Book 2 Basic Semiconductor Devices for Electrical Engineers

Professor C.R. Viswanathan Electrical and Computer Engineering Department University of California at Los Angeles

Distinguished Professor Emeritus

Chapter 1 Introductory Solid State Physics Introduction

An understanding of concepts in semiconductor physics and devices requires an elementary familiarity with principles and applications of quantum mechanics. Up to the end of nineteenth century all the investigations in physics were conducted using Newton’s Laws of motion and this branch of physics was called classical physics. The physicists at that time held the opinion that all physical phenomena can be explained using classical physics. However, as more and more sophisticated experimental techniques were developed and experiments on atomic size particles were studied, interesting and unexpected results which could not be interpreted using classical physics were observed. Physicists were looking for new physical theories to explain the observed experimental results.

To be specific, classical physics was not able to explain the observations in the following instances:

1) Inability to explain the model of the atom. 2) Inability to explain why the light emitted by atoms in an electric discharge tube contains sharp spectral lines characteristic of each element. 3) Inability to provide a theory for the observed properties in thermal radiation i.e., heat energy radiated by a hot body. 4) Inability to explain the experimental results obtained in photoelectric emission of electrons from solids.

Early physicists Planck (thermal radiation), Einstein (photoelectric emission), Bohr (model of the atom) and few others made some hypothetical and bold assumptions to make their models predict the experimental results. There was no theoretical basis on which all their assumptions could be justified and unified. In the 1920s, a revolutionary and amazing observation was made by De Broglie that particles also behave like waves. Einstein in 1905 had formulated that light energy behaves like particles called photons to explain the photoelectric results. De Broglie argued that energy and matter are two fundamental entities. The results of the photoelectric experiments demonstrated that light energy behaves also like particles and therefore matter also should exhibit wave properties. He predicted that a particle with a momentum of magnitude will behave like a plane wave of wavelength, equal to

풑 = 흀 (1.1) ℎ where is Planck’s constant and numerically equal to 휆 푝 풉 = 6.625 10 Joules - sec. −34 This led to the conclusionℎ that both particles푥 and energy (light) exhibited dual properties of behaving as particles and as waves.

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Since Planck’s constant is extremely small, the De Broglie wavelength is correspondingly small for particles of large mass and momentum and therefore, the wave properties are not noticeable in our daily life. However, subatomicℎ particles such as electrons have such a small mass and their momentum is small to make the wavelength large enough to observe interference and diffraction effects in the laboratory.

A particle moving with a momentum is represented by a plane wave of amplitude ( , ) given by: 풑��⃑ 훹 푟⃑ 푡

( , ) = = ( ) 푝��⃑ ∙ �푟�⃑ (1.2) 푖�푘�⃑ ∙��푟⃑ − 휔 푡� 푖 ℏ − 휔 푡 where is the propagation vector of magnitude equal to , = , = 2 times the frequency 훹 푟⃑ 푡 퐴 푒 퐴 푒 2 휋 ℎ of the wave and , the radius vector is equal to 풌��⃑ 푘 휆 ℏ 2 휋 휔 휋 푟⃑ = + +

and , and are the unit vectors푟⃑ along푎⃑ the푥푥 three푎⃑푦 coordinate푦 푎⃑푧 푧 axes. The radius vector denotes the position where the amplitude is evaluated. The wavelength of the plane wave according to equation 풂��⃑풙 �풂�⃑풚 풂��⃑풛 (1.1), is given by

= = (1.3) | | | | ℎ 2 휋 ( , ) is also called the wave-function or state휆 function푝 . The푘 expression for the plane wave given in equation (1.2) shows that the amplitude of the plane wave varies sinusoidally with time and position . For훹 illustrating푟⃑ 푡 the plane wave properties we will consider a plane wave travelling from negative infinity to positive infinity. Figure (1.1a) shows the amplitude of the plane wave at as a sinusoidal풕 function of 풙 time at some value = . Figure (1.1b) shows the amplitude varying as a sinusoidal function of position at some time = . 풕 풙 풙ퟏ 풙 풕 풕ퟏ

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Plane wave

흍

흍 Amplitude, Amplitude,

a) Amplitude as a function of time, t at x = x1

흍

흍 Amplitude, Amplitude,

b) Amplitude as a function of position, x at t = t1

Figure (1.1): (a) Plane wave travelling from negative infinity to positive infinity, a) the amplitude of the plane wave as a function of time, t at some value = , (b) the amplitude of the plane wave as a function of x at some time = 푥 푥1 The dynamic behavior of a physical system such as a single푡 푡or1 a collection of particles can be deduced from , the wave-function of the system. In quantum mechanics, the wave-function is obtained by solving an equation called Schrödinger equation. Another consequence of applying quantum mechanical훹 principles to electrons is that the momentum and position of the particle say an electron, cannot be measured or known accurately at the same time to any precision that the measuring equipment is capable of. Either the momentum or the position alone can be measured to any accuracy the measuring equipment is capable of; but both of them cannot be measured at the same time to the accuracy or precision that each property can be measured separately. This is called the Heisenberg Uncertainly Principle. That is, the more precisely position or momentum is measured, the other can be measured only less precisely. The physics underlying the uncertainty principle is that the act of

- 4 - measurement of one property disturbs the physical state of the system making measurement of the other property less accurate. In quantum mechanics, there are other examples of pairs of physical properties that obey the Heisenberg Uncertainty Principle. For example, energy and time is another pair that obeys the Uncertainty Principle.

Mathematically the uncertainty principle says that the product of the uncertainty in simultaneous measurement of position and momentum of a particle has to be higher than a minimum value of the order of Planck’s constant , . The uncertainly principle is not observable in the size and mass of particles in our daily life where the uncertainties of the properties are too small to 푥 observe because of the small value of Planck’s풉 ∆푃 ∆constant.푥 ≥ ℎ When we deal with atomic particles such as electrons the uncertainty effects become significant.

When the wave-function is determined by solving Schrödinger equation for an electron that is confined to a narrow region of space along the -axis, say between = 0 and = , the result shows that the - component of momentum can have only discrete values given by 푥 푥 푥 퐿푥 푥 =

푥 푥 ℎ 푃 푛 ( = ±1, ±2, ±3, . . . ) where is called a quantum number, and is an integer 퐿푥 . Thus the - component of momentum, , can have only discrete values and are said to be discretized or quantized 풙 and hence풏 is called a quantum number. If the electron 푛is constrained to move푒�푒 in a restricted푥 region in 풙 the - plane, then 2 quantum푷 numbers and will be needed to quantize the and components 풏풙 of the momentum. In the case when the electron is confined to a small three dimensional region, the 푥 푦 풏풙 풏풚 푥 푦 electron state will be specified by three quantum numbers and and , as discussed in the next section. 풏풙 풏풚 풏풛

Free Electron Model

The electric current in a conducting solid such as a metal is explained in solid state physics using the Free Electron Model. According to this model, each atom of the solid contributes one (or two in some cases) electron to form a sea of electrons in the solid and these electrons are free to roam around in the solid instead of being attached to the parent atom. If an electric field is applied in the solid, the electrons will move due to the force of the electric field and each electron will carry a charge – coulomb thus giving rise to the flow of an electric current. Thus free electrons in the solid are treated quantum mechanically as equivalent to electrons in a box of the same dimensions as the solid. We푞 will now derive some properties of the free electrons in a conducting solid by using free electron model.

Let us now consider an electron of mass that is confined to move within a three dimensional box. Let the box be rectangular with dimensions , and . As shown in Figure (1.2), we will choose 푚 the origin of our Cartesian coordinate system at one corner and the , and axes to be along the 퐿푥 퐿푦 퐿푧 three edges of the box. 푥 푦 푧 We assume that the potential energy to be constant and that it does not vary with position inside the box. Since force is equal to the gradient of the potential energy, there is no force acting on the

- 5 - electron because the gradient is zero. The electron is said to be in a force free region. Let the potential energy be taken as zero. A particle moving in force-free region has a constant momentum and is therefore represented by a plane wave as discussed earlier. The wave-function for the particle is therefore given by

( ) ( , ) = = = ( ) (1.4) 푖�푘�⃑ ∙푟⃑− 휔 푡� 푖 푘�⃑ ∙푟⃑ −푖𝑖 −푖𝑖 where 훹 푟⃑ 푡 퐴 푒 퐴 푒 푒 휓 푟⃑ 푒 ( ) = ( ) �⃑ ( ) 푖 푘 ∙푟⃑ can be written as, in rectangular coordinates,휓 푟⃑ 퐴 푒 흍 �풓 ⃑ ( ) = ( , , ) = 푖푘푥푥 푖푘푦푦 푖푘푧푧 where 휓 푟⃑ 휓 푥 푦 푧 퐴푥 푒 퐴푦 푒 퐴푧 푒 =

퐴 퐴푥퐴푦퐴푧

Figure (1.2): Electron in a box

We can also write

( , , ) = ( ) ( ) ( ) (1.5)

Where 휓 푥 푦 푧 휓푥 푥 휓푦 푦 휓푧 푧 ( ) = (1.6) 푖푘푥푥 휓푥 (푥) = 퐴푥 푒 (1.7) 푖푘푦푦 휓푦 (푦) = 퐴푦 푒 (1.8) 푖푘푧푧 휓푧 푧 퐴푧 푒 - 6 -

The boundary condition on the wave-function is that the wave-function is the same in amplitude and phase when is incremented by , is incremented by and is incremented by i.e.

푥 푦 푧 ( , ,푥 ) = ( + 퐿, ,푦 ) = , +퐿 , 푧 = ( , , +퐿 ) (1.9)

This boundary휓 푥 condition푦 푧 is휓 called푥 the퐿푥 Periodic푦 푧 Boundary휓 �푥 Condition푦 퐿푦 .푧 � 휓 푥 푦 푧 퐿푧 The above boundary condition requires

( ) = ( + ) (1.10)

휓푥 (푥) = 휓푥 푥 +퐿푥 (1.11) and 휓푦 푦 휓푦 �푦 퐿푦 � ( ) = ( + ) (1.12)

By applying the boundary condition given 휓in푧 Equation푧 휓(1.6)푧 to푧 Equation퐿푧 (1.10), we obtain ( ) = (1.13) 푖푘푥푥 푖푘푥 푥+퐿푥 This implies that 퐴푥 푒 퐴푥 푒 = 2 (1.14) where is a positive or negative integer.푘 Therefore,푥퐿푥 휋 푛푥 푥 푛 = (1.15) 2 휋 and 푘푥 퐿 푥 푛푥

( ) = 2 휋 푛푥 (1.16) 푖 퐿푥 푥

The -component of the momentum, equal휓푥 to푥 , is퐴 therefore푥 푒 quantized with values, 풙 푥 = 0, ± ℏ풌, ± , ± , ± . (1.17) ℎ 2ℎ 3ℎ 4ℎ Thus the -component of the momentum푝푥 is quantized퐿푥 퐿 푥with the퐿푥 same퐿 푥quantum∙∙∙ 푒�푒 number . The푥 kinetic energy is related to the momentum. The kinetic energy due to the motion풏풙 of the particle along the -axis is also quantized. Let us denote the kinetic energy due to motion along the - axis as . 푥 푥 퐸1 ħ ħ = = = (1.18) 2 2 2 2 2 푝푥 푘푥 푛푥 1 2 Thus , and are all quantized with퐸 the2 same푚 quantum2푚 number2 푚 퐿푥 .

풑풙 풌풙 푬ퟏ 풏풙

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Example

Let us now calculate , and for an electron in a state = 1 in a box of dimension 10 Å × 20 Å × 1 10 , 30 Å along the , and푥 axes푥 respectively.1 Remembering 푥 is 푘 푝 퐸 푛 −10 × = 푥 푦 푧= = = 6.28 ×Å 10 푚 × 2 휋 2 휋 1 2 휋 9 −1 −10 −9 푘푥 퐿푥 푛푥6.62610 10 10 푚 = ħ = × 6.628 × 10 = 6.26 × 10 2 −34 푥 9 −25 푝푥 푘푥 퐾 푔� − 푚⁄ = = = 2.41 × 10 = 1.50 푠 2 . ×2 휋 푝푥 푝푥 −19 −31 퐸 1 2 푚 2 푥 9 11 10 퐽퐽퐽퐽𝐽 푒푒

, and are related to , and components of momentum through the De Broglie relation. By using풌풙 풌 풚the boundary풌풛 condition푥 for푦 and푧 , it can be shown = and = where 2 휋 푛푦 2 휋 푛푧 풚 and are quantum numbers similar푘푦 to 푘. 푧 푘푦 퐿푦 푘푧 퐿푧 풏 The kinetic풏풛 energy due to motion along the풏풙 and axes are denoted by and

푦 ħ 푧 ħ 푬ퟐ 푬ퟑ = 2 2 = 2 2 (1.19) 푘푦 푛푦 2 2 and 퐸 2푚 2푚퐿푦

ħ ħ = = (1.20) 2 2 2 2 푘푧 푛푧 2 In Figure (1.3), is plotted as a function퐸 of3 . Since2푚 is2푚 quantized퐿푧 , is also quantized. Since the adjacent values of lie so close to each other, the plot looks like a continuous curve. We say ퟏ 풙 풙 ퟏ are 푬qusai-continuous. The total wave풌 -function풌 is obtained as푬 a product function of 풙 ퟏ , and풌 is푎 given�푎 푬 by 풌풙 푎�푎 푬1 풙 풚 풛 흍 흍 푎�푎 흍 ( , , ) = ( ) ( ) ( ) (1.21)

( ) 휓 푥=푦 푧 휓푥 푥 휓푦 푦 휓푧 푧 (1.22) 푖 푘푥푥+푘푦푦+푘푧푧 푥 푦 푧 = 퐴 퐴 퐴 푒 (1.23) 푖푘�⃑ ∙푟⃑ where is called the propagation vector and퐴 푥is 퐴given푦 퐴 by푧 푒 ��⃑ 풌 = + + (1.24)

푥 푥 푦 푦 푧 푧 푘�⃑ = 푎⃑ 푘 푎⃑ +푘 푎⃑ 푘 + (1.25) 2 휋 푛푥 2 휋 푛푦 2 휋 푛푧 푎⃑푥 퐿푥 푎⃑푦 퐿푦 푎⃑푧 퐿푧

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where = radius vector, and , and are the unit vectors along the , and axes respectively. , the total kinetic energy is equal to 풂��⃑풙 ��풂�⃑풚 풂��⃑풛 푥 푦 푧 ′ 풓��⃑ 푬 ħ ħ ʹ = + + = = + + (1.26) 2 2 2 2 2 2 푘 푛푥 푛푦 푛푧 2 2 2 1 2 3 푥 푦 푧 where is the magnitude퐸 of the퐸 propagation퐸 vector퐸 .2 The푚 number2푚 �퐿 , 퐿and 퐿 are� called quantum numbers, and once you assign a particular set of three integers to these three quantum numbers, you 풌��⃑ 풏풙 풏풚 풏풛 have specified푘 the momentum states for the particle i.e., the value of momentum the particle will have in that state. We therefore denote the wave-function by the subscripts , and .

풙 풚 풛 ( 풏) 풏 풏 = 푛푥푥 푛푦푦 푛푧푧 (1.27) 푖 2 휋 퐿푥 + 퐿푦 + 퐿푧 휓푛푥푛푦푛푧 퐴푥 퐴푦 퐴푧푒

Figure (1.3): Energy vs.

The state = , = and = , represents a state in퐸 1which푘 푥the particle is moving with a momentum equal to 풏풙 ퟏ 풏풚 ퟐ 풏풛 −ퟏ �풑�⃑ = 2 ħ + = h + (1.28) a�⃑x 2a�⃑y a�⃑z a�⃑x 2a�⃑y a�⃑z x y z x y z Since , and determine푝⃑ the휋 momentum�L L state,− L and� can �beL only Lintegers,− L �we see that the value

changes풏풙 by풏풚 from풏풛 one state to the next and similarly the and values change by and 풑풙 ℎ ℎ ℎ respectively. 풚 풛 퐿푥 풑 풑 퐿푦 퐿푧

It should not be surprising that adjacent values of , differ by . The maximum uncertainty in ℎ the -component of position is = since the particle 푥can be anywhere in the box. Therefore the 푝 퐿푥 uncertainty in the -component of momentum is required from the Uncertainty Principle to satisfy the 푥 condition푥 ∆푥 퐿 푥

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ℎ ∆푝푥 ≥ The adjacent values differing by are consistent with the퐿푥 Uncertainty Principle. 풉 Consider a three dimensional푳풙 space (imaginary of course!) in which the three axes are ,

and as shown in Figure (1.4). Each set of integers for , and generates a point in this space 풑풙 풑풚 (called the momentum space) and each point represents a particular momentum state. Such points in 풑풛 풏풙 풏풚 풏풛 momentum space are called representative points or phase points.

Figure (1.4): Momentum space with , and axes

The momentum space comprises an infinite number of푝푥 phase푝푦 points푝푧 each separated from its neighbor by along the axis or along the axis or along the axis. The origin of this space is 풉 풉 풉 풙 풚 풛 located at 푳=풙 , and hence풑 represents푳풚 the state풑 with no 푳kinetic풛 energy.풑 The vector connecting the origin of the momentum space to a point with coordinate numbers , and represents the 풑� ퟎ = = momentum vector of the state with -component , -component풏풙 풏풚 풏 풛 , and - 풉 풉 component = . 푥 풑풙 푳풙 풏풙 푦 풑풚 푳풚 풏풚 푧 풉 풑풛 푳풛 풏풛

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Figure (1.5): ABCDEFGH represents the elementary volume in the momentum space containing one momentum state.

We obtained three quantum numbers because we restricted the motion of the particle to be within the small box of volume . If we constrain the particle to only a small region in one dimension, say along the axis, we get one quantum number and the momentum along the axis is 푳풙푳풚푳풛 quantized. Similarly, if we restrict the particle to a small area in the plane, we will get two quantum 풙 numbers and and the푥 momentum along the and axes풏 will be quantized. If we restrict푥 the 푥� particle to a three dimensional space, three quantum numbers , and will be needed to specify 풏풙 풏풚 푥 푦 the momentum state. 풏풙 풏풚 풏풛 Let us consider a specific state characterized by a particular set of quantum numbers , and , . This state is represented by a point A in the momentum space with coordinates 풏 and풙 풏 풚 풉 풉 풉 풏as풛 shown in Figure(1.5). The rectangular volume contained in the momentum space푳풙 풏within풙 푳풚 풏 풚coordinate푳풛 풏풛 numbers ( , , ), ( + 1, , ), ( , , + 1, ), ( , , + 1), ( + 1, + 1, ), ( , +1, + 1), ( +1, +1, + 1) are the corner has eight adjacent states, one at each of 푛푥 푛푦 푛푧 푛푥 푛푦 푛푧 푛푥 푛푦 푛푧 푛푥 푛푦 푛푧 푛푥 푛푦 푛푧 its corners, and no states inside the rectangular volume. The rectangular volume has sides 푛푥 푛푦 푛푧 푛푥 푛푦 푛푧 = , = and = , and has a volume equal to Since each corner is shared by eight풙 ퟑ . ∆풑 풉 풉 풉 퐡 풚 풛 adjacent푳풙 ∆풑 rectangular푳풚 ∆ 풑volumes,푳풛 we can think that each rectangular퐋퐱퐋퐲퐋퐳 volume equal to contains only ퟑ 풉 one momentum state. The density of momentum states in the momentum space is 푳the풙푳풚 푳reciprocal풛 of this volume. Refer to Figure (1.5).

According to quantum mechanics, the electron state is completely specified by four quantum numbers which are the three quantum numbers such as , and obtained by constraining the particle to a small three dimensional volume and a fourth quantum number called the spin quantum 풏풙 풏풚 풏풛 number. The spin quantum number can either be or - for electrons. Thus for a specific set of integers 1 1 -2 11 - 2

for , and , we have 2 quantum states one with numbers , and and , and the other with 1

, 풙 and풚 and풛 - . 푛푥 푛푦 푛푧 2 풏 풏 풏 1 푛푥 푛푦 푛푧 2 In the momentum space, in a rectangular volume equal to 3 there are two electron states. ℎ Another way of saying this is that each electron state is contained within퐿푥퐿푦퐿푧 a volume = of the 3 3 ℎ ℎ momentum space where the volume of the box is .. We can write that the2 퐿number푥퐿푦퐿푧 of2 푉states in unit volume of the momentum space� is equal to 푳and풙푳풚 th푳풛is is called the density of electron states in 2푉 momentum space. If we take the potential energy 3to be zero (by suitably choosing the zero reference ℎ for the energy), then the total energy , is equal to the kinetic energy. If we describe a sphere of radius in the momentum space with the origin at the center of the sphere, all states lying on the surface of 퐸 the sphere will have the same energy. This leads us to conclude that the constant energy surface, i.e., a surface풑 generated by connecting all states with the same energy, is a spherical in the momentum space.

Suppose we want to determine the number of electron states having momentum components between and + , and + , and + . We consider an elementary rectangular volume of sides with its corner at the coordinates in the momentum space as 푝푥 푝푥 �푝푥 푝푦 푝푦 �푝푦 푝푧 푝푧 �푝푧 shown in Figure (1.6), and count the number of states within this volume. Instead of counting, we can �푝푥�푝푦�푝푧 푝푥 푝푦 푎�푎 푝푧 simply multiply the volume of the elementary rectangular momentum space by the density of states 2푉 to obtain the number of states. 3 ℎ The magnitude of the momentum , is given by

풑 = + + 2 2 2 푥 푦 푧 Usually, we are more interested in finding푝 the� 푝number푝 of states푝 having the magnitude of momentum between and + . In the momentum space if we describe two spheres around the origin, one with the radius equal to and the other with radius equal to + as shown in Figure (1.6), all the states in 푝 푝 𝑑 the interspace between the surfaces of these two spheres correspond to momentum magnitude between and +푝 . The volume of the interspace between푝 𝑑 the two spheres is equal to 4 . If ( )ퟐ we multiply푝 this푝 volume𝑑 by the density of states in the momentum space, we obtain 흅풑 풅� the 2푉 number of states with the magnitude of momentum3 in the range between and + . The number �ℎ � 푍 푝 𝑑 of states having magnitude of momentum between and + as 푝 푝 𝑑 푝 2 푝 𝑑 ( ) = 4 � 2 푍 푝 𝑑 3 휋푝 𝑑 = ℎ (1.29) 8 휋� 2 3 ℎ 푝 𝑑

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Figure (1.6): Constant momentum (magnitude) and constant energy surface

Fermi-Dirac Statistics

When we have a single electron, this electron will occupy the state with the lowest energy. If we have a very large number of electrons in the box, all these electrons cannot occupy the same lowest energy state. According to a famous principle in physics called Pauli exclusion principle, only one electron can occupy a given quantum state i.e., a state specified by the assignment of four quantum numbers. We start therefore filling the states, starting from the lowest energy state, until we have exhausted all the electrons. The last electron fills the state with the highest energy among all occupied states. This kind of distribution of electrons among the states is called the Fermi-Dirac statistics or distribution.

According to Fermi-Dirac statistics, the probability that a given state is occupied by an electron is given by the following function:

( ) = (1.30) 1 퐸−퐸퐹 푘푘 This function given by Equation (1.30)푓 퐸is called푒 the Fermi+1 function and the parameter is called the Fermi energy. This function is plotted in Figure (1.7). 푬푭 At absolute zero temperature, ( ) is 0 for > and is 1 for < . This means that at = 0 , states with energy less than are occupied and states with higher energy are empty. 퐸 퐸퐹 퐸 퐸퐹 represents0 the maximum value of 풇the푬 energy of occupied states. The Fermi energy acts as a 푇 퐾 퐸퐹 퐸퐹 - 13 - threshold energy separating the occupied states and the vacant states. The Fermi function exhibits a discontinuity at = 0 in going from a value of 1 to 0 at = . 0 As the temperature푇 퐾 is raised, the electrons in states퐸 close퐸퐹 to the Fermi energy get excited by thermal energy to states with energy higher than the Fermi energy. This leads to a gradual variation with energy of the Fermi function from 1 to 0 around . Since the thermal energy is of the order of which is very much smaller than , the region of transition from 1 to 0 is very small compared with 퐹 as shown in Figure (1.7). Even at higher temperature,퐸 states with energy less than the Fermi energy 푘by� 퐹 퐹 several , have a probability of퐸 1 for occupation by an electron. Similarly, states with energy greater퐸 than the Fermi energy by several have a probability of nearly 0 for occupation by an electron. The 푘� probability of a state with energy equal to , being occupied by an electron is . 푘� 1 퐹 Determination of Fermi energy 퐸 2

We denote the number of electrons having the magnitude of momentum between and + as . is obtained by multiplying Equation (1.29) and Equation (1.30). 푝 푝 𝑑 �푁푝 �푁푝 = ( ) ( ) = (1.31) 8 휋 푉 1 2 푝 3 퐸−퐸퐹 ℎ 푘푘 Let us now determine� the푁 distribution푍 푝 𝑑 of 푓electrons퐸 as a funct푒 ion of+ 1energy푝 𝑑 . We want to find the number of electrons having energy between and + . Let us denote this number by . In Figure (1.6) the surfaces of the two spheres with radius and + are constant energy퐸 surfaces one 퐸 with energy and the other with + . Therefore퐸 퐸 is𝑑 the same as , but needs� 푁to be 푝 푝 𝑑 expressed in terms of instead of in terms of . Since , the total energy equal to the kinetic energy, 퐸 퐸 𝑑 �푁퐸 �푁푝 �푁퐸 when the potential energy is taken as zero, 퐸 푝 퐸 = 2 2 푝 Differentiating, 퐸 푚

= = (1.32) 2푝 푑� 푝 푑� 𝑑 =2푚 푚 (1.33) and 푝 𝑑 푚 𝑑 = 2 (1.34)

Substituting these in Equation (1.31), we obtain 푝 √ 푚 퐸 ( ) = = = 3 1 (1.35) 2 2 8 휋 푉 1 2 4 휋 푉 2푚 퐸 퐸 푝 3 퐸−퐸퐹 3 퐸−퐸퐹 ℎ 푘푘 ℎ 푘푘 �푁 �푁 푒 + 1 푝 𝑑 푒 + 1 𝑑

- 14 -

Figure (1.7): Fermi function (also called the Fermi-Dirac distribution function)

This can also be written in the form

= ( ) ( ) (1.36) ( ) ( ) where is the Fermi function and � 푁is퐸 given푓 by퐸 푍 퐸 𝑑 푓 퐸 푍 퐸 ( ) ( ) = 3 (1.37) 2 1 4 휋 푉 2푚 2 3 ( ) is called the density of states (in energy)푍 퐸 and can beℎ interpreted퐸 as the number of quantum states in unit energy interval i.e., ( ) is the number of states having energy between and + . 풁 푬 Figure (1.8) shows 풁a plot푬 풅 of� the density of state function, ( ). This figure gives퐸 the퐸 distribution𝑑

of states as a function of energy. We see that the density of states푍 is퐸 proportional to 1. We multiply ( ) by the Fermi function ( ) or divide Equation (1.35) by to obtain , the distribution2 of 푑푁퐸 퐸 e푍lectrons퐸 as a function of energy푓 퐸 . represents the number𝑑 of electrons푑� with energy between 푑푁퐸 and + and is plotted in Figure (1.9). At = 0 , no electron has energy larger than . However, 푑� 𝑑 퐸 at high temperatures, a few electrons occupy states0 with larger energy. The higher the temperature, the 퐹 more퐸 electrons𝑑 there are with higher and higher푇 energy.퐾 This process is referred to as thermal퐸 excitation of electrons to higher energy states.

- 15 -

Figure (1.8): Distribution of states as a function of energy

The Fermi energy, , can be obtained by a simple treatment as shown below: At absolute zero temperature, the total number of occupied states in the momentum space shown in Figure (1.6) are 푭 those which lie within a sphere푬 on whose surface the energy is equal to and this should be equal to the total number of electrons in the box. A sphere bounded by the constant surface has a radius 퐸퐹 푁 given by 퐸퐹 풑푭 = 2 2 퐹 푝 퐹 퐸 The volume of the sphere is equal to . Multiplying푚 this by the density of states, we obtain 4 휋 3 2푉 3 3 푝퐹 ℎ = 4 휋 3 2 푉 퐹 3 푁 = 3 푝 ℎ

8 휋 푉 3 3 퐹 3 ℎ 푝 = (2 ) (1.38) 3 3 8 휋 푉 2 2 3 We used the relation between and in the above3 ℎ equation.푚 We퐸퐹 can now rearrange this equation to express as 푝퐹 퐸퐹 퐸퐹 2 = 3 3 3 ( ℎ ) 푁 3 퐹 2 푉 퐸 �8 휋 2푚 �

- 16 -

2 = (1.39) 3 3 3 ( ℎ ) 3 �8 휋 2푚 2 푛�

Figure (1.9): Electron distribution as a function of energy

where = is the number of electrons per unit volume (i.e. density) in the box. The Fermi energy is 푁 dependent only on the density and not on the total number of electrons. As we will see later on, 푛 푉 electrons in metallic solids that are free to roam around in the solid are modeled as electrons in a box of the same dimensions as the piece of solid. We can determine the Fermi energy of the solid, knowing , the density of electrons. Whether a solid is large or small in size, the Fermi energy is the same independent of size because it depends only on the density of electrons. 푛

______Example

Let us calculate the Fermi energy of a solid that has 1022 cm-3 free electrons. These electrons are free in the sense that they can move from one region of the solid to another region. Therefore they can be treated as electrons in a box of the same dimension as the solid. = 10 22 −3 푛 = 10 𝑐 28 −3 = 9.11 ×푚 10𝑚𝑚 −31 푚 = 6.63 × 10 퐾 푔� −34 ℎ 퐽퐽퐽퐽𝐽 − 푠𝑠

- 17 -

Substituting these values in the expression for Fermi energy given in Equation (38) we obtain the value of Fermi energy as

3 × 10 2 = 3 × (6.63 × 10 ) 28 8 × × (2 × 9.11 × 10 ) −34 2 퐸퐹 � 3 � −31 2 휋= 2.71 × 10 = 1.69 −19 퐽퐽퐽퐽𝐽 푒푒

According to classical physics, all motions stop at = 0 . But according to quantum mechanics, even at = 0 electrons move with speeds distributed 0between 0 and a maximum value 푇 퐾 , where is the speed0 of electrons with energy equal to . We can calculate the speed of the 푇 퐾 electron퐹 with퐹 energy (kinetic) , through the relation = 퐹 . The maximum velocity that an 푣 푣 1퐸 푣 electron can have at = 0 is given by 2 퐸 퐸 2 푚 푣 0 푇 퐾 = (1.40) 1 2 퐹 퐹 퐸 2 푚 푣 = (1.41) 2 퐸퐹 퐹 We can determine the average speed푣 or momentum� 푚 or energy of a collection of electrons in a box since we know the distribution functions , for the electrons. In the example below we 푑푁푝 푑푁퐸 calculate the average speed of electrons in a box at absolute zero temperature. 푑푑 푑�

Example

, the speed, is equal to and hence the average speed at = 0 0K is given by 푝 푣 푚 푇 < > = < >

∞ 푝 푣 = ∞푝 푚

∫0 푚 푑푁푝 푝 ∞ ∫ 0 (푑푁) ( ) = ∞푝 ( ) 0 푍 푝 푓( 퐸) 𝑑 ∫ 푚 To evaluate this integral we take advantage of∫0 the푍 property푝 푓 퐸 𝑑that ( ) is equal to 1 when is between 0 and and is equal to 0 when is greater than at absolute zero temperature. 푓 퐸 푝 푝퐹 푝 푝퐹 - 18 -

< >= 푝퐹 푝 2 ∫0 푚 푝 푑� 푃퐹 2 푣 ∫0 푝 푑� = × 4 푝퐹 3 3 4푚 푝퐹 = = 3 푝퐹 3 Thus we see that the average speed of the electrons4 푚 is of4 the푣퐹 maximum speed . 3 퐹 If we want to localize the position of the electron4 in the box, by representing푣 it as a wave packet, we superimpose a large number of plane waves of suitable amplitude and phase and of different values with an average value of . The velocity of the electron is the group velocity of the wave packet and is equal to 푘 푘0 = (1.42) ħ 푑� 1 푑� determined at = . Referring to Figure (1.10)푑�, the velocity푑� of the electron for the state = , is related to the slope of the energy vs plot at = . Since we have 푘 푘0 푘 푘0 퐸 푘 푘 ħ푘0 = 22 2 푘 the velocity of the electron in the state = 퐸is equal to 푚 푘 푘0 ħ = = = (1.43) ħ 1 푑� k0 푒푒 푔 This is the usual result that velocity is momentum푣 푣 divided 푑by� �푘=. We푘0 considered푚 the potential energy to be constant inside the box. However, the potential energy inside the solid will be periodically varying. When we assume that the potential energy inside the box is varying푚 periodically, an interestingly different vs diagram is obtained and the velocity is not equal to the momentum divided by the mass.

In퐸 the 푘treatment of electrons in a box, the potential energy was assumed to be constant and hence there was no force on the electron. Hence the electrons were said to be free. When we apply this model to treat the case of electrons in a metallic solid, we call it the free electron model of the solid. We found that the energy levels are discrete (quantized). However, since the dimensions of the solid are large in comparison with atomic distances, the discrete energy levels are very close to each other and hence for all practical purposes can be considered to be continuous. This is particularly true when we consider higher energy states where the quantum numbers are large and the separation in energy between adjacent states expressed as a fraction of the energy of the state becomes extremely small. For this reason, the energy is said to be “quasi-continuous”. For the same reason the corresponding momentum , and as well as the magnitude of | | are quasi-continuous. This model is applicable to describe the behavior of conduction electrons in metallic solids. You will see in the next chapter that, 푘푥 푘푦 푘푧 푘 in a metallic solid, each atom of the solid contributes an electron or two to form a sea of electrons that belong to the whole solid and are free to move in the solid like electrons in a box. This model does not

- 19 - explain why some solids are conductors and some others are not. We have to use a new model called the “Band Theory of Solids” to understand the conducting properties of solids.

Figure (1.10): vs digram

퐸 푘 Band Theory of Solids

In a typical solid with free electrons, the electrons are subjected to an internal force due to the positively charged ions of the atoms. The potential energy varies periodically due to the periodic arrangement of the atoms in the solid. This force acting on the electrons is called the internal force as distinct from an externally applied force such as that due to an electric field arising from the application of a voltage between the two ends of the solid.

To illustrate the concept of “allowed” and “forbidden” bands of energy, let us assume a one- dimensional solid as shown in Figure (1.11). The positive ions are assumed to be situated periodically along the -axis with an inter-atomic distance of ‘a’ units. The electrons can move only along the -axis in either direction (i.e., in only one dimension). The potential energy of the electron due to the attractive electrostatic푥 force of the positive ions varies periodically with as shown in Figure (1.12). Due to 푥the periodic potential energy, the electron states are allowed to have energy values only in certain ranges and not in others. The range of energy values in which the electron푥 has states is called an allowed band of energy. The allowed bands are separated by forbidden (or disallowed) bands of energy in which the electrons have no states and the forbidden bands are called energy gaps, (see Figure 1.13).

In the free electron model, the electron energy varies quasi-continuously from 0 ∞ as shown in Figure (1.14). In this figure, the kinetic energy, which is the same as the total energy since the potential energy is constant and can be taken as zero by a proper choice of the zero reference푡� for energy, is plotted as a function of . Motion is possible only along one dimension. varies quasi- continuously from ∞ + ∞. 푘 푘 − 푡� - 20 - free electron Periodic location of positive ions - q

+ + + + + + a

One –dimensional solid

Figure (1.11) Positive ions and free electrons in a one-dimensional solid

Figure (1.12) Electrostatic potential energy variation with distance

______allowed band

Energy ______

forbidden band (energy gap) ______allowed band

0

Figure (1.13) Bands of allowed and forbidden energy

Let us assume that the potential energy varies periodically with the periodicity and is given by

( ) = ( + ) 푎 (1.44)

푈 푥 푈 푥 푎

- 21 -

When the existence of the periodic potential energy is treated quantum mechanically, it is shown in solid state physics books that the energy varies with as shown in Figure (1.15). Notice that the energy of the states is split into bands. The solid line in the Figure (1.15) represents the energy according to the band theory. Let us now examine the essential features푘 of the band theory.

Figure (1.14): diagram for free electrons. Quasi-continuous distribution of energy as a function of the magnetitude of the momentum. Notice that the energy values extend from 0 to ∞ 퐸 − 푘

- 22 -

Figure (1.15): A plot of energy as a function of ( ) in the band theory in the one-dimensional approximation. Notice that the energy values are broken into bands of allowed and forbidden values. 푘 퐸 푣� 푘 As the magnitude of is increased from 0, the energy increases symmetrically with for both positive and negative values of until becomes equal to ± . At this value of , the energy has a 푘 휋 푘 discontinuity and if is increased beyond푘 푘 then the energy again푎 increases. The discontinuity푘 in 휋 ± energy at is called푘 the energy gap or the푎 bandgap. The continuous range of energy for values of 휋 + between - 푎 and is called the first energy band. It can be noticed that another energy gap occurs푘 휋 휋 at - 2 and푎 + 2 is called푎 the second band of energy. In general, the bandgap occurs at = ± 휋 휋 푛� where is an integer. Thus, according to the band theory of solids, the energy values are split into 푎 푎 푘 푎 allowed and forbidden bands of energy. The forbidden band of energy is also referred to as bandgap and the allowed푛 band is referred to as energy band. At the bottom and at the top of the energy band, the energy plot has zero slope. In the middle of the band, the energy has a parabolic dependence on , i.e., is proportional to as in the free electron model. Notice that, in the region of the band, the dotted line and the solid line ov2erlap showing that the behavior of electrons in this region of the band is similar푘 to the free electron behavior푘 .

It is shown in text books in solid state physics that in a band, two states whose value differs by where is a positive or negative integer are equivalent states if they also have the same 푘 2 푛 휋 energy. Referring푎 to the푛 Figure (1.16), A and B are equivalent states because the values differ by 2 휋 and they have the same energy. Similarly, C and D are equivalent states. 푘 푎 Since the velocity of the electron is the group velocity of the wave-packet as given by Equation (1.42), we notice that the electron velocity varies in the bandgap due to the variation of the slope in the versus plot i.e., in the band. The electron has zero velocity in the bottom and in the 푑� top of the band since the slope is zero. It has maximum velocity in the middle of the band where the 퐸 푘 푑� inflection point occurs.

- 23 -

Figure (1.16): One dimensional energy band diagram illustrating equivalent states

Effect of an Applied Electric Field

We will now examine the behavior of the electrons in a band under the application of an external field. Let us assume that the electric field is switched on at time = 0. There is a force on the electron equal to – . In a short interval of time t, the electron moves a distance t and gains an 푡 energy equal to 퓔 풒� 훿 푣 훿 = t (1.45) + The energy of the electron is changed훿� by −푞, andℰ 푣 it훿 should be now in a new state . The differential in energy is given by 훿� 푘 훿�

훿� = (1.46) 푑� The velocity was seen earlier to be equal훿 to� the group�푑� �velocity훿� of the wave packet i.e.

= = (1.47) 1 푑� Hence can be expressed in terms of the푣 velocity푣� of ℏthe푑� electron as 훿� = ` (1.48)

Equating this expression for with that훿 given� inℏ equation푣 훿� (1.45), we obtain after rearrangement 훿� = (1.49) 푑� −푞 ℰ ℏ 푑� - 24 -

Thus we see that is equal to the external force on the electron due to the applied electric field. We 푑� can infer, therefore, that the effect of an external force is to make the value change with time at a rate ℏ 푑� given by 푘

= (1.50) 푑� 푒𝑒푒𝑒�푒 푓푓�푓� What do we mean by a change푑� in value? ℏThe electron changes or makes a transition from a state of some value to another of different value. Under the application of the external field (force) the electron goes from one state to the푘 next higher state and from that state to the next higher state and푘 thus keeps on making these 푘transitions as long as the external force is applied. In other words, the electron changes its푘 state under the application푘 of an external force. 푘 Referring to Figure (1.16),푘 an electron in the state will jump to the next higher state due to the external field. Then it will jump to a next higher state and keep on moving to higher states until it reaches the state . According to Equation (1.45), the velocity푃 of the electron in any given state푄 is proportional to the slope of the energy versus plot known푘 usually as – plot. In the upper푘 left half of the bandgap, the 퐴 plot is concave downwards which means that the slope (velocity) decreases as is increased. As the electron changes its state푘 progressively from 퐸 to 푘 , the electron velocity is decreasing i.e., the electron퐸 − 푘 is slowing down. Since the slope is zero at , the electron velocity becomes 푘 푃 퐴 zero. The value cannot increase to a value higher than since there is an energy discontinuity. 휋 퐴 However, 푘the state is equivalent to and therefore when푎 the electron is in the state = , it is 휋 equivalent to being 퐵in the state = 퐴 . Now, can increase at the rate from the state푘 푎 , where 휋 푑� = . Since is negative, the velocity− 푎 is negative.푘 The electron starts푑 �to move in the opposite퐵 휋 푑� direction and the magnitude of the velocity in the negative direction increases from zero when it is in 푘 − 푎 푑� ’ the state to a maximum value when it reaches the state P . Then it starts to slow down and reaches zero velocity in the state denoted by . As the force continues to act on the electron, the electron starts to move in퐵 the positive direction and its velocity increases until it reaches the state . When the electron goes from state to a higher푂 state, it starts to slow down and comes to rest when it reaches the state . This cycle keeps on repeating as long as the force continues to act on the푃 electron. The electron changes states in푃 the band in a푘 cyclic fashion. When a band is completely full, all the electrons are continually퐴 changing states in a cyclic fashion. At any given instant, there are as many electrons traveling in the positive direction as those in the negative direction. This is because electrons occupying states between = 0 (state ) and = (state A) correspond to electrons traveling in the positive 휋 direction while electrons occupying states between = 0 (state ) and = (state ) travel in the 푘 푂 푘 푎 휋 negative direction. The average velocity is therefore zero and hence a completely filled band does not 푘 푂 푘 − 푎 퐵 contribute to electrical conductivity.

When the electron is in a state between state and state , the effect of the external field is to decelerate the electron (i.e., as time progresses its velocity decreases). Normally, when we consider free electrons, the electron will accelerate under the action푃 of the external퐴 field. It is as though an electron in states between and has a negative mass. Similarly, the electron behavior in states between and is as though the electron has a negative mass. We define therefore, an effective 푃 퐴 푃′ 퐵 - 25 - mass *, which, when multiplied by the acceleration, gives the force on the electron. Since states in the upper half of the band correspond to deceleration of electrons, they have a negative effective mass while 풎those in the bottom half of the band have a positive effective mass.

The acceleration of the electron is given by

= = = = (1.51) 2 2 푑푑 푑 1 푑� 1 푑 퐸 1 푑 퐸 푑� 2 But the force -q is equal푎 to 푑� , Hence푑� �ℏ 푑�� ℏ 푑�푑� ℏ 푑푘 푑� 푑� ℇ ℏ 푑� = = ( ) = × (1.52) 2 2 2 1 푑 퐸 푑� 1 푑 퐸 1 푑 퐸 2 2 2 2 2 2 ℏ 푑푘 푑� ℏ 푑푘 ℏ 푑푘 The term relates acceleration푎 to� ℏthe applied� force just−푞 asℰ mass relates total퐹𝐹𝐹 force and 2 1 푑 퐸 acceleration.2 We2 thus define the effective mass * as ℏ 푑푘 풎 = (1.53) 2 ∗ ℏ 푑2퐸 푚 푑푘2 The curvature of the plot is positive in the lower half of the band and hence * is positive. 2 푑 퐸 The curvature is negative2 in the upper half of the band, and hence the effective mass is negative. 푑푘 퐸 푣� 푘 푚 What is the philosophical interpretation of the effective mass? The effective mass relates the acceleration of the electron to the external force. If one were to include all the forces internal and external, then indeed the relation between the acceleration and the total force will be through the real mass of the electron. We are relating only the external force to the acceleration and hence we need to define an effective mass.

The effective mass is a useful concept since it enables one to determine the behavior of electrons by treating them as though they are free from any internal force (i.e., constant potential energy) and replacing the true mass by the effective mass * in the expressions describing its dynamic behavior. In the 3-dimensional solid, the effective mass * is a second-order tensor which relates the external force along one direction to the acceleration along푚 another direction. For example, in a general case, unless the external force is applied along one of the푚 three principal axes of the crystal, the resulting acceleration will not be in the same direction as the force. However, for the purposes of our discussion, we will assume that the effectiveness * is a scalar quantity i.e., the external force and the acceleration are in the same direction. 푚

Conductors, Insulators and Semiconductors

Let us now consider a material in which the upper most occupied band is partially filled while all the lower bands are completely filled. There is no contribution to the electrical conductivity from the lower bands since they are full as stated earlier. On the other hand, the electrons in the partially filled (highest) band move continuously to higher states under the action of the eternal force. In a time interval t, the electrons would have changed to new states that differ in the value by 푘 ∆ - 26 - 푘

= t (1.54) 푒𝑒푒𝑒�푒 푓푓�푓� One would expect that the electrons∆푘 will be continuallyℏ ∆ changing states in this fashion but this does not occur. The electrons collide with scattering centers such as impurities and other atoms of the crystal that have been displaced from their normal atomic sites. The scattering causes them to give up their excess energy and return to lower energy states. This process is called a relaxation process. The electrons are never able to go to very high energy states due to the relaxation process. According to solid state theory, it is possible to assume that, under steady state conditions, the distribution of electrons in the presence of the electric field , is the same as what one would obtain if the field acted on the electrons for a time period . .is called the relaxation time. Referring to Figure (1.17), we find that under steady state conditions, allℇ the electrons have shifted to new states separatedℇ from the original � 풄 states by an amount 휏 흉 푘 = ( ) (1.55) 1 All the states below the dotted line∆푘 in the ℏenergy−푞 axisℇ have휏� equal number of electrons going in the positive direction as in the negative direction. Only those electrons lying above the dotted line travel in the positive direction without an equal number traveling in the opposite direction. There is thus a net flow of electrons in the positive direction. These electrons contribute to electrical conductivity. We can conclude therefore that only partially filled bands contribute to electrical conductivity.

Figure (1.17) Electron distribution in the presence of an electric field

We can distinguish three classes of materials: conductors, insulators and semiconductors. An insulator is one in which the top-most occupied band is completely filled as shown in Figure (1.18). The top most filled band is called the valence band since the electrons in this band are the ones which give rise to the chemical bonding. The next higher band is called the conduction band, since any electron excited to this band will give rise to electrical conductivity due to the fact that the conduction band

- 27 - becomes partially filled. The difference in energy between the top of the valence and the bottom of the conduction band is the energy gap. Insulators’ energy gap which is also called the bandgap, is very large and hence even at very high temperatures only a negligible number of electrons is excited into the conduction band and the material does not carry electrical current. On the other hand, if the bandgap is small, then, even at moderately low temperatures, electrons will be excited from the valence band into the conduction band and we will get electrical conductivity from both the partially filled valence and conduction bands. Such materials are called semiconductors. At absolute zero temperature, the semiconductor is an insulator since there are no electrons in the conduction band. However, as the temperature is raised, electrons are thermally excited from the valence band to the conduction band and the material starts to conduct electricity. Conductors are materials in which the highest occupied band is only partially filled. The band structure for conductors is shown in Figure (1.19). The metallic solids have this feature, i.e., the highest occupied band is partially filled, and hence is a good conductor.

Fig. (1.18): Band Structure of an insulator

Figure (1.19): Band structure of a conductor

- 28 -

Concept of Holes

We saw in the last section that a semiconductor has a small bandgap with its valence band completely filled while its conduction band is empty at very low temperatures. As the temperature is raised by heating the semiconductor, electrons are thermally excited from the valence band to the conduction band. This gives rise to a partially filled conduction band and the electrons in the conduction band can give rise to electrical conduction. Similarly, the remaining electrons in the valence band can give rise to electrical conduction since the valence band is now only partially filled. However, it is a very difficult task to determine the contribution of the large number of electrons in the valence band. A simple technique can be employed to calculate the contribution of the valence band electrons to electrical conduction.

For the sake of simplicity, let us assume that only one electron is excited from the valence band to the conduction band, leaving a vacant state in the valence band. The electron in the conduction band gives rise to electrical conduction as though it is a free electron with an effective mass, , appropriate to the state it occupies in the conduction band. The electron will occupy only a low energy � state in the bottom of the band and therefore its effective mass is positive. On the other hand푚 the vacant state in the valence band is in the top of the band and the effective mass is negative. An electron

with an effective mass , can occupy the vacant state in the valence band, and make the valence band fully occupied. Assume that we introduce two fictitious particles one with a charge – coulomb −푚� and a mass and the other with a charge coulomb and a mass . The particle with the negative charge and negative effective mass fills the vacant state making the valence band completely푞 filled. � � However, the− 푚particle with the positive charge푞 and positive effective mass푚 will now be free to move around the crystal and contribute to the electrical conductivity. This fictitious particle with a positive charge and a positive effective mass is called a hole. It behaves like a free particle and carries a charge. It therefore contributes to electrical conductivity. For every electron excited from the valence band, there is hole in the valence band. We obtain electrical conductivity due to both electrons and holes.

The number of electrons in the conduction band is small in comparison with the total number of states in the conduction band. Hence, the electrons occupy a narrow range of states in the bottom of the conduction band. We can therefore assume that all these electrons have the same effective mass and calculate the electrical conductivity due to them by using the free electron model (electrons in a box) except that we will use the effective mass in the place of the true mass. Similarly, all the holes are nearly at the top of the valence band and occupy a narrow range of states in the � valence band. Therefore they all have the same effective mass,푚 , corresponding to the curvature of the top of the valence band. These holes can be treated as holes in a box. Thus, we can calculate the � electrical conductivity of the semiconductor using free electron 푚and free hole models.

- 29 -

Chapter 2 Semiconductor Material Electronic Properties

Intrinsic Semiconductor

We will briefly review, in this chapter, the essential aspects of semiconductor physics that will be relevant to the discussion of device physics. A semiconductor is a material that has an electrical conductivity much less than a metallic conductor but is much more conducting than an insulator. Furthermore, its electrical conductivity is zero at absolute zero degree temperature and increases with temperature. The atoms in a semiconducting crystal are held together by covalent bonds. In order to understand what a covalent bond is consider two hydrogen atoms which are initially infinitely apart. On bring these two atoms closer, the electron in each atom is simultaneously subjected to the attractive forces of both the nuclei and the potential energy of each electron is lowered. The lowering of the potential energy with decrease in the distance between the two nuclei corresponds to an attractive force. The reason the two nuclei do not collapse on each other is because of a repulsive force that arises between the two nuclei (Coulomb repulsion) at very close inter-nuclear distance. The two nuclei are in stable equilibrium at a separation where the attractive force is exactly balanced by the repulsive force. The two electrons have a very high probability of being found midway between the two nuclei and therefore the electrons spend most of the time midway between the two nuclei. This will not violate Pauli’s exclusion principle because each atom has only one 1 electron. The electron in one atom can have opposite spin to the electron in the other atom and therefore these two electrons can simultaneously exist in the same place. It is as though each atom푠 has two electrons in the 1s orbit. However if we bring two helium atoms together, a covalent bond will not result because of the fact that Pauli’s exclusion principle is violated since each atom already has a filled 1 state (usually called 1 orbital). Therefore we can conclude that for covalent bonds to exist, each atom should have a partially- filled orbital. 푠 푠

The same covalent bond gives rise to the bonding force in many of the solids like carbon, nitrogen, and oxygen. The covalent bond is directional and therefore these atoms crystallize in structures which satisfy the directionality of the covalent bond. An atom with electrons outside the close-shell structure requires (8- ) electrons to complete the s and sub-shell. Since two neighboring atoms each contribute one electron to the bond between them, an atom with푁 valence electrons requires (8- ) nearest neighbors푁 and the crystal structure that results푝 has to provide (8- ) nearest neighbors which are symmetrical situated with respect to the first atom. For example,푁 carbon (diamond), silicon,푁 and germanium, each has two electrons and two electrons. Therefore푁 they require 4 nearest neighbors and this is obtained in the diamond structure. Figure (2.1) shows the crystal structure of diamond in which each atom is surrounded푠 by four covalently푝 bonded neighboring atoms. Silicon and Germanium also have similar crystalline structure.

It might be asked how Pauli’s exclusion principle is not violated when we have the -orbitals filled in each of the carbon atoms. This is explained by assuming a rearrangement of the states of the four electrons. One of the 2 electrons goes into a 2 state and thereby, we have half-filled푠 -orbitals and half filled -orbitals. Such a rearrangement is called hybridization. The four new orbitals thus 푠 푝 푠 푝 - 30 - obtained are ( ) hybrids. Figure (2.2) shows how the energy bands are formed by hybridization of and orbitals. 3 푠푝 푠 푝 The covalent crystals exhibit great hardness, low electrical conductivity at low temperature and in pure state. These usually have strong binding but some of the crystals (semiconductors) have weaker binding. These semiconductors exhibit electronic conductivity at high temperatures.

Figure (2.1): The diamond crystal structure

Figure (2.2): Energy band formation in diamond by hybridization of and orbitals Band Structure of Semiconductors 푠 푝

Some of the common semiconductor materials are made up of atoms of Group IV elements like germanium and silicon; these materials are referred to as elemental semiconductors. Semiconductor materials such as gallium arsenide or indium antimonide are made up of compounds of atoms of Group

- 31 -

III and Group V elements and therefore these materials are known as compound semiconductors. The energy band structure of some common semiconductors is given in Figure (2.3). In this figure, the energy is plotted as a function of the wave vector (proportional to the momentum) (in two crystallographic directions viz., [111] and [100]. In all the semiconductors, the states in the lower bands are completely filled with electrons and the states in the higher bands are empty at very low temperatures. The electrons in the lower bands are the ones that participate in covalent bonding and as such, they are not free to move around in the crystal. Since the electrons in the lower bands are the ones in the covalent bond, the lower bands are called the valence band. The states in the higher band are empty; therefore, there are no electrons to move around in the crystal. At very low temperatures there is no broken bond and the valence band is completely filled.

Figure (2.3): Energy band structure of some typical semiconductors

According to the principles of solid state physics, electrons in a completely filled band do not contribute to electrical conductivity and only the electrons in a partially filled band contribute to electrical conductivity.

When a covalent bond is broken and an electron is freed to move around in the solid, this electron is in one of the states in the higher bands. The missing electron in the broken covalent bond is represented by a vacant state in the valence band. Electrons in the higher bands are called free electrons since they are free to move around in the solid.

As the temperature of the crystal is increased, some electrons break loose from the covalent band and become free to move around in the crystal. That is, some electrons are thermally excited to the states in the higher bands. The higher bands (which were empty at low temperatures) become partially occupied at higher temperatures and thus, these electrons give rise to electrical conductivity. For this reason, the higher bands are called the conduction band.

At higher temperatures, the valence band (which was fully occupied by electrons at low temperatures) become partially occupied and hence, gives rise to electrical conductivity. When there is a vacancy in an otherwise completely filled valence band, the collective behavior of the rest of the electrons in the valence band is equivalent to that of a fictitious particle called the hole, which has a positive effective mass and a positive charge. The holes are free to move around in the solid and thus

- 32 - called “free holes.” An electron in the conduction band is free to move around in the crystal and a hole in the valence band is free to move around in the crystal as shown in Figure (2.4).

Figure (2.4): Generation of electron-hole pairs

In Figure (2.5), the lowest energy in the conduction band is denoted and the highest energy in the valence band is denoted . The minimum energy needed to excite an electron from the valence 퐸� band to the conduction band is obviously - and is called the energy gap, . We notice that there is 퐸� a difference between elemental semiconductors and compound semiconductors. In silicon and 퐸� 퐸� 퐸푔 germanium, the minimum energy in the conduction band, , occurs at a momentum value different from that at which the maximum energy in the conduction band, , occurs. In order to excite an � electron from the top of the valence band to the bottom of퐸 the conduction band, not only does energy � need to be given, but also some momentum needs to be imparted.퐸 A photon does not have momentum and can excite an electron in the valence band to a state with the same momentum in the conduction band. This is referred to as vertical transition. The minimum energy needed to excite an electron to the conduction band is = - . However in silicon and germanium, to excite an electron to the conduction band with the minimum energy, additional momentum also needs to be given. Hence, in 퐸푔 퐸� 퐸� these materials, a phonon is also involved to supply the needed change in the momentum between the initial and final states. For this reason, these materials are called indirect semiconductors. In gallium arsenide and other materials, in which the bottom of the conduction band, , also occurs at zero momentum values, it is possible to excite an electron optically from the top of the valence band to the � bottom of the conduction band directly (or vertically) without the need of a퐸 phonon. These materials are called direct semiconductors for this reason. The direct and indirect transitions are illustrated in Figure (2.5).

- 33 -

Figure (2.5): Illustration of direct and indirect gap semiconductors

We will now simplify our discussion by denoting the conduction and the valence band by two horizontal lines—one for and another for with the understanding that the conduction band states are above the line and the valence band states are below the line as illustrated in Figure (2.5). In � � this figure, is the bottom퐸 of the conduction퐸 band, and is the top of the valence band. The � � difference - 퐸 is the energy gap where there are no electron states.퐸 퐸� 퐸� In a퐸 �crystal,퐸� on which no external radiation is incident, or in which no externally applied electric fields are present, the excitation energy for the electron to jump from the valence band to the conduction band comes from the thermal energy. For this reason, this process is called the thermal excitation process and the electron is said to be thermally excited from the valence band to the conduction band. In a pure crystal, which contains no impurities, thermal excitation from the valence band (breaking up of covalent bonds) is the mechanism for generating electrons and holes. (Usually in literature, electrons in valence bands are referred to as valence electrons and electrons in conduction bands as conduction electrons.) Since in a thermal excitation process the number of electrons in the conduction band is a function of the intrinsic properties of the crystal such as the band-gap, effective mass, etc., a semiconducting material in which free electrons (and holes) are only obtained by thermal excitation from the valence band to the conduction band is called an intrinsic semiconductor. It is customary to denote the number of electrons per unit volume (concentration or density) in the conduction band as n and the number of holes per unit volume in the valence band (concentration or density as p. In an intrinsic semiconducting material

= (2.1) i where the subscript is purposely used to denote푛 푖that the푝푖 material is intrinsic. The main requirement for a crystal to be a semiconductor is that the bonds between the atoms of the crystal should be covalent and also comparatively a small amount of energy should be needed to break up the bond. Table (2.I) gives the energy gap of some of the typical Group IV elemental semiconductors and Group III-V intermetallic compound semiconductors.

- 34 -

Material Chemical Symbol Energy Gap (eV) Diamond (Carbon) C 5.3 Silicon Si 1.17 Germanium Ge 0.72 Gallium Phosphide GaP 2.25 Gallium Arsenide GaAs 1.34 Indium Phosphide InP 1.27 Indium Antimonide InSb 0.18

Table (2.1): Energy Gap in typical semiconductdors

Figure (2.6): An electron in a state with energy E in the conduction band has a kinetic energy equal to . The kinetic energy is small compared with the band gap energy . This figure is not drawn to scale 퐸 − 퐸� 퐸� − 퐸� Let us now calculate the density (concentration) of electrons in the conduction band. When an electron is in a state which is at the bottom of the conduction band i.e., a state with energy , it has no velocity and therefore no kinetic energy. Hence, we can say that the electron in this state has only � potential energy. Therefore the potential energy of the electron is taken as equal to . The퐸 kinetic energy of the electron in a state in the conduction band with energy is therefore equal to as � shown in Figure (2.6). We have already seen that the conduction electrons can be considered퐸 nearly � free, and therefore we can apply the techniques of free electron theory퐸 i.e., treat them as electrons퐸 − 퐸 in a box. The density of states in the conduction band close to the bottom of the band i.e., close to , can be approximated by the density of states of free electrons or electrons in a box with one assumption. � The mass of the electron is to be replaced by the effective mass determined by the curvature of퐸 the conduction band near . Since only states that are nearly at the bottom of the conduction band are occupied, treating these electrons as free electrons subject to the assumption of the same effective � mass for all electrons is퐸 valid. The momentum of the conduction electrons can be readily written from the box model as 푝 = 2 ( ) (2.2)

The density of states for electrons in the푝 conduction� 푚 band� 퐸 is− then퐸� given by

- 35 -

( ) ( ) = 3 ( ) 1 (2.3) 푐 2 4 휋 푉 2푚 2 3 where is the volume of the semiconduct푍� 퐸 or solid ℎand the subscript퐸 − 퐸� c in ( ) has been used to denote the density of states in the conduction band. The density of states for electrons is shown in Figure (2.7). � 푍� 퐸 The number of electrons , with energy between and + per unit volume of the crystal, is then given by �푛퐸 퐸 퐸 𝑑 ( ) ( ) ( ) ( ) = = 3 1 (2.4) 푍푐 퐸 푓 퐸 푑� 4휋 2푚(푐 2 퐸−퐸 푐 )2푑� 퐸 3 퐸−퐸퐹 푉 ℎ 푘푘 Where ( ) is the Fermi�푛 function. 푒 + 1

푓The퐸 concentration (density) of free electrons can now be obtained by integrating Equation (2.4) between the limits and where is the top of the conduction band i.e., the maximum energy of the conduction band. 퐸� 퐸푡𝑡 퐸푡𝑡 = 퐸푡푡� 푛 � �푛퐸 Typically, the concentration of free electrons퐸푐 is very low; hence the states in the lower portion of the conduction band will be occupied. Two further assumptions can then be made: one, can be assumed to be infinite i.e., the upper limit of the integral can be assumed to be ∞ instead of 퐸푡𝑡 without any appreciable error, and two, the effective mass is assumed to be the same for all the 퐸푡𝑡 electrons since the curvature of the band is the same for all the states in the bottom of the conduction � band. Hence can be considered constant and taken outside푚 the integral. Under these conditions, the density of electrons is given by 푚� ∞ ( ) = (2 ) 1 3 (2.5) 푐 2 4휋 2 (퐸−퐸 푑 �) 퐸−퐸 3 � 푐 퐹 ℎ 퐸 푘푘 This integral has to be evaluated푛 numerically as푚 it stands.∫ However,푒 + 1some simplifications can be made. Let us assume that >> i.e., the Fermi energy lies in the band gap at least a few below the conduction band minimum . Since the lower value that can have is , >> , and hence - � 퐹 >> . The equation퐸 for− the퐸 electron푘� density can now be simplified by neglecting the unity푘� term in the � � � 퐹 denominator to yield 퐸 퐸 퐸 퐸 퐸 퐸 퐸 푘� 푛 ∞ = (2 ) ( ) (2.6) 3 1 퐸퐹−퐸 4휋 2 2 � 푘푘 � 3 � 푐 � Let us define a new variable푛 =ℎ 푚 . Then ∫the퐸 above퐸 − integral퐸 푒 equation𝑑 can be transformed as 퐸−퐸푐 휂 푘푘 ∞ ( ) = ( ) 1 1 퐸퐹−퐸 3 퐸퐹−퐸푐 퐸푐−퐸 퐶 2 2 � 푘푘 � 2 � 푘푘 � 퐸−퐸 � 푘푘 � 퐸 ∫퐸푐 퐸 − 퐸� 푒 𝑑 푘� 푒 ∫ � 푘푘 � 푒 � �푘푘� - 36 -

= (2.7) 3 퐸퐹−퐸푐 1 2 � 푘푘 � 2 −휂 Hence 푘푇 푒 ∫ 휂 푒 𝑑 ∞ = (2 ) ( ) (2.8) 3 퐸퐹−퐸푐 3 1 4휋 2 � 푘푘 � 2 2 −휂 3 � 0 The integral in the right hand푛 sideℎ of the푚 above푒 equation is푘 just� equal∫ 휂to 푒 . The𝑑 equation for can then 휋 be simplified to √ 2 푛

= 2 3 (2.9) 퐸푐−퐸퐹 푐 2 2 휋푚 푘푘 −� 푘푘 � 2 The above equation is usually written푛 as � ℎ � 푒

= (2.10) 퐸푐−퐸퐹 −� 푘푘 � where 푛 푁�푒

= 2 3 (2.11) 2 휋푚푐푘푘 2 2 This quantity is called the effective푁� density� of ℎstates� in the conduction band. The reason why is called so is as follows:� The factor in the expression for , gives the probability that a 푁 퐸푐−퐸퐹 state located at level will be occupied,− since� 푘푘 it� can be shown that when , 푁� 푒 푛 퐸� 퐸� − 퐸퐹 ≫ 푘� ( ) = (2.12) 퐸푐−퐸퐹 1 −� 푘푘 � � 퐸푐−퐸퐹 푘푘 If there are states located at 푓 =퐸 , then푒 we will+ 1 have≈ 푒for the same expression as derived above.

We 푁can� similarly calculate퐸 the 퐸number� of holes in the valence푛 band per unit volume of the crystal. While we can calculate the density of states in the valence band by methods similar to what we did for the conduction band, we must use 1 ( ) as the probability that a hole will exist in a state of energy , which is the probability that an electron will not exist in that state. � − 푓 퐸 � We퐸 will consider the holes as free particles in a box. A hole at the top of the valence band has no velocity and therefore has no kinetic energy. As we go to lower energy states in the valence band, the velocity of the hole increases and therefore the kinetic energy increases. It is therefore possible to write the kinetic energy of the hole as , where is the energy of the top of the valence band. The density of states expression is the same as for electrons except for kinetic energy we write and is � � given by 퐸 − 퐸 퐸 퐸� − 퐸 ( ) ( ) = 3 ( ) 1 (2.13) 푣 2 4휋 푉 2푚 2 3 The density of states in the valence푍 �band퐸 is plotted inℎ Figure (2.퐸8�). − 퐸 - 37 -

We can calculate as we did for the electrons, the number of holes per unit volume of the crystal. We again assume that the Fermi Energy is at least a few above , the valence band maximum. Then proceeding exactly similar to the case of electrons, we can show 푘� 퐸�

= 2 3 = (2.14) 퐸퐹−퐸푣 퐸퐹−퐸푣 푣 2 2 휋푚 푘푘 −� 푘푘 � −� 푘푘 � 2 where is the effective 푝density of� statesℎ in the� valence푒 band, and푁� is푒 defined as � 푁 = 2 3 (2.15) 2 휋푚푣푘푘 2 2 When the Fermi energy is a few푁� below� ℎ and also� a few above , the material is called a non-degenerate semiconductor. When this condition is not satisfied, the material is called a 퐹 � � degenerate semiconductor. 퐸 푘� 퐸 푘� 퐸

We denote the Fermi energy in an intrinsic material . Since = according to Equation (2.1), we can write 퐸푖 푛푖 푝푖 = (2.16) 퐸푐−퐸푖 퐸푖−퐸푣 −� 푘푘 � −� 푘푘 � And also 푁�푒 푁�푒

= × = = = (2.17) 퐸푐−퐸푖 퐸푖−퐸푣 퐸푐−퐸푣 퐸푔 2 −� 푘푘 � −� 푘푘 � −� 푘푘 � −푘푘 This푛 is푖 an important푛푖 푝푖 result.푁 �This푁� shows푒 that the푒 electron density푁� 푁 (as�푒 well as the hole푁 density)� 푁�푒 in an intrinsic material depends on the width of the energy gap (band gap). This result helps us to determine the location of , the Fermi energy in an intrinsic material. If we multiply both sides of Equation (2.16) by and rearrange terms we get 퐸푖 퐸푖 푘푘 = (2.18) 2퐸푖 퐸푐+퐸푣 푣 푘푘 푁 � 푘푘 � 푐 Taking logarithms of both sides and multiplying푒 by 푁 , 푒we get 푘푘 2 = ln + (2.19) 푘푘 푁푣 퐸푐+퐸푣 푖 푐 We notice first that represents퐸 2an energy�푁 � level exactly2 in the middle of the band gap. 퐸푐+퐸푣 Secondly, if the effective mass of the electrons and that of the holes are nearly equal, the first term in 2 the right hand side of the above equation is very nearly equal to 0. Hence we conclude that , the Fermi energy in the intrinsic material, is nearly at the middle of the band gap. 퐸푖

- 38 -

Example

Let us now calculate the effective density of states in a semiconductor at room temperature ( = 300 0K). Let us assume that the effective mass of the electron is the same as , the mass of an electron in vacuum. 푇 푚� 푚0 = 9.11 × 10 −31 � 0 푚 =≈ 1푚.38 × 10 푘𝑘 −23 퐽퐽퐽퐽� 푘 퐾 = 6.63 × 10 𝑑𝑑𝑑 −34 퐽퐽퐽퐽� ℎ 퐾 Substituting these values in the expression for we get 𝑑𝑑𝑑 푁� 2 × × 9.11 10 × 1.38 10 × 300 = 2 3 (6.63− 31× 10 ) −23 2 휋 푥 푥 � −34 2 푁 =�2.51 × 10 = 2.51 × 10 � 25 −3 19 −3 If we assume similarly that the effective mass푚𝑚𝑚 of the hole is the same as that𝑐 of the electron in vacuum, we will get the same value for . 푚� 푁�

We can now determine , the density of electrons or holes, in the intrinsic semiconductor. By taking the square root of both sides of Equation (2.17) we can write 푛푖

= (2.20) 퐸푔 − 2푘푘 In order to express the carrier density푛 at푖 any� temperature푁�푁� 푒 in terms of the number for the density of states that were calculated for =300 0 , we rearrange the terms to yield

푇 퐾 = 2.51 × 10 3 3 (2.21) 퐸푔 푐 푣 4 2 19 푚 푚 � −2푘푘 −3 푖 2 Where is the mass of푛 an electron in vacuum.� This푚0 expression� �300 �can푒 more readily𝑐 be used to calculate , the intrinsic carrier density at any temperature. 푚0 푛푖 Example

Let us now calculate , the intrinsic electron concentration in silicon at room temperature. Substituting the following values 푛푖 = 0.29

� 0 푚 - 39 - 푚

= 0.57

( =푚300� ) =푚 10.12 We get for 퐸푔 푇 퐾 푒푒 푛푖 2 × 300 × 1.38 × 10 × 9.1 10 = 4 3 × (0.29 × 0.57) (6.63 × 10−23) −31 3 휋 푥 2 푛푖 � 푥 � . × . × −34 2 � × × × . × −19 1 12 1 6 10 −� −23� 2 300 1 38 10 푒 = 2.54 × 10 9 −3 We must point out that the above calculation is in𝑐 error since it does not take into account the fact that there are six conduction bands in silicon and hence the effective density of states that we calculated must be multiplied by a factor 6. Hence the value for has to be multiplied by a factor 6. 푁� For our purpose, we approximate as 푛푖 √ 푛푖 = 1.00 × 10 10 −3 We will use this value for for silicon푛 푖at room temperature throughout𝑐 this book. 푛푖

Figure (2.7): Density of states for electrons in the conduction band

Figure (2.8): Density of states for holes in the valence band

- 40 -

Extrinsic Semiconductor

The properties of the intrinsic (i.e., pure) semiconductor can be altered by adding a small amount of impurity atoms to occupy the lattice sites originally occupied by the atoms of the intrinsic material. The semiconductor is then said to be extrinsic, since its properties depend on the externally added impurities: the impurities are said to be substitutionally added since they substitute the original atom in the lattice site.

Consider, for example, an element belonging to Group III in the Periodic Table which is trivalent, such as boron, aluminum, gallium or indium. Such an atom has three valence electrons and so, when it substitutes for a tetravalent atom of the semiconductor, the site into which it goes has only three of the four bonds completed and has the fourth bond incomplete. The boron atom is in the neutral state. This is illustrated in Figure (2.9 A), with boron as the impurity atom. As electron from one of the covalent bonds of a nearby silicon atom can break free from its bond and jump into the incomplete bond of the boron atom. When that happens, all the four bonds surrounding the boron atom are completed, and an electron ( a negative charge) is bound to the boron atom as shown in Figure (2.9 A). The boron atom is now negatively ionized as shown in Figure (2.9 B). In this process, we have a bound electron (at the site of the boron atom) and a broken bond in a nearby semiconductor atom. The broken bond represents a free hole. Since the boron atom (and all other group III elements) accepts an electron, releasing a hole in the valence band, boron and other group III elements are called acceptor impurities. When the acceptor atom accepts an electron, it becomes ionized. Each ionized acceptor atom gives rise to a free hole.

If instead of a group III element, we substitute an atom of the elements of group V which is penta-valent, such as phosphorous, we will then have one electron more than the number needed to complete the four covalent bonds (tetra-valent bond) as shown in Figure (2.10 A). This fifth electron requires very little energy to get freed from the parent phosphorous atom; when it is freed, we have a free electron in the crystal and an ionized phosphorous atom as shown in Figure (2.10 B). We can think of the removal of the fifth electron as a positive charge bound to the site of the phosphorous atom. The bound positive charge in the phosphorous atom donates a free electron. It is called a donor impurity. A donor atom donates an electron when it is ionized. Each ionized donor atom gives rise to a free electron.

In Table (2.2), we give the ionization energy (energy required to free an electron from the donor atom, or to free a hole from the acceptor atom) for some of the common impurity atoms in germanium and silicon.

Table (2.2 A): Ionization Energy in eV of Donor Atoms

Impurity Si Ge P 0.044 0.012 4 As 0.049 0.013 Sb 0.039 0.096 Bi 0.067

- 41 -

Table (2.2 B): Ionization Energy in eV of Acceptor Atoms

Impurity Si Ge B 0.045 0.01 4 Al 0.057 0.01 Ga 0.067 0.11 In 0.11

Figure (2.9): Substitution of trivalent boron for one of the tetravalent silicon: A)Boron atom is neutral and an electron is missing from the covalent bond. B) An electron breaks loose from the covalent bond of a neighboring silicon atom and jumps into the covalent bond of the boron atom in which an electron was missing. The boron atom is now negatively charged. The broken covalent bond in the neighboring silicon atom can be thought of as a free hole.

From Table (2.2), it can be seen that the ionization energy is approximately of the same order for all the different impurities in a given semiconductor. In the case of a neutral donor atom, a single (the fifth) electron is orbiting around a positive charge in a space characterized by the dielectric constant of the semiconductor. It is like a hydrogen atom, which can be described by the Bohr model. We can therefore substitute the value of permittivity of the semiconductor for the permittivity of free space in the expression for the ionization energy of the hydrogen atom to obtain the ionization energy of the donor atom. We can treat the acceptor atom similarly, where we assume a positive charge to be revolving around a fixed negative charge. It is therefore usual to assume that the energy level of the acceptor atoms is typically about 0.05 above the top of the valence band, and that the energy level

푒푒

- 42 - of the donor atoms is typically about 0.05 below the bottom of the conduction band in silicon. This is illustrated in Figure (2.11). 푒푒 In a neutral material, the charge neutrality condition should hold. The magnitude of negative charge per unit volume should be equal to the amount of positive charge per unit volume. If in a material there are donor atoms in unit volume of the crystal, the amount of positive charge per unit volume is equal to 푁퐷 ( + ) + and the magnitude of negative charge per unit푞 volume푝 푁 is퐷 equal to

Equating the two we get 푞� = + (2.23) + Where is the number of positively ionized푛 donor푝 impurities푁퐷 per unit volume. We have free electrons due to two+ reasons: (1) thermal excitation of the valence electrons, and (2) donor ionization. In the 퐷 above equation,푁 the first term on the right hand side represents the density of electrons due to thermal excitation from the valence band and the second term represents the density of electrons due to ionization of donor atoms.

Since both electrons and holes carry a charge, they are called charge carriers or, more simply, carriers. In a material in which we have donor type of impurities we have more electrons than holes per unit volume since, from the above equation, it can be inferred that n is larger than p. For this reason, the electrons are called majority carriers, and the holes are called minority carriers. For the same reason, this material is called a n-type semiconductor.

If we have a material with only acceptor type of impurities of density , then charge neutrality requires 푁퐴 = + (2.24) − where is the number of negatively ionized푝 acceptor푛 atoms푁퐴 per unit volume. Free holes are created due to two− reasons: (1) thermal excitation of valence electrons, and (2) acceptor ionization. As before, 퐴 the first푁 term on the right hand side of the above equation represents the density of holes due to thermal excitation and the second term the density of holes due to ionization of acceptor atoms. Holes are majority carriers in this material, and electrons are minority carriers. This type of material is called p- type material. To denote the carrier density and other parameters in an extrinsic material, we use a subscript p in p-type material, and a subscript n in n-type material.

In practice, the purest semiconductors prepared in the laboratory have an impurity ration of 1 in 1010 or 1011. This corresponds to approximately 1012 or 1011 impurity atoms per cubic centimeter. Therefore, the acceptor and donor impurities that are added should be greater than 1012 per cubic centimeter to sensibly alter the properties of this material.

- 43 -

Figure (2.10): Substitution of a pentavalent phosphorus atom for one of the tetravalent silicon: a) Phosphorus atom is neutral and there is an extra electron orbiting around the phosphorous atom similar to the electron in the hydrogen atom b) The orbiting electron breaks loose due to the imparting of a small amount of energy. The phosphorous atom is positively ionized. The electron is now free to move around in the crystal and is called the free electron.

Figure (2.11): Impurity energy levels in silicon and generation of free carriers.

- 44 -

Carrier densities in extrinsic semiconductors

The expression for the thermal equilibrium density of free electrons and free holes in an extrinsic semiconductor is the same as what we derived earlier for an intrinsic semiconductor, since in these derivations we did not use any intrinsic property of the material. Denoting푛 the thermal equilibrium푝 carrier densities with a subscript 0,

= = (2.25) 퐸푐−퐸푖 퐸푖−퐸푣 −� 푘푘 � −� 푘푘 � as long as the Fermi energy,푛0 푁, is� 푒at least a few 푎 �below푎 푝 0 and푁 a� few푒 above i.e., the material is non-degenerate. 퐸퐹 푘� 퐸� 푘� 퐸� When the impurity concentration , is approximately equal to or larger than in an -type material, or when the impurity concentration is equal to or greater than in a -type material, 퐷 � is closer to or by less than a few , 푁and the material is said to be degenerate material.푁 In푛 these 퐴 푉 퐹 cases the majority carrier density is given by Equation푁 (2.5) for electrons or a푁 similar푝 one for holes and퐸 � � not by the simplified퐸 퐸 expressions in Equation푘� (2.25).

The expression for the center densities in Equation (2.25) is the same for both the intrinsic and the non-degenerate extrinsic material. The difference between the values of the carrier concentrations in intrinsic and extrinsic semiconductors arises due to the difference in the positions of the Fermi level in the two cases. Also we note, as before, that in the non-degenerate semiconductor,

= = 퐸푔 (2.26) −�푘푘� 2 This relationship is called the law푛0 of푝 mass0 action푁�푁� which푒 is valid푛 even푖 in the extrinsic material as long as it is non-degenerate.

While and are the impurity concentrations in and type materials respectively, and are respectively the ionized impurity densities. The probability that an acceptor atom will be+ 퐷 퐴 퐷 ionized− is the probability푁 푁 that an electron will occupy a state with푛 energy푝 equal to and this is given푁 by 퐴 the Fermi푁 Dirac statistics. Therefore1, 퐸퐴 = (2.27) − 푁퐴 퐴 퐸퐴−퐸퐹 푘푘 Similarly the probability that donor atom푁 will be ionized1+푒 is equal to the probability that an electron will not occupy a state with energy equal to the donor energy. Hence2,

= 1 (2.28) 1 + 퐸 −퐸 퐷 퐷 퐷 퐹+1 푘푘 푁 푁 � − 푒 �

1 In reality the expression for should include a factor in the denominator called the degeneracy factor but we will ignor that here for the purpose− of simplicity. 퐴 2 Again we have neglected the푁 degeneracy factor but it will not introduce any serious error.

- 45 -

It is possible to have both types of impurities simultaneously present in the same region of the material. Then the material becomes or dependent on the type of impurity which is higher in density. For example, if the donor density is larger than the acceptor density, the material will be type and if the opposite is true, the material푛 will푝 be type. If the densities of acceptor and donor atoms are equal, then the electron and hole densities will be equal and the material is like an intrinsic material.푛 This type of material is called a compensated material.푝

In the fabrication of semiconductor devices, regions of and type are formed in the same wafer by first starting with a particular type of semiconductor with the lightest doping. Then the wafer is covered with an oxide layer by oxidizing the wafer in a furnace at푛 high푝 temperature. Next windows are cut in the oxide layer and impurities of the opposite type to the type of the starting wafer are diffused. If the wafer initially had an impurity concentration of acceptors, then donor type of impurities will be diffused through the windows in the oxide such that in a region close to the window there are more donor atoms than acceptor atoms i.e., > . This region will therefore have a net donor type of impurities of density - , and hence will be -type. The majority carrier density will be equal to the 퐷 퐴 net donor density. We have to use the 푁net donor푁 density in the calculation of the carrier densities 퐷 퐴 instead of the total donor푁 푁 density. If on the other푛 hand, is greater than , the material will be - type and we will use the net acceptor density - in the calculation of the carrier density instead of 퐴 퐷 the total acceptor density. 푁 푁 푝 푁퐴 푁퐷 n-type material:

Let us consider a non-degenerate n-type semiconductor. Using the subscript for the -type material and the subscript 0 for the thermal equilibrium, we have 푛 푛 = + + Hence 푛푛0 푁퐷 푝푛0

= 1 + (2.29) 퐸푐−퐸푖 퐸퐹−퐸푣 −� 푘푘 � 1 −� 푘푘 � 퐸 −퐸 � 퐷 퐷 퐹+1 � 푘푘 The above expression is 푁difficult푒 to solve, and푁 has� to− be푒 solved numerically.� 푁 푒 At absolute zero temperature, no donor atom is ionized and no free carriers exist. Therefore, = 0. The Fermi energy lies halfway between and , since a slight increase in the temperature will give rise to an equal 푛0 number of electrons in the conduction band and ionized donor atoms. As the푛 temperature is gradually � 퐷 raised, more and more퐸 donor퐸 atoms are ionized, and so the Fermi level moves downward. When the Fermi level is between and , the material is said to be in the freeze-out region. In this range of temperatures, only some of the donor atoms are ionized. This is referred to as partial ionization of the � 퐷 donor atoms. When the퐸 Fermi 퐸energy is close to , such that the electrons in the conduction band are essentially due to ionization of donor atoms, the material is said to be in the extrinsic region. When the 퐷 Fermi level goes way below as the temperature퐸 is increased further, the material is said to be in the exhaustion region, since in this case the donor energy lies above the Fermi energy and all the donor 퐷 atoms are ionized. No increase퐸 in electron density can arise due to further ionization of donor atoms, since all the donor atoms are already ionized. Any further increase of electron density can arise only due to thermal excitation of the valence electrons and this happens when the temperature is increased to a

- 46 - much higher value. The Fermi level decreases with temperature and asymptotically approaches the intrinsic Fermi level at high temperature ( becomes equal to ). At high temperatures the material becomes intrinsic. It can be seen, then that if we want to take advantage of the properties 푖 푛0 푛0 brought about by external퐸 impurities in the semiconductor,푛 we must푝 limit the operating temperature so that the material is used in the extrinsic, or exhaustion range. The variation of Fermi level with temperature is shown in the upper half of Figure (2.12) for an -type material for various concentrations of impurity atoms. The effect of increasing the concentration of impurities is to extend the extrinsic and exhaustion range of temperatures to higher values. 푛

The minority carrier densities in a semiconductor can be obtained from the law of mass action (Equation (2.26)). In an -type material the thermal equilibrium minority carrier density is therefore obtained as 푛 = 2 (2.30) 푛푖 푝푛0 푛푛0

Fig. (2.12): Variation of Fermi Level with Temperature for - and -type semiconductor

푛 푝 From Equation (2.23) (charge neutrality expression in an -type material), we get

= + = =푛 + 2 (2.31) + + 푛푖 Rearranging terms we obtain푛푛0 the following푁퐷 푝 quadratic푛0 푛푛 equation:0 푁퐷 푛푛0

= 0 (2.32) 2 + 2 The solution to this equation is푛 푛0 − 푛푛0푁퐷 − 푛푖

- 47 -

= + + 2 2 (2.33) 푁퐷+��푁퐷� +4 푛푖 This solution to the quadratic푛푛0 equation corresponding2 to the negative sign before the radical is neglected since it will not give physically meaningful solution. It can be seen that this expression for approaches and in the two limits of small and large respectively. When >> , as in -type 푛푛0 semiconductors at moderate+ and low temperatures, + + 푛푖 푁퐷 푁퐷 푁퐷 푛푖 푛 (2.34) + and 푛푛0 ≈ 푁퐷 = = 2 2 (2.35) 푛푖 푛푖 + When >> as at very high temperatures, 푝푛0 푛푛0 푁퐷 + 푛푖 푁퐷

푛푛=0 ≈ 푛=푖 2 (2.36) 푛푖 푛0 푖 푝 Example푛푛0 푛

Let us calculate the carrier densities in a -type material in which the donor density is 10 cm-3. Assume all the donor atoms are ionized at room temperature. 14 푛

× -3 = + + 2 2 = 10 cm 푁퐷+��푁퐷� +4 푛푖 14 28 20 10 +√10 +4 10 14 and 푛푛0 2 2 ≈

-3 = = = 10 2 20 cm 푛푖 10 6 14 푝푛0 푛푛0 10 -type material:

푝 Let us now consider a -type material. Using the subscripts for the type material and 0 for thermal equilibrium in the charge neutrality equation (Equation (2.24)) we get 푝 푝 푝 = + (2.37) − where is the density of ionized acceptors.푝푝0 But 푁퐴 푛푝0 − 푁퐴 = (2.38)

1 − 퐸 −퐸 퐴 퐴 퐴 퐹+ 1 푘푘 푁 푁 �푒 � - 48 -

Therefore

= = + (2.39) 퐸퐹−퐸푣 퐸푐−퐸퐹 퐴 −� 푘푘 � 푁 −� 푘푘 � 퐸 −퐸 푝0 � 퐴 퐹+ 1 � 푘푘 This equation has to be solved푝 numerically푁 푒 as before �and푒 the variation� 푁 of the푒 Fermi level with temperature is plotted in the lower half of Figure (2.12). For convenience we are plotting the Fermi energy variation with temperature for both and type materials in the same figure. In the pure -type material, the Fermi level lies between and at low temperatures, and approaches the intrinsic Fermi level as the temperature is increased. 푛The -푝type material is also characterized by freeze-out,푝 퐴 � extrinsic, exhaustion and intrinsic ranges퐸 of temperatures.퐸 푝 Using the law of mass action, transforming Equation (2.37) into a quadratic equation and solving, we get

= − − 2 2 (2.40) 푁퐴 +��푁퐴 � +4 푛푖 As before, if >> as in a -type semiconductor푝푝0 at moderate2 and low temperatures, − 푁 퐴 푛 푖 푝 (2.41) − and 푝푝0 ≈ 푁퐴 = 2 2 (2.42) 푛푖 푛푖 푝0 − When as at very high temperatures,푛 푝푝0 ≈ 푁퐴 − 푛 푖 ≫ 푁 퐴 (2.43)

and 푝푝0 ≈ 푛푖 = = 2 (2.44) 푛푖 푝0 푖 푛 푝푝0 푛 Determination of the Fermi Energy

We will now derive an expression for the Fermi energy location in the band gap for an extrinsic semiconductor. As stated before the material is assumed to be a non-degenerated semiconductor. n-type material.

In the non-degenerate n-type semiconductor, the majority carrier density can also be expressed from Equation (2.25) as

= 퐸푐−퐸퐹 −� � 푘푘 푛푛0 푁�푒 - 49 -

Dividing both sides by and taking logarithms of each side and rearranging terms, we get

푁� = + (2.45) 푛푛0 Since 퐸퐹 푘� 𝑙 � 푁푐 � 퐸� + 푛0 퐷 푛 ≈ 푁 = + + (2.46) 푁퐷 퐸퐹 푘� 𝑙 �푁푐 � 퐸� Since is less than in the non-degenerate semiconductor, the term ln + is negative. Hence is + 푁퐷 퐷 � 푁푐 퐹 below푁 by an amount푁 equal to + . This is illustrated in Figure (2.13� A).� At room temperature퐸 푁퐷 all the donor퐶 atoms can be assumed to be ionized and hence = . 퐸 푘� 𝑙 �푁푐 � + Therefore, 푁퐷 푁퐷

= + (2.47) 푁퐷 퐸퐹 푘� 𝑙 �푁푐 � 퐸퐶 p-type material

The majority carrier density in the non-degenerate -type semiconductor can be expressed using Equation (2.25) as 푝 = 퐸푣−퐸퐹 � � 푘푘 Proceeding as we did in the case of the -푝type푝0 material,푁�푒 we get 푛 = + (2.48) 푁푉 퐹 � 퐸 푘� 𝑙 �푝푝0� 퐸

But = . Hence, − 푝0 퐴 푝 푁 = + (2.49) 푁푣 − The Fermi Energy is above the valence퐸 band퐹 in푘 �a 𝑙-type�푁 퐴semiconductor� 퐸� by an amount equal to

as illustrated in Figure (2.13 B). At room푝 temperature all the acceptor atoms are assumed to 푁푉 − be푘� ionized.𝑙 �푁퐴 � Hence, = − Therefore, 푁퐴 푁퐴

- 50 -

= + (2.50) 푁푣 퐸퐹 푘� 𝑙 �푁퐴� 퐸�

Figure (2.13): Location of the Fermi energy in an extrinsic semiconductor: (A) -type material (B) -type material. 푛 푝 Example

Let us now determine the location of the Fermi energy in an -type material with a net impurity concentration of = 5 × 10 cm-3 at room temperature. Let us assume that the effective mass of the 푛 electron is the same as that of 15the electron in vacuum. We had calculated in an earlier example that 퐷 under these assumptions푁 is equal to 2.51 x 1019 cm-3. All the donor atoms can be assumed to be ionized 퐶 at room temperature. 푁

-23 = 1.38 x 10 x 300 Joules

푘 � = 0.02589 eV

= + 퐷 퐹 푁 퐶 퐸 푘� 𝑙5�× 10� � 퐸 = 0.0258 ln 푁 + 2.51 × 1015 19 퐶 = 0�.22 + � 퐸

− 푒푒 퐸퐶 - 51 -

The Fermi energy is located in the band gap 0.22 eV below the conduction band.

Carrier Densities in terms of

Starting from the expression퐸 for푖 the thermal equilibrium carrier densities and in terms of the Fermi energy as given in Equation (2.25), it is easy to show (left as a homework exercise) that 푛 푝

= (2.51) 퐸퐹−퐸푖 � 푘푘 � and 푛푥0 푛푖푒

= (2.52) 퐸퐹−퐸푖 −� 푘푘 � where stands for the intrinsic Fermi energy푝푥0 and the푛푖 푒subscript stands for in a -type material and for in a -type material. Therefore 퐸푖 푥 푛 푛 푝 푝 = + (2.53) 푛푥0 and 퐸퐹 푘� 𝑙 � 푛푖 � 퐸푖

= + (2.54) 푛푖 In an -type material, 퐸퐹 푘푇 𝑙 �푝푥0� 퐸푖

푛 = = (2.55) + Hence 푛푥0 푛푛0 푁퐷 = + = + + (2.56) 푛푛0 푁퐷 It is trivial to show that starting퐸퐹 from푘 �Equation𝑙 � 푛 (2.54)푖 � and퐸푖 putting푘� 𝑙=� 푛,푖 we� will퐸 푖get the same result. Similarly in a -type material, 푥 푛푖 푝 = + = + = − (2.57) 푛푖 푛푖 푁퐴 퐹 푖 − 푖 푖 The location of the퐸 Fermi푘 energy� 𝑙 � with푝푝0� respect퐸 to 푘 is� illustrated𝑙 �푁퐴 � in Figure퐸 (2.14퐸 − A)푘 for� 𝑙 type� 푛 푖material� and in Figure (2.14 B) for type material. 퐸푖 푛 푝 Example

Let us determine the location of the Fermi energy with respect to in a -type semiconductor with = 10 cm-3 at room temperature. At room temperature it is reasonable to assume that all the 퐸푖 푝 acceptor 15atoms are ionized, i.e., . 퐴 푁 − 푁퐴 ≈ 푁퐴 - 52 -

= + 푖 퐹 푛 푖 퐸 푘� 𝑙 �10퐴� 퐸 = 0.0259 푁 + 1010 15 푖 = 0.297𝑙 � +� 퐸

The Fermi energy is 0.2976 eV below in the− bandgap푒푒. This is퐸 illustrated푖 in Figure (2.14). 퐸푖

Figure (2.14): Fermi energy in terms of the intrinsic Fermi energy: (A) -type material (B) -type material

푛 푝 Electric Current in Semiconductors

We will now study the flow of electric current in semiconductors. According to thermodynamic treatments, the Fermi energy is equal to the sum of chemical potential (energy) and the internal electrostatic potential energy. The electrochemical potential Ϛ for electrons is given by = q

where is the chemical potential and is theϚ electrostaticµ − 휓 potential. Thus we see that the Fermi energy is the electrochemical p[otential in a solid in which equilitbrim conditions exist i.e., no externally applied electricµ field or no radiation is present.휓 When we apply a voltage across a piece of semiconductor, the

- 53 - thermal equilibrium conditions are disturbed. Even then, as long as there are no excess carriers, the Fermi energy is equal to the electrochemical potential. The electrochemical potential is different at the two ends of the semiconductor across which the voltage is applied. The difference is equal to the amount of the applied voltage multiplied by the electron charge. Hence, the Fermi level at one end of the semiconductor is shifted in energy with respect to the Fermi level at the other end by times the applied voltage as shown in Figure (2.15). Since the material is uniform i.e., the carrier density is the same everywhere, the conduction band and the valence band vary in direct accordance with푞 the variation in the Fermi energy. In other words, the conduction band and the valence band are shifted from one end of the crystal to the other by the amount – where is the applied voltage. The electric field is the negative gradient of the electrostatic potential . 푞� � = - 휓 (2.58)

In one dimension ℰ⃗ 훻�⃗휓 = (2.59) 푑푑 The electrostatic potential is the potentialℇ energy− divided푑� by charge. We saw earlier that , the bottom of the conduction band, is the potential energy of the electron. The electrostatic potential� is obtained by dividing by – ,휓 and is therefore equal to 퐸

푞 = = (2.60) 퐸푐 퐸푖 퐸푔 and 휓 − 푞 − 푞 − 2푞

= = = (2.61) 푑푑 1 푑�푐 1 푑�푖 The energy band diagram in the presenceℇ of− an 푑electric� 푞field푑� is represented푞 푑� in Figure (2.15). The electric field is shown as being in the – direction in this figure. Hence the gradient, according to Equation (2.61) is also negative and hence the bands are tilted downward to the right. The slope is constant since the electric field is constant. 푥

We saw earlier that in the presence of an electric field, the values and the momentum values change by an amount proportional to the external force under steady state. 푘 푝 = = (2.62) 퐹𝐹𝐹 −푞ℇ Where is the scattering relaxation time훥� and Forceħ is휏 �the externallyħ 휏� applied force due to the electric field. The change in momentum correspondingly is 휏� = = (2.63)

훥� ħ훥� −푞ℇ 휏�

- 54 -

Figure (2.15): Energy band diagram in the presence of an electric field

Before the application of the electric field, the average momentum was zero since for every electron having a particular value of momentum, there was another electron having the opposite momentum. In the presence of the electric field, all the electrons have the same net change in momentum given by Equation (2.63). Since all the electrons suffer the same net change in momentum, the average momentum of the electrons is equal to the net change, and is given by

= (2.64)

Since the momentum is given by = * 푝 where푎� −*푞 ℇis 휏the� effective mass of the electron and is its velocity, the average velocity, which is also called the drift velocity is given by 푝 푚 푣 푚 푣 푑 = = 푣 (2.65) 푝푎� 푞ℇ 휏푐 ∗ ∗ Equation (2.65) shows that the drift (average)푣푑 푚velocity −is proportional푚 to the electric field . This is usually expressed as ℇ = (2.66)

velocity per The proportionality constant is called the푣 mobility푑 − of휇 ℇthe electron and is given in units of unit electric field. Typically we express mobility in units of . Since we will be discussing the µ drift velocity of electrons as well as that of holes, we denote2 − the1 parameters−1 with a subscript for electrons and for holes. Therefore, for electrons 𝑐 � 푠𝑠 푛 푝 = (2.67) and 푣푑� −휇푛 ℇ

= (2.68) 푞 휏푐� 휇푛 푚푛

- 55 -

For holes we must bear in mind that the charge is positive and therefore the externally applied force is . Hence,

푞 ℇ = = (2.69) 푞ℇ 푑� 𝑐 푛 and 푣 푚푝 휏 휇 ℇ

= (2.70) 푞 휏푐� 푝 휇 푚푝

Figure (2.16): Motion of an Electron with and without an Electric Field

In the above equations, and are the effective masses for electrons and holes. The relaxation time due to scattering is also different for electrons and holes; hence different subscripts 푚푛 푚푝 are also used to denote the relaxation time. Figure (2.16) shows on the left the random motion of the � electron in the absence of an electric휏 field and the average velocity is zero. On the right, the motion of the electrons has a drift velocity superimposed on the random motion in the presence of the electric field. If the charge carriers viz., electrons and holes, have a non-zero drift velocity, then an electric 푑� current flows. The particle flux density푣 can be determined as follows: Since all the particles are moving with the same velocity , the particles passing through a unit cross sectional area in the next one second will be only those that are within a distance numerically equal to and they are equal to 푣푑� = 푣 푑 � (2.71)

The electric current density is obtained푃𝑃�푃𝑃� by multiplying퐹𝐹� 𝑑𝑑𝑑� the particle푛 flux푣푑 �density by the charge – and hence the electric current density due to electron flow is 푞 = = (2.72)

The current flow due to the electric퐽 field푛 is− called푞�푣 drift푑� current.푞 푛 휇 Similarly,푛ℇ the electric current density due to hole flow is given by

= = (2.73)

The total drift current density is =퐽 푝 + −, 푞Hence�푣푑� 푞 푛 휇푝ℇ 퐽 퐽푛 퐽푝

- 56 -

= +

The total electric current density is proportional퐽 푞 푛 휇 to푛ℇ the applied푞 푛 휇 푝electricℇ field. This is called Ohm’s Law. The electrical conductivity is the electric current density per unit electric field and is given by

= 퐽 Therefore 휎 ℇ = ( + ) (2.74)

휎 푞 푛 휇푛 푝 휇푝 The resistivity ρ is the reciprocal of .

σ = = ( ) 1 1 It must be pointed out that the drift current휌 is 휎essentially푞 푛 휇determined푛+푝 휇푝 by the majority carriers and the contribution of the minority carriers is negligible since the majority carrier density is orders of magnitude larger than the minority carrier density.

Interpretation of and

The mobility defines휏𝑐 the ease휏𝑐 with which the carriers move in the semiconductor under the application of an external field. It depends primarily on the effective mass of the carrier and the relaxation time . can be interpreted as the average or mean time between scattering or collisions suffered by the electron with a similar interpretation for . Since the random motion of the electron � 𝑐 is due to thermal휏 velocity,휏 the average or mean distance traveled by the electron between collisions, 𝑐 which is denoted , is given by 휏

� 푛 = (2.75)

Where is the average of the magnitude�푛 of thermal푣푡ℎ 휏 𝑐velocity. The parameter is called the mean free path. 1 can be interpreted as the probability per unit time that an electron will be scattered. 푡ℎ 푛 Collisions푣 or scattering of the electrons can be due to several mechanisms such� as lattice (phonon) 𝑐 scattering⁄ 휏 and impurity scattering. Therefore when several scattering mechanisms are simultaneously present, the probabilities of scattering due to each of the mechanisms are added.

= + (2.76)

1 1 1 푐� 푛 푝ℎ 푛 푖𝑖 where is the probability per unit 휏time of scattering휏 휏due to lattice vibrations or phonons and

1 is휏 푛the푝ℎ probability per unit time of scattering due to impurities. Therefore,

1 휏푛 푖𝑖

- 57 -

= + (2.77)

1 1 1 µ푛 µ푛 푝ℎ µ푛 푖𝑖 Here and are respectively the mobilities that the electron would have had if only scattering due to lattice vibrations alone was present or if only scattering due to impurities alone was 푛 푝ℎ 푛 푖𝑖 present. The twoµ scatteringµ mechanisms have different temperature dependence. For example,

3 − 2 and µ푛 푝ℎ 훼 푇 1

3 2 µ푛 푖𝑖 훼 �푇 � Similar arguments apply for hole mobility. 푁푖𝑖

Figure (2.17): Electron mobility versus temperature in silicon (From Sze)

Figure (2.17) gives the variation of electron mobility with temperature for different donor concentrations. The mobility at low temperatures is limited by impurity scattering. The impurity scattering probability decreases with an increase in temperature and the mobility therefore increases with temperature. At higher temperatures the lattice or phonon scattering probability becomes significant and the mobility starts to fall off with temperature since the scattering probability due to lattice vibrations increases with temperature. The mobility thus exhibits a peak at some temperature where the influence of the two scattering mechanisms is comparable. The peak shifts to the right and falls off also in height as the impurity concentrations increase since the probability of scattering due to impurity scattering increases with donor concentration.

Figure (2.18) gives the variation of mobility with impurity concentration. The mobility is high and fairly constant with impurity density until about 1016 cm-3 after which it falls to a lower value. The figure

- 58 - gives both the electron and the hole mobility variation with the total impurity concentration. In compensated materials the mobility should be determined from this figure using the total density of impurities (the sum of donor and acceptor impurity densities). The resistivity at room temperature for silicon for different donor or acceptor concentrations is given in Figure (2.19). Once we know the resistivity, we know the impurity concentration. Similarly when we know the impurity concentration, we know the resistivity.

Figure: (2.18) Electron and Hole mobility variation with impurity concentration in silicon at = 300 0K (From Mueller and Kamins) 푇

- 59 -

Figure (2.19): Resistivity variation with dopant concentration in silicon. (From Sze)

Diffusion

An electric current can also flow in a semiconductor due to a variation of the density of electrons and holes. The process by which the concentration variation produces carrier flow is called diffusion. In a diffusion process the particles flow from regions of high concentration to regions of low concentration. In Figure (2.20), we have assumed an electron concentration that increases with ; hence electrons tend to flow from right to left. The particle flux density is proportional to the concentration gradient. The direction of the flow is opposite to the gradient. 푥

= (2.78)

Where is the proportionality푃𝑃� const푃𝑃�ant.푓 It𝑓� is called𝑑𝑑𝑑� the electron− diffusion퐷푛∇푛 constant. The negative sign in the equation for the particle flux density indicates that the flux of particles is in the opposite direction to 푛 the concentration퐷 gradient. is the electron concentration gradient. Since the dimension of the left – hand side of the above equation is and the dimension of the concentration gradient is , ∇푛 we infer that the dimension of the diffusion−2 constant,−1 , is . Since each electron carries a −4 𝑐 푠𝑠 𝑐 charge equal to – coulombs, the electric current density due to2 diffusion−1 is given by 퐷푛 𝑐 푠𝑠 푞 = q (2.79)

In one dimension this becomes equal to 퐽푛 퐷푛∇푛

= q (2.80) 푑� When one calculates the current퐽푛 flow due퐷 to푛 hole푑� diffusion noting that each carries a charge + , we get for the hole current density 푞 = q (2.81) 푑� The electric current due to the density퐽푝 − gradient퐷푝 푑is� called the diffusion current. We saw earlier that the electric current cue to the presence of the electric field is called the drift current. The electric current density caused by electron flow is due to both diffusion and drift and is given by

= q n + q (2.82) 푑� and similarly the hole electric current퐽푛 density is µgiven푛ℇ by 퐷푛 푑�

= q p q (2.83) 푑� The total electric current density퐽푝 is the sumµ푝 ℇof −electron퐷푝 and푑� hole electric current densities and is given by

= + (2.84)

퐽 퐽푛 - 60퐽푝 -

In our discussions, we will be concerned with the diffusion current due to the minority carriers only. The diffusion current due to majority carrier density gradient will be seen to be balanced by a drift current arising from a self-induced electric field.

Einstein Relation

The mobility and diffusion constants are related as follows:

= (2.85) 푞 and µ푛 푘푘 퐷푛

= (2.86) 푞 This relationship is called Einstein relationshipµ푛 푘푘. Although퐷푛 this relationship can be derived, it is sufficient for our purposes to assume the relationship. It is easy to remember this relationship if one bears in mind the dimensions of mobility has the dimension . Since has the dimension 2 −1 −1 푘푘

of volt it is easy to remember that the mobility times is the𝑐 diffusion� 푠𝑠 constant. 푞 푘푘 푞

Figure (2.20): Variation of electron concentration with distance

Generation Recombination Process

We saw that free electrons and holes were created by thermal excitation of electrons from the valence band into the conduction band. The process of creating electrons and holes is called generation. Generation can be either due to thermal excitation i.e., by imparting to the electron in the valence band an amount of thermal energy sufficient to be excited to the conduction band, or by other means such as

- 61 - optically exciting the valence band or by bombardment with high energy particles. Electrons in the valence band are those that are localized in the covalent bonds. As shown in figure (2.21), when a covalent bond is broken the free electron is the electron in the conduction band and the vacancy represented in the broken bond is the hole in the valence band.

Figure (2.21): Illustration of connection between free electrons and free holes in the semiconductor and electrons in the conduction band and hole in the valence band

A free electron moving in the crystal can recombine with a free hole resulting in the loss of both the electron and hole. This process is called the recombination of an electron and a hole.

There are two mechanisms by which generation-recombination process takes place: 1) band to band generation or recombination; 2) generation or recombination through an impurity or a defect. In the first process electrons and holes recombine directly by an electron in the conduction band jumping back to a vacant state in the valence band. Physically, the recombination process is one in which the free electron wanders to the site of a broken bond and the covalent bond gets completed when the electron jumps into the broken bond. This results in the loss of a free electron and a free hole.

In the second process electrons and holes recombine through the intermediary of an impurity or trap which has an energy level in the band gap. Such impurities or traps are called generation- recombination centers or - centers. Recombination takes place by the recombination center first capturing an electron or a hole and then subsequently capturing a carrier of the opposite charge. This gives rise to the annihilation푔 푟or recombination of an electron and a hole. The recombination centers also generate electrons and holes. The generation takes place by the center emitting an electron or a hole first and subsequently emitting a carrier of the opposite charge. Generation requires imparting of energy to the solid and recombination results in the release of energy. In band to band recombination energy is released in the form of light or photons. In recombination through - centers energy is released as thermal (phonon) energy. 푔 푟 These processes of generation and recombination always require the involvement of both an electron and a hole each time the process takes place and it is customary to say that the recombination

- 62 - of an electron-hole pair or the generation of an electron-hole pair occurs. In silicon the generation- recombination process occurs mostly through - centers and the band to band process is highly improbable. 푔 푟 Independent of the actual process, direct or indirect, the rate of recombination , equal to the number of electron-hole pairs lost due to recombination per unit volume in unit time, is proportional to the density of electrons and the density of holes. During the proportionality constant as 푅, we have

= 푟 (2.87)

In thermal equilibrium, the generation푅 rate 푟, 푛which푝 is equal to the number of electron-hole pairs generated per unit volume in unit time, should be equal to the recombination rate . What do we mean by thermal equilibrium? When there is no externally퐺 applied electric, magnetic or electromagnetic field, the solid is in thermal equilibrium with its surroundings. Under that condition, the temperature푅 is constant everywhere within the solid and the only way electrons are excited into the conduction band is by thermal excitation and this process is called thermal generation. In thermal equilibrium, no electric current flows. We denote the carrier densities and with subscript 0 such as and to denote thermal equilibrium carrier densities. Denoting the recombination rate in thermal equilibrium , and 0 0 denoting the thermal generation rate , we푛 have 푝 푛 푝 푅푡ℎ 푡ℎ 퐺= = = (2.88) 2 If we increase the density 퐺of푡 oneℎ type푅푡ℎ of carriers,푟 푛0 푝the0 probability푟 푛푖 of recombination for the other type of carriers increases. Hence the density of the second type of carriers decreases. This is the basis of law of mass action.

The thermal generation rate is the rate at which electron-hole pairs are generated thermally at a given temperature, and therefore has the same value whether thermal equilibrium conditions hold 푡ℎ no or not. 퐺

The recombination-generation process involving a - center has been modeled by a theoretical treatment developed by Schockley, Read and Hall. This model is called the Schockley-Read Hall model or S-R-H model. According to the S-R-H model, the constant 푔is 푟given as,

= 푟 (2.89) 1 푛+푛1 푃+푝1 + where 푟 푁푡휎푝푣푝 푁푡휎푛푣푛

= density of traps or g-r centers,

푁푡, = capture cross-sections for electrons and holes respectively,

푛 푝 휎 휎 , =thermal velocity of electrons and holes respectively and is equal to , where is the 3 푘 � ∗ 푛 푝 ∗ effective푣 푣 mass of the carrier, � 푚 푚

= , (2.90) 퐸푐−퐸푡 �− 푘푘 � 1 � 푛 푁- 6푒3 -

= , (2.91) 퐸푡−퐸푣 �− 푘푘 � = the trap energy level. 푝1 푁� 푒 퐸푡 The capture cross-section s relate to the ability (probability) of the - centers to capture the charge carriers. and are short-hand notations for the expression given in the above equations. However, one can assign a physical휎 meaning′ to them. is equal to the number푔 푟of free electrons per 1 1 unit volume in the푛 semiconductor푝 when the Fermi energy is at the same level as the trap energy level. 1 Similarly, is the number of holes per unit volume if the푛 Fermi energy is at the same level as the trap level. 푝1 Let us define the terms involving electron and hole capture as follows:

= (2.92) 1 and 휏푛0 푁푡휎푛�푛

= (2.93) 1 푝0 Then 휏 푁푡휎푝�푝

= (2.94) ( ) ( ) 1 The recombination rate and can푟 then푛 be+ 푛 written1 휏푝0+ as푝 + 푝1 휏푛0

푡ℎ 푅 퐺 = ( ) ( ) 푛 푝 푅 푛+ 푛1 휏푝0+ 푝+ 푝1 휏푛0 = (2.95) ( ) 2( ) 푛푖 푡ℎ The net recombination rate is the dif퐺ference푛 between+ 푛1 휏푝 0+ and푝+ 푝1 .휏 In푛0 regions where there is thermal equilibrium, the product is equal to and hence the net recombination rate is zero. In regions 푅 퐺푡ℎ where there is no thermal equilibrium, if 2the product is more than , a net recombination will 푛� 푛푖 occur. If the product is less than , then a net generation of electron2 -hole pairs will occur. Let us 푛� 푛푖 now illustrate these points by considering2 an extrinsic semiconductor. 푛� 푛푖

Extrinsic semiconductor

We will consider an -type semiconductor although the discussion will be similar for a p-type semiconductor. Let us consider the thermal equilibrium recombination rate first. 푛

= ( ) ( ) 푛푛0 푝푛0 푡ℎ 푅 푛푛0+- 64푛1 - 휏푝0+ 푝푛0+ 푝1 휏푛0

= (2.96) 2 푛푖 We neglected the term involving in the denominator푛푛0휏푝0+푛1휏 푝of0 +the푝1 above휏푛0 equation since . Expressing and in terms of , 푝푛0 푝푛0 ≪ 푛푛0 1 1 푖 푛 푝 푛= 2 푛푖 퐸 −퐸 �퐸 −퐸 � 푡ℎ 푡 1 − 푡 푖 푅 푘푘 푘푘 푛푛0휏푝0+푛푖휏푝0푒 +푛푖휏푛0푒 = (2.97) 푝푛0 퐸푡−퐸1 �퐸푡−퐸푖� 푛푖 − 푝0 푝0 푘푘 푛0 푘푘 휏 +푛푛0�휏 푒 +휏 푒 � Equation (2.97) can be rewritten as

= (2.98) 푝푛0 where 푅 휏푝

= (2.99) 1 1 푛1 푝1 푝 푝0 푝0 푛0 휏 휏 +푛푛0휏 +푛푛0휏 We can infer from Equation (2.98) that can be interpreted as the probability per unit time that a hole, 1 (i.e., a minority carrier) will combine. Using휏푝 this interpretation one can show that is the average time that a minority carrier spends before it is lost due to recombination. Hence it is called the minority 푝 carrier lifetime. Equation (2.99) tells us that the minority carrier lifetime depends 휏on the energy of the - center is near the middle of the band gap.

푔 푟 When the trap levels are close to the middle of the gap, then the minority carrier lifetime becomes 휏푝 = (2.100)

The thermal generation rate, is equal휏푝 to 휏푝� and hence, 퐺푡ℎ 푅푡ℎ = 2 푛푖 푡ℎ 푛푛0 휏푝0+ 푛1휏푝0+푝1 휏푛0 퐺 = (2.101) 푝푛0 We can rewrite the above equation as 휏푝

= (2.102)

Which gives the interpretation that 푝the푛0 thermal퐺푡 ℎequilibrium휏푝 minority carrier density is the thermal generation rate times the lifetime.

- 65 -

Till now we considered a - center with a discrete energy level. However, generally the - centers are distributed over a range of energies. Then the minority carrier lifetime is obtained by adding the contributions of all the - centers.푔 푟 푔 푟 푔 푟 Excess carriers

Suppose that excess carriers over and above the thermal equilibrium densities are created by illuminating the semiconductor with light or by electrical injection as we will see later on in our study of - junctions. This corresponds to a non-thermal equilibrium situation. The carrier densities and can be written as 푝 푛 푛 푝 = + (2.103)

푛 = 푛푛0 + ∆푛 (2.104)

Where and are thermal equilibrium푝 carrier푝푛0 densities∆푝 and and are called excess carrier densities. When excess carrier densities are created optically, equal number of excess holes and 푛0 푛0 electrons푛 are created푝 and hence = . When excess minority ∆carriers푛 ∆ are푝 inject ted as in a - junction, excess majority carriers will flow into this region to preserve charge neutrality and hence again = . The recombination rate∆ 푛in the∆ 푝presence of excess carriers is therefore given by 푝 푛

∆ 푛 ∆ 푝 = ( ) ( ) 푛 푝 푛+ 푛(1 휏푝0+)(푝+ 푝1 휏) 푛0 푅= (2.105) ( ) ( ) 푛푛0+∆푝 푝푛0+∆푝 푝 The net recombination rate, , is the푅 difference푛푛0+∆푝 +between푛1 휏푝0+ the푝푛0 recombination+∆푝+ 푝1 휏푛0 rate and the generation rate. The latter is the thermal generation rate 푈 =

( )( ) 푡ℎ = 푈 푅 − 퐺 ( ) ( ) 2 푛푛0+∆푝 푝푛0+∆푝 푝 푛푖 푛푛0+∆푝+ 푛1( 휏푝0+ 푝푛)0+∆푝+ 푝1 휏푛0 − 푛푛0 휏푝0+ 푛1휏푝0+푝1 휏푛0 = (2.106) ( ) ( 2 ) 2 푛푛0푝푛0+ 푛푛0+푝푛0 ∆푝+∆푝 푛푖 We will now distinguish푛푛 0two+∆ 푝limiting+ 푛1 휏 푝cases,0+ 푝 푛one0+∆ called푝+ 푝1 low휏푛0 injection− 푛푛0 휏 and푝0+ the푛1휏 other푝0+푝 1called휏푛0 high injection. The low injection conditions are obtained when .

Under low injection conditions, term is negligib∆푝 le≥ and푛푛0 and are also negligible in comparison with . Hence, 2 ∆푝 푝푛0 ∆푝 푛푛0 ( ) = ( ) ( ) ( ) 2 푛푛0+푝푛0 ∆푝+ 푛푛0푝푛0 푛푖 푈 푛푛0+∆푝+ 푛1 휏푝0+ 푝푛0+∆푝+ 푝1 휏푛0 − 푛푛0+푛1 휏푝0+푝1 휏푛0

- 66 -

( ) = ( ) 2 ( ) 2 푛푛0∆푝 +푛푖 푛푖 푛푛0+푛1 휏푝0+푝1 휏푛0 − 푛푛0+푛1 휏푝0+푝1 휏푛0 = ( ) 2 ∆푝 푛푖 푛1 푝1 푝0 푛0 푛0 1 푝0 1 푛0 휏 +푛푛0+푛푛0휏 − 푛 +푛 휏 +푝 휏 = = (2.107) ∆푝 ∆푝 푛1 푝1 푝0 푝0 푛0 푝 The net recombination is due to휏 excess+푛푛 0carriers.휏 +푛푛 0Therefore휏 휏 is the rate of excess carrier recombination and is proportional to . In other words, is the probability per unit time that an excess carrier will be 푈 1 lost due to recombination.∆푝 Therefore the excess휏푝 carrier lifetime is also which is minority carrier lifetime. 휏푝 Under high injection > and hence the term ( + ) is negligible in comparison

with . Then 푛0 푛0 1 2 ∆푝 푛 푛 푛 ∆푝 ∆ 푝 = + (2.108)

Also is negligible in comparison with푛 R. Under푛푛0 this∆ 푝condition≈ ∆푝 퐺푡ℎ = ( + + ) +2( + + ) ∆푝 푈 푛푛0 ∆푝 푛1 휏푝0 푝푛0 ∆푝 푝1 휏푛0 = +2 ∆푝 푝0 푛0 =∆푝�휏 =휏 � (2.109) ∆푝 ∆푝 Where is the lifetime of the excess carrier 휏at푝 0high+휏푛 injection.0 휏ℎ푖 Under high injection conditions, the excess carrier lifetime becomes the sum of and . At intermediate injection, the lifetime is ℎ푖 between휏 the high injection value and the low injection value. 휏푝0 휏푛0

Depletion Region

We will see in our study of - junctions with an applied voltage that it is possible to have regions within the semiconductor under non-thermal equilibrium condition, in which the carrier densities n and p are much less than푝 the푛 thermal equilibrium carrier densities. Under this condition ( 0) the recombination rate is negligible for

푛 ≈ 푝 ≈ 푅 = 0 (2.110) 푛 푝 푁푡�0휎0 The thermal generation rate is equal푅 to 푛+푛1+푝+푝1 ≈ 퐺푡ℎ - 67 -

= ( + ) +2 ( + ) 푛푖 퐺푡ℎ 푛 푛1 휏푝0 푝 푝1 휏푛0 = +2 ( ) 푛푖 +푛푖 퐸푡−퐸푖 − 퐸푡−퐸푖 ≈ 푛0 푛1휏푝0 푝1휏 푘푘 푘푘 = 휏푝 0 푒 휏 푛 0 푒 (2.111) 푛푖 Where is defined as the generation lifetime and2휏푔 equal to

휏푔 = + (2.112) 퐸푡−퐸푖 −�퐸푡−퐸푖� 1 푘푘 푘푘 The generation rate is maximum휏푔 2 only�휏푝0 when푒 the trap휏푛 level0푒 is in� the middle of the band gap i.e., when = = . Another way of stating this is that only those traps which are within a few above 퐸푡 or below will be effective as generation centers. When = , is a minimum and equal to 푛1 푝1 푛푖 푘� 퐸푖 푡 푖 푔 = 퐸 퐸 휏 (2.113) 휏푝0+휏푛0 Thus we conclude that the thermal휏푔 generation2 rate is in the non-thermal equilibrium 푛푖 depletion region, while in the neutral region it is equal to .2 휏푔 푝푛0 휏푝 Continuity Equation

Let us consider an elementary volume in an -type semiconductor. In this elementary volume minority carriers are being thermally generated at a rate and let us further assume that minority carriers are also generated at a rate ∆� due to푛 external light sources. Minority carriers are 퐺푡ℎ also lost due to recombination at rate . In addition, if there is an outward flow of current out of this 퐺𝑜� elementary volume, then the minority carriers will be lost if the minority carriers are holes, and minority carries will increase in number if the minority푅 carriers are electrons. Since we are considering an -type semiconductor, the minority carriers are holes in our example. 푛 The total outward flow of current out of a volume surrounded by a surface is equal to

= (2.114)

푝 𝑣� ����⃗ In an elementary volume , the outward∫푠푠𝑠𝑠� flow퐽���⃗ ∙ of���� �current푑⃗ ∫ reduces∇ ∙ 퐽푝 ∙ to𝑑 . The outward flow represents the charge flow,∆� and hence the number of holes lost in unit∇ time∙ 퐽���푝⃗ ∆is� equal to . 1 Hence we can write the rate of increase of the number of holes in the elementary volume as 푞 ∇ ∙ 퐽���푝⃗∆�

- 68 -

1 = ( + + ) (2.115) 훿푝 푡ℎ 𝑜� 푝 Dividing the above equation훿� ∆by� ∆we� get−푅 퐺 퐺 − 푞 ∇ ∙ �퐽�⃗

∇=� + + (2.116) 훿� 1 This equation is called the continuity훿� −equation.푅 퐺푡 ℎFrom퐺 our𝑜� earlier− 푞 ∇ discussion,∙ 퐽���푝⃗ we know that

= (2.117) ∆푝 푡ℎ Hence the continuity equation reduces to 푅 − 퐺 휏푝 ( ) = = + (2.118) 훿� 훿 ∆푝 ∆푝 1 𝑜� 푝 But 훿� 훿� − 휏푝 퐺 − 푞 ∇ ∙ 퐽���⃗

= (2.119)

In one dimension, the continuity equation퐽푝 reduces푞휇푝푝ℰ to− 푞퐷푝∇푝 ( ) = + (2.120) 훿 ∆푝 ∆푝 1 dJp 𝑜� The expression for the hole current훿 �density− is 휏given푝 by퐺 − 푞 dx

= (2.121) 훿푝 Let us assume that the electric field 퐽푝 is zero.푞휇 푝푝ℰ − 푞퐷푝 훿푥 ℰ = 훿푝 Then 퐽푝 −푞퐷푝 훿푥 ( ) = + + (2.122) 2 훿 ∆푝 ∆푝 δ p 𝑜� 푝 2 A simple example: 훿� − 휏푝 퐺 퐷 δ x Let us assume that light is incident on the semiconductor to generate excess carrier uniformly within the semiconductor. Growth of excess carriers

Let the light be switched on a time t = 0. Since the excess carriers are uniformly generated everywhere in the semiconductor, there is no concentration gradient, and hence the term due to the current flow can be dropped in the continuity equation. The continuity equation governing the growth of the excess carriers is then given by

- 69 -

= + (2.123) 푑∆푝 ∆푝 𝑜� The solution to this equation can be readily푑� seen −to be휏푝 퐺

( ) = 1 푡 = 1 푡 (2.124) −휏푝 −휏푝 𝑜� 푝 0 Where = is the∆ excess푝 푡 carrier퐺 density휏 � at− t =푒 , i.e.,� when∆푝 steady� − state푒 conditions� have been reached. Equation (2.124) describes that the excess carrier density grows exponentially with time, from ∆푝0 퐺𝑜�휏푝 ∞ zero to its steady state value.

Decay of excess carriers

Let us assume that the light has been on for a long time, and that steady state conditions have been reached. Now, let the light be turned off at time t = . The decay of excess carrier density will be governed by putting = 0 in the continuity equation, which will then be ∞ 퐺𝑜� = (2.125) 푑∆푝 ∆푝 The solution to this equation is 푑� − 휏푝

( ) = 푡 (2.126) −휏푝 = 0 Where is the excess carrier density just∆푝 when푡 the∆ light푝0푒 was turned off, i.e., at . ∆From푝0 the above treatment, we can conclude that the time taken to reach steady푡 state during the growth and the decay of excess carriers is determined by the minority carrier lifetime.

- 70 -

Summary

A semiconductor behaves like an insulator at very low temperatures, but it exhibits increasing electrical conductivity as the temperature is increased. The electrical conductivity of a semiconductor at room temperature is much lower than that of a conductor but very much higher than that of an insulator. As the temperature is raised, in a semiconductor, some of the electrons get enough thermal energy to break loose from the covalent bond and become free electrons. Similarly, the broken bonds represent free holes. In terms of a band picture, at very low temperatures the valence band is completely full and the conduction band is empty. Additionally, the band gap (the separation in energy between the conduction and valence bands) is small. As the temperature is increased, the electrons get excited from the valence band to the conduction band. The electrons in the conduction band are the free electrons while the vacant states in the valence band are the fee holes. Since an electron has a charge of – coulombs and a hole has a charge of coulombs, the charge moves with them when they move. Hence electrons and holes are called charge carriers, or simply carriers. 푞 푞 An intrinsic semiconductor is one in which electrons and holes are generated only by thermal excitation from the valence band to the conduction band. The free electron density is equal to the free hole density ( = ). The electron has no velocity when it occupies a state at the bottom of the conduction band, and hence has no kinetic energy. Hence , the bottom of the conduction band, is the 푖 푖 potential energy푛 of푝 the electron. An electron that occupies a state with energy has a kinetic energy is 퐶 . Similarly, the potential energy of the hole is and퐸 its kinetic energy is . In other words, the kinetic energy of the hole increases as we go deeper into the valence band.퐸 퐸 − 퐸퐶 퐸푉 퐸푉 − 퐸 Free holes and free electrons can be treated as particles in a box by assuming the appropriate effective masses for the electrons and holes. The probability of occupation of a state by an electron is given by the Fermi function, ( ). The probability of occupation of a state by a hole is the probability that an electron will not occupy that state, (1 ( )). 푓 퐸 In deriving the expression for the density− 푓 of퐸 electrons and holes, a numerical integration is necessary. This procedure can be simplified by assuming that the Fermi energy is at least a few below the conduction band and at least a few above the valence band. When the assumption is not valid in a material, the semiconductor is said to be degenerate. The Fermi function approximates푘� the Boltzmann distribution function. Hence this approximation푘� is also called the Boltzmann approximation.

The electron (and hole) density in an intrinsic semiconductor is determined by the band gap energy and the effective masses. It is also a strong function of temperature. The Fermi energy is approximately at the middle of the band gap in an intrinsic material. If the effective masses of the holes 퐸푔 and the electrons are assumed to be equal to the mass of the electron in vacuum, then the Fermi energy lies exactly at the middle of the band gap. The Fermi energy in an intrinsic material is denoted by , and hence it is customary to denote the middle of the bandgap as . 퐸푖 An extrinsic semiconductor is one in which controlled 퐸amount푖 of trace impurities are substitutionally added to increase one type or the other type of carriers. When elements of group V are added as impurities in silicon, the electron density is increased without a corresponding increase in the free hole density. These impurities are called donors, and the material is called an -type semiconductor. In an -type semiconductor, electrons are called majority carriers and holes are called minority carriers since electron density is much larger than hole density. On the other푛 hand, when 푛 - 71 - impurity atoms belonging to an element in group III of the periodic table are added, the hole density is larger than electron density. This type of material is called a -type semiconductor. In this type of material, holes are called majority carriers and electrons are called minority carriers. 푝 In an -type semiconductor, free electrons arise due to two reasons: thermal excitation from the valence band, and thermal excitation from the donor atoms (ionization of the donor atoms). The holes in an -type푛 material arise only due to thermal excitation of electrons from the valence band. The positive charge in an -type semiconductor arises due to free holes and positively ionized donor atoms while the negative푛 charge arises due to free electrons. 푛 In a -type semiconductor, free holes arise due to two reasons: thermal excitation of electrons from the valence band, and ionization of acceptor atoms. The electrons arise only due to thermal excitation from푝 the valence band. The positive charge in a -type material arises due to free holes, while the negative charge in an -type material arises due to electrons and negatively ionized acceptor atoms. In a neutral semiconductor, ( -type or -type) the net charge푝 density is zero. When both acceptors and donors are present in the 푛same region, the net impurity density (the donor density minus the acceptor density) determines the type푛 of semiconductor.푝 Such a material is called a compensated material.

The expression for the carrier density in an extrinsic semiconductor can be derived similarly to the expression for an intrinsic material. For a high density of impurities, the carrier density can only be obtained by numerical; integration. However, when the impurity density is not that high, the material can be considered to be non-degenerate and the expression for the carrier density is of the same form as what is obtained in the intrinsic case under non-degenerate assumption. In a non-degenerate semiconductor, = . This is called the law of mass action. When the electron density is increased by adding donor impurities,2 the hole density decreases. Similarly, when the hole density is increased by 푖 adding acceptor 푛impurities,� 푛 the electron density decreases. The number of ionized impurities increases with temperature. At moderate and low temperatures, the majority carrier density is equal to the ionized net impurity atoms.

In an -type material the Fermi energy lies in the upper half of the band gap. As the donor density is increased, the Fermi energy moves closer to the conduction band. In a -type material the Fermi energy 푛lies in the lower half of the band gap. As the acceptor density is increased, the Fermi energy moves closer to the valence band. 푝

At high temperatures, all the impurity atoms are ionized and an increase in carrier density arises only by thermal excitation from the valence band. At still higher temperatures, the electron and hole densities become equal and the material becomes intrinsic.

An electric current in a semiconductor can arise due to either a drift process or a diffusion process. In the drift process, an externally applied electric field drives the charge carriers. The resulting current is called the drift current. The drift current can be expressed in terms of a parameter called the mobility. The drift current is generally due to majority carriers. In the diffusion process, carriers diffuse from regions of high concentration to regions of low concentration. The resulting current is called the diffusion current. The diffusion current is expressed in terms of a constant called the diffusion constant. The diffusion current is due to minority carriers only.

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The diffusion constant is related to the mobility by a factor . This relationship is called the 푘푘 Einstein relation. The mobility is limited by scattering from impurity atoms or scattering from phonons. 푞 The two processes have different temperature dependence.

In a semiconductor, electrons and holes are constantly being created (generated) and lost due to recombination. Recombination is the process in which an electron in the conduction band jumps back to a vacant state in the valence band. In this manner, both an electron and a hole have been annihilated. In steady state, the generation rate is equal to the recombination rate.

There are two mechanisms for generation-recombination to occur. One is what is called the band to band generation-recombination. The other is called generation-recombination through a trap or a generation-recombination ( - ) center. In silicon, the second is the dominant mechanism. The average time that a minority carrier spends in its band before it is lost due to recombination is called the minority carrier lifetime. The thermal푔 푟 equilibrium minority carrier density is given by the product of the thermal generation rate and the minority carrier lifetime. The minority carrier lifetime is inversely proportional to the density of - centers. When excess carriers are generated, the process is called injection of carriers. The average time an excess carrier spends in its band before it is lost due to recombination is called excess푔 carrier푟 lifetime. The steady state excess carrier density is given by the product of the excess carrier generation rate and the excess carrier lifetime.

When the excess carrier density is much less than the thermal equilibrium majority carrier density, it is called a low injection process. When the excess carrier density is comparable to or higher than the thermal equilibrium majority carrier density, then it is called a high injection process. Under low injection, the excess carrier lifetime is the same as the minority carrier lifetime. The growth and decay of excess carriers is governed by the excess carrier lifetime when the generation process is switched on or off respectively.

The flow and growth (or decay) of excess carriers is determined by solving the continuity equation.

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Glossary

= number of electrons with energy between + per unit volume of the crystal

�푛퐸 = elementary volume 퐸 푎�푎 퐸 𝑑

𝑑 = electron diffusion constant

퐷푛 = hole diffusion constant

퐷푝 = energy

퐸 = acceptor energy level

퐸퐴 = bottom of the conduction band, also the potential energy of an electron in the bottom of the conduction band 퐸퐶 = donor energy level

퐸 퐷 = Fermi energy

퐸퐹 = bandgap equal to

퐸푔 = Fermi energy in an퐸 intrinsic퐶 − 퐸푉 material, usually referred to as the intrinsic Fermi energy

퐸푖 = trap level or energy level of the g-r center

퐸푡 = top of the conduction band or the maximum energy of the conduction band

퐸 푡𝑡 = electric field

휀( ) = Fermi function; the probability of occupation by an electro

푓 퐸 = generation rate

퐺 = rate at which minority carriers are generated by an external light source

퐺𝑜� = thermal generation rate

퐺 푡ℎ = Planck’s constant

ℎ = total current density

퐽 = electric current density due to electron flow

퐽푛 = electric current density due to hole flow

퐽푝 = Boltzmann constant

푘 = average distance travelled by the electron between collisions

�푛 = the mass of an electron in vacuum

푚0 = effective mass of an electron in the conduction band

푚퐶, = effective masses for electrons and holes respectively

푛 푝 푚 푚 - 74 -

= effective mass of hole in the valence band

푚� = effective mass of the carrier ∗ 푚 & = thermal equilibrium carrier densities

푛0 푝0 = number of free electrons per unit volume in the semiconductor when the Fermi energy is at the same level as the trap energy level 푛1 = electron density

푛 = density of electrons or holes in the intrinsic semiconductor

푛푖 = thermal equilibrium electron density in the n-region

푛0 = thermal equilibrium electron density in the -type of semiconductor where is or

푛푥0 = effective density of states in the conduction푥 band 푥 푛 푝

푁퐶 = density of traps or - centers

푁푡 = effective density of푔 states푟 in the valence band

푁푉 = density of acceptor atoms

푁퐴 = density of ionized acceptors, number of negatively ionized atoms per unit volume − 푁퐴 = density of donor atoms

푁 퐷 = density of ionized donors, number of positively ionized atoms per unit volume + 푁 퐷 = density of free holes

푝P = momentum of carriers = number of holes per unit volume if the Fermi energy is at the same level as the trap level

푝1 = thermal equilibrium hole (minority carrier) density in an -type semiconductor

푝푛0 = thermal equilibrium hole (majority carrier) density in a 푛-type semiconductor

푝푝0 = thermal equilibrium hole density in -type material where푝 is or

푝푥 0 = charge 푥 푥 푛 푝

푞 = proportionality constant for recombination rate

푟 = recombination rate

푅 = recombination rate at thermal equilibrium

푅 푡ℎ = time

푡 = temperature in degrees Kelvin

푇 = net recombination rate, the recombination rate minus the generation rate

푈 - 75 -

= velocity

푣 = average velocity, also called the drift velocity

푣푑 = drift velocity of electrons

푣푑� = drift velocity of holes

푣푑�, = thermal velocity of electrons and holes respectively

푣푛 푣푝 = volume of a semiconductor solid

� ( ) = density of states for electrons in the conduction band

푍퐶 퐸 = excess electron density

∆ 푛 = excess hole density

∆푝 = change in momentum

∆ 푝 = excess carrier density at =

∆푝 0 = elementary volume 푡 ∞

∆ � = ideality factor

휂 = mobility of the carriers

휇 = electron mobility

푛 휇 = mobility that an electron would have if scattering were due only to impurities

푛 푖𝑖 휇 = mobility that an electron would have if scattering were due only to lattice vibration

휇푛 푝ℎ = hole mobility

휇푝 = resistivity, the reciprocal of

휌 = electrical conductivity 휎

휎 = capture cross-section

휎 , = capture cross-sections for electrons and holes respectively

휎푛 휎푝 = scattering relaxation time or average time between collisions

휏퐶 = average time between scattering or collisions suffered by the electron

휏퐶� = average time between scattering or collisions suffered by the hole

휏퐶 � = hole lifetime, i.e. average time that a minority carrier spends before it is lost due to recombination 휏푝 = electrostatic potential

∅ = electrostatic potential

휓 - 76 -

Problems

1. Derive an expression for the density of holes in a non-degenerate semiconductor using a method similar to that used in the text for the derivation of an expression for the density of electrons.

2. Given that = 10 at = 300 , calculate at = 200 and at = 150 assuming = 1.1210 at− all3 temperatures0 (i.e., neglecting variation0 of with temperature).0 푛푖 𝑐 푇 퐾 푛푖 푇 퐾 푇 퐾

퐸푔 푒푒 퐸푔

3. A piece of silicon is doped with 2 10 phosphorus atoms per . What are the majority and minority carrier concentrations at room15 temperature? 3 푥 𝑐 4. Suppose that the hole concentration in a piece of silicon at room temperature is 10 , find 5 −3 𝑐

(a) the electron concentration (b) the location of the Fermi energy

5. A sample of silicon is first doped with 10 boron atoms, and then doped with 4 10 arsenic atoms. 15 −3 15 𝑐 푥 (a) −What3 is the type of the semiconductor? 𝑐 (b) Find the location of the Fermi energy.

6. In an n-type semiconductor, the temperature is lowered such that only half the donor atoms are ionized. Neglecting the degeneracy factor, show that

7. In an n-type semiconductor say n-type, at higher temperatures the electrons come from ionization of the donor atoms as well as from excitation of electrons from the valence band to the conduction band. If is the donor density, show that the intrinsic carrier density, at the temperature at which the +electron density is twice that of the hole density, is 2 . 푁퐷 + √ 푁퐷 8. Find the resistivity of the same in Problem 4. Take the mobility values from the figure in the text.

9. Using the mobility curves in the text, find the resistivity of the sample in the Problem 5.

10. A piece of n-type silicon has a resistivity of 5 at room temperature (27 ). Find the thermal equilibrium concentration at 17 . 0 훺 − 𝑐 퐶 0 퐶

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11. Two scattering mechanisms are present in a semiconductor. The mobility, if the first mechanism alone is present is 500 . The mobility, if the second mechanism alone is present, is 900 . What 2is the−1 mobility−1 of the carrier in the sample? 𝑐 � 푠 2 −1 −1 𝑐 � 푠 12. If an intrinsic piece of silicon is illuminated such that carriers are generated at the rate of 10 uniformly everywhere in the sample, and if the lifetime of the carriers is 10 s, 18 dete−3rmine−1 the change in the electrical conductivity. −6 𝑐 푠

13. Write the expression for the total electron current in a semiconductor. From this write the relation between the concentration gradient and ( ) under conditions of no current 푑� flow. 푑� 푛 푥

14. Show that the probability that a hole will survive without recombining for seconds is given by

푡 assuming that it was created at time = 0 and that is the probability푡 that a hole will −휏푝 1 푒recombine in unit time. 푡 휏푝

15. Show that the average time that a hole lasts without recombining is seconds. Assume that the hole was created at time = 0. (This is why is called the hole lifetime.) 휏푝

푡 휏푝 16. Consider an n-type silicon sample with = 10 . Calculate the location of the Fermi level at a) 300 K, and b) 200 K. 16 −3 푁푑 𝑐 0 0

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Chapter 3 - Junction

A - junction is a device structure in푃 which푁 a single crystal of a semiconductor say silicon, has one region doped with donor impurities to make it -type while the rest of the semiconductor is doped with acceptor푝 푛 impurities to make it -type. For example, Figure (3.1) shows the impurity variation from one end of the crystal to the other. This figure is called푛 the impurity profile. The ordinate is the net impurity concentration - which푝 is positive in the -type region and negative in the -type region. The abscissa is , position. 푁퐷 푁퐴 푛 푝 Let us now푥 consider an isolated -type semiconductor. The net impurity concentration - is negative. The density of states in the two bands is plotted in Figure (3.2 A). The Fermi distribution 퐷 퐴 function is plotted for a -type material 푝in Figure (3.2 B). The Fermi energy is close to the valence푁 band푁 as shown in Figure (3.2 B). The probability of occupation by electrons i.e., the Fermi function, is nearly zero in value for states in푝 the conduction band whereas the probability of occupation by holes i.e., 1 minus the Fermi function, is much larger for states in the valence band. The distribution of electrons in the conduction band and the holes in the valence band is obtained by multiplying the density of states and the probability of occupation, and is given in Figure (3.2 C). The area under these distribution curves gives the number of carriers in the conduction and the valence band. As is to be expected in a -type semiconductor, a small density of electrons and a large density of holes are obtained. 푝 Let us now similarly consider an isolated -type semiconductor. The net impurity concentration - is positive. Again the density states in two bands is plotted in Figure (3.3 A). The probability functions for the occupation of states by electrons푛 and holes are plotted in Figure (3.3 B). The 퐷 퐴 푁distribution푁 of carriers in the two bands is given in Figure (3.3 C). We see now in the -type material a small density of holes and a large density of electrons are obtained. 푛 Before going further we will now show that the gradient in Fermi energy is zero when no electric current flows. Let us consider the electric current due to electron flow. The electric current is given by

= + (3.1) 푑� But we know that under thermal equilibrium,퐽푛 푞 푛 휇푛 ℰ 푞 퐷푛 푑�

= (3.2) 퐸퐹−퐸푖 푘푘 Where we have used the subscript 0 to푛 denote0 푛 푖thermal푒 equilibrium densities, and

= = (3.3) 푑푑 1 푑퐸푖 ℰ − 푑� 푞 푑�

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Figure(3.1): Impurity profile in a - junction. This junction is called an abrupt junction since the net impurities change from one type to the other abruptly. 푝 푛

Figure (3.2): Carrier distribution in an isolated p-type semiconductor. A) Plot of the distribution of density of states in the valence and in the conduction band. B) Plot of ( ) (Fermi function) and 1 ( ). c) Distribution of electrons and holes in the respective bands. 푓 퐸 − 푓 퐸

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Figure (3.3): Carrier distribution in an isolated n-type semiconductor. A) Plot of the distribution of density of states in the valence and in the conduction band. B) Plot of ( ) (Fermi function) and 1 ( ). C) Distribution of electrons and holes in the respective bands. 푓 퐸 differentiating the− equation푓 퐸 for we get

푛 = (3.4) 푑푛0 푛0 푑퐸퐹 푑퐸푖 Substituting the above values of and 푑in� Equation푘푘 � (3.1)푑� −we 푑get� � 푑� 푑� ℰ = + (3.5) 1 푑퐸푖 푛0 푑퐸퐹 푑퐸푖 Recalling Einstein relation which 퐽is푛 푞 푛 휇푛 푞 푑� 푞 퐷푛 푘푘 � 푑� − 푑� �

= (3.6) 푞 we can write as 휇푛 퐷푛 푘푘 푛 퐽 = + = (3.7) 푑퐸푖 푑퐸퐹 푑퐸푖 푑퐸퐹 Similarly 퐽푛 푛0 휇푛 푑� 푛0 휇푛 푑� − 푛0 휇푛 푑� 푛0 휇푛 푑�

= (3.8) 푑퐸퐹 퐽푝 푝0 휇푝 푑�

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When is equal to zero, is zero, which means that under conditions of no current flow, 푑퐸퐹 is constant and does not vary with distance. Under that condition is also zero. Another way of 푑� 퐽푛 stating this is that under thermal equilibrium conditions, no current flows and that the Fermi energy 퐹 퐽푝 퐸does not vary with .

푥 P-N junction Under Thermal Equilibrium

The - junction is in thermal equilibrium when no external voltage is applied. Hence no electric current flows through the junction and the Fermi energy is constant and does not vary with distance. Let us define the푝 metallurgical푛 junction as the plane at which the semiconductor changes from -type to - type and assume that it is located at = 0. is closer to on the -side and closer to on the - side. (Recall the discussion on the location of the Fermi energy in the band gap of extrinsic 푝 푛 퐹 푉 퐶 semiconductor in chapter 2.) This means푥 that퐸 in order for 퐸 to be constant푝 with , has퐸 to decrease푛 with as we pass through the metallurgical junction from the -side to the -side. On the -side far 퐹 퐶 away from the junction, there is no electric field and hence퐸 is constant i.e., does푥 퐸not vary with distance푥 and is separated from by an amount determined by푝 the net ionized푛 acceptor concentration푝 퐶 . is below the intrinsic Fermi energy level , by an amount퐸 given by 퐹 − 퐸 퐴 퐹 푖 푁 퐸 퐸= − (3.9) 푁퐴 Similarly on the -side far away from the퐸푖 junction,− 퐸퐹 푘 is� constant𝑙 � 푛푖 �and is separated from by an amount determined by the net ionized donor concentration . is above the intrinsic Fermi energy 푛 퐸퐶 퐸퐹 level by an amount given by + 푁퐷 퐸퐹 푖 퐸 = + (3.10) 푁퐷 However, close to the junction,퐸 퐹 −(potential퐸푖 푘 energy)� 𝑙 � and푛푖 � therefore and vary gradually from their respective values on the -side to their values on the -side. This region where the potential 퐶 푖 푉 energy varies is called the transition region.퐸 In this region there is an electric퐸 field since퐸 and the electrostatic potential varies. Outside푝 the transition region the electric푛 field is zero. Since the electric 퐶 field lines emanate (flow out) from the positive charges and terminate on negative charges,퐸 the transition region comprises positive charges on one side and negative charges on the other side. The charge density in the transition region is non-zero whereas the charge density is zero outside the transition region. For this reason the transition region is called the space-charge region since it has a net charge density. The region outside the transition region is called the neutral region since it has no net charge density. How does a space charge arise? In the transition region on the -side, electron density is much less than the thermal equilibrium majority carrier density and therefore a charge density equal to the magnitude of the electrical charge times the ionized donor atom density arises푛 in this region. Similarly hole density is less than the thermal equilibrium majority carrier density value in the transition region on the -side. Thus there is a region on the -side close to the metallurgical junction with a net negative charge density equal to the electron charge times the ionized acceptor atom density. Electric lines of force emanate푝 from the positive charge on푝 the -side and terminate on the negative charge on the -side. 푛 푝 - 82 -

Another way of looking at the origin of the transition region is as follows: Again let us assume that the - junction is formed by joining a -type semiconductor and a -type semiconductor and fusing them. As soon as the two pieces of semiconductor have been joined together, electrons, which are large푝 in푛 number in the -region diffuse to푝 the -region where there are푛 only a few of them. Thus there is a diffusion current. The diffusion of electrons charges the -region negative with respect to the -region and therefore the푛 potential energy of the푝 electrons on the -side raises relative to the -side i.e., a potential energy barrier arises between the and regions.푝 This give rise to an electric field in the transition푛 region due to the gradient of the potential energy since the푝 potential energy changes 푛with from the -side to -side. This electric field causes푛 a drift푝 flow of electron current in the direction from the -side to the -side i.e., in the opposite direction to the diffusion current. The potential energy 푥 barrier in 푝the transition푛 region reduces the diffusion current and hence fewer electrons diffuse and therefore푝 the potential푛 energy rises less. Ultimately a steady state condition is reached in which the potential energy barrier is of such a magnitude that the drift current is exactly equal and opposite to the diffusion current. Hence no net current flows through the junction. In the transition region on the -side there is a net positive charge density due to the ionized donor atoms since electrons left this region to go initially to the -side. Similarly in the transition region on the -side, there is a net negative charge푛 density due to the majority carriers viz., holes having been neutralized by the electrons that came from the -side. We discussed푝 this model as though electrons initially 푝went from the -side to the -side. We would have arrived at the same conclusion if we had assumed that holes initially diffused form the - side푛 to the -side. 푛 푝 푝 Let푛 us now define the electrostatic potential as

∅ = + = + (3.11) 퐸퐶 퐸푖 퐸푔 The constant is for the purpose∅ − of푞 referencing𝑐𝑐𝑐𝑐 our electrostatic− 푞 − potential2푞 𝑐𝑐𝑐𝑐 to a particular reference or zero value. Since is constant with distance, choose the constant as + . Then becomes 퐸푔 퐸퐹 퐹 퐸 ( ) 2푞 푞 ∅ = + = (3.12) 퐸푖 퐸퐹 퐸퐹−퐸푖 , the electrostatic potential, varies∅ with distance− 푞 푞 in the transition푞 region. Since , the thermal equilibrium density of electrons, is given by ∅ 푥 푛0

= (3.13) �퐸퐹−퐸푖� 푘푘

We can express as 푛0 푛푖 푒 푛0 = 푞∅ (3.14) 푘푘 Similarly, since , the thermal equilibrium density푛0 of푛 푖holes,푒 is given by 푝0 = (3.15) �퐸푖−퐸퐹� 푘푘 we can express as 푝0 푛푖 푒 푝0 - 83 -

= 푞∅ (3.16) −푘푘 Thus both the electron and the hole densities vary with distance in the transition region 푝0 푛푖푒 . By taking the logarithm of both sides of Equations (3.14) and (3.16), we can express the electrostatic potential in terms of the thermal equilibrium carrier densities. 푥

= ln (3.17) 푘푘 푛0 and also ∅ 푞 �푛푖 �

= ln (3.18) 푘푘 푝0 These two equations relate the potential variation∅ − with푞 thermal�푛푖 � equilibrium carrier density variation. , the electrostatic potential, defined as above, can also be used in the neutral semiconductor. In a neutral -type semiconductor, , the thermal equilibrium electron density, is equal to , the thermal ∅equilibrium majority carrier density which is equal to , the ionized donor density, Substituting for 푛 푛0 푛푛0 in Equation (3.14) we find that , the electrostatic potential+ in the neutral semiconductor, is given+ by 푁퐷 푁퐷 0 푛 ∅ = ln + (3.19) 푘푘 푁퐷 where we have used a subscript in to denote∅푛 that푞 we talking� 푛푖 � about the electrostatic potential in a neutral -type semiconductor. Similarly we can define an electrostatic potential for a -type material as 푛 ∅ 푛 = ln 푝 − (3.20) 푘푘 푁퐴 where we have used a subscript in to denote∅푝 the− electrostatic푞 � 푛푖 � potential in a neutral p-type semiconductor. 푝 ∅

Example

Let us now calculate the electrostatic potential for a -type silicon which has a concentration of 10 cm-3 acceptor atoms at room temperature. We assume that at room temperature all the impurity ∅ 푝 atoms16 are ionized, i.e., = . Let us take as equal to 1 × 10 . − 10 −3 푁퐴 푁퐴 푛푖 𝑐 = ln − 푘� 푁퐴 ∅푝 − � � 푞 푛푖 - 84 -

10 = .02585 × ln 1 × 1016 10 = −0.02585 × 13.81�= 0.357�

Let us now go back to the space-charge− region. Differentiating− given� in Equation (3.14) we obtain

0 = 푛 (3.21) 푑푛0 푞 푑∅ This yields 푑� 푛0 푘푘 푑�

= (3.22) 푞 which can be written as �푛0 푛0 푘푘 �∅

= (3.23) 푘푘 푑푛0 We can integrate Equation (3.23) to obtain�∅ the 푞potential푛0 difference between two points say A and B in a semiconductor. The potential difference is equal to

∅퐴−퐵 = = = (3.24) 퐴 푘푘 퐴 푑�0 푘푘 푛퐴 We can use this result∅ 퐴to− obtain퐵 ∫ an퐵 �expression∅ 푞 for∫퐵 the푛0 potential푞 𝑙 barrier�푛퐵� that exists between the -side and the -side in a - junction. This potential barrier is called the built-in voltage since it is already in existence in a - junction under thermal equilibrium. This built-in voltage which is denoted by푛 is the potential푝 difference푝 푛 between the neutral -side and the neutral -side and according to the above equation, is equal 푝to푛 푽풃풃 푛 푝 = = = − + (3.25) 푘푘 푛푛0 푘푘 푛푛0푝푝0 푘푘 푁퐴 푁퐷 푏� 2 2 Equation (3.25) can be� rewritten푞 as𝑙 �푛푝0� 푞 𝑙 � 푛푖 � 푞 𝑙 � 푛푖 �

= + + − (3.26) 푘푘 푁퐷 푁퐴 푏� 푖 푖 We could have arrived at the� same result푞 � 𝑙by �a different푛 � 𝑙 route� 푛 also.�� We know and , the electrostatic potentials of the neutral and the region respectively. The built-in voltage is the ∅푝 ∅푛 electrostatic potential difference between the neutral and the regions. 푝 푛 �푏� Hence 푛 푝

= (3.27)

Since is negative from Equation (3.20) we� can푏� write∅ 푛 − ∅푝 ∅푝

- 85 -

= + = + + − (3.28) 푘푘 푁퐷 푁퐴 푏� 푛 푝 푖 푖 which is the same result that� we obtained∅ before.�∅ � 푞 �𝑙 � 푛 � 𝑙 � 푛 ��

Example

Let us now determine the built-in potential for a p-n junction in which the n-region is uniformly doped with a net donor concentration of 2 x 1016 cm-3 and the p region is uniformly doped with a net acceptor concentration of 1 x 1015 cm-3. Assume room temperature and that all the impurity atoms are ionized. From Equation 25 we find 2 10 1 10 = 0.02585 × (1 ×1610 ) 15 푥 푥 푥 푏� 10 2 � = 0.02585𝑙 �× 26.02 � = 0.673

An abrupt - junction is one in which the net acceptor� density is uniform in the -region and the net donor density is uniform in the -region. The impurities change from acceptor to donor type abruptly as we go across푝 푛 the metallurgical junction. In Figure (3.4 A) the abrupt junction is푝 shown with the metallurgical junction at = 0. The푛 and the regions are connected externally by a copper wire. The junction is therefore in thermal equilibrium. In Figure (3.4 B) is shown the band-bending i.e., the bending of , and , that푥 occurs in 푝the transition푛 region as we go from the -side to the -side. The built-in voltage is the difference between the potential energy in the -side and that in the -side 퐶 푖 푉 divided by 퐸. In퐸 other words,퐸 it is equal to the amount of band-bending divided by푝 . In Figure푛 (3.4 C) the space charge distribution is shown in the transition region. The charge푝 density in the transition푛 region is due to the 푞ionized impurity atoms that are not compensated by equal number of majority푞 carriers. Hence the charge density in the space charge region is = on the -side and = + on the -side. Since the - junction is in thermal equilibrium, is constant and does not vary with . On 퐴 퐷 the other hand varies with in the depletion region. 휌The electron−푞 푁 density푝 and the hole휌 density푞 푁 in the 퐹 specie푛 -charge region 푝are푛 given by 퐸 푥 퐸푖 푥

= 퐸퐹−퐸푖 푘푘 and 푛 푛푖 푒

= 퐸퐹−퐸푖 − 푘푘 Hence and vary with position, , in the 푝depletion푛푖 푒 region. The electron density decreases from , the thermal equilibrium majority carrier concentration on the -side, to , the thermal equilibrium 푛 푝 푥 푛푛0 minority carrier concentration on the -side. The electron density in the portion of the transition (space 푛 푛푝0 charge) region lying on the -side is less than , the thermal equilibrium majority carrier density in the 푝 푛 푛푛0 - 86 - neutral -region. Hence this region has a net charge density equal to a and is depleted of electrons. By a similar argument, the portion of the transition region on the -side is+ depleted of holes and the 퐷 charge density푛 in this region Is . The carrier densities are smaller푁 than the thermal equilibrium 푝 majority carrier densities in respective+ portions of the entire space-charge region. Hence the space- 퐴 charge region is said to be depleted−푞푁 of charge carriers and the space-charge region is usually referred to as the depletion region. The product of the electron and the hole densities anywhere in the transition region is equal to the square of the intrinsic carrier concentration since the junction is in thermal equilibrium, i.e., the law of mass action is obeyed. At the two edges of the space-charge region the charge density gradually approaches zero as shown in Figure (3.4 C) since the carrier densities change gradually. However for ease of calculation and analysis, we approximate the space charge density distribution as a rectangular distribution in which the charge density becomes zero abruptly at the two edges as shown in Figure (3.4 D). We assume here that the negative charge density due to ionized acceptors exists between = and = 0 and the positive charge density due to ionized donors exists between = 0 and = . The electric field variation in the space region is shown in Figure (3.4 푥 −푥푝 푥 E). Outside the space charge region, the electric field is zero. Inside the space charge region, the electric 푛 field varies linearly푥 and the푥 magnitude푥 reaches a maximum value at the metallurgical junction. The electric field is the negative gradient of the electrostatic potential. Therefore the line integral of the electric field yields the electrostatic potential difference between two points. The area under the plot of the electric field in Figure (3.4 E) is equal to the built-in voltage . The linear variation of the electric field in the space-charge region is due to the assumption that the impurity density is constant in each 푏� region and that it changes from one type to the other abruptly at� the metallurgical junction.

- 87 -

Figure (3.4): (A) An abrupt p-n junction with the metallurgical junction at = 0. (B) Bending of the band at the junction. The built-in voltage is the difference in the potential energy between the two sides divided by the electron charge . (C) The actual space charge region with푥 the charge density gradually approaching zero at the two edges. (D) Approximation by a rectangular charge density in which the density abruptly goes to zero푞 at the edges. (E) Electric field variation in the space charge region.

Space Charge Region

We will now calculate the potential distribution and the width of the space charge region in the abrupt junction. There are two ways by which one can determine the space charge region characteristics. One is using Poisson equation. Another is to use Gauss theorem. Both the approaches are the same from the physics point of view. In certain cases one method may be more direct while in some other cases the other method may be more direct. Hence we will consider both the methods. Poisson Equation

The Poisson equation is given by

= (3.29) 2 푑 ∅ 휌 2 Let us assume that the metallurgical junction푑푥 is at −=휖0푠 and the space charge region is approximated by the rectangular density shown in Figure (34 D) bounded by = on the side and = on the 푥 side. The charge density in the various regions can then be written as 푥 −푥푝 푝 푥 푥푛 푛 For < , = 0

For 푥 < −푥<푝 0 휌 = − 푝 For − 푥 0 < 푥 < , 휌 = −푞푁퐴 + For >푥 , 푥 푛 휌 = 0− 푞푁퐷 In the space-charge region, under steady푥 state푥 conditions,푛 all the impurity휌 atoms are ionized even at low temperatures3 i.e., = and = . This is in contrast to the neutral region where the impurity atoms are partially ionized+ at low temperatures.− 푁퐷 푁퐷 푁퐴 푁퐴 Let us consider the space charge region (the depletion region) on the side. = = 푝 − The Poisson equation is given by 휌 − 푞푁퐴 − 푞푁퐴

= (3.30) 2 푑 ∅ 푞푁퐴 2 3 푑푥 휖푠 The theoretical reason for this is that the electric field in the depletion region ionizes all the impurity atoms even at low temperatures

- 88 -

Integrating the above equation we get

= + (3.31) 푑∅ 푞푁퐴푥 where is an integration constant. 푑� 휖푠 퐶1

1 퐶Since the electric field in the neutral region is 0, must be equal to 0 at = . Applying 푑∅ this boundary condition we obtain 푝 푑� 푥 −푥

= (3.32) 푞푁퐴푥푝 1 푠 And substituting this value of in the equation퐶 for 휖 we get 푑∅ 1 퐶 푑� = + (3.33) 푑∅ 푞푁퐴 Again integrating 푑� 휖푠 �푥 푥푝�

= + + (3.34) 푞푁퐴 2 푞푁퐴 Where is an integration constant.∅ 2휖푠 푥 휖푠 푥푝푥 퐶2 Starting퐶 with2 the charge density in the -side, which is equal to

푛 = = + We get 휌 푞 푁퐷 푞 푁퐷

= (3.35) 2 푑 ∅ 푞푁퐷 2 Integrating, 푑푥 − 휖푠

= + (3.36) 푑∅ 푞푁퐷푥 ′ 푠 1 Where is an integration constant. Using푑 �the boundary− 휖 condition퐶 that the electric field is zero ′ 푑∅ 1 in the neutral퐶 n region and hence is zero at = , we find �− 푑� � 푑∅ 푛 푑� 푥 푥 = ′ 푞푁퐷푥푛 퐶1 Hence, 휖푠 ( ) = + = (3.37) 푑∅ 푞푁퐷푥 푞푁퐷푥푛 푞푁퐷 푥푛−푥 Integrating 푑� − 휖푠 휖푠 휖푠

- 89 -

= + + (3.38) 푞푁퐷 2 푞푁퐷 where is an integration constant. ∅ − 2휖푠 푥 휖푠 푥푛푥 퐶3 퐶Realizing3 that given by Equations (3.34) and (3.38), should have the same value at = 0, we put = . Therefore, the potential at – = , is given, from Equation (3.34), by ∅ 푥 퐶2 퐶3 ∅ 푥 −푥푝 = = 2 2 + = 2 + (3.39) 푞푁퐴푥푝 푞푁퐴푥푝 푞푁퐴푥푝 Similarly, the potential ∅ at�푥 = −푥 is푝 obtained� 2 from휖푠 −Equation휖푠 (3.38)퐶2 as − 2휖푠 퐶2 ∅ 푥 푥푛 ( = ) = + = + (3.40) 2 2 2 푞푁퐷푥푛 푞푁퐷푥푛 푞푁퐷푥푛 The total potential difference∅ 푥 between푥푛 the− neutral2휖푠 − side휖 푠and the퐶 neutral2 2 휖side푠 is given퐶2 by = ( = 푛) = 푝

�푏� ∅ 푥 푥푛 − ∅ 푥 −푥푝 = + � + � 2 2 2퐷 푛 2퐴 푝 푞푁 푥 2 푞푁 푥 2 푠 퐶 − �− 푠 퐶 � = 휖 + 휖 (3.41) 2 2 푞푁퐷푥푛 푞푁퐴푥푝 The amount of positive charge2 in휖푠 the -side2휖푠 of the space-charge region, per unit area of the junction, is equal to . Similarly the amount of negative charge in the -side of the space-charge region, per unit area of the junction, is equal to푛 . Since the positive charge in the side of the 푞푁퐷푥푛 푝 depletion region should be equal to the magnitude of the negative charge in the side, we get −푞푁퐴푥푝 푛 = 푝 (3.42)

Using this equality we can express as 푁퐴푥푝 푁퐷푥푛 �푏� = + = + 2 2 2 2 푞푁퐷 2 푞푁퐴 푞푁퐷 2 푞푁퐷푥푛푥푝 �푏� 푥푛 푥푝푥푝 푥푛 휖=푠 휖푠 + = 휖푠 +휖푠 푞푁퐷푥푛 푞푁퐷푥푛 푁퐷푥푛 2휖푠 �푥푛 푥푝� 2휖푠 ( �푥푛 ) 푁퐴 � = 1 + = 2 2 푞푁퐷푥푛 푁퐷 푞푁퐷푥푛 푁퐴+푁퐷 Therefore 2휖푠 � 푁퐴� 2휖푠 푁퐴

= (3.43) ( ) 2휖푠푉푏� 푁퐴 푛 Similarly it can be shown that 푥 � 푞 푁퐷 푁퐴+푁퐷

- 90 -

= (3.44) ( ) 2휖푠푉푏� 푁퐷 푝 퐴 퐴 퐷 The total depletion region width is푥 equal to� 푞 푁 푁 +푁

푥 푑 = +

푑 푛 푝 =푥 푥 푥 + ( ) 2휖푠푉푏� 푁퐴 푁퐷 �푞 푁퐴+푁퐷 ��푁퐷 �푁퐴� = ( ) 2휖푠푉푏� 푁퐴+푁퐷 푞 푁퐴+푁퐷 �푁퐷푁퐴 � ( � ) � = (3.45) ( ) 2휖푠푉푏� 푁퐴+푁퐷 It must be pointed out that the temperature� 푞 푁퐷푁 dependence퐴 of arises due to the temperature dependence of as expressed in Equation (3.28). In that equation we expressed the built-in voltage as 푑 sum of two components and . We have to remember that these푥 two components are not the �푏� potential drop across the depletion region on the side and the potential drop across the depletion ∅푛 ∅푝 region on the side. In Equation (3.41) we again expressed as sum of two components. However in this equation, the first component expresses the potential푛 drop across the depletion region in the side 푏� while the second푝 component is the potential drop across the� depletion region in the side. 푛 Let us now express the potential as a function of distance in the depletion region.푝 The electrostatic potential given in Equation (3.34) is measured with respect to some arbitrary zero reference by an appropriate choice of . Let us measure the potential with respect to the neutral - ∅ side i.e., choose the electrostatic potential on the neutral -side as zero, and denote this potential by . 퐶2 푝 can be obtained from proper choice of . Using Equation (3.34) we obtain for the region 푝 < < 0 as 휓 휓 ∅ 퐶2 휓 푝 − 푥 푥 = + + (3.46) 푞푁퐴 2 푞푁퐴 ′ where ʹ is the new constant which sho휓 uld 2make휖푠 푥 equal to휖푠 zero푥푝 푥at =퐶2 . By putting this condition in the above equation we obtain 퐶2 푥 −푥푝

= 2 (3.47) 퐴 푝 ′ 푞푁 푥 By substituting this in the above equation for 퐶 2we get 2휖푠

휓 (3.48) = + + 2 = + 퐴 푝 푞푁퐴 2 푞푁퐴 푞푁 푥 푞푁퐴 2 휓 2휖푠 푥 휖푠 푥푝푥 2휖푠 2휖푠 �푥푝 푥� at = 0 is obtained as

휓 푥 - 91 -

( ) (3.49) = 0 = 2 푞푁퐴푥푝 Equation (3.48) describes the potential휓 푥 variation in 2the휖푠 depletion region on the side, the potential being measured with respect to the neutral region. Since we are measuring the potential, , 푝 with respect to the neutral region, the value of at = given by Equation (3.49) is equal to the 푝 potential drop across the segment of the transition region lying on the -side. 휓 푝 휓 푥 −푥푝 Using Equation (3.38) we can express the potential variation in 푝the depletion region on the side with respect to the same zero reference also by using the same constant , and show that the 푛 potential ( ) in the depletion region on the n side i.e., for 0 < < as ′ 퐶2 푥 푥 푥푛 휓 ( ) (3.50) = + + 2 퐴 푝 푞푁퐷 2 푞푁퐷 푞푁 푥 This is equivalent to putting the휓 constant푥 − in2휖 Equation푠 푥 (3.휖푠38푥) 푛as푥 equal to2휖 푠 퐶3 2 푞푁퐴푥푝 We could have started also with the zero2 휖reference푠 for the potential as that on the neutral region and derived a similar set of expressions for the potential in the depletion regions on the two sides of the metallurgical junction. 푛

We summarize in Table (3.1), the step by step solution of Poisson equation in the space charge region of an abrupt p-n junction. The solutions to the Poisson equation with two different zero references are also given in this table.

Table (3.1). The summary of the step by step solution of Poisson equation in the space charge region of an abrupt p-n junction

Region Region

< < 0 0 < <

−푥푝 푥 푥 푥푛 - 92 -

2 퐴 퐷 � ∅ 푞푁 푞푁 2 푠 − 푠 �푥 휖 휖 + + ʹ 퐴 퐷 �∅ 푞푁 푥 푞푁 푥 퐶1 − 퐶1 𝑑 휖푠 휖푠 Boundary condition = 0 = = 0 = �∅ �∅ 푎� 푥 푥푝 푎� 푥 푥푛 𝑑 𝑑 = ʹ = Integration constant 푞푁퐴푥푝 푞푁퐷푥푛 퐶1 퐶1 휖푠 ( 휖푠 ) + �∅ 푞푁퐴 퐷 푛 푝 푞푁 푥 − 푥 푠 �푥 푥 � 푠 𝑑 휖 휖

+ + + + 2 2 푞푁퐴 2 푞푁퐴 퐷 2 퐷 푝 2 푞푁 푞푁 푛 3 ∅ 푠 푥 푠 푥 푥 퐶 − 푠 푥 푠 푥 푥 퐶 휖 휖 휖 휖 Boundary condition at = zero: ( = 0) 2 3 2 퐶 퐶 퐶 ∅ 푥 Zero reference = = 0

푝 ∅�푥 −푥 � = Integration Constant 2 2 2 2퐴 푝 푞푁퐴푥푝 푞푁 푥 2 퐶 푠 푠 휖 휖

+ + + 2 = 1 + 2 2 2 2 퐷 퐷 2퐴 푝 퐴 2푝 2퐴 푝 푞푁 2 푞푁 푞푁 푥 푞푁 �푥 푥� 푞푁 푥 푥 푛 ∅ � � − 푠 푥 푠 푥 푥 푠 휖푠 휖푠 푥푝 휖 휖 휖

Zero reference ( = ) = 0

푛 ∅ 푥 푥 Integration constant = = 2 2 2 2 푞푁퐷푥푛 푞푁퐷푥푛 퐶2 − 퐶3 − 휖푠 휖푠 + ( ) = 1 22 2 2 2 2 2 2 ∅ 푞푁퐴 푥 푞푁퐷푥푛 푞푁퐷 2 푞푁퐷푥푛 푥 푝 푛 푠 �푥 푥 � − 푠 − 푠 푥 − 푥 − 푠 � − 푛� 휖 휖 휖 휖 푥

Gauss Theorem Approach

The origin of the electric field in the depletion region is more intuitively understood if the depletion region is examined using Gauss Theorem. We will now derive the same result using Gauss theorem. As before let us assume that the depletion region charge is rectangular as shown in Figure (3.5 A).

- 93 -

If we were imagine a cylindrical section of the depletion region, with unit area of cross-section, as shown in Figure (3.5 B), then the total negative charge contained between – and = 0 in the cylindrical section is equal in magnitude to the total positive charge in the cylinder between = 0 an 푥푝 푥 . This equality of charge magnitude is required according to Gauss theorem due to the fact that the electric field is zero outside the depletion region i.e., in the neutral and regions. The electric푥 field at 푛 any푥 point in the depletion region can be obtained by partitioning the cylinder with a plane at and considering the charge contained in either part of the cylinder. For example,푛 푝 in Figure (3.5 C), the left part of the푥 cylinder contained between = and is shown. The electric field at is the charge푥 contained in the left part of the cylinder divided by , the permittivity of the semiconductor. 푥 푥푝 푥 푥 휖푠 ( ) = = (3.51) 푞푁퐴�푥−�−푥푝�� 푞푁퐴�푥+푥푝� In the above discussionℰ 푥 we assume− that휖 푠 is located between− 휖 –푠 and 0 i.e., in the -side of the depletion region. The field lines come from the right to the left and hence the field is negative. in 푥 푥푝 푝 Equation (3.51) is also negative. 푥 The electric field, ( ), at is also equal to the charge contained in the right part of the cylinder between and as shown in Figure (3.5 D), divided by , the permititivity. Hence, ℰ 푥 푥 푥 푥푛 ( ) ( 휖푠) ( ) = + = + (3.52) 푞푁퐴 푥−0 푞푁퐷 푥푛−0 푞푁퐴푥 푞푁퐷푥 Since = ℰ ,푥 , this expression− � 휖푠 for , the electric휖푠 field,� can− be� seen휖푠 to be휖 푠the� same as the one in Equation (3.51). It is always convenient to consider the charge contained in that part of the cylindrical 퐷 푛 퐴 푝 section푁 in푥 which푁 the푥 impurity atoms are of ℰthe same type. Hence Equation (3.51) is the preferred choice to express ( ) for < < 0.

If ℰ was푥 chosen−푥 푝to be푥 between 0 and i.e., in the -side of the depletion region, then we can write ( ) as the charge contained in the right part of the cylinder divided by . 푥 푥푛 푛 ℰ 푥 ( ) 휖푠 ( ) = (3.53) 푞푁퐷 푥푛 − 푥 We can obtain the potential variationℰ 푥 as a− function휖푠 of by taking the line integral of ( ). If we take the zero reference for the potential as that at the neutral -region, then 푥 ℰ 푥 ( ) = (푝) 푥 휓 푥 − ∫−푥푝 ℰ 푥 𝑑 = 푥 푞 푁퐴�푥 + 푥푝� 푝 푠 = ∫−푥 +휖 𝑑 2 푥 푞 푁퐴 푥 휖푠 2 푝 � 푥 푥�−푥푝 = + 2 2 푝 푞 푁퐴 푥 푥 2 휖푠 � 2 푥푝푥 − � 2 − 푥푝�� - 94 -

= + + 2 2 푞 푁퐴 푥 푥푝 휖푠 2 푝 2 = � +푥2 푥 + � 푞 푁퐴 2 2 2휖 푠 푥 푥푝푥 푥푝 = � + � (3.54) 푞 푁퐴 2 푝 This is the same result that we obtained2휖푠 in� Equation푥 푥 (3.48)� by solving Poisson equation.

Figure (3.5): The charge distribution in the depletion region of an abrupt p-n junction: A) rectangular profile of the charge distribution. B) a cylindrical section of unit cross-sectional area of the entire depletion region. C) the left part of the cylindrical section obtained by partitioning the cylindrical section with a plane at . We have chosen x to be in the -side of the depletion region. D) the right part of the cylindrical section. 푥 푝 A review of the p-n junction under thermal equilibrium

Let us now examine the - junction under thermal equilibrium i.e., when no external voltage is applied across the junction. Free carriers are present in the two neutral regions with their densities equal to the thermal equilibrium푝 values.푛 In the depletion region, the carrier densities decrease from the thermal equilibrium majority carrier density values at one end of the depletion region to the thermal equilibrium minority carrier density value at the other end. Therefore there is a diffusion of electrons

- 95 - across the depletion region from the -side to the -side, and a diffusion of holes from the -side to the -side. The electric current due to the diffusion action is called the diffusion current. 푛 푝 푝 푛 The electric field in the depletion region drives the minority carriers entering the depletion region from one neutral side to the opposite neutral side. For example, electrons entering the depletion region (due to the random thermal motion) from the neutral -region where they are minority carriers will be driven by the depletion region electric field to the neutral -region. The holes will similarly be driven from the neutral -region to the neutral -region. The 푝resulting electric current is called the drift current. The direction of the drift current balances out exactly the푛 diffusion current and hence the net current through the junction푛 is zero. 푝

This process of diffusion and drift can be understood by reference to Figure (3.6 A). This figure shows in addition to the band diagram of the - junction under thermal equilibrium, the electron and hole distributions in the two neutral regions. The carrier distribution functions are rotated by 90 degrees from the way they are drawn in Figure (3.2 C) 푝and푛 Figure (3.3 C) since the variation of energy is vertical in this figure. The electrons in the -region lying above the dotted line represent those that have the kinetic energy larger than the barrier height and therefore will be able to diffuse to the -region. 푛 푝

Figure (3.6): Energy band diagram of a p-n junction and the carrier distribution under (A) thermal equilibrium (B) a forward bias and (C) a reverse bias.

- Junction under Forward Bias

푃 푁 When an external voltage source such as a battery is connected across the - junction such that the positive terminal of the battery is connected to the -side and the negative terminal is connected to the -side as shown in Figure (3.7), the - junction is said to be forward푝 푛 biased. The potential energy of the electrons on the -side is increased relative푝 to the potential energy of the electrons on the 푛-side by an amount equal to where푝 푛 is the externally applied voltage. The 푛 푝 푞�퐹 �퐹 - 96 - junction is said to be forward biased under this condition. The band diagram under forward bias is illustrated in Figure (3.6 B). The amount of band-bending is equal to the difference between (the potential energy of the electrons) in the neutral region and that in the neutral region: The potential 퐶 energy barrier height, is now ( ). The barrier height is hence reduced by an amount퐸 under forward bias. In the neutral regions far away from푝 the depletion region, the carrier푛 densities are at their 푏� 퐹 퐹 thermal equilibrium values and푞 the� −Fermi� energy is located in the band gap at a level determined푞� by the net impurity (acceptor and donor) concentrations. The Fermi energy in the neutral region (far away from the depletion region) is shifted upward by from the Fermi energy in the neutral region (far away from the metallurgical junction). 푛 푞�퐹 푝 The expression for the depletion region width under forward bias can be derived exactly the same way as we did for the thermal equilibrium case either by solving the Poisson equation or by using Gauss theorem. The expression for the depletion region width will have essentially the same form excerpt the potential barrier height will be replaced by ( - ). Using the same notation for the depletion region width as in the thermal equilibrium case, we have �푏� �푏� �퐹

= ( ) 1 (3.55) ( ) 2휖푠푁퐷 2 푥푝 �푞푁퐴 푁퐴+푁퐷 �푏� − �퐹 � = ( ) 1 (3.56) ( ) 2휖푠푁퐴 2 and 푥푛 �푞푁퐷 푁퐴+푁퐷 �푏� − �퐹 � = +

푑 푛 푝 푥 푥 ( 푥 ) = ( ) 1 (3.57) 2휖푠 푁퐴+푁퐷 2 We denote that the depletion region width is smaller� under푞푁퐷 푁forward퐴 � bias푏� − conditions�퐹 � than under thermal equilibrium. The potential barrier is less under forward bias conditions, and hence less charge is needed to sustain in a smaller potential barrier. The depletion region has a smaller width.

- 97 -

Figure (3.7): A - junction under forward bias. is the voltage applied across the -

푝 푛 �퐹 푝 푛

Example

Calculate the depletion region width in an abrupt - junction with a net impurity density of = 10 on the -side at room temperature (a) in thermal equilibrium and (b) at a forward applied 푝 푛 푁퐴 voltage16 bias−3 of 0.4 V. Take as equal to 10 . Using Equation (3.25) we get 𝑐 푛 10 −3 푛푖 𝑐 × = 0.0259 × = 0.656 16 15 10 10 20 (a) Thermal equilibrium: �푏� 𝑙 � 10 � � . × . × × = 1 . × ×( −14 15 ) 2푥11 9 8 854 10 10 2 −19 16 16 15 푝 1 6푥10 10 10 +10 푥 � = 3.627 × 10 × � × 0.656 .15 3 10 16 16 = 0.0886 × 10 �cm10 푥1 1푥10 √ −4 Similarly

= 3.627 × 10 × × 0.656 × .16× 3 10 15 16 푥 푛 = 0.886 × 10 cm� 10 1 1 10 √ = (0.886 + 0.−08864 ) × 10 cm −4 푑 = 0.975 × 10 cm 푥 −4 (b) Forward Bias: The barrier height is now – equal to 0.656 0.4= 0.256 V

�푏� �퐹 − = 3.627 × 10 × × 0.256 × .15× 3 10 16 16 푥 푝 = 0.553 × 10 cm�10 1 1 10 √ Similarly −4

= 3.627 × 10 × × 0.256 × .16× 3 10 15 16 푥 푛 = 0.553 × 10 cm�10 1 1 10 √ = (0.553 + 0.553−4 )x10 cm −4 푑 푥 - 98 -

Referring to Figure (3.6 B), the electrons that are distributed above the dotted line have adequate energy to clear the barrier, and diffuse to the other side. This will result in diffusion current. This diffusion current is larger than in thermal equilibrium since there are more electrons with kinetic energy adequate to go over the smaller barrier height. On the other hand, the drift current of electrons will be the same since it does not depend on the barrier height. Similarly more holes will be able to diffuse than in the thermal equilibrium case causing a larger hole diffusion current while the hole drift current will be the same as in thermal equilibrium.

Due to the diffusion of electrons from the -side to the -side, the (electron) minority carrier density in the -side increases from its thermal equilibrium value; hence, this process is called minority carrier injection. Similarly, holes are also injected into푛 the neutral푝 region from the region under this process. 푝 푛 푝 We saw earlier that under conditions of no current flow, the electron densities at two different locations and are related by the following expression:

푥퐴 푥퐵 ( ) = ( ) (3.58) 푞휓퐴� 푘푘 and 푛 푥퐴 푛 푥퐵 푒

( ) = ( ) (3.59) 푞휓퐴� − 푘푘 Where is the electrostatic potential푝 difference푥퐴 푝 between푥퐵 푒 and . This is called the Boltzmann relation. 휓퐴� 푥퐴 푥퐵 In the case of a - junction under thermal equilibrium, the condition of zero net current is realized, due to two large, equal and opposite currents. One of these currents is due to diffusion, and the other to drift. However,푝 푛 when a forward bias is applied, a current flows across the junction due to the minority carrier injection process. We will assume that the current through the device under forward bias is very small in comparison with the drift and drift components of current flowing under thermal equilibrium. This is tantamount to assuming that the drift and diffusion components are nearly equal under forward bias. We can therefore assume that the carrier densities are still given by the Boltzmann relation (Equation (3.58)) even under forward bias.

As before, we will choose the origin = 0 at the metallurgical junction. With the boundary of the depletion region on the - side at = , and that of the - side at = , as shown in Figure 푥 (3.8). The width of the neutral region is the distance between the boundary of the depletion region 푝 푥 −푥푝 푛 푥 푥푛 ( = ) and the ohmic contact to the neutral region. The ohmic contact is what is used to apply a 푛 voltage to the device and send 푛a current. In our example푊 we have taken the location of the ohmic 푛 contact푥 푥 to the neutral region as = + 푛. Similarly, the ohmic contact to the neutral region is located at = , where is the width of the neutral region. 푛 푥 푥푛 푊푛 푝 The푥 electron−푥푝 − densities푊푝 at 푊=푝 and = are related푝 by the Boltzmann distribution, and therefore, 푥 푥푛 푥 −푥푝

- 99 -

( ) = = ( = ) − 푞 푉푏�−푉퐹 푘푘 Similarly, the hole density 푛at푝 �푥= −is given푥푝� by 푛푛 푥 푥푛 푒 푥 푥푛 ( ) ( = ) = ( = ) − 푞 푉푏�−푉퐹 푘푘 The minority carrier density푝 푛under푥 thermal푥푛 equilibrium푝푝 푥 is− 푥푝 on푒 the side, and on the side. The extra (also called excess) minority carrier densities푛푝� at =푝 and at 푝푛=� under푛 forward bias are given by 푥 푥푛 푥 −푥푝 ( = ) = ( = )

and ∆푝 푥 푥푛 푝푛 푥 푥푛 − 푝푛� = = =

Due to the fact that we have∆ excess푛�푥 minority−푥푝 �carriers푛푝 in� 푥the neutral−푥푝� regions,− 푛푝� equal amounts of excess majority carriers flow into the neutral regions from their respective ohmic contacts to maintain charge neutrality. Therefore, the majority carrier densities are also increased from their thermal equilibrium values, i.e.,

= = + = = + = (3.60) and 푝푝�푥 −푥푝� 푝푝� ∆푝�푥 −푥푝� 푝푝� ∆푛�푥 −푥푝� ( = ) = + ( = ) = + ( = ) (3.61)

We distinguish푛푛0 two푥 cases:푥푛 one called푛푛� high∆ 푛injection,푥 푥 and푛 the푛 other푛� called∆푝 low푥 injection.푥푛 When

= is > ,

or when ∆푛�푥 −푥푝� 푝푝� ( = ) is > ,

we call it a high injection condition. When ∆푝 푥 푥푛 푛푛� = < , i.e., ( = ) ( see Eq. (3.60) ) and when ∆푛�푥 −푥푝� 푝푝� 푝푝 푥 −푥푝 ≈ 푝푝� ( = ) < , i.e., ( = ) ( see Eq. (3.61) ) we have a low and moderate∆푝 푥 injection푥푛 condition.푛푛� We will푛 푛restrict푥 our푥푛 discussion≈ 푛푛� to low and moderate injection cases only. Therefore,

- 100 -

( = ) and ( = ) (3.62)

The expression for (푛푛= 푥 ) can푥푛 be≈ simplified푛푛� as 푝푝 푥 −푥푝 ≈ 푝푝� 푛푝 푥 −푥푝 = = ( = ) −푞푉푏� 푞푉퐹 푘푘 푘푘 푝 푝 푛 푛 푛 � 푥 − 푥 � = 푛 푥 푥 푒 푒 −푞푉푏� 푞푉퐹 푘푘 푘푘 푛� = 푛 푒 푒 (3.63) 푞푉퐹 푘푘 In the above equation, we made use of the Boltzmann푛푝 relation� 푒 for the thermal equilibrium carrier densities, i.e.,

= −푞푉푏� 푘푘 Similarly, the expression for ( = ) 푛can푝� be obtained푛푛� 푒 as 푝푛 푥 푥푛 ( = ) = (3.64) 푞푉퐹 푘푘 = = The excess minority carrier densities at푝 푛 푥 and푥푛 푝푛 are�푒 obtained as 푥 푥푛 푥 −푥푝 ( = ) = = 1 (3.65) 푞푉퐹 푞푉퐹 푘푘 푘푘 and ∆푝 푥 푥푛 푝푛�푒 − 푝푛� 푝푛� �푒 − �

= = = 1 (3.66) 푞푉퐹 푞푉퐹 푘푘 푘푘 These excess minority∆푛�푥 carrier−푥 푝charges� 푛 will푝�푒 now diffuse− 푛푝� into the푛푝 �two�푒 respective− � neutral regions. Before we study the diffusion process, we will discuss another concept called the Quasi-Fermi Level.

- 101 -

Figure (3.8): The depletion and neutral regions in a forward biased - junction

Example 푝 푛

Let us calculate the excess carrier density in the neutral and regions of a - junction, at the boundary of the depletion region. Assume = 10 on the -side, and = 5 x10 on the -side. 푝 푝 푛 Assume a forward bias voltage of 0.4 V. 16 −3 16 −3 푁퐴 𝑐 푝 푁퐷 𝑐 푛 10 and = × = 10 2 20 16 −3 푛푖 10 4 −3 16 푝푝� ≈ 𝑐 푛푝� 푝푝0 10 𝑐 = 5 × 10 and = = = 2 × 10 2 × 20 15 −3 푛푖 10 4 −3 15 푛푛0 𝑐 푝푛0 푝푝0 5 10 𝑐 ( = ) = 1 푞푉퐹 푘푘 ∆푝 푥 푥푛 푛푝� �푒 − � = 0.0259 푘� . � ( ) . 푞 [ ] = = 10 0 4 1 10 × 5.1 × 10 1 5.1 × 10 4 0 0259 4 6 10 −3 ∆푝 푥 푥푛 �푒 − � − ≈ 𝑐 . . = = 2 10 × 0 4 1 × 10 5.1 × 10 = 1.02 × 10 4 0 0259 4 6 11 −3 푝 ∆푛�푥 −푥 � 푥 �푒 − � 푥 𝑐

Quasi-Fermi Level

In thermal equilibrium, we expressed the minority carrier density in terms of Fermi energy. However, under forward bias it will not be possible to do so. We therefore define a parameter called the quasi-Fermi energy, to enable us to write the expression for the minority carrier density in the same form as in the thermal equilibrium case.

We define a quasi-Fermi energy level, , for electrons such that the electron density can be written as 퐸퐹푛 ( ) = 퐸퐹푛−퐸푖 푘푘 Similarly, the hole density can be written 푛as푝 푛푖푒

- 102 -

( ) = 퐸퐹푝−퐸푖 − 푘푘 where is the quasi-Fermi energy level푝 for푛 holes.푛푖푒 The quasi-Fermi levels for electrons and holes are plotted in푝 Figure (3.9). Since we are considering low and moderate injection conditions, the majority 퐸퐹 carrier densities are nearly the same as the thermal equilibrium values, and the majority carrier quasi- Fermi levels are nearly the same as Fermi energy under thermal equilibrium conditions. We make a further assumption that the quasi-Fermi levels in the depletion region of the forward biased p-n junction is at the same level as the majority carrier quasi-Fermi level. This is illustrated in Figure (3.9). At distances far from the junction the excess carrier densities would become zero, as discussed in the next section. Hence, the quasi-Fermi level for the minority carriers would be at the same level at the thermal equilibrium Fermi energy. For example, at x = - , is at a higher level than the Fermi energy under thermal equilibrium, and gets lower and lower and approaches the thermal, and gets lower and lower 푥푝 퐸퐹푛 and approaches the thermal equilibrium value as one proceeds further into the neutral p region.

Fig. (3.9): Quasi-Fermi Levels in a forward biased - junction 푝 푛 Minority Carrier Diffusion

The diffusion of the excess minority carriers in the neutral regions can be analyzed by setting up the continuity equation. Let us first consider the neutral n region. The continuity equation is given by

= (3.67) 휕Δ푝 Δ푝 1 푑퐽푝 We assume that there is no electric field휕 �in the −neutral휏푝 − n 푞region,푑� and hence, the hole current is only due to diffusion.

= = 휕� 휕Δ푝 Substituting this into the continuity equation,퐽푝 − we푞퐷 get푝 휕 � −푞퐷푝 휕�

- 103 -

( ) ( ) = + (3.68) 2 휕 Δ푝 Δ푝 휕 Δ푝 푝 2 Under D.C. conditions (i.e. when does휕� not vary− 휏 푝with time,퐷 this휕푥 equation is simplified by setting the left hand side as equal to zero, and rewriting it as: 훥� ( ) = 2 휕 Δ푝 Δ푝 or 2 퐷푝 휕푥 휏푝 ( ) = (3.69) 2 휕 Δ푝 Δ푝 2 By considering the dimensions of and , it can휕 푥be seen that퐷푝휏 푝the term has units of square length. We define a parameter as given by 퐷푝 휏푝 퐷푝휏푝 푝 퐿 = (3.70) 2 The equation of continuity is now written as 퐿푝 퐷푝휏푝 ( ) = 2 휕 Δ푝 Δ푝 2 2 Let us define a new variable ’ = – 휕, 푥(i.e. we choose퐿푝 the origin of the horizontal axis at = .) The continuity equation, written in terms of ’, is given by 푥 푥 푥푛 푥 푥푛 ( ) 푥 = (3.71) 2 휕 Δ푝 Δ푝 2 2 The general solution to this equation is 휕푥 퐿푝

( ) = −푥′ + −푥′ (3.72) 퐿푝 퐿푝 ′ ’ = 0 A and B are integration constants thatΔ will푝 be푥 determined퐴푒 by the퐵 푒boundary conditions at and ’ = . The boundary condition at ’ = 0 is given by 푥 푥 푊푛 푥 ( = 0) = ( 1) 푞푉퐹 ′ 푘푘 according to Equation (3.65). Δ푝 푥 푝푛� 푒 − The excess carrier density at an ohmic contact is by definition zero. Hence, the boundary condition at ’ = is

푥 푊 푛 ( = ) = 0 ′ Applying these boundary conditions to determineΔ푝 푥 and푊 푛 , we can arrive at the solution to the continuity equation, as done in the next chapter, and obtain 퐴 퐵

- 104 -

( ) ′ ( ) = ( ) 푊푛−푥 (3.73) sinh ( 퐿푝 ) ′ 푊푛 Δ푝 푥 Δ푝 � sinh 퐿 Where ( ) is defined as ( = 0) However, for the present we푝 limit our discussion to two limiting approximations ′ Δ푝 � Δ푝 푥 >> and <<

푊푛 퐿푝 푊푛 퐿푝 Wide Base Case ( >> )

When >> 푊, 푛we call퐿푝 it a wide base case. Let us consider Equation (3.72), which gives the solution to the continuity equation. If we evaluate at ( ’ = ), we get 푊푛 퐿푝 Δ푝 푥 푊푛 ( ) = −푊푛 + 푊푛 퐿푝 퐿푝 ( ) = Where is evaluated at Δ푝 푊푛. Since 퐴푒>> , the퐵 first푒 term on the right hand side is negligible and hence ′ Δ푝 푊푛 Δ푝 푥 푊푛 푊푛 퐿푝

( ) = 푊푛 퐿푝 푛 But our boundary condition at theΔ 푝ohmic푊 contact퐵 requires푒 ( ) to be zero and 푊푛 is a large quantity. In order to satisfy the boundary condition, we set = 0. The solution to the continuity퐿푝 푛 equation then becomes Δ푝 푊 푒 퐵

( ) = −푥′ 퐿푝 ’ = 0 ′ If we apply the boundary condition at Δ푝 ,푥 we can evaluate퐴푒 A and obtain 푥 ( ) = ( ) −푥′ (3.74) ′ 퐿푝 where ( ) is ( = ) = ( 1), as stated earlier. 푞푉퐹 Δ푝 푥 Δ푝 � 푒 ′ 푘푘 ΔThe푝 � plotΔ of푝 푥 as �a function푝푛� 푒 of distance− as given by Equation (3.74), is given in Figure (3.10 A). We find that the excess minority carrier density decays exponentially with distance. What is the physical interpretation? A holeΔ푝 that is injected from the left at = 0 diffuses to the right, and as it travels in the neutral region, it recombines with an electron. Hence,′ the excess minority carrier density decreases with x’. We can now explain why it decays exponentially.푥 푛 Let ( ) be the probability that an excess hole injected at = 0, will recombine in an elementary distance′ ’ between ’ and ’ + ’. Then ( ’) is the probability′ per unit distance that an excess minority푃 푥 carrier𝑑′ (hole) will recombine and be lost. The number of푥 excess holes that recombine (and therefore be lost)𝑑 in the interval푥 ’푥 is equal𝑑 to the푃 product푥 ( ), i.e., the number of excess hole

𝑑 Δ푝 푥 - 105 - density at ’, multiplied by ( ’) ’. The difference in excess hole density at ’ and at ’ + ’ is what is lost due to the recombination in the interval ’. Hence 푥 푃 푥 𝑑 푥 푥 𝑑 ( ) = ( +𝑑 ) ( ) = ( ) ( ) ′ ′ ′ ′ ′ Assuming ( ’) does�� Δnot푝 vary푥 �with distanceΔ푝 푥 and𝑑 is equal′ − toΔ 푝 ,푥 this equation−Δ푝 can푥 be푃 rewritten푥 𝑑′ as 푃 푥 ( ( )) 푃� = , ( ′) 푑 Δ푝 푥 ′ and directly integrated to obtain Δ푝 푥 −푃�𝑑′

( ) = � ’ (0) ′ −� 푥′ ’ = 0 The constant is obtained as Δ푝 by푥 applying퐴′ 푒the boundary condition at : ’ 퐴 ∆ 푃 ( ) = ( ) 푥 ′ −��푥 By comparing the above to Equation (3.74),Δ푝 we푥 see thatΔ푝 � 푒 = 1 � 퐿푝 This means that is the probability per unit distance푃 that a hole will recombine. That is, is the 1 푑�′ probability that an퐿푝 excess hole will recombine in an interval ’ 퐿푝 Using this argument, it can be shown that is the average𝑑 distance an excess hole (minority carrier) will travel before it recombines. (This is left as a homework problem.) For this reason, is 퐿푝 called the minority carrier diffusion length. 퐿푝 Figure (3.10 A) gives the profile of the excess minority carrier density. As mentioned before in order to maintain charge neutrality, excess majority carrier density equal to the excess minority carrier density will be present in the neutral region, with a profile exactly equal to that of the minority carrier density. This is illustrated in Figure (3.10 B). 푛 Due to the concentration variation of the excess minority carrier density, a diffusion current will flow. The resulting electric current density is given by

( ) ( ) = q ′ 푑 Δ푝 푝 푝 ′ 퐽 푥 − 퐷 (푑)� = q −푥′ (3.75) Δ푝 � 퐿푝 푝 The diffusion current also decays exponentially with distance.퐷 퐿푝 푒 While there is a diffusion current due to the exponential decay of the minority carrier density, there will be no similar majority carrier diffusion current arising from the exponential variation of the excess majority carrier density. The reason for this is that the excess majority carrier density is

- 106 -

maintained by the presence of a small electric field, whose value is proportional to the ratio of ( ) and ( ’) . ( ’) is equal to + ( ) where is the donor density in the region. The drift current′ Δ푛 푥 of electrons due to this electric field, given′ by = ( ) ( ), is exactly balanced by the 푥 푛 푥 푁퐷 Δ푛 푥 푁퐷 푛 ′ ′ excess electron diffusion current, given by 푑𝑑𝑑= . In the푛 case of minority carriers, there is a 퐽 푞푑�Δ푛푥 휇 ℇ 푥 drift current due to this electric field, but it is negligibly small compared to the diffusion current. This is 퐽푑𝑑� 푞퐷푛 푑�′ the reason why we take into account the minority carrier diffusion current, and not the majority carrier diffusion current.

Figure (3.10): Plot of the excess carrier density in the neutral n region for a wide base case: (A) Plot of the excess minority carrier (hole) density, and (B) Plot of the excess majority carrier (electrons) density

In Figure (3.11), the minority carrier diffusion current is plotted as a function of distance in the neutral region. For each hole lost due to recombination, there must also be the loss of an electron. Therefore, electrons must flow into the region in the opposite direction from the ohmic contact to replace 푛the electrons lost due to recombination. Since >> , all the injected holes are lost, due to 푛 recombination, and the hole diffusion current becomes zero at large distances. The (recombinating) 푊푛 퐿푝 electron current density flowing from the ohmic contact into the region should be equal to the injected hole current density at ’ = 0. The recombinating electron current density falls off as 푛 푥 - 107 -

(1 −푥′) with a decrease in . The total electric current density (due to the injection of minority carriers퐿 in푝 the region) is the sum′ of injected hole diffusion current density and the recombining electron− 푒 current density and is푥 constant in the region as shown in Figure (3.11). The total electric density is equal푛 to the injected hole current density at ’ = 0, where the recombining electron current density is zero, and is therefore obtained by putting푛 ’ = 0 in Equation (3.75), as 푥 ( ) 푥 = q = ( 푞푉 1) (3.76) 푝 푛� 퐹 Δ푝 0 q퐷 푝 푘푘 푝 푝 The above equation gives us the퐽 current퐷 density퐿푝 flowing퐿 in푝 the 푒p-n junction− due to the injection of minority carriers (holes) in the region.

푛

Figure (3.11): Injected hole diffusion current density and the recombining electron current density in the neutral n region

We can similarly treat the injection of minority carriers (electrons) in the region. We will find that the injected minority carrier (electron) density decays exponentially with distance in the neutral 푝 region, and the minority carrier diffusion length, , will be given by = in the neutral 푝 region. The current density through the junction, due to injection of electrons (minority carriers) in the 퐿푛 퐿푛 퐷푛휏푛 푝 region, will be given by � 푝

= ( 푞푉 1) 푛 푝� 퐹 q퐷 푛 푘푘 푛 The total electric current density through퐽 the junction퐿푝 will푒 be the− sum of in the equation above and in Equation (3.76) and is given by 퐽푠 퐽 푝

- 108 -

= + = [ + ]( 푞푉 1) 푛 푝� 푝 푛� 퐹 푞퐷 푛 푞퐷 푝 푘� (3.77) 퐽 퐽푛 퐽푝 퐿푛 퐿푝 푒 − This is usually written as

= ( 1) 푞푉퐹 푘푘 where 퐽 퐽푠 푒 −

= + (3.78) 푞퐷푛푛푝� 푞퐷푝푝푛� 푠 푛 푝 is called the saturation current density퐽 . 퐿 퐿

퐽푠 Figure (3.12) shows the various components of current that flow in the forward biased p-n junction. is the injected hole diffusion current density. is the hole current density that is flowing in the neutral p region, in order to inject holes in the n region. is the electron current density in the 퐽푝2 퐽푝3 neutral region to replace the electrons lost due to recombination with holes in the neutral region. 퐽푛1 Similarly, is the injected electron diffusion current density in the neutral region. is the electron 푛 푛 current density flowing in the neutral region to inject electrons in the region. is the recombining 퐽푛2 푝 퐽푛3 hole current density in the neutral region. The total electric current density, , is the sum of all these 푛 푝 퐽푝1 various components. 푝 퐽푡 The current flowing through a forward biased junction is obtained by multiplying the current density by the area of the junction

퐴 = ( 1) (3.79) 푞푉퐹 푘푘 where 퐼 퐼푠 푒 − = = [ + ] (3.80) 푞퐷푛푛푝� 푞퐷푝푝푛� 푠 푠 푛 푝 is called the saturation current.퐼 The current퐴� in퐴 the forward퐿 biased퐿 junction is plotted in Figure (3.13). At values of larger than a few , it can be seen from Equation (3.79) that the term 1 is negligible in 푠 comparison퐼 with the exponential term, and hence, I is given by �퐹 푘� −

( ) (3.81) 푞푉퐹 푘푘 The current increases exponentially with the forward퐼 ≈ 퐼푠 bias푒 voltage . �퐹

- 109 -

Figure (3.12): Various components of electric current density in the forward biased p-n junction. The arrows indicate the direction in which the electrons or hole flow.

Figure (3.13): Forward Bias current through the junction

- 110 -

Narrow Base Case

When << , we call it a narrow base junction.

푊Starting푛 퐿푝 from Equation (3.72), we can expand the exponential terms and retain only the first two terms:

( ) = 1 + 1 + (3.82) ′ ′ ′ 푥 푥 By applying the boundary conditionsΔ푝 푥 at ’ =퐴 0� and− at퐿푝 �’ = 퐵 �, we obtain퐿푛�

푥 ( ) 푥 푊푛 = 1 + Δ푝 0 퐿푝 and 퐴 2 � 푊푛� ( ) = 1 Δ푝 0 퐿푝 Substituting these values in Equation (3.82),퐵 we obtain2 � − 푊푛�

( ) = (0) 1 (3.83) ′ ′ 푥 The excess minority carrier (hole)Δ푝 density푥 inΔ the푝 neutral� − region푊푛� is plotted in Figure (3.14 A). As stated before, a neutralizing excess majority carrier density ( )(= ( )) also exists and is plotted 푛 in Figure (3.14 B). The injected hole diffusion current density is obtained′ from′ Equation (3.83) as Δ푛 푥 Δ푝 푥 ( ) ( ) = ′ 푑 Δ푝 푝 푝( ) ′ 퐽 푥 = −푞퐷 푑 푥 푞퐷푝Δ푝 0 푊푛 = ( 푞푉 1) 푝 푛0 퐹 푞퐷 푝 푘푘 푊푛 푒 −

Figure (3.14): Excess carrier density in the neutral region of a forward biased - junction for a narrow base approximation. (A) excess minority carrier (hole) density, and (B) excess majority carrier (electron) density푛 푝 푛

- 111 -

First, we notice that the injected hole diffusion current density is constant and independent of ’. This means that any hole current injected at ’ = 0 from the region reaches the ohmic contact. There is no recombination of minority carriers in the region. Since is very small in comparison with 푥the diffusion length , the probability of an injected푥 hole recombining푝 in the region is negligible. 푛 푊푛 Hence the injected hole current density reaches the ohmic contact without any attenuation. 퐿푝 푛 Secondly, we notice that the expression for the injected hole diffusion current density is similar to that for the wide base case, except that is replaced by .

If we now assume that the neutral 퐿푝 region can also 푊be푛 approximated to be narrow base, i.e., , we get, for the injected electron diffusion current density, 푝 푊푝 ≪ 퐿푛 = 1 퐹 푛 푝0 푞푉 푞퐷 푛 푘푘 퐽푛 �푒 − � The total current through the junction is given 푊by푝

= + 1 퐹 푛 푝0 푝 푛0 푞푉 푞퐷 푛 푞퐷 푝 푘푘 퐼 퐴 � 푝 푛 � �푒 − � 푊= 푊 1 푞푉퐹 푘푘 where is the saturation current. Thus, we see퐼푆 that�푒 the −form� of the expression for the current under forward bias of a narrow base - junction is similar to that for a wide base junction. 퐼푆 It should be noted that푝 it푛 is possible that in a - junction, one side can be approximated as narrow base, and the other as wide base. In such a case, the current through the junction will be expressed as an appropriate combination of narrow base푝 푛 and wide base expressions.

- Junction under Reverse Bias

푃 푁 When an external voltage is connected such that the positive terminal is connected to the -side and the negative terminal to the -side of the junction, then the junction is said to be reverse-biased. The potential energy of the electrons on the -side is increased relative to that of the electrons on 푛the - side by an amount equal to 푝 where is the externally applied reverse bias voltage. The band diagram in a - junction under reverse bias 푝is illustrated in Figure (3.6 C). The height of the potential 푅 푅 energy푛 barrier as before is the 푞difference� between� on the neutral -side and on the side and therefore equal푝 푛 to ( + ). The band-bending, i.e., the barrier height is thus increased from the 퐶 퐶 thermal equilibrium value of to ( + ). 퐸 푝 퐸 푛 푞 �푏� �푅 In the neutral regions푞 far�푏� away푞 �from푏� the�푅 junction, the carrier densities are at their thermal equilibrium values and the Fermi energy is located in the band gap at a level determined by the net impurity (acceptor and donor) concentrations. Therefore, the Fermi energy in the neutral region (far away from the junction) is shifted downward by from the Fermi energy in the neutral region (far away from the junction). 푞�푅 푝 - 112 -

The expression for the depletion region width under reverse bias can be derived exactly the same way as we did earlier either by solving the Poisson equation or by using Gauss theorem. The expression for the depletion region width will have essentially the same form except the potential barrier height will be replaced by ( + ). Using the same notation for the depletion region width as earlier, we have �푏� �푏� �푅

= ( + ) 1 (3.84) ( ) 2휖푠푁퐷 2 푥푝 �푞푁퐴 푁퐴+푁퐷 �푏� �푅 �

= ( + ) 1 (3.85) ( ) 2휖푠푁퐴 2 and 푥푛 �푞푁퐷 푁퐴+푁퐷 �푏� �푅 � = +

푑 푛 푝 푥 ( 푥 ) 푥 = ( + ) 1 (3.86) 2휖푠 푁퐴+푁퐷 2 Thus we see that the depletion region� width푞푁퐷푁 under퐴 �reverse푏� � bias푅 � increases with the reverse voltage. The increase is physically understandable because a larger space charge (and hence a wider space charge region) is needed for a larger potential energy barrier.

Example

Calculate the depletion region width in an abrupt - junction with a net impurity density of = 10 on the -side and = 10 on the -side. Assume room temperature and a reverse 푁퐴 bias16 voltage−3 of 6 . We determined the15 depletion−3 푝region푛 width for this diode under thermal equilibrium 퐷 case 𝑐and under a forward푝 bias 푁of 0.4 in 𝑐the example worked푛 earlier in the section on forward bias. We can therefore use� the value, 0.656 for from the previous example. � �+ �푏�= 0.656 + 6 = 6.656

푏� 푅 � � 10 � = 3.627 × 10 × × 6.656 10 × 1.115× 10 3 푥푝 � 16 16 √ = 0.282 × 10 cm −4 Similarly

- 113 -

10 = 3.627 × 10 × × 6.656 10 × 1.116× 10 3 푥푛 � 15 16 √ = 2.81 × 10 cm −4 = (2.821 + 0.282) × 10 cm = 3.103 × 10 cm −4 푥푑 −4

Due to the increase in the barrier height, the hole (minority carrier) density in the neutral region at the boundary of the depletion region, i.e., = , is reduced from its thermal equilibrium value since every hole that enters the depletion region will be pulled towards the side by the electric 푛 푛field in the depletion region. Another way of stating this푥 is 푥that while there was a balance between the drift and the diffusion currents in thermal equilibrium, under reverse bias condition푝 the diffusion current is reduced due to the increase in the barrier height. Thus only the drift current remains. Yet another way of looking at this is by considering the carrier density at the edges of the depletion region using the Boltzmann relation in the following manner:

( ) ( = ) = = = = (3.87) −푞�푉푏�+푉푅� −푞 푉푏�−푉퐹 −푞푉푅 푘푘 푘푘 푘푘 We see푝 that푥 when푥푛 is푝 much�푥 larger−푥푝 than�푒 , at = 푝푝 is0 nearly푒 zero. Using푝 the푛0 above푒 relation we see that the excess hole density at = is negative and is given by 푞�푅 푘� 푝 푥 푥푛 푥 푥푛 ( = ) = = 1 (3.88) −푞푉푅 −푞푉푅 푘푘 푘푘 We can now ∆set푝 up푥 the 푥continuity푛 푝푛 equation0푒 which− 푝푛 will0 be푝 the푛0 �same푒 as Equation− � (3.68). Again setting up a new variable ’, as we did in the forward bias case we obtain the same equation as Equation (3.71) which is given below 푥 ( ) = 2 푑 ∆푝 ∆푝 ′2 2 Solving this equation under the boundary 푑condition푥 퐿that푝 the excess carrier density is equal to 0 at x’ = and is equal to 1 at = 0, we get for the wide base diode, −푞푉푅 푘푘 ′ 푊푛 푝푛0 �푒 − � 푥 ( ) = −푞푉 1 푥′ (3.89) 푅 − ′ 푘푘 퐿푝 and for the narrow base diode, ∆푝 푥 푝푛0 �푒 − � 푒

( ) = −푞푉 1 푥′ 1 (3.90) 푅 − ′ 푘푘 퐿푝 푥′ 푛0 푊푛 ∆푝 푥 푝 �푒 - 114− - � 푒 � − �

When we consider the electron density in the neutral region, we get similar results. The excess minority carrier densities in the neutral and regions are negative and are plotted in Figure (3.15). 푝 We will find that the diffusion current푛 푝is of the same form as we got for the forward-biased junction except that will be replaced by – . The current through the diode will be similar in form as the forward bias current but opposite in direction. The current is called the reverse current, . �퐹 �푅 퐼푟𝑟 = 1 (3.91) −푞푉푅 푘푘 In the case of wide base diode, 퐼푟𝑟 퐼푠 �푒 − �

= = + (3.92) 푞퐷푛푛푝0 푞퐷푝푝푛0 푠 푠 is the same as what we obtained퐼 under퐴퐽 forward퐴 � bias퐿 푛for a wide 퐿base푝 �diode. In the case of the narrow base diode, 퐼푠

= = + (3.93) 푞퐷푛푛푝0 푞퐷푝푝푛0 푠 푠 is again the same as what we obtained퐼 퐴 under퐽 forward퐴 � 푊 bias푝 for a narrow푊푛 � base diode.

The퐼푠 current in the reverse biased junction is plotted in Figure (3.16). At values of larger than a few , it can be seen from Equation (3.91) that the exponential term is negligible in comparison with the 푅 term -1 and hence, approaches or saturates to a value � 푘� 퐼 = (3.94)

and is independent of the reverse voltage. This 퐼is the− reason퐼푠 why is called the saturation current. 퐼푠

Figure (3.15): The Plot of the excess (depleted) carrier density in the reverse biased junction.

- 115 -

Figure (3.16): Reverse current in the - junction.

푝 푛

Example

Let us now calculate the current through a wide-base diode at a forward bias of 0.5 and at a reverse bias of 0.5 : The diode is assumed to be an abrupt - junction with = 10 on the -side and � = 10 16 −3 with on the n side. Let us assume room temperature. The퐴 minority carrier diffusion � 2 -1 푝 푛 푁 𝑐 푝 constant is assumed15 − 3to be 10 and 25 in the neutral n and p regions respectively and let the 퐷 minority푁 carrier 𝑐diffusion length be 5 × 10 and 10-2 cm respectively in the neutral and regions. -3𝑐 2 푠 Let the area of the junction , be 10 cm . −3 푛 푝 퐴 = = + 푞퐷푛푛푝0 푞퐷푝푝푛0 푠 푠 퐼 퐴 퐽 = 퐴25� 퐿푛 퐿푝 � 2 −1 퐷푛 = 10𝑐 푠 −2 푛 퐿 = = 𝑐10 20 10 4 −3 16 푛 푝 0 =1010 𝑐 2 −1 퐷푝 = 5 10𝑐 푠 −3 퐿푝 = 푥 = 10𝑐 20 10 5 −3 15 Substituting these values in the equation for푝 푛 0 we get10 𝑐 = 퐼3푠 .60 × 10 −14 푠 . = 3.60 10 퐼 × . 1 퐴= 9.41 × 10 0 5 −14 −6 퐹 0 02589 퐼 푥 �푒 - 116 - − � 퐴

= 3.60 10 × . 1 = 3.60 × 10 5 −14 − −14 푅 0 02589 Interpretation퐼 of Reverse푥 Current� 푒 − � − 퐴

We saw in the last section that the current through the diode is given by

= + 푞푉 1 (3.95) 푝 푛0 푛 푝0 퐹 퐷 푝 퐷 푛 푘푘 for a wide base diode where 퐼 is the푞 �externally� 퐿푝 applied퐿푛 voltage.� �푒 We− can �give a physical interpretation for this reverse current. When a reverse voltage is applied, the term in the above equation should 퐹 be replaced by – . When � is large than a few , the exponential term becomes negligible. The 푅 퐹 resulting current, , denoted by , is the leakage� or saturation current -� and is given by �푅 �푅 푘� 퐼 퐼푅 퐼푆 = = + = (3.96) 퐷푝푝푛0 퐷푛푛푝0 푅 푆 푠� 푠� where and are the퐼 saturation−퐼 current−푞� components� 퐿푝 due퐿푛 to� the flow−퐼 holes− 퐼and electrons respectively. Recall from our discussion in Chapter 2 that the minority carrier thermal generation rate in 퐼푠� 퐼푠� the neutral region of the semiconductor is given by

푛 = (3.97) 푝푛0 푡ℎ Also 푔 휏푝

= (3.98)

We can therefore write 퐿푝 �퐷푝휏푝 = = = × (3.99) 퐷푝푝푛0 퐷푝퐿푝푝푛0 퐿푝푝푛0 2 푝 푡ℎ The reverse current can therefore퐿푝 be expressed퐿푝 as 휏푝 퐿 푔

= = = (3.100)

Where and are 퐼the푅 thermal−퐼푆 generation−퐼푠� − rates퐼푠� of electrons−푞 퐴 � 푔and푝퐿 holes푝 − 푔in푛 the퐿푛 neutral� and regions respectively. 푔푛 푔푝 푝 푛 What is the physical interpretation of this equation? A minority carrier that enters the depletion region will be swept away to the other side due to the electric field in the depletion region. This gives rise to the reverse current. For example, an electron that enters the depletion region from the side where it is a minority carrier will be pushed to the side by the electric field in the depletion region. Similarly a hole will be pushed from the side to the side. Since a minority carrier will travel on푝 an average a distance equal to the diffusion length before푛 it recombines, carriers generated in the neutral region farther than a diffusion length away푛 from the edge푝 of the depletion region will not reach the depletion region. It is only those minority carriers that are generated within a diffusion length that will

- 117 - produce the reverse current. The first term in the above equation represents the current flowing through the junction due to minority carriers (holes) generated in the neutral n region within a diffusion length from the depletion region and similarly the second term represents the current due to the minority carriers (electrons) generated in the neutral region within a diffusion length from the � depletion퐿 region. Such a physical interpretation is not readily possible for the reverse current in the 푛 narrow base diode. 푝 퐿

Non-Ideal Current Characteristics

The current characteristics that we discussed so far are called the ideal current characteristics. We will now discuss additional components of current that flow in a - junction and these additional components of current are called non-ideal currents. In our treatment we neglected the component of current due to generation of electron-hole pairs in the depletion region푝 푛in the reverse current and the component of current due to recombination of electron-hole pairs in the depletion region in the forward current. Reverse Bias

In the depletion region of the reverse-bias junction, the product is less than . Recall from our discussion in Chapter 2, that under this condition, there is a net generation in the 2depletion region equal 푖 to 푛 푝 푛

= (3.101) 푛푖 푡ℎ Each electron-hole pair that is generated in the푔 depletion2휏 푔region will be acted upon by the electric field in the depletion region such that the electron will be propelled to the neutral region and the hole will be propelled to the neutral region. This results in an electric current through the junction equal to the flow of the charge of an electron for each electron-hole pair that is generated.푛 Since each electron is carrying a charge of coulombs,푝 and the electron-hole pairs are generated throughout the depletion region of volume equal to where is the area of the junction and is the width of the depletion region, the reverse current푞 is given by 퐴 푥푑 퐴 푥푑 = + 1 = 1 −푞푉푅 −푞푉푅 푝 푛0 푛 푝0 푖 퐷 푝 퐷 푛 푘푘 푛 푘푘 퐼 푅 푞 퐴 � 퐿푝 퐿 푛 � � 푒 − � − 푞 � 푥 푑 2 휏 푔 퐼 푠 � 푒 − � − (3.102) 푛푖 푑 Since푞� 푥, the2휏 depletion푔 region width, is dependent on the applied reverse voltage, the term due to generation in the depletion region has a voltage dependence as shown in Figure (3.17). 푥푑

- 118 -

Figure (3.17): Non-ideal current characteristics under a reverse bias

When the junction is forward biased the product in the depletion region is larger than . Hence there is a net recombination in the depletion region. This gives rise to an additional component2 in 푖 the forward current. The term due to recombination푛 푝 in the depletion region is of the form 푛

= (3.103) 푞푉퐹 2푘푘 This expression for the current can be derived퐼푟𝑟 as shown퐼푠 푟𝑟 below:푒 The rate of recombination is given by the Schockley Read Hall Model discussed in chapter 2 and is repeated below:

= (3.104)

Where n and p are the electron and hole densities푅 respectively푟 푛 푝 and r is given by = (3.105) ( ) ( ) 1 Where 푟 푛+푛1 휏푝0+ 푝+푝1 휏푛0 1 =

휏푝0 푁푡휎푝푣푝 1 =

휏푛0 - 119푁푡 -휎 푛푣푛

= density of traps or g-r centers.

푁푡, = capture cross-sections for electrons and holes respectively,

푛 푝 휎 휎 , =thermal; velocity of electrons and holes respectively and is equal to , and is the 3 푘푘 ∗ 푛 푝 ∗ effective푣 푣 mass of the carrier, � 푚 푚

= exp 퐸퐶−퐸푡 1 퐶 푘푘 푛 =푁 �− � (3.106) 퐸푡−퐸푖 푖 푘푘 = 푛 exp푒𝑒 � � 퐸푡−퐸푉 1 푉 푘푘 푝 = 푁 �− � (3.107) 퐸푡−퐸푖 where = the trap energy level 푛푖푒𝑒 �− 푘푘 � The value퐸푡 of at the edge of the depletion region near the neutral -side is equal to the majority carrier density and has a value equal to . Hence the product of and at the edge of the 푛 푞푉퐹 푛 depeletion region near the neutral region is푘푘 given by 푛푛0 푝 푝푛0푒 푛 푝 푛 = = (3.108) 푞푉퐹 푞푉퐹 푘� 2 푘푘 Similarly by considering the푛 hole푝 and푛 푛electron0푝푛0푒 densities푛 at푖 푒 the edge of the depletion region near the neutral region it can be shown that the product is the same as at the edge of the depletion region near the neutral -side. We now assume that the product has the same value throughout the interior of the푝 depletion region and is given by푛 Equation푝 (3.108). This is pictorially shown in Figure (3.18). Hence the expression for푛 the net recombination rate can 푛be푝 written as

푞푉퐹 = = = 2 푘푘 (3.109) ( ) ( 2 ) ( 푛푖)�푒 ( −1� ) 푛�−푛푖 푡ℎ The net recombination푈 푅 rate− 퐺 is a maximum푛+푛1 when휏푝0+ the푝+ denominat푝1 휏푛0 or푛+ is푛 a1 minimum.휏푝0+ 푝+푝 Near1 휏푛0 the edge of the depletion region close to the neutral region, is large and in the region close to the neutral region, is larger. and are both small only in the middle of the depletion region when and are nearly equal as shown in Figure (3.18). Therefore푛 only푛 a small region in the middle of the depletion region푝 is effective푝 in푛 producing푝 recombination. Using Equation (3.108) and using the fact that푛 and푝 are nearly equal in this region, we can write 푛 푝

= = = (3.110) 푞푉퐹 2푘푘 푛 푝 �푛� 푛푖푒

- 120 -

푞푉퐹 = 2 푘푘 (3.111) 푛푖 �푒 −1� 푞푉퐹 푞푉퐹 2푘푘 2푘푘 푈 �푛푖푒 +푛1�휏푝0+�푛푖푒 +푝1�휏푛0 Substituting the expression for and in terms of ( ) , we obtain

푛1 푝1 퐸푡 − 퐸푖 푞푉퐹 = 2 푘푘 푛푖 �푒 ( −)1� ( ) 푞푉 퐸 −퐸 퐸 −퐸 퐹 푡 푖 − 푡 푖 2푘푘 푘푘 푘푘 푈 푛푖푒 �휏푝0+휏푛0�+푛푖푒 휏푝0+푛푖푒 휏푛0

푞푉퐹 = 푘푘 (3.112) 푛푖�푒 ( −1�) ( )

푞푉퐹 퐸푡−퐸푖 퐸푡−퐸푖 2푘푘 푘푘 − 푘푘 When | | > , the denominator푒 �휏푝0 is+ large휏푛0�+ and휏푝0 hence푒 is+ nearly휏푛0푒 zero. Only when the trap is in the middle of the gap i.e., , becomes not negligible. 퐸푡 − 퐸푖 푘� 푈 퐸푡 ≈ 퐸푖 푈 푞푉퐹 푘푘 (3.113) 푛푖�푒 −1�

푞푉퐹 2푘푘 푈 ≈ �휏푝0+휏푛0� �푒 +1�

Figure (3.18). Electron and hole density variation in the depletion region of a forward biased - junction. The solid lines represent the carrier density variation under forward bias. The dotted lines represent the carrier density variation in the depletion region of a - junction under thermal푝 푛 equilibrium. In thermal equilibrium the product is equal to . Under forward bias the product is 2 푝 푛 assumed to be equal to throughout the depletion region푖 since the product has this value at 푞푉퐹 푛� 푛 푛� 2the푘푘 boundary of the depletion region on either side. 푛푖 푒 푛� The term 1 in both the numerator and the denominator is negligible in comparison with the exponential terms and can therefore be omitted.

- 121 -

푞푉퐹 = 푞푉퐹 (3.114) 푘푘 2푘푘 푛 푖푒 푛푖푒 푞푉퐹 푈 ≈ 2푘푘 �휏푝0+휏푛0� When an electron and a hole recombine,푒 �휏푝0 an+휏 푛electron0� flows from the neutral region into the depletion region and similarly a hole flows into the depletion region from the neutral region. This constitutes a current flow through the junction in the same direction as the ideal forward푛 current that we considered in the previous section. Hence the additional component of current due푝 to recombination in the depletion region is given by

= = 푞푉퐹= (3.115) 푞푉퐹 deff 푖 2푘푘 퐴 푥 푞 푛 푒 2푘푘 푟𝑟 deff 푆 푟𝑟 where is the small퐼 region퐴 in 푥the middle푞 푈 of the depletion�휏푝0+휏푛0 region� 퐼 where the푒 electron and hole densities are both minimum and equal. 푥deff The total forward current including recombination in the depletion region is given by

= = + 푞푉 + 푞푉 = 푞푉 + 푝 푛0 푛 푝0 푎 퐹 퐹 퐷 푝 퐷 푛 푘푘 2푘푘 푘푘 퐹 퐿푝 퐿푛 푆 푟𝑟 푆 푖𝑖𝑖 퐼 퐼 푞� � � 푒 퐼 푒 퐼 푒 (3.116) 푞푉퐹 2푘푘 If one were to 퐼calculate푆 푟𝑟 푒 and for modern devices as illustrated in the worked out example below, the former will be much smaller than the latter. Hence, when the forward bias is not 푆 푖𝑖𝑖 푆 푟𝑟 that large, the forward current is퐼 dominated퐼 by the recombination current and depends on the voltage as . At much larger forward bias voltages, the forward current is due mainly to the ideal junction 푞푉퐹 2푘푘 current and depends on the bias voltage as . At intermediate voltages, the forward current is due to 푒 푞푉퐹 푘푘 both the mechanisms and the bias voltage dependence푒 is approximated by 푞푉퐹 . The parameter is called the ideality factor and is usually used as a parameter to define how good휂𝜂 a junction is. 푒 휂

Example

Consider a p-n junction with the following characteristics where and are the net impurity concentration on the n and p sides respectively. A is the area of the junction. 푁퐴 푁퐷 = 10 , = 10 , = 10 , = 30 , = −4 2 16 −3 15 −3 2 −1 4 , = 10퐷 , = 10 퐴 푛 푝 퐴 2 −1𝑐 푁 −4 𝑐 −5푁 𝑐 퐷 𝑐 푠 퐷 The𝑐 ideal푠 current휏푛 prefactor 푠 휏푝 and the recombination푠 current prefactor can now be calculated using the values for the universal constants and and assuming room temperature i.e., = 300 . 푆 푖𝑖𝑖 푆 푟𝑟 Taking the intrinsic carrier density퐼 as equal to 10 , the thermal equilibrium퐼 minority densities 푞 푘 푇 퐾 can be calculated as 10 −3 푛푖 𝑐

- 122 -

= = = 10 (3.117) 2 20 푛푖 10 4 −3 16 and 푝푛0 푁퐷 10 𝑐

= = = 10 (3.118) 2 20 푛푖 10 5 −3 15 The minority carrier diffusion lengths푛 are푝0 calculated푁퐴 10as 𝑐

= = 10 4 = 6.32 10 cm (3.119) −5 −3 푝 푝 푝 퐿 =�휏 퐷 =√ 10 푥 30 = 5.48푥 x10 cm (3.120) −4 −2 Substituting these values퐿푛 in the� 휏equation푛퐷푛 for√ 푥 we get 퐼푆 푖𝑖𝑖 × × = 1.6 10 × 10 × + . × 4 . × 5 −19 −4 4 10 30 10 푆 푖𝑖𝑖 −3 −2 퐼 = 1.6푥 10 × (5.7 × 10�6 32+ 510.47 × 105 48 ) 10 � −23 6 7 = 9.66푥 × 10 Amperes −16 For calculating we use the following typical parameters. Let the density of the traps, be 10 and = = 10 . Assume = = where 퐼푆 푟𝑟 푁푡 15 −3 −16 푛 푛 푛 푝 푡ℎ 𝑐 휎 휎 𝑐 × . × 푣 ×푣 푣 = = = 1,17 × 10 (3.121) . −23 3푘푘 3 1 38 10 300 5 푡ℎ −31 And we have assumed푣 the effective� 푚 mass� of9 the11 푥carriers10 to be the same as for electrons in vacuum. Since we will be using mostly as the unit of length we take as 1.17 10 . Let us assume tht the width of the region wwhere the recombination is effective, , is 3 x107 cm.− 1 is equal to due 𝑐 푣푡ℎ 푥 𝑐 푠 to our assumption of equality of and as well as and . Therefore−5 푥diff 휏푛0 휏푝0 휎푛 1.휎6푝 10 푣10푛 푣3푝 10 10 + = −19 1,71− 4 10 −5 10 푥 푥 푥 푥 푥 휏푛0 휏푝0 −6 = 2.81 10 Amperes푥 −12 Thus we see that is larger than 푥 and therefore dominates at low bias voltages. At higher bias voltages, the ideal current dominates because of the absence of the factor 2 in the denominator of 푆 푟𝑟 푆 푖𝑖𝑖 the exponent of the퐼 voltage dependent퐼 term.

______In the discussion so far we considered recombination generation in the bulk of the semiconductor. Normally the depletion region intersects a surface and the g-r centers (called surface- states or interface states) on the surface of the depletion region can also provide a recombination

- 123 - current under forward bias and a generation current under reverse bias in addition to the previously discussed components of current. The net recombination rate per unit area of the surface, is given by

푠 = 푈 2 (3.122) 푛�−푛푖 푠 푛+푛1 푝+푝1 + Where is the number of g-r centers per푈 unit area푁푠�휎 푝of푣 푝the푁 surface푠�휎푛푣푛 (surface density), and all other parameters have the same meaning as before for the bulk recombination. Under forward bias the 푁푠� product in the surface region is much larger than and hence a surface recombination current flows in 푛 푝 addition to the bulk recombination current. The surface2 recombination current also has the same 푛푖 voltage dependence as the bulk recombination current i.e., the current is proportional to . 푞푉퐹 2푘푘 When the junction is under reverse bias, the product is negligible in comparison with 푒 and hence the net generation rate under the assumption that only the surface states whose energy is in2 the middle 푖 of the gap contribute to the generation, is given푛 푝 by 푛

= = = 2 (3.123) 푛푖 푛푖 푛푖푠 푛1 푝1 1 1 푈 푁푠�휎푝푣푝+푁푠�휎푛푣푛 − 푁 휎푝푣푝+푁 휎푛푣푛 − 2 where s equal to 푠� 푠�

= 2 1 1 + is called the surface recombination velocity. Recall푁 that푠�휎푝 we푣푝 defined푁푠�휎푛푣 푛a minority carrier lifetime in the bulk to describe the generation rate. We now define a surface recombination velocity for the surface to describe the generation at the surface. The electron-hole pairs created at the surface depletion region will give rise to another component of reverse current which we will call the surface generation current. This component of surface generation current will add to the bulk generation current under reverse bias.

A typical current voltage plot is given in Figure (3.19) in which the current is plotted on a logarithmic scale while the voltage is plotted on a linear scale. The slope at small values of forward bias voltage is half that at larger values of bias voltage. The recombination component is larger than the ideal and hence at small voltages the recombination current dominates 푆and푟𝑟 the slope is . 퐼 푞 푆 푖𝑖𝑖 At higher 퐼voltage the ideal component dominates and the slope is . It is customary to express the 2푘푘 푞 current as 푘푘

= (124) 푞푉퐹 휂𝜂

Where is called the ideality factor. The more퐼퐹 ideal퐼 푆a diode푒 is, closer is to unity. 휂 휂

- 124 -

Figure (3.19): Plot of the forward current in a typical p-n junction as a function of the bias voltage. The current is on a logarithmic scale. Notice how the slope increases by a factor 2 between the lower and higher voltages.

Capacitive Effects

The - junction behaves like a capacitor for small signal (applied) voltages. There are two mechanisms that give rise to capacitive effects. One arises due to the dependence of the space charge in the depletion푝 region푛 (and hence the depletion region width) on the applied voltage. The capacitance is called Junction Capacitance or Depletion Capacitance. The other is due to the minority carrier charge stored in the neutral regions. This capacitance is called Storage Capacitance or Diffusion Capacitance. We will examine both these now: = + and �푄 푞푁퐷�푥푛 = − and they flow from the external voltage source�푄 into− the푞푁 diode.퐴�푥푝 The two elementary charges are equal in magnitude but opposite in sign. When the step voltage is removed, the elementary charges flow back to the external voltage source. Hence the diode behaves like a capacitor of capacitance per unit area equal to

= + �푄 퐶 ��푎 - 125 -

We will now derive an expression for the depletion capacitance. At any point in the depletion region, the increase in the electric field, d due to the small step voltage, is obtained from Gauss theorem as 푥 ℰ = + �푄 dℰ It is independent of . The incremental change in the applied휖푠 voltage across the diode can be obtained from the line integral of across the depletion region. Since is constant with , the line 푎 integral is 푥 �� dℰ dℰ 푥 = ×

Substituting the expression for in terms of푎 wed get �� +푥 dℰ dℰ �푄 = × + �푄 ��푎 푥d Dividing both sides of the equation by , we obtain for the휖푠 depletion capacitance per unit area of the junction, + �푄 = = (3.125) + 푑푄 휖푠 푑 We find that this is exactly the same expression퐶 that푑푉 푎we would푥d have had if we have a parallel plate capacitor with the semiconductor as the dielectric between the parallel plates of unit area and a spacing of between the plates. Thus we see that the junction behaves like a parallel plate capacitor. What we calculated was the expression for the capacitance for unit area of the junction. The capacitance of a junction푊 of area A is

= = (3.126) 휖푠 What we derived for the junction퐶 capacitance퐽 퐴 퐶푑 is valid퐴 푥 ford a junction even if it did not have uniform impurity concentrations i.e., if the diode was not an abrupt junction. We can arrive at the expression for the capacitance of an abrupt junction by an alternate method also. The total positive charge in the depletion region, per unit area of the junction is given by + 푄 = = + + 푁퐷푁퐴 푄 푞 푁퐷푥푑 푞 푥d Where we have used the relation 푁퐷 푁퐴 =

Differentiating this expression for with respect푁퐷푥n to 푁 we퐴푥 pcan get + 푎 푄 =� 휖푠 퐶푑 which is the same as what we obtained earlier. It is left푥 asd an exercise for the student to show this. - 126 -

Figure (3.20): a) Flow of elementary charges and into the junction diode due to the application of a step voltage: b) The widening of the +space charge− region due to the step voltage: and are the elementary increase in the� space푄 charge�푄 on the n and p sides respectively. 𝑑n ______𝑑p ______Example

Consider a junction with = 10 on the p side and = 3 10 on the side. Let us calculate the junction capacitance15 for a− reverse3 voltage of 5 . Assume the15 area−3 of the junction to 퐴 퐷 be 10 . We saw that푁 the expression𝑐 for in terms of 푁 is given푥 as 𝑐 푛 −3 2 � 푑 푅 𝑐 푥 ( ) � = ( + ) 1 2휖푠 푁퐴+푁퐷 2 푥푑 � 푞푁퐷푁퐴 �푏� �푅 � 10 × 3 × 10 = 0.0259 × ln = 0.625 15 10 15 푏� 20 Substituting the values� for and other parameters,� we get � �

�푏� . × . × ( )× = (5 + 0.625) 1 = 3.13 x 10 cm . × −×14 × 15 2 211 9 8 84 10 3+1 10 −4 −19 15 15 푥푑 � 1 6푥10 3 10 10 � - 127 -

. × . × = =3.36 x 10 F . −14 11 9 8 84 10 −9 −2 −4 and 퐶푑 3 13 x 10 𝑐 = 10 × 3.36 × 10 = 3.36 × 10 = 3.36 −3 −9 −12 퐶퐽 퐹 푝� Diffusion or Storage Capacitance

We saw earlier that under a forward bias minority carriers are injected into the neutral region on either side of the depletion region. The excess minority carrier density decayed with distance in the neutral regions either exponentially as in the case of the wide base diode or linearly as in the case of the narrow base diode. Although the excess carriers were diffusing and therefore constantly moving until they recombined (wide base) or until they reached the ohmic contact (narrow base), at any given time they had a density distribution given by Figure (3.10) and (3.14). We can think as though the minority carriers equal to the area under the curve in these two figures are stored in the neutral regions. The minority carrier charge stored in the neural regions can be obtained by integrating the excess minority carrier density in the neutral region and by multiplying this by the electron charge and the area, , of the junction. The charge due to excess holes stored in the neutral region is therefore given by 푞 퐴 푛 = ( ) 푊푛 ′ ′ 푝� where we have used the subscript s푄 to denote푞 퐴that� 0we areΔ푝 considering푥 𝑑 stored charge.

Wide Base Diode

Let us consider the wide base diode. Substituting the expression for the excess hole density in the wide base case which is

( ) = ( ) −푥′ ′ 퐿푝 into the expression for and integratingΔ푝 we푥 get, Δ푝 � 푒 푄푝� = ( ) −푥′ = ( ) (3.127) 푊푛 퐿푝 ′ We can relate this to the푄 current푝� 푞through퐴 Δ푝 the� diode.∫0 The푒 current𝑑 due퐴 to�Δ hole푝 � injection퐿푝 is ( ) ( ) = = ′ 푝 푝 푝 �Δ푝 푥 푞 퐴 퐷 Δ푝 � 퐼 −푞 퐴 퐷 ′ � ′ Dividing the expression for , by and rearranging,�푥 푥 we=0 get 퐿푝 푄푝� 퐼푝 - 128 -

= 2 = (3.128) 퐿푝 푝� 푝 푝 푝 Thus we see that the stored charge푄 in the퐼 neutral퐷푝 region퐼 휏 is the injected minority carrier current times the life time. The stored charge is dependent on the applied forward bias, , since

�퐹 ( ) = 1 푞푉퐹 푘푘 Let us assume that a step voltage Δ 푝is applied� in푝 푛series0 �푒 with − as� shown in Figure (3.21 A). Due to this step voltage the stored charge will increase by an amount as shown in Figure (3.21 B). This ��퐹 �퐹 extra charge , flowed into the neutral region from the external source through the neutral 𝑑푝� region. In order to maintain charge neutrality the electron density in the neutral region also has to 𝑑푝� 푛 ��푎 푝 increase and the increase in the negative charge will be exactly equal and oppoissite to , and 푛 has to flow from the external source through the neutral region as shown in Figure (3.21 C). 𝑑푛� 𝑑푝 Therefore the diode appears to the external source as a capacitor of capacitance equal to 푛

= (3.129) 푑푄푝� Hence we can obtain the diffusion capacitance퐶diff 푝as 푑푉퐹

= = (3.130) 푑�퐼푝 휏푝� 푑퐼푝 But 퐶diff 푝 푑푉퐹 휏푝 푑푉퐹

= 1 푞푉퐹 푘푘 which becomes equal to 퐼푝 퐼푠� �푒 − �

푞푉퐹 푘푘 Under the approximation . Differentiating퐼푝 ≈ 퐼푠 �this푒 with respect to we get 퐹 퐹 � � ≫ 푘 � = = � (3.131) 푑퐼푝 푑퐼푝 푞퐼푝 Therefore the diffusion capacitance becomes푑푉푎 equal푑 to푉퐹 푘푘

= (3.132) 푞퐼푝 The DC current through the diode 퐶alsodiff increases푝 휏푝 by푘푘 a small increment due to the small step voltage and therefore the diode appears to the step voltage as a resistor of resistance equal to 𝑑푝 = 푎 푓� �� 푟 푝 - 129� - 퐼 or a conductor of conductance equal to

= (3.133) 푑퐼푝 Using Equation (3.132), we can write the diffusion푔푓� capacitance푑푉푎 as

= (3.134)

So far we considered only the small signal퐶diff circuit푝 휏eleme푝푔푓�nts due to the hole injection in the neutral region. We have similar diffusion capacitance due to electron injection in the neutral region so that we can write for the total diffusion capacitance as 푛 푝

= + = + (3.135) 푞퐼푝 푞퐼푛 where is the component 퐶ofdiff forward푝 bias휏푝 푘푘 current휏 due푛 푘푘 to electron휏푝푔 푓injection� 휏푛 in푔 푓the� neutral region and and are the forward conductance due to hole and electron injection respectively. Similarly the 퐼푛 푝 forward conductance will be equal to 푔푓� 푔푓� = + = = + 푞 푞퐼퐹 푔푓 �퐼푝 퐼푛� 푔푓� 푔푓� 푘� 푘�

Figure (3.21): The effect of the application of a step voltage on a forward biased junction diode a) The step voltage, is applied in series with the forward bias. B) The increase in the injected minority carrier distribution in the neutral n-region is shown by the shaded area. C) The increase in the majority 푎 carrier distribution�� in the neutral n-region to maintain the charge neutrality is shown by the shaded area.

- 130 -

Narrow Base Diode

Similar to the wide base diode case, the minority carrier distribution increases with the application of a step voltage in the case of the narrow base diode. Hence we get stored charge effects and the diode gives rise to diffusion capacitance. It is left as an exercise for the student to show that the stored charge due to the injection of minority carriers in the neutral region is given by

( ) 푛 = q A Δ푝 � 푊푛 Similarly the diffusion capacitance can be shown푄푝� to be 2

= 2 2 푛 푝 diff 푝 푊 푞퐼 퐶 푝 In the next chapter, it will be shown that 2 is the transit퐷 time푘� for the holes in the neutral -region i.e., 푊푛 the time taken for the hole to traverse the2퐷 푝-region. A diffusion capacitance arises also due푛 to the injection of electrons in the neutral p-region. The total diffusion capacitance is the sum of both the diffusion capacitances and is given by 푛

= + (3.136) 2 2 푞퐼푝 푊푛 푞퐼푛 푊푝 diff If one side, say the -side, is approximated퐶 as a wide푘푘 2base퐷푝 and푘푘 the2 퐷other푛 side ( -side) is approximated as a narrow base, the diffusion capacitance is given by 푝 푛 = + (3.137) 2 푞퐼푝 푊푛 휏푛푞퐼푛 diff The expression for the small signal conductance퐶 and푘푘 resistance2퐷푝 푘푘is the same as what we obtained earlier for the wide base case.

Small Signal Equivalent Circuit

The diode behaves like a capacitor due to the junction capacitance under reverse and forward bias voltages and due to the diffusion capacitance under forward bias. Reverse-biased Diode

Under reverse bias, the current in an ideal diode depends on the reverse voltage for small values of voltage but is independent of the voltage for larger values of reverse voltage. Hence we can conclude that the conductance under reverse bias is zero or the resistance is infinite. However when the non- ideal components are considered, the reverse current depends on the reverse voltage. Hence there is a non-zero conductance. However, the conductance is very small. Hence the diode behaves like a large resistor. The junction also behaves as a capacitor due to the junction capacitance whose value depends

- 131 - on the reverse voltage. Hence the equivalent circuit for the reverse biased diode is a large resistor in parallel with the junction capacitance as shown in Figure (3.22 A). In circuit representation, the diode is denoted by the symbol of an arrow and a vertical line as depicted in this figure. The direction of the arrow is the direction of the forward current flow. Hence the side containing the arrow is the -side and the side containing the vertical line is the -side. 푝 푛

Figure (3.22). Equivalent circuit of a diode under A) reverse bias, and B) forward bias

Forward-biased Diode

We saw earlier that under forward bias the current through the diode current increases with the forward voltage and diode behaves like a small signal resistor of resistance equal to

= 푘� 푟푓 Since is approximately equal to 0.0258 V at room temperature,푞퐼퐹 a current of 1 mA under forward bias 푘푘 . will give푞 rise to a small resistance of = 25.8 . Thus we see that a forward biased diode offers 0 0258 very low resistance to a small signal voltage.−3 In addition, the diode gives rise to capacitive effects 10 �ℎ푚 through the junction capacitance and the diffusion capacitance. Therefore the response of the diode to a small signal voltage can be represented by the equivalent circuit shown in Figure (3.22 B). The

- 132 - capacitors and the resistor are all in parallel because the incremental charge or the incremental current all flow independent of each other and only depend on the small signal voltage. The resistance has a small value. Junction Breakdown

The reverse current in an ideal diode remains constant with increasing reverse voltage until at a certain reverse voltage called the breakdown voltage the current suddenly increases to an extremely large value as shown in Figure (3.23). If current limiting resistors were not included in the circuit the current will become so large as to destroy the junction. Under this condition, the junction is said to have suffered a breakdown and the reverse voltage at which the breakdown occurs is called the junction breakdown voltage. There are three mechanisms that cause the junction to break down and they are: a) Thermal mechanism, b) Avalanche breakdown and c) Tunneling breakdown.

Although the reverse current, which is called the diode leakage current, is extremely small at room temperature, it is very sensitive to temperature and increases very rapidly with temperature. The power dissipated in the junction, is the product of the leakage current and the reverse voltage. At large enough reverse voltage, the power dissipation can become excessive and hence can cause a heating of the junction. This increases the junction temperature and hence the leakage current increases further and this in turn increases the junction power dissipation. This cycle of increased power dissipation and increase in temperature keeps on recurring until the current becomes very large and the diode is destroyed. This is called the thermal breakdown. The thermal breakdown used to be a big problem in the days when germanium instead of silicon was used as the semiconducting material for fabricating junction devices. However with modern silicon devices, in which adequate heat sinks are used as part of the device design, thermal breakdown is not usually a problem.

In modern devices the breakdown due to the other two mechanisms limits the maximum reverse voltage that can be applied across the junction. The avalanche mechanism can be explained by reference to Figure (3.24). In this figure, the band diagram of a junction across which a larger reverse voltage has been applied is shown. The depletion region is wide and also has a large electric field. An electron entering the depletion region is accelerated by the large electric field, and within a short distance its kinetic energy increases to a sufficiently large value such that it is able to create an electron- hole pair by impact ionization. The electron loses most of its kinetic energy in the impact ionization process. It starts to accelerate again and within a short distance again causes the generation of another electron-hole pair. Thus by the time the electron reaches the other end of the depletion region, it has generated many more electrons. This process is called avalanche multiplication. The ratio of the number of electrons coming out of the depletion region to the electrons entering the depletion region is called the multiplication factor. The holes can also cause avalanche multiplication. At sufficiently large reverse voltage the multiplication factor becomes infinite and the current increases infinitely to cause a junction breakdown.

The tunneling mechanism can be explained with reference to Figure (3.25 A). In this figure a junction in which the impurity concentrations on the two sides are very high is shown. The depletion region width is very narrow because of this. In order to go to the conduction band the electrons in the valence band need not thermally get excited but can tunnel through a triangle barrier to the conduction band as shown in Figure (3.25 B).

- 133 -

Figure (3.23): Breakdown in a junction diode. The voltage at which the current rapidly increases is defined as the breakdown voltage �푏�

Figure (3.24): The avalanche breakdown mechanism

- 134 -

Figure (3.25): The tunneling breakdown mechanism: A) the band diagram under tunneling conditions and B) the triangular barrier through which the electron tunnels through

A detailed theoretical model is available to treat the avalanche process and derive an expression for the avalanche breakdown voltage. Similarly a detailed theoretical model is can be used to derive an expression for the tunneling breakdown voltage. However for our purpose we will assume that there is a certain critical electric field which when exceeded causes the junction to breakdown independent of the mechanism. We can derive an expression for the breakdown voltage in terms of the critical electric field. For an abrupt junction the maximum electric field is equal to

= = 푞푁퐷 푞푁퐴 ℰ푚𝑚 푥푛 푥푝 But the total voltage across the junction is equal∈ to푠 ∈푠 + + = 2 푥푛 푥푝 Substituting for and from the previous�푅 � 푏equation� ℰ푚 we𝑚 get 푥푛 푥푝 + + = 푛 2 푝 푅 푏� 푚𝑚 푥 푥 � � = ℰ + ℰ푚𝑚 ℰ푚𝑚∈푠 ℰ푚𝑚∈푠 2 � 푞푁퐷 푞푁퐴 � = 2 + (3.138) ℰ푚𝑚 ∈푠 1 1 The breakdown voltage is the value of at2푞 which�푁 퐷 푁becomes퐴� equal to , the critical electric field. Thus we obtain the expression for the breakdown voltage as �푏� �푅 ℰ푚𝑚 ℰ𝑐𝑐 - 135 -

= 2 + (3.139) ℰ푐𝑐� ∈푠 1 1 There are two points that� must푏� be mentioned2푞 �푁퐷 with 푁respect퐴� − to�푏 the� junction breakdown. One is that the avalanche mechanism is the dominant mechanism in diodes in which the breakdown voltage is larger than and the tunneling mechanism is the dominant one in diodes in which the breakdown 6퐸푔 voltage is less푞 than . For silicon diodes avalanche process is dominant in diodes in which the 4퐸푔 breakdown voltage is above 6.6 V and tunneling is the dominate process for diodes with a breakdown 푞 voltage less than 4.4 V. Secondly the avalanche breakdown process has a positive temperature coefficient i.e., the breakdown voltage increases with temperature whereas the tunneling breakdown process has a negative temperature coefficient with the breakdown voltage decreasing with increasing temperature. Silicon diodes which have a breakdown voltage between 4.4 and 6.6 volts have both mechanisms present and hence have very little temperature dependence of the breakdown voltage. Usually diodes which are in the breakdown mode are used a voltage reference at a voltage of and these are called Zener diodes. �푏� The expression we derived for the breakdown voltage is for planar diodes i.e., ones in which the junction is planar. However in practical diodes the sides and corners of the junction are curved and hence the actual breakdown voltage is lower than what we derived due to the fact that the electric field is higher in curved junctions than in planar junctions for a given applied voltage.

Example

Given that the critical electric field is 3 10 for a diode with equal to 10 on the p side and equal to 10 on the n side,5 calculate−1 the breakdown voltage. 15 −3 퐴 16 −3 푥 �𝑐 푁 𝑐 푁퐷 𝑐 10 = = 0.0258 ln = 0.654 1031 푘� 푁퐴푁퐷 �푏� 𝑙 � 2 � � 20� � Hence 푞 푛푖 (3 10 ) × 11.9 × 8.84 10 = × (10 + 10 ) 0.654 5 22× 1.6 × 10 −14 푥 = 325 0.654 푥324 −15 −16 �푏� −19 − − ≈ �

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Summary

• In thermal equilibrium, a potential barrier exists across the - junction such that the diffusion current is balanced by a drift current in the opposite direction. The height of the potential 푝 푛 barrier is denoted by . • A depletion region is formed on either side of the metallurgical junction, and the charge density �푏� in the depletion region is due to ionized acceptor atoms on the -side, and to donor atoms on the -side. 푝 = + 푛 Where = width of the depletion region on the -side, and = width of the depletion region 푑 푛 푝 on the -side and is the total width of푥 the depletion푥 푥 region. 푥푛 푛 푥푝 • The positive charge in the depletion region on the -side is equal in magnitude to the negative 푝 푥푑 charge in the depletion charge on the -side ( = ) 푛 • The depletion region occurs more on the lightly doped side. The electric field exists only in the 푛 푁퐴푥푝 푁퐷푥푛 depletion region. The neutral and regions do not have any electrical field.

푛 푝 Forward Bias

• Under forward bias, the height of the potential barrier across the junction is reduced to , where = bias voltage. • The expression for the depletion region widths is the same as the one for thermal equilibrium, �푏� − �퐹 �퐹 with replaced by . The depletion region width decreases under forward bias. • Under forward bias, excess minority carriers are injected into the two neutral regions. The value �푏� �푏푖 − �퐹 of the excess minority density at the boundary of the deletion region and the neutral region is

( ) = ( 1) …………… -side 푞푉퐹 푘푘 푛� Δ푝 (� ) =푝 (푒 − 1) …………….푛 side 푞푉퐹 푘푘 • A net current flows across theΔ푛 junction� 푛 due푝� to푒 the −diffusion of the injected푃 excess minority carrier’s density. • Under ideal conditions.

= ( 1), 푞푉퐹 푘푘 Where 퐼퐹 퐼푆 푒 −

= + 퐷푝푝푛0 퐷푛푛푝0 푆 When퐼 푞� � 퐿 푝 퐿푛 �

푊푛 ≫ 퐿푝 푎�푎 푊푝 ≫ 퐿푛 - 137 -

and

= + 퐷푝푝푛0 퐷푛푛푝0 푆 When퐼 푞� � 푊푛 푊푝 �

The first term in the expression for represents푊푛 ≪ 퐿푝 푎 the�푎 diffusion푊푝 ≪ 퐿푛 current, and the second one represents the drift current. 퐼퐹 Reverse Bias

• Under reverse bias, the height of the potential barrier increases to + , where = reverse voltage. �푏� �푅 �푅 • The expression for the depletion region under reverse bias is the same as the one for thermal equilibrium, with replaced by + . The depletion region width increase under forward bias. �푏� �푏� �푅 • The net current under reverse bias has the same form as the forward current, with replaced by . �퐹 −�푅 = 1 −푞푉푅 푘푘 Again, as in the forward bias case,퐼푅 the 퐼first푆 � 푒term represents− � the diffusion current, and the second term the drift current. When >> , the reverse current is only due to the drift current. 푞�푅 푘� is called the saturation current.

퐼푆 is called the leakage (or reverse) current.

• It can be shown that퐼푅 is due to the generation of minority carriers within a diffusion length from the boundary of the depletion region, in the two neutral regions. 퐼푆 Non-ideal Current

• Current due to recombination of electron-hole pairs also contributes to the current in the - junction under forward bias. This component of current has a voltage dependence given 푝 by , and is dependent on the number of traps (g-r centers) per unit volume. 푛 푞푉퐹 • Similarly,푘푘 an additional component due to the generation of electron-hole pairs in the depletion 푒 region flows in the junction under reverse bias.

Capacitance

• The depletion region behaves like a parallel plate capacitor with spacing equal to the depletion region width, with respect to applied small signal voltage. The capacitance is called the junction capacitance.

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• Injected minority carriers give rise to a stored charge in the two neutral regions. In the one- sided abrupt junction, + 푝 − 푛 =

Where 푄푠 퐼휏푝 = stored charge

푄푠 = minority carrier lifetime

휏푝= current through the diode

A capacitance effect arises due퐼 to the stored charge, and this is called the diffusion capacitance. • When the diode is switched from ‘on’ to ‘off’, there is a delay in the reverse current attaining the steady-state value. This delay time is related to the minority carrier lifetime.

Breakdown Voltage

• As the reverse voltage is increased at sufficiently large voltages, when the maximum electric field in the depletion region becomes equal to the critical electrical field, the reverse current increases enormously. This voltage is called the Breakdown Voltage. • There are three mechanisms: 1) Thermal instability 2) Tunneling process 3) Avalanche process • The breakdown voltage due to tunneling has a negative temperature coefficient. • The cylindrical and spherical junctions have lower breakdown voltage than planar junctions.

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Glossary

A = area of the junction

= an integration constant ′ A퐴 and B = two points in a semiconductor A and B = integration constants that will be determined by the boundary conditions at = 0 and = ′ 푥 C = 푥capacitance푊푛 = an integration constant

퐶1 = an integration constant ′ 퐶1 = an integration constant ′ 퐶1 = an integration constant

퐶2 = an integration constant ′ 퐶1 = an integration constant

퐶3 = depletion capacitance per unit area

퐶푑 = diffusion or storage capacitance

diff 퐶 = diffusion capacitance due to stored holes

퐶diff 푝 = junction capacitance

퐶퐽 = step voltage

𝑑푎 = electron diffusion constant

퐷푛 = hole diffusion constant

퐷푝 = energy at the bottom of the conduction band, al;so the potential energy of electrons

퐸퐶 = Fermi energy

퐸퐹 = quasi-Fermi energy level for electrons

퐸퐹� = quasi-Fermi energy level for holes

퐹� 퐸 = energy gap or bandgap energy

퐸푔 = intrinsic Fermi energy level

퐸푖 = trap energy level

퐸푡 = energy at the top of the valence band

퐸 푉 = electric field

ℰ - 140 -

= critical electric field

ℰ𝑐𝑐 = maximum electric field in the space charge region

ℰ푚( 𝑚) = the electric field at

ℰ(푥) = Fermi function 푥

푓 푥 = small signal forward conductance

푔푓 =forward conductance due to electron injection

푔푓� = forward conductance due to hole injection

푔푓� = thermal generation rates of electrons in the p region

푔푛 = thermal generation rates of holes in the n region

푔푝 = thermal generation rate

푔 푡 ℎ = current flowing through the junction

퐼 = current through a forward biased junction

퐼퐹 = forward current due to electron flow

퐼푛 = forward current due to hole flow

퐼푝 = current through a reverse biased junction

퐼푅 = component of forward bias current due to recombination in the space-charge region

퐼푟𝑟 = reverse current in the - junction

퐼푟𝑟 = saturation current 푝 푛

푆 퐼 = ideal current prefactor

퐼푆 푖𝑖𝑖 = saturation current component due to the flow of electrons

퐼푆� = saturation current component due to the flow of holes

푆� 퐼 = prefactor in the expression for current due to recombination in the space-charge region

퐼푆 푟 𝑟 = total electric current density through the junction

퐽 = electric current density due to electron flow

퐽푛 = electron current density in the neutral n region to replace the electrons lost due to recombination with holes in the neutral region 퐽푛1 = (injected) electron diffusion current density푛 in the neutral p region

퐽푛2 = electron current density flowing in the neutral n region to inject electrons in the p region

푛3 퐽 = excess electron diffusion current

퐽푛 diff - 141 -

= drift current density due to electron flow

퐽푛 drift = electric current density due to hole flow

퐽푝 = recombining hole current density in the neutral p region

퐽푝1 = (injected) hole diffusion current density

퐽푝2 = hole current density that is flowing in the neutral p region to inject holes in the n region

퐽푝 3 = saturation current density

퐽푆 = the total electric current density

퐽푡 = Boltzmann constant

푘 = minority carrier diffusion length (of holes) in the n region

퐿푝 = minority carrier diffusion length (of electrons) in the p region

퐿푛 = effective mass of the carrier ∗ n푚 = electron density = electron density in the semiconductor if the Fermi energy at trap level

푛1 = thermal equilibrium density of electrons

푛0 = thermal equilibrium majority carrier (electron) density

푛푛0 = electron density in the n region

푛푛 = electron density in an intrinsic semiconductor

푛푖 = electron density in the neutral p region

푛푝 = thermal equilibrium minority carrier (electron) density

푛푝0 = acceptor density

푁퐴 = ionized acceptor density − 푁퐴 = donor density

푁퐷 = ionized donor density + 푁퐷 = net impurity concentration

푁퐷 − 푁 퐴 = density of traps or g-r centers

푁 푡 = hole density

푝 = hole density in the semiconductor if the Fermi energy is at trap level

푝1 = thermal equilibrium density of holes

푝0 - 142 -

= hole density in the neutral n region

푝푛 = thermal equilibrium minority carrier (hole) density

푝푛0 = hole density in the p region

푝푝 = thermal equilibrium majority carrier (hole) density

푝푝(0 ) = probability that an excess hole, injected at = 0, will recombine in an elementary distance ′ ′ between and + ′ 푃 푥 �푥 ′ ′ ′ ′ 푥 ( ) =𝑑 probability per풙 unit푥 distance�푥 that an excess minority carrier (hole) will recombine and be lost ′ 푃 푥 = probability that an excess minority carrier will recombine and that does not vary with x

푃q0 = total electric charge, expressed in coulombs Q = charge

= stored charge due to excess electrons

푄푛� = stored charge due to excess holes

푄푝 � = stored charge r푄 푆 = proportionality constant in the recombination rate = small signal forward resistance

푟R푓 = recombination rate s = surface recombination velocity

t = time

T = Temperature

= net recombination rate per unit area of the surface

푈푠 , = thermal velocity of electrons and holes respectively and is equal to 3푘푘 푛 푝 ∗ 푣 푣 = thermal velocity � 푚

푣푡ℎ = applied voltage

�푎 = built-in voltage, potential difference between the neutral n-side and the neutral p-side

�푏� = breakdown voltage

�푏� = externally applied forward bias voltage

�퐹 = externally applied reverse bias voltage

�푅 = width of the neutral n region

푊푛 - 143 -

= width of the neutral p region

푊 푝 = distance measured with the origin at the metallurgical junction

푥 = distance measured with the origin at = ′ 푥 = total depletion region width = + 푥 푥푛

푑 푛 푝 푥 = part of the depletion region where푥 the푥 electron and hole are both minimum and equal

푥푑 eff = boundary of the depletion region on the n-side, or the width of the depletion region on the n-side 푥푛 = width of the depletion region on the p-side

푥푝 = boundary of the depletion region on the p-side

−푥 푝 = excess electron density

∆푛( = ) = excess minority carrier densities at = under forward bias

∆푛 푥 − =푥 excess푝 hole density 푥 −푥푝

∆푝(0) = excess carrier (hole) density at = 0, i.e., at = ′ ∆푝( ) = evaluated at = 푥 푥 푥푛 ′ ∆푝(푊푛= ) ∆ 푝 = excess minority푥 carrier푊푛 density at = under forward bias, same as (0)

∆푝 푥 푥 =푛 permittivity of a semiconductor 푥 푥푛 ∆푝

∈ 푠 = ideality factor

ힰ = electron mobility

휇푛 = hole mobility

휇푝 = charge density

휌 , = capture cross-sections for electrons and holes respectively

휎푛 휎 푝 = generation lifetime

휏푔 = minority carrier (or excess carrier) lifetime in the p-region

푛 휏 = 1 휏푛 0 = minority푁푡휎푛� carrier푛 (or excess carrier) lifetime in the n-region 휏푝 = 1 푝0 휏 = electrostatic푁푡휎푝�푝 potential

∅ = potential difference between two points A and B

∅퐴−퐵 - 144 -

= electrostatic potential in an n-type material

∅푛 = electrostatic potential in a p-type material

∅푝 = electrostatic potential difference

흍(x) = electrostatic potential difference at some point x 흍 = electrostatic potential difference between 휓퐴� 푥퐴 푎�푎 푥퐵

Problems

1. Calculate the electrostatic potential or as the case may be, at room temperature for each of the following impurity concentrations: a) 10 boron atoms; b) 10 gallium ∅푛 ∅푝 atoms and 10 phosphorous atoms. 15 −3 16 −3 𝑐 𝑐 2. Redo problem17 abov−e3 at =100 0K. (Assume that all the impurity atoms are ionized at this 𝑐 temperature.) 푇 3. Consider a -type silicon sample with = 10 . Calculate at a) 300 0K and b) 200 0K 4. An abrupt - junction has an impurity concentration16 −3 of10 boron atoms on the -side 푛 푁퐷 𝑐 ∅푛 and 10 antimony atoms on the -side. Calculate the15 built−-3in potential of the junction at 푝 푛 𝑐 푝 room temperature.17 −3 𝑐 푛 5. Calculate the depletion region width on the -side and that on the side as well as the total depletion region width for the case described in the previous problem. Do you expect your 푛 푝 answer to depend on the temperature? (The impurity atoms are completely ionized in the depletion region even though in the neutral region the impurity atoms will only be partially ionized at low temperatures.) If so, why? 6. Consider an abrupt - junction with the net impurity concentration of 4 × 10 on the -side and 6 × 10 on the -side. Determine the built-in potential. 15 −3 푝 푛 7. For the sample in the16 previous−3 problem, calculate the depletion region width under𝑐 thermal 푛 𝑐 푝 equilibrium at =300 0K 8. For the sample in the previous problem, 푇 (a) What is the maximum electric field , in the depletion region? (b) At what values of in the depletion region will the electric field be of ? ℰ푚𝑚 1 9. Consider a one sided abrupt - n junction in which the acceptor density on the -side is 푥 2 ℰ푚𝑚 10 and the donor density+ on the n-side is 10 . Calculate the depletion region 푝 푝 wid18th in− thermal3 equilibrium. 15 −3 𝑐 𝑐 10. An abrupt - junction has 10 donor atoms on the n-side and 10 acceptor

atoms on the -side. Determine15 , −3, and 16 −3 푝 푛 𝑐 𝑐 푝 �푏� 푥푝 푥푛 푥푑

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11. Consider an abrupt junction with = 10 on the -side and = 10 on the -side. Assume that the edge of the depletion16 region−3 occurs at and . Assume15 −3 to be 푁퐷 𝑐 푛 푁퐴 �푚 10 at room temperature. 푝 −푥푝 푥푛 푛푖 10 −3 = 𝑐(a) What is the electron density at in thermal equilibrium? 푥푝 (b) What is the hole density at the same point under the same condition? 푥 − 2 12. The equations in the text give the potential variation in the -side and in the -side respectively for an abrupt junction, taking the zero reference for the potential as that at the neutral region. 푝 푛 Write down similar equations for the potential taking the zero reference for the potential as 푝 that at the neutral region. 13. Consider a linear junction where the impurity concentration varies as = and is a 푛 constant. Derive an expression for the electric field and the potential variation in the depletion 푁 −푎� 푎 region assuming the boundary of the depletion region to be at = and = . 푊 푊 14. A diffused - junction has a uniform doping of 2 × 10 on the -side. The impurity 푥 − 2 푥 2 concentration varies linearly from the -side to the -side14 with−3 = 10 . At zero bias the 푝 푛 𝑐 푛 depletion region on the -side is 0.8 m. Find the depletion region width,19 built−4-in potential and 푛 푝 푎 𝑐 maximum electric field in the depletion region. 푝 휇 15. Show that, for a narrow base diode, the excess carrier density variation in the neutral region is given by 푛 ( ) = (0) 1 ′ ′ 푥

∆푝 푥 ∆푝 � − 푊푛� 16. Write down the expression for the ideal diode current when the region is approximated as a narrow base and the region is approximated as a wide base. 푛 17. Consider an abrupt - junction in which the acceptor density on the -side is 10 and 푝 the donor density on the -side is 6 10 . Calculate the depletion region widths16 − on3 the 푝 푛 푝 𝑐 and on the side for a forward bias voltage15 of− 30.6 V. 푛 푥 𝑐 푝 18. Assume that the area of the junction for the device in the last problem is 10 . Calculate 푛 the current through the device under a forward bias voltage of 0.5 V assuming−4 that2 the diode is 𝑐 a wide base diode. Take the minority carrier lifetime in the p and in the n regions to be 10 and 10 seconds respectively. Use the mobility curve in chapter 2 in your text. −6 19. Consider−5 as abrupt junction with = 10 on the n side and = 2 10 on the p side. Let the area of the junction be 10 16 . Given−3 the diffusion constant to be15 20 and−3 10 푁퐷 𝑐 푁퐴 푥 𝑐 and the lifetime to be 80 and 10−3 s in2 the n and p regions respectively, determine the 𝑐 ideal2 reverse−1 current for a reverse voltage of a) 0.06 and b) 6V. 𝑐 푠 휇 20. Given that = 1015 3, = = 10 and = 107 1 for both the electrons and the holes in− the diode in the previous−16 2 problem, calculate the− forward current due 푁푡 𝑐 휎푛 휎푝 𝑐 푣푡ℎ 𝑐 푠 to recombination in the depletion region for a forward bias of 0.4 V. Assume that the effective depletion region width in which the recombination is effective is = where is the 푥푑 depletion region width. 푥deff 3 푥푑

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21. Given that non-ideal reverse current is only due to generation in the depletion region occurring in the bulk of the device (i.e., neglecting surface generation) calculate the reverse current at a reverse bias of 6V. 22. Consider an abrupt - junction with the metallurgical junction located at = 0 and the boundaries of the depletion region located at = on the -side and at = on the - 푝 푛 푥 side. 푥 −푥푛 푛 푥 푥푝 푝 (a) What is the ratio of the electric field at = and that at = where and are constants of value less than 1. 푥 −푟1푥푛 푥 푟2푥푝 푟1 (b) What is the built-in voltage at room temperature if = 10 in the -side 푟2 and = 10 on the -side? Assume = 1 10 16 . −3 푁퐴 𝑐 푝 23. An abrupt p-n junction is fabricated15 −3 with = 2 10 on the n10-side− and3 = 푁퐷 𝑐 푛 푛푖 푥 𝑐 5 10 on the -side. Let the area of the junction16 − be3 10 . 푁퐷 푥 𝑐 푁퐴 15(a) −What3 fraction of the forward current is due to the minority−4 2 carrier injection in the 푥 𝑐 푝 퐴 𝑐 neutral -side? Assume that and on the -side to be 40 and 0.02 respectively and that and on the n-side to be 10 2 −1 and 푝 퐷푛 퐿푛 푝 𝑐 푠 0.005 respectively. 2 −1 𝑐 퐷푝 퐿푝 𝑐 푠 24. An abrupt - junction has 10 donor atoms on the -side and 10 acceptor 𝑐 atoms on the -side. If the area15 of the−3 junction is 10 and the minority16 carrier−3 lifetime is 푝 푛 𝑐 푛 𝑐 0.4 and 1 s on the and side respectively, find the−4 diode2 current at a) 0.6 and b) -1 volt bias 푝 𝑐 applied across the device, 휇 푛 푝 25. An abrupt - junction has 10 donor atoms on the n-side and 10 acceptor atoms on the -side. Find the current15 through−3 the diode for a forward bias of 0.519 volt− given3 that the 푝 푛 𝑐 𝑐 diffusion lengths in the and regions are 10 and 3 × 10 respectively. Assume the 푝 area of the junction to be 2 × 10 . Take the−2 diffusion constants−4 from the curve in chapter 푛 푝 𝑐 2 of the text. −4 2 𝑐 26. An abrupt - junction has 10 donor atoms on the -side and 10 acceptor atoms on the -side. Calculate15 the junction−3 capacitance a) at a forward bias19 of −0.43 volt and b) at 푝 푛 𝑐 푛 𝑐 a reverse bias of 10 volt. 푝 27. In an abrupt - junction = 5 10 , = 10 , and the area of the junction is 10 . Given that the minority carrier16 diffusion−3 constant17 and−3 lifetime are 20 and 푝 푛 푁퐷 푥 𝑐 푁퐴 𝑐 2 −103 2 on the side and 8 and 2 10 on the -side, calculate the current2 −1 𝑐 𝑐 푠 through−5 the diode at room temperature2 −1 under −6 푥 푠 푝 𝑐 푠 푥 푠 푛 (a) A forward bias voltage of 0.5 .

�

- 147 -

Chapter 4 Bipolar Transistor

In this chapter, we will be discussing a device called bipolar junction transistor or a bipolar transistor. The principle of the bipolar transistor action is based on the minority carrier injection into a neutral region using a forward-biased - junction, and the collection of the injected minority carriers by placing a second - junction (which is reverse-biased) very close to the forward-biased - junction. In order to illustrate this principle, let us푝 푛first consider an ideal - junction. Under forward bias, the current through the 푝junction푛 comprises two components: one due to electron injection into푝 the푛 side, the other due to hole injection into the side. The forward biased푝 푛 - junction represents a low- impedance junction: for a small voltage applied across the junction, a large amount of current flows푝 through that junction. On the other hand,푛 if a - junction is under푝 reverse푛 bias, the current through the junction is due to minority carriers that enter the depletion region, and does not depend on the reverse voltage (barrier height). The same current will푝 essentially푛 flow independent of the reverse voltage. In this sense, it is a constant current source or a high impedance current source: even for a large change in the reverse bias, the increase in the current is negligibly small.

Two types of transistor structure are possible in the bipolar transistors, and are shown in Figure (4.1). They are called - - and - - transistors. Let us consider the - - transistor. Assume that the first - junction is forward-biased, and the second junction is reverse-biased as shown in Figure (4.1A). The first region풏 풑is 풏called 풑the풏 emitter풑 , the region the base, and푛 the푝 푛 second region the collector푛. The푝 emitter-base junction is forward biased, and therefore minority carriers (electrons) are injected into the푛 base region. These injected minority푝 carriers diffuse in the base region푛 towards the collector-base junction. If the collector-base junction is close to the emitter-base, then a substantial fraction of the injected minority carriers will reach the collector-base depletion region without being lost in recombination. The carriers reaching the collector-base depletion region are driven by the electric field in the collector-base depletion region into the neutral collector region. Thus, the injected minority carriers in the base region give rise to a current through the collector lead, and this current is called the collector current. The collector current is like a current from a constant current source since it is due to minority carriers that enter a reverse-biased junction. If a large load resistance is connected between the collector and the base, then a large amount of output power will be delivered to the load resistance. Let us look at the band diagram of the device as given in Figure (4.2). The energy barrier for minority carrier flow in the emitter-base junction is reduced because it is forward-biased, giving rise to electron injection from the emitter into the base. The energy barrier in the collector-base junction is increased because it is reverse-biased. Most of the electrons injected into the base will reach the collector-base junction if the base region is narrow. Let us consider the circuit shown in Figure (4.3). A small signal voltage, is connected between the emitter and the base in series with the forward-bias voltage source . A load 푠 resistor ( ) is connected between the collector and the base, in series with the collector voltage 푣supply� 퐵� ( ). Let an alternating current ( ) flow in the emitter lead due to the small signal voltatge.� Let and 퐿 be the푅 corresponding collector and base currents. The collector current ( ) flows through the load 퐶퐶 푒 � resistance.� The power that the small� -signal source, has to deliver to the transistor is given by � �푏 �� = (4.1) 2 where is the input resistance of푃 the�푃 forward𝑃 퐼𝐼𝐼-biased �-� junction,푟퐸 and is given by 푬 풓 - 148 -푝 푛

= (4.2) 푘푘 with as the DC emitter current. The power푟퐸 output푞퐼퐸 is given by

푰 푬 = (4.3) 2 We can now calculate the power gain, which푃�푃 is𝑃 the퐼𝐼𝐼 ratio of the�� power푅퐿 output to the power input.

= (4.4) 2 푖푐 푅퐿 2 If we design the transistor so that the collector푃�푃𝑃 current퐺𝐺� ( ) is very푖푒 푟퐸 nearly equal to , then the power gain is equal to �� �푒

푅퐿 푃�푃𝑃 퐺𝐺� ≈ However, we already noticed that is the resistance of a forward푟퐸 -biased - junction and hence is a very small quantity. By choosing the load resistance to be large, we can obtain a large power gain. 퐸 This is the principle of a bipolar transistor.푟 푝 푛 푅퐿

Figure (4.1): Two alternate structures for a bipolar transistor: A) - - transistor B) - - transistor.

푛 푝 푛 푝 푛 푝

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Figure (4.2): The energy band diagram for the two - junctions connected back-to-back.

푛 푝

Figure (4.3): Application of a small signal voltage source between the emitter and the base of a - - transistor. 푛 푝 푛

Figure (4.4): The various components of current, flowing in an - - transistor

Let us now consider the various components of current flowing through푛 푝 푛the device. The DC current flowing through the emitter (which is denoted ) comprises two components: one due to electron injection from the emitter into the base ( ), and the other due to hole injection from the 퐸 base into the emitter ( ). The electrons injected into퐼 the base diffuse toward the collector region and 퐼퐸� result in a collector current. However, not all the electrons injected into the base are able to reach the 퐼퐸� - 150 - collector-base depletion region, since some of the injected electrons recombine in the neutral-base region. Hence the electron current ( ) that reaches the collector-depletion region and flows through the collector lead is smaller than , the electron current injected from the emitter into the base. 퐼퐶� In the collector, in addition퐼퐸 �to , we also have a leakage current that normally arises in a reverse-biased - junction. Let us denote the leakage current . The collector current (denoted ) 퐶� is therefore equal to the sum of (the퐼 leakage current), and (the current due to the injection of 퐶�퐶 퐶 electrons from the푝 푛 emitter into the base). 퐼 퐼 퐼퐶�퐶 퐼퐶� Holes have to flow into the base region to participate in the recombination process. This hole current flows through the base lead. Similarly, the holes injected into the emitter form the base also flow through the base lead. Additionally, the collector leakage current, , also has to flow through the base lead, The flow of the various components of current in an - - transistor is shown in Figure 퐶�퐶 (4.4). In this figure, the arrows indicate the direction of the conventional퐼 electric current, and not the direction of the particle flow. The electron (particle) flow is opposite푛 to푝 the푛 conventional electric current flow for and . We can now write the emitter, base and collector currents as equal to

퐼 퐸 � 퐼 퐶 � = + (4.6)

퐼퐸 = 퐼퐸� + 퐼퐸� (4.7) = 퐼퐶 = 퐼퐶� 퐼퐶�퐶+ (4.8) We define the current gain of퐼퐵 the device퐼퐸 − as퐼퐶 equal퐼 to퐸� the− collector퐼퐶� 퐼 퐸current� − 퐼 퐶due�퐶 to electron injection from the emitter into the base, divided by total emitter current. The current gain is usually denoted , and is given by 휶ퟎ = (4.9) 퐼퐶� The subscript in implies that the gain is evaluated훼0 퐼 퐸for very low (zero) frequency input signals. It must be noted that we are considering an - - transistor, and hence the collector current due to 0 minority carrierퟎ (electron)훼 injection in the base from the emitter is denoted with a subscript . On the other hand, if we were to consider a - - 푛 transistor,푝 푛 the collector current arising from minority carrier injection from the emitter into the base will be due to holes and will be denoted with a subscript푛 in place of . 푝 푛 푝 푝 The푛 current gain, , can be related to two other parameters, which are (the emitter injection efficiency), and (the base transport factor). The emitter injection efficiency is defined as 0 the ratio of the emitter current훼 due to minority injection from the emitter into the휸 base and the total 푻 emitter current. Hence it is휶 equal to 훾

= = (4.10) 퐼퐸� 퐼퐸� The base transport factor, , is definedγ as the퐼퐸 ratio퐼퐸 of�+ the퐼퐸푝 collector current and the minority carrier current injected from the emitter into the base, i.e., the fraction of the injected minority carrier current � that reaches the collector. 훼This relationship is expressed by

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= (4.11) 퐼퐶� It is readily seen that is given by the product훼� of 퐼퐸 �and , that is

0 � 훼 = = 훾 훼= (4.12) 퐼퐶� 퐼퐸� 퐼퐶� We can now write the collector current훼0 as 퐼equal퐸 to퐼퐸 퐼퐸� 훾 훼�

= + = +

퐼 퐶 = 퐼 퐶 � 퐼 퐶�퐶+ 훼�퐼=퐸� 퐼퐶�+퐶 (4.13) and similarly, the base current can be expressed훼�훾 퐼퐸 in terms퐼퐶�퐶 of the훼 0emitter퐼퐸 퐼current퐶�퐶 as = = (1 ) (4.14)

In principle, the emitter퐼퐵 injection퐼퐸 − efficiency퐼퐶 is made− 훼 0very퐼퐸 close− 퐼퐶 to� 퐶unity by making the emitter impurity concentration very large in comparison to the base impurity concentration. Similarly, the base transport factor can be made very close to unity by making the width of the neutral base region (i.e., the distance which the minority carriers that are injected from the emitter into the base have to travel to reach the collector-base depletion region) very small compared with the diffusion length of the minority carriers in the base. Under these conditions, the current gain will be very close to unity.

Example

Let us now calculate the power gain of an - - transistor, based on our discussions so far. Let us assume that the transistor is carrying a DC emitter current , equal to 1 mA. Let the load resistance be equal to 259 . Let the current gain 푛 푝be푛 taken as 0.99. Assume a small signal voltage is applied 퐸 퐿 between the emitter and the base giving rise to an emitter 퐼current and and a collector current . 푅 퐾훺 훼0 푣�푠 = R �푒 �� 2 푃�푃=𝑃(푂0.𝑂𝑂�99 ) R�c L 2 = (0.99 ) ×� 푒259 L10 2 3 �푒 = 푥 r 푊 2 But 푃�푃𝑃 퐼𝐼𝐼 �e e =

푒 푘� 푟 . =푞 퐼퐸 0 0259 −3 = 2510.9 훺

- 152 - 훺

= 25.9 2 (0.99 e) 259 10 푃 �푃𝑃 퐼=𝐼𝐼 � 푥 2 25.9 3 �푒 푥 푥 푃�푃𝑃 퐺𝐺� 2 = 0.99 � e 10푥 2 4 = 0.98 푥 10 = 9800 4 푥 From the above example, we see that a large power gain is obtained by choosing a large value of . From this rudimentary discussion it appears that infinite power gain can be obtained by choosing to be infinitely large. This is not so. Later on, we will see that our initial assumption that the collector- 퐿 퐿 base푅 junction acts as a current source with infinite impedance is in error, and that the output impedance푅 of the collector-base junction (the ratio of the incremental change in the collector voltage to the incremental change in the collector current) is large but not infinite. Hence the maximum power that can be drawn from the device is what is obtained with equal to the output impedance of the collector-base junction. 푅퐿 The bipolar device is denoted by the symbol shown in Figure (4.5A) for - - transistors, and that in Figure (4.5B) for - - transistors. The lead with an arrow is the emitter and the direction of the arrow indicates (as in a - junction) the direction of the forward-bias current. 푛 푝 푛 푝 푛 푝 푝 푛 D.C. Characteristics

Let us now investigate the characteristics of the bipolar transistor in a quantitative manner. We make the following assumptions: 1) The impurity concentration in the emitter, base and collector, regions are uniform. This means that there is no spatial variation of the impurity concentration within any one of the three regions. 2) The injection of minority carriers in the base corresponds to a low-level injection. 3) There is no recombination or generation in the depletion region of the emitter-base junction, or in the collector-base junction. In other words, we treat the two junctions as ideal junctions. 4) There is no series resistance in the bulk regions of the device.

Ideal Transistor Characteristics

The characteristics of a device under the assumption of ideal junctions are called ideal transistor characteristics. Let us examine the base region. Referring to Figure (4.6), the metallurgical junctions on the emitter-base side and on the collector-base side are located at = 0 and = , respectively. However, the active neutral region of the base is obtained by subtracting (the width of the 푥 푥 푊퐵� depletion-region occurring in the base region due to the emitter-base junction), and (the width of 𝑃 the depletion-region occurring in the base region due to collector-base junction)푥 from the width . 푥𝑃 We choose a new coordinate system, , where the origin is chosen at the edge of the neutral base퐵 � ′ 푊 푥 - 153 - region on the emitter side. The edge of the depletion region on the collector side occurs at = . is then equal to ′ ′ ′ 푥 푊 푊 = (4.15) ′ From now on, we will use 푊to denote푊 퐵the� −width푥𝑃 of− the푥 𝑃neutral base region. The minority carriers injected from the emitter at ′ = 0 diffuse in the neutral base region and travel toward the 푊 collector-based junction. This gives rise′ to a diffusion current. In order to derive an expression for the diffusion current due to injection of minority푥 carriers in the base, we first set up the continuity equation in the base region. Since we are considering an - - transistor, electrons are the minority carriers in the base. The continuity equation for electrons in the base region is given by 푛 푝 푛 = (4.16) 휕휕� 1 𝛥 Here, stands for the excess minority휕� 푞 carrier훻 ∙ 퐽푛 in− the휏푛 base region due to minority carrier injection from the emitter, is the minority carrier current density, and is the minority carrier lifetime. If we 훥assume� a one-dimensional case, this continuity equation reduces to 퐽푛 휏푛 = ( ) (4.17) 휕휕� 1 휕 𝛥 ′ where 휕� 푞 휕푥 퐽푛 − 휏푛

= + (4.18) 휕휕� ′ Let us assume that there is no 퐽electric푛 푛 field�휇푛 inℰ the neutral푞퐷푛 휕 base푥 region. Then the expression for the current density has only the diffusion term, and the gradient of becomes equal to

퐽푛 = (4.19) 2 휕퐽푛 휕 𝛥 ′ ′2 Substituting this expression for the gradient휕푥 of 푞in퐷 the푛 휕continuity푥 equation, we get 퐽푛 = (4.20) 2 휕휕� 휕 𝛥 𝛥 ′2 We can solve the continuity equation휕� first퐷푛 under휕푥 DC− conditions.휏푛 We assume that sufficient time has been allowed to elapse after the application of the DC bias so that steady-state conditions can be assumed.

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Figure (4.5): The circuit symbol for the bipolar transistor. A) - - transistor and B) - - transistor. 푛 푝 푛 푝 푛 푝

Figure (4.6): The definition of the neutral base region. The region between the metallurgical junction on the emitter side and that on the collector side represents the total base region. However, the neutral base region width is obtained by subtracting appropriate segments of the depletion region widths on the two sides. Steady State (DC Conditions)

Under steady-state conditions, partial of with respect to time is equal to zero. Hence, the continuity equation can be rearranged to obtain 훥�

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= = (4.21) 2 휕 𝛥 𝛥 𝛥 ′2 2 where = is the diffusion length,휕푥 which we퐷푛 휏came푛 across퐿푛 in our study of the - junction. The solution to the continuity equation is seen to be 퐿푛 �퐷푛휏푛 푝 푛

= −푥′ + 푥′ (4.22) 퐿푛 퐿푛 The boundary conditions for the excess훥 carrier� 퐴density푒 in the퐵 푒base region are: At = 0 ′ 푥 ( = 0) = = 1 (4.23) 푞 푉퐵� ′ 푘푘 And at = , the boundary condition훥� 푥 4 can be approximated훥푛0 푛푝 as0 � 푒 − � ′ ′ 푥 푊 = 0 (4.24) = 0 By applying the boundary condition at in훥 Equation� (4.22), we get ′ 푥 = A + B = 1 (4.25) 푞 푉퐵퐸 푘푘 Similarly, by applying the boundary훥푛 0condition at = 푛푝, �we0 �get푒 − � ′ ′ 푥 푊 0 = −푊′ + 푊′ (4.26) 퐿푛 퐿푛 Solving for the constants A and B, (left as a homework퐴 푒 problem)퐵푒 we obtain

= −푊′ (4.27) 퐿푛 푒 0 −푊′ 푊′ 퐵 훥푛 퐿푛 퐿푛 푒 − 푒 = 푊′ (4.28) 퐿푛 푒 0 푊′ −푊′ 퐴 훥푛 퐿푛 퐿푛 Substituting these expressions for A and B, we obtain푒 − 푒

( ) ( ) ( ) ( ) 푊′−푥′ 푊′−푥′ ′ ′ ( ) = + − = 푊 −푥 (4.29) 퐿 퐿 푛 푛 sinh (퐿 ) ′ 푒 푒 푛 푊′ − 푊′ 푊′ 푊′ 푊′ 0 0 − + 0 훥� 푥 훥푛 퐿푛 퐿푛 훥푛 퐿푛 퐿푛 훥푛 sinh 퐿푛 4 A more exact boundary condition푒 than− the푒 one we have used푒 is − 푒

( = ) = 1 −푞 푉퐶� ′ ′ However, the resulting expression for ( ) will look more푝� cumbersome.푘푘 Very little error results due to our simpler assumption that ( = 훥)�= 푥0 ′ 푊 푛 �푒 − � ′ ′ 훥� 푥 훥� 푥 푊 - 156 -

This expression for is the most general expression for excess carrier density in the base region. It may be recalled from our discussion of the - junction that the excess carrier density in the neutral region decreased linearly with훥� distance when the width of the neutral region was small compared to the minority carrier diffusion length (narrow푝 푛base diode) and decayed exponentially with distance when the width of the neutral region was large compared to the diffusion length (wide base diode). It is easy to show that Equation (4.29) reduces to an exponential or linear relation under appropriate assumptions.

As we will see later on, in order to make (the base transport factor) as close to unity as possible, (the neutral base width) has to be kept very small in comparison with (the minority 훼� carrier diffusion′ length). When , the expression for ( ) can be simplified as 푛 푊 ′ 퐿 푾 ≪ 푳풏 ( ) 훥� 푥 ( ) 푊′−푥′ = (1 ) (4.30) ′ �푛0 퐿 ′ 푛 푥 푊′ 0 ′ 훥� 푥 ≈ 퐿 훥푛 − 푊 ( ) 푛 given by Equation (4.29) is plotted in Figure (4.7) as a function of for various values of ′. It can 푊 be noticed′ that when , a linear plot is obtained; when ′, an exponential plot is 푛 푥 푥 퐿푛 obtained. We can now calculate′ the diffusion current in the base due′ to the injected minority carriers by 푛 푛 substituting the expression푊 ≪ for퐿 ( ) given by Equation (4.29) in푊 the≫ expression퐿 for electron diffusion current. The electron diffusion current′ density is given 훥� 푥 ( ) ( ) ( ) = = 푊′−푥′ (4.31) cosh ( 퐿 ) ′ 푑𝑑 퐷푛�푛0 푛 ′ 푊′ 푛 푛 푛 퐽 푥 푞퐷 휕푥 −푞 퐿 sinh 퐿푛 Base Transport Factor:

The electron current density at = 0, is obtained by putting = 0 in the above equation. ′ ′ 푥 ( ) 푥 ( ) ( = 0) = 푊′ = 푊′ (4.32) cosh ( 퐿 ) �푛0 coth 퐿 ′ �푛0 푛 푛 푊′ 퐽푛 푥 −푞 퐷푛 퐿푛 −푞 퐷푛 퐿푛 The emitter current due to electrons injected fromsinh the emitter퐿푛 into the base ( ) is equal to

퐸� ( 퐼 ) = ( = 0) = 푊′ (4.33) �푛0coth 퐿 ′ 푛 The electron current density퐼퐸� at 퐴 =퐽푛 푥 is obtained by− putting푞 퐴 퐷 푛 = 퐿in푛 Equation (4.31) ′ ′ (푥 =푊′ ) = 푥 푊′ (4.34) ( ) ′ �푛0 1 푊′ 퐽푛 푥 푊′ −푞 퐷푛 퐿푛 This is the collector current density arising due to minority carriersinh injection퐿푛 in the base. The collector current ( ) is equal to

푰푪� - 157 -

= ( = ) = (4.35) ( ) ′ �푛0 1 푊′ 퐼퐶� 퐴 퐽푛 푥 푊′ −푞 퐴 퐷푛 퐿푛 The difference between and represents the decrease in current duesinh to reco퐿푛 mbination in the neutral base region. The base transport factor is the ratio of to , and is given by 퐼퐸� 퐼퐶� ( ) 퐼퐶� 퐼퐸� = = = (4.36) ( ′ )′ 퐼퐶� 퐽푛 푥 = 푊 1 ( ) ′ 푊′ � 퐼퐸� 퐽푛 푥 =0 훼 cosh 퐿푛 When becomes negligibly small in comparison with , ( ) 1 and becomes equal to 1 ′ . 푊 This can be seen from a comparison of the electron current푛 density at = 0 and� that at = . 푊′ 퐿 𝑐�ℎ 퐿푛 ≈ 훼 When , ′ ′ 푥 푥 푊′ 푛 푊 ′ ≪ 퐿 ( = 0) = (4.37) ′ 푞 퐷푛�푛0 푛 ′ ( = ) is also equal to 퐽 . 푥Therefore, −= 1. 푊This reflects the fact that there is no ′ ′ 푞 퐷푛�푛0 recombination푛 in the base region. ′ � 퐽 푥 푊 − 푊 훼

Figure (4.7): A plot of the excess minority carrier density in the base region for various values of ′. 푊 Emitter Injection Efficiency: 퐿푛

In order to determine , the emitter injection efficiency, we must get an expression for , the current injected into the base by the emitter as well as , the current injected into the emitter by the γ 퐼퐸� base, is already determined in Equation (4.33) where is given by 퐼퐸� 퐼퐸� 훥푛0 - 158 -

= 1 = 1 (4.38) 푞�퐵� 2 푞�퐵� 푘푘 푛푖 푘푘 Therefore 훥푛0 푛푝�0 �푒 − � 푁퐴� �푒 − �

= coth ( ) ′ 푛 퐸� 퐷 0 푊 퐼 −푞 퐴 푛 훥푛 푛 = 퐿 coth ( 퐿 ) 1 (4.39) 2 ′ 푞�퐵� 푛 퐷 푛푖 푊 푘푘 The component of the emitter− current푞 퐴 퐿 푛due푁퐴 to� hole injection퐿푛 �from푒 the− base� into the emitter can similarly be obtained by solving the continuity equation for holes in the neutral emitter region. Let 퐸� us assume that the width퐼 of the neutral emitter region, , is large compared with the minority carrier (hole) diffusion length ( ) in the emitter. Then we can use the wide base approximation in the 퐸 emitter region. We will get an expression for the injected푊 hole current density which decays 𝑃 exponentially (because of퐿 our assumption that is much large than ). We determine the expression for the hole current in the emitter just at the boundary of the emitter-base depletion region, 푊푛� 퐿푝� and take that as the current injected into the emitter by the base. Hence is given by

퐼퐸� = 1 퐵� 푝� 푞� 퐸� 퐷 푛�0 푘푘 퐼 −푞 퐴 푝� 푝 �푒 − � = 퐿 1 (4.40) 2 푞�퐵� 푃� 퐷 푛푖 푘푘 The emitter injection efficiency, , can be− 푞obtained퐴 퐿푝� as푁 퐷� �푒 − �

훾 = = + 퐼퐸� 퐼퐸� 훾 퐼퐸 퐼퐸� 퐼퐸� ( ) ′ = 퐷푛 1 푊 퐿푛 푁퐴 � coth 퐿푛 ( ) ′ 퐷푝� 1 퐷푛 1 푊 퐿푝� 푁퐷� + 퐿푛 푁퐴� coth 퐿푛 = (4.41) ( ) 1 퐷 ′ 푝퐸 퐿푛 푁퐴� 푊 1 + 퐷 퐿 푁 푡𝑡ℎ 퐿 It is seen that when is very large in comparison with푛 푝� , 퐷 approach� 푛1. When , ′ coth ( ) . Therefore퐷� 퐴� 푛 ′ 푁 푁 훾 푊 ≪ 퐿 푊 퐿푛 ′ 퐿푛 ≈ 푊 = (4.42)

1 퐷 ′ 푝� 푊 푁퐴� 훾 1 + �� 퐷 �� 퐿 �� 푁 �� 푛 푝� 퐷�

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Example

Let us calculate the base transport factor given: = 1 = 10 ′ −4 푊 = 10 휇� = 10 𝑐 −3 By substituting these values in the expression퐿푛 for 휇�, we obtain𝑐 � = =훼 = 0.995 (4.43) . 1 ( ) 1 � 푊′ 훼 cosh 1 005 퐿푛

Example

Let us now calculate the emitter injection efficiency, given

10 19 −3 푁퐷� =≈ 10 𝑐 17 −3 푁퐴� = 1.0 𝑐 ′ 푊 = 10 휇� 퐿푝=� 1 휇� 2 퐷푝�= 10 𝑐 ⁄푠𝑠 2 By substituting these values in the expression퐷푛 for , we𝑐 obtain⁄푠𝑠 1 훾 1 = = 0.9999 1 1 1 1 1 + 1 + 10 10 100 10 훾 ≈ �� � � � � �� 4

DC Current-Voltage Characteristics

The DC current-voltage characteristics of the bipolar transistor can be measured in either of two modes: the common base mode and the common emitter mode. Common Base Mode

In the common base mode, the collector current is measured as a function of the collector-base voltage for a specified value of the emitter current using the experimental configuration shown in Figure

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(4.8). The emitter-base voltage ( ) is adjusted to a suitable value, such that a specified emitter current is obtained. The collector current is measured for different values of the collector-base voltage, , 퐵� while adjusting to keep constant� at the specified value. Thus a family of curves can be obtained, 퐶� each curve corresponding to a specific value of emitter current, as shown in Figure (4.9). The -axis� in 퐵� 퐸 this figure is the� collector-base퐼 voltage, , and the –axis is the collector current, . When is positive5, the collector-base junction is reversed-biased. Since is very close to unity, the collector푥 퐶� 퐶 퐶� current is very nearly equal to the emitter� current. When푦 is negative, the collector퐼 -base junction� is 0 forward-biased, and the collector also injects minority carriers 훼into the base just as the emitter does. 퐶� Hence as the forward bias on the collector is increased, the� collector current decreases and ultimately becomes zero. When the collector-base junction is forward-biased, the transistor is said to be saturated, or said to be operating in the region of saturation. When the collector-base junction is reverse-biased, the excess carrier density at = (the edge of the depletion region of the collector-base junction) is negligibly small. The excess carrier′ density′ profile in the neutral base region under this condition is shown in Figure (4.10A). On the푥 other푊 hand, when the collector-base junction is forward-biased, the excess carrier density at = is increased due to injection of minority carriers from the collector into the base. The excess carrier′ density′ profile in the neutral base region under this condition is shown in Figure (4.10B). The slope푥 of the푊 excess carrier density profile is less now and hence the collector current is reduced.

Figure (4.8): Experimental measurement of characteristics.

퐼퐶 − �퐶�

5 Remember we are dealing with an n-p-n transistor

- 161 -

Figure (4.9): The collector current-voltage characteristics of an - - transistor in the common base mode. 푛 푝 푛

Figure (4.10): The excess carrier density in the base region A) when the collector-base junction is reverse-biased, the device is said to operate in the active mode. B) When the collector-base junction is forward-biased, the device is said to operate in the saturation mode.

- 162 -

The vertical dotted line drawn through = 0 in Figure (4.9) separates the active region (when the collector-base is reverse-biased) from the saturation region. It will be noticed that the collector 퐶� current (for a given emitter current) increases slightly� with the collector voltage in the active region. This causes the output impedance ( ) of the device to be finite, and hence our earlier assumption

푑퐼푐 that the collector-base junction behaves−1 like a constant current source is not strictly valid. This will be � 푑�퐶� � further discussed in the section on output impedance.

= + (4.44)

If we increment by , increases by퐼퐶 . 훼Hence0퐼퐸 퐼 퐶 � 퐶 퐸 퐸 퐶 퐶 퐼 훥 퐼 퐼 훥 � = (4.45) 𝛥퐶 Therefore, can be obtained from the characteristics훼0 � illustrated퐼퐸 in Figure (4.9). The훼 characte0 ristic curve corresponding to = 0, represents the collector-base junction leakage current, . The letters CB in the subscript refer to the fact that it is the collector-base junction 퐸 current, and the letter o refers to the fact that the퐼 emitter is open, i.e. there is no emitter current. At 퐶�퐶 some large퐼 value of , the current is seen to increase suddenly with collector voltage in all the curves. This is due to the breakdown of the collector-base junction. The breakdown phenomenon will be 퐶� discussed in the section� on voltage limitations. The collector leakage current, , is given by the expression for the leakage current for a reverse-biased - junction and is equal6 to 퐼퐶�퐶 푝 푛 = + −푞푉 1 (4.46) 푛 푝0 푝� 푛�0 퐶� 퐷 푛 퐷 푝 푘푘 퐶�퐶 퐼 −푞� � 퐿푛 퐿푝� � �푒 − � Common Emitter Mode

In the common emitter mode, the collector current-voltage characteristics are measured as a function of the collector-emitter voltage, keeping the base current constant. The experimental set-up to obtain these characteristics is illustrated in Figure (4.11). As in the previous method, the collector current is measured at different values of the collector-emitter voltage while adjusting the base-emitter voltage to keep the base current constant at a specified value. As before, a family of curves is obtained, each curve corresponding to a particular value of the base current ( ) as shown in Figure 퐶 퐶� (4.12). The voltage represents the sum of the voltage drop across the collector-base퐼 − junction� and 퐵 that across the base-emitter junction. 퐼 �퐶� = + , Hence, when becomes less than�퐶� �퐶� becomes�퐵� negative, and the collector-base junction becomes forward-biased. The device becomes saturated under this condition. In Figure (4.12), the �퐶� �퐵� �퐶�

6 Although the base width is small in comparison with the minority carrier diffusion length, we use the wide base expression for leakage current since minority carriers generated in the base within a diffusion length from the collector base junction will contribute to the leakage current.

- 163 - portion of the curve lying to the left of the dotted line marked = 0 corresponds to the saturation region of operation, while the portion to the right corresponds to the active region of operation. As 퐶� expressed before, the collector current is equal to � 퐼퐶 = +

But 퐼퐶 훼0퐼퐸 퐼퐶�퐶 = (1 )

Referring terms in this equation to obtain퐼퐵 , we− get훼 0 퐼퐸 − 퐼퐶�퐶 퐸 =퐼 + (4.47) ( ) ( ) 퐼퐵 퐼퐶�퐶 Substituting this expression for in the퐼퐸 equation1−훼 0for ,1 we−훼 get0 퐼퐸 퐼퐶 = + + 퐼퐵 퐼퐶�퐶 퐶 0 1−훼0 1−훼0 퐶�퐶 퐼 = 훼 � + � 퐼 + (4.48) 훼0 훼0 If we give an increment in the base current,�1−훼 the0� collector퐼퐵 �1 current−훼0� 퐼퐶 increases�퐶 퐼퐶 by�퐶 . We now define a current gain, in the common emitter mode as equal to 훥�퐵 훥�퐶 0 훽 (4.49) 𝛥퐶 From the last equation for , 훽0 ≡ 𝛥퐵 퐼퐶 = (4.50) 훼0 since does not change due to a change훥 in�퐶 . Hence�1− 훼0� 훥�퐵

퐶�퐶 퐵 퐼 퐼 = (4.51) 훼0 Hence, can be written in terms of as 훽0 1−훼0

퐼 퐶 = +퐼 퐵 + = + ( +1) (4.52) = 0, The collector current퐼 퐶with the훽0 base퐼퐵 open,훽0퐼 퐶i.e.,�퐶 with퐼 퐶�퐶 is 훽denoted0퐼퐵 훽0, and from퐼퐶� this퐶 equation we can see that 퐼퐵 퐼퐶�퐶 = ( +1) (4.53)

Again, the letters CE refer to the fact that퐼퐶� 퐶the current훽0 is flowing퐼퐶�퐶 between the collector and the emitter, and the letter o signifies the fact that the base is open, i.e., no base current is flowing. The leakage current in the common emitter mode, , is larger than the leakage current in the common base mode, , by a factor ( + 1). 퐼퐶�퐶 퐼퐶�퐶 훽0 - 164 -

Figure (4.11): Experimental measurement of characteristics.

퐼퐶 − �퐶�

Figure (4.12): The collector current-voltage characteristics of a n-p-n transistor in the common emitter mode.

- 165 -

Different Modes of Operation

The bipolar transistor can be used in active, cut-off or saturation modes by biasing the emitter-base and collector-base junctions suitably as shown in Table 1.

Mode E-B Junction Bias C-B Junction Bias active forward reverse cut-off reverse reverse saturation forward forward inverse reverse forward Table 1: Modes of Operation

Low Frequency Small Signal Response and the Equivalent Circuit

Till now we discussed the DC characteristics of the transistor. Let us now consider the response of the device to a small signal voltage input. The response of the bipolar transistor to a small signal input can be represented in terms of an equivalent circuit. The equivalent circuit will be different for common base and common emitter modes of operation. Common Base Mode

Consider the circuit shown in Figure (4.13). is the DC collector supply voltage. A small signal voltage source ( ) is connected between the emitter and the base in series with the DC bias voltage �퐶퐶 ( ). We assume that the frequency of the small signal voltage source is very low and that the injected 푠 minority carrier 푣density� everywhere in the base is varying in step with voltage variation. Since is �퐵� varying sinusoidally with time, the injected carrier density in the base varies sinusoidally as shown in 푠 Figure (4.14). The top line in this figure shows the minority carrier density in the base when the푣� sinusoidal function ( ) is at the most positive excursion. Assuming = , the base emitter voltage at its most positive excursion is + . The middle line in this figure represents the carrier 푠 푠 푠 distribution when the푣� same function is zero. The bottom line represents푣� the푣 carrier푠𝑠 휔 �density when is at 퐵� 푠 its most negative excursion, with the instantaneous� 푣 base-emitter voltage equal to . The 푣�푠 excess carrier density at = 0 is given by 퐵� 푠 ′ � − 푣 훥� 푥 ( ) = ( = 0) = 1 (4.54) 푞 푉퐵� + 푣�푠 ′ 푘푘 0 푝𝑝 Denoting the total emitter current훥푛 as 훥�, we푥 obtain 푛 �푒 − �

( 퐼)퐸� ( ) = 1 = (4.55) 푞 푉퐵� + 푣�푠 푞 푉퐵� + 푣�푠 푞푉퐵� 푞푣�푠 푘푘 푘푘 푘푘 푘푘 퐼퐸� 퐼퐸� �푒 − � ≈ 퐼퐸� 푒 퐼퐸� 푒 푒 Expanding as an exponential series and retaining only the leading terms, we obtain 푞푣�푠 푘푘 푒 (1 + ) (4.56) 푞푉퐵� 푠 푘푘 푞�� 퐼퐸� ≈ 퐼퐸� 푒 푘푘 - 166 -

therefore comprises a DC term and a sinusoidally varying term, and can expressed as

퐼 퐸 � = + (4.57) where is the DC current and is the sinusoidally퐼퐸� 퐼퐸 varying횤̃푒 component of current. By comparing Equations (4.56) and (4.57), we obtain 퐼퐸 횤̃푒

= (4.58) 푞푉퐵� 푘푘 which is the DC emitter current when no sinusoidally퐼퐸 퐼퐸� 푒varying voltage is applied.

= (4.59) 푞푣�푠 푘푘 횤̃푒 퐼퐸 푒

Figure (4.13): Common base mode amplifier connection. is the small signal sinusoidal input voltage. is the load resistor in the collector circuit. 푣�푠 푅퐿

Figure (4.14): Modulation of the injected minority carrier density in the base in step with a small sinusoidal signal voltage.

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Figure (4.15): Equivalent circuit of the input circuit in the common base mode.

The ratio of the sinusoidal input voltage, , to the sinusoidal emitter current (input current) defines an equivalent resistor which has a conductance equal to . The input circuit is equivalent to a 푣�푠 푒 푞 횤̃ resistance (or conductance) and is shown in Figure (4.15)�. 푘푘 The� 퐼 ratio퐸 is the small signal input 횤̃푒 conductance, , and = = is called the input resistance.��푠 Due to ( ) varying sinusoidally 1 in step with , 𝒊a small signal collector푖� 퐸 current flows in step with , the total collector current is 품 푔푖� 푟 푟 훥� 푥 given by 푣�푠 푣�푠 퐼퐶� = + = + + (4.60)

The DC current in the collector퐼퐶� is 훼0퐼퐸� 퐼퐶�퐶 훼0퐼퐸 훼0횤̃푒 퐼퐶�� = +

and the small signal sinusoidal collector current퐼퐶 훼is 0퐼퐸 퐼퐶�퐶 = (4.61)

The small signal current gain is equal to 횤̃� 훼0횤̃푒 = = (4.62) 횤̃�표� 횤̃푐 The output voltage is 횤̃푖� 횤̃푒 훼0

= = = = (4.63) 푞 where 푣�𝑜� 횤̃�푅퐿 훼0횤̃푒푅퐿 훼0 푘푘 퐼퐸 푣�푠 푅퐿 푔푚푣�푠 푅퐿

= = (4.64) 푞 훼0 is called the transconductance. In our treatment,푔푚 훼 0 푘푘= 퐼퐸 where푟퐸 is the sinusoidally varying small signal emitter-base voltage. Transconductance relates the output (collector) current to the input 푣�푠 푣�푏� 푣�푏� (emitter-base) voltage. Since flows through a reverse-biased - junction, is (nearly) independent of the value of , the load resistor. This is equivalent to a constant current source, . The output � 푝 푛 � circuit behaves as though a current횤̃ source ( ) exists between the횤 ̃ collector and the emitter. 푅퐿 � The equivalent circuit for the output circuit is shown in Figure (4.16). Combining the횤 equivalent̃ circuits 푔푚푣�푠 푔푚푣�푏� - 168 - for the input and the output sides, the equivalent circuit for the transistor is shown in Figure (4.17). The rectangular box enclosed by the dotted line outlines the transistor. The voltage gain equals

= = (4.65) ���표� 푔푚��푠 푅퐿 The power gain equals ��푖� ��푠 푔푚 푅퐿

= = = (4.66) 횤̃�표�����푡 2 2 푅퐿 횤̃푖���푖� 훼0 푔푚 푅퐿 훼0 푔푖� 푅퐿 훼0 푟퐸

Figure (4.16): Equivalent circuit of the output circuit in the common base mode

Figure (4.17): Equivalent circuit for the bipolar transistor in the common base mode.

Figure (4.18): Circuit connection for common emitter mode amplifier. is the small signal sinusoidal input voltage, and is the load resistor. 푣�푠 Common Emitter Mode 푅퐿

Consider the circuit shown in Figure (4.18). In the common emitter mode, the small signal sinusoidal input voltage is connected in series with the DC bias base-emitter voltage across the base and the emitter. is the collector supply voltage. The load resistor is connected between the collector

�퐶퐶 푅퐿 - 169 - and the emitter in series with the collector supply voltage. Thus the emitter is common to both the input and the output circuit. The input current is the base current. The total base current ( ) is given by

= (1 ) = (1 ) + (1 ) =퐼퐵� + (4.67) ( ) (1 ) , The퐼퐵 base� current− 훼 thus0 퐼contains퐸� − 퐼퐶 a� DC퐶 current− 훼0 equal퐼퐸 to − 훼0 횤̃푒 − 퐼퐶 � and퐶 a small퐼퐵 signal횤̃푏 current ( ) equal to (1 ) . The input current is equal to 퐼퐵 − 훼0 퐼퐸 − 퐼퐶�퐶 횤 푏̃ − 훼 0 횤 푒̃ = = (1 ) (4.68)

But 횤̃푖� 횤̃푏 − 훼0 횤̃푒 = 푣�푠 횤̃푒 Hence 푟퐸

= (1 ) = (4.69) 푞퐼퐸 1−훼0 푏 0 푠 푠 퐸 The ratio is called the input 횤conductancẽ −, 훼 . Therefore푘푘 푣� 푣� 푟 횤̃푖� 푠 𝒊 �� 품 = = (1 ) = (4.70) 횤̃푖� 푞퐼퐸 1−훼0 푖� 푠 0 퐸 Recalling that we defined as푔 equal to�� , we can− 훼 write푘푘 the input푟 resistance as 푘푘 퐸 푟 푞퐼퐸 = = = (4.71) ( ) ( ) 1 푘푘 푟퐸 The input circuit therefore behaves푟푖� as 푔though푖� there푞퐼퐸 1 were−훼0 a resistance,1−훼0 , between the base and emitter terminals. The output current is equal to 푟푖�

= = = = = (4.72) 푞퐼퐸 ��푠 As before, the output circuit횤̃𝑜� behaves횤�̃ 훼as0 though횤̃푒 훼 a 0current푘푘 푣� 푠source,훼0 푟퐸 , exists푔푚푣� between푠 the collector and the emitter. The equivalent circuit is shown in Figure (4.19). In this instance, the current gain is 푚 푠 equal to 푔 푣�

= = = 푞퐼 = = (4.73) ( )퐸 횤̃�표� 횤̃푐 푔푚��푠 훼0 푘푘 ��푠 훼0 푞퐼퐸 푖� 푏 푖� 0 0 Thus the current gain in the횤 ̃common횤̃ emitter횤̃ mode 1−훼. The0 푘푘 subscript��푠 1 −0 훼in the 훽 and refers to the fact we are considering a very low frequency input signal, i.e., at nearly zero frequency. We are 0 0 0 neglecting capacitive effects. The voltage gain equals훽 훼 훽

= = (4.74) ���표� 푔푚��푠푅퐿 ��푖� ��푠 푔푚푅퐿 - 170 - and the power gain equals

= = (4.75) 횤̃�표� ���𝑢 훼0훽0 횤̃푖� ��푖� 훽0푔푚푅퐿 푟퐸 푅퐿

Figure (4.19): Equivalent circuit of the transistor in the common emitter mode.

Voltage Limitation

The maximum collector voltage that can be applied is limited by one of the following physical mechanisms: thermal, punch-through, the breakdown of the collector-base junction in the common base mode, or the collector breakdown in the common emitter mode. In a practical device, the limit on the magnitude of the collector voltage that can be applied is due to that mechanism which yields the lowest breakdown voltage. In some devices the punch-through voltage may be the limiting mechanism while in some other devices the collector breakdown maybe the limiting mechanism. We will now examine the three voltage limiting mechanisms.

Punch-Through Voltage,

When the reverse collector�푝�-base junction voltage ( ) is increased, the depletion region widens, and hence the width of the neutral base region decreases. This is illustrated in Figure (4.20). 퐶� is the width of the base region between the metallurgical� junctions on the emitter side, and that on the collector side. The width of the segment of the emitter-base depletion region occurring on the base 푊퐵� side is denoted . Similarly, the width of the segment of the collector-base depletion region occurring on the base side is푝� denoted . The width of the neutral base region is denoted , and is equal to 푥 ′ 푝� 푥 = 푊 (4.76) ′ However, we know from our study of the푊 - junctions푊퐵� − that푥푝� − 푥푝� 푝 푛

- 171 -

( ) ( ) = (4.77) ( ) 2 휖푠 푉푏𝑏−푉퐵� 푁퐷� 2 휖푠 푉푏𝑏−푉퐵� 푝� Since and 푥 � 푞 푁퐴� 푁퐷�+푁퐴� ≈ � 푞 푁퐴�

퐷� 퐴� 푁 ≫ 푁 ( ) = (4.78) ( ) 2 휖푠 푉푏𝑏+푉퐵� 푁퐷� 푝� Hence 푥 � 푞 푁퐴� 푁퐷�+푁퐴�

( ) ( ) = (4.79) ( ) ′ 2 휖푠 푉푏𝑏−푉퐵� 2 휖푠 푉푏𝑏+푉퐵� 푁퐷� 퐵� where and are푊 the built푊-in voltages− � of푞 the푁퐴 emitter� −-base� 푞and푁퐴 collector� 푁퐷�+-푁base퐴� junctions. , and are the impurity concentrations in the emitter, base and collector regions, respectively. �푏𝑏 �푏𝑏 푁퐷� 푁퐴� 푁퐷�As the collector-base applied voltage is increased, which is the third term on the right- hand side of Equation (4.79), representing the portion of the collector-base depletion region lying on the �퐶� 푥푝� base side increases. Thus, decreases as is increased. At some value of voltage (called the punch-through voltage) ′ becomes zero. can be determined by putting = 0 in the above 푊 �퐶� �퐶� ≡ �푝� equation and solving for . Hence,′ ′ 푊 �푝� 푊 �푝� ( ) ( ) = (4.80) 2 푞 푁퐴� 푁퐴�+푁퐷� 2 휖푠 푉푏𝑏−푉퐵� 푝� 퐵� 푏𝑏 Under the punch through� conditions,2 휖 the푠푁퐷 depletion� �푊 region− extends� 푞all푁 the퐴� way �from− the� collector to the emitter, and the base current (or voltage) has no control on the collector current. The collector current that flows under this condition is a space-charge limited current, and is called the punch-through current. The punch-through phenomenon is important only in transistors in which the metallurgical base region which is small.

Figure (4.20): The neutral base region.

- 172 -

Example

Given that the impurity concentration in the emitter, base and collector regions are = 10 , = 10 and = 10 , and the width of the base region is = 2 , determine19 −3 푁퐷� 𝑐 the punch-16through−3 voltage, given 15the emitter−3 -base junction forward voltage ot be 0.5 V. The built-in 푁퐴� 𝑐 푁퐷� 𝑐 푊퐵� 휇� voltage, , of the emitter base junction is

�푏𝑏 10 = ln = 0.0258 = 0.893 1035 푘� 푁퐷�푁퐴� 푏𝑏 2 20 � � 푖 � � � � 2 ( 푞0.5) 푛2 × 11.9 × 8.84 × 10 × 0.393 = = 0.227 1.6 10 × 10−14 휖푠 �푏𝑏 − 푝� −19 16 푥 ≈ � 퐴� � 휇� The built-in voltage of the푞 collector푁 -base junction is 푥 10 = ln = 0.0258 = 0.655 0.66 1031 푘� 푁퐴�푁퐷퐶 �푏𝑏 � 2 � � 20� � ≈ � Substituting these values into푞 Equation푛푖 80, we obtain 1.6 × 10 × 10 × (1.1 10 ) = (2 × 10 0.227 × 10 ) 0.66 2 × 11.9− 19× 8.8416 × 10 × 1016 −4 −4 2 푝� 푥 � = 2.−6314.33 150.66 262.7 − −

− ≈ � Collector- Base Breakdown Voltage,

In the common-base mode, the maximum퐵� 퐶voltage�퐶 that can be applied on the collector is limited by the breakdown characteristics of the collector-base junction. Recall our discussion on avalanche breakdown. The breakdown voltage of the collector-base junction is calculated the same way as the - junction breakdown voltage. That is, the collector-base junction voltage is determined by the applied voltage needed to attain critical electrical field in the depletion region, and is therefore a function of 푝the푛 dopant impurity concentrations in the base and collector regions.

represents the maximum collector voltage that can be applied in the common-base configuration unless punch-through occurs earlier, i.e., at a lower collector voltage. can be 퐶�퐶 calculated퐵� by using the expression for the breakdown voltage for the - junction in terms of the critical 퐵�퐶�퐶 electric field, , discusse in Chapter 3. 푝 푛 𝑐𝑐 ℰ = + (4.81) 2 ℰ푐𝑐� 휖푠 1 1 In the avalanche process,퐵�퐶� 퐶the carriers2푞 entering�푁 퐷the퐶 depletion푁퐴� � region− �푏𝑏 get multiplied due to impact ionization in the depletion region. The multiplication factor is given by

푀

- 173 -

= (4.82) 1 휂 푉 푀 1− � � where is the voltage applied across the junction, is푉 푏the� breakdown voltage of the junction, and is an empirical parameter. By putting = , we can see that the multiplication factor increases to 푏� infinity� at breakdown. The symbol that is used to denote� the common-base collector breakdown voltage휂 푏� is . The 0 is in the subscript implies� that� the emitter is open. Equation (4.82) can be푀 rewritten in terms of as 퐵�퐶�퐶 퐶�퐶 퐵 � = (4.83) 1 휂 푉퐶� 푀 1− � � where is the voltage across the collector-base junction.퐵푉퐶� 퐶

�퐶� Example

For the transistor discussed in the previous example, let us calculate given that = 200,000 V/cm. Substituting the values for the various quantities, we have 퐵�퐶�퐶 ℰ𝑐𝑐 × × . × . × = + 0.66 5 2× . × −14 �2 10 � 11 9 8 84 10 1 1 −19 15 16 퐵�퐶�퐶 2 =1 614410 .6 0.66 �10143.910 � −

Collector-Emitter Breakdown Voltage,− ≈ �

In the common-emitter mode of operation,퐵� recall퐶�퐶 that the base current is given by = (1 ) (4.84)

When the base is open, is 0, and퐼 hence퐵 − 훼0 퐼퐸 − 퐼퐶�퐶 퐵 퐼 = (4.85) ( ) 퐼퐶�퐶 This is the leakage current that flows between퐼퐸 the collector1−훼0 and the emitter when the base lead is open. However, at larger collector voltages, the collector multiplication factor arises, and hence the current through the collector-base junction increases by a factor , i.e., it gets multiplied by .

= ( +푀 ) 푀 (4.86) = When the base is open, . Hence퐼퐶 푀 훼0퐼퐸 퐼퐶�� 퐶 퐸 퐼 퐼 = (4.87) 푀 퐼퐶�퐶 As the collector-emitter voltage is increased,퐼 퐸 becomes1−훼0 larger푀 than 1 (as seen from Equation (4.82)), and hence increases. Ultimately when = 1, becomes infinite, and breakdown occurs. The 푀 퐼퐸 훼0푀 퐼퐸 - 174 - collector voltage at which becomes infinite is denoted . The notation in the subscript is similar to what we had for ; 0 indicates that the base is open. The condition for this breakdown to occur 퐸 퐶�퐶 is given by 퐼 퐵� 퐵�퐶�퐶 = 1 = = (4.88) α0 훼0 휂 휂 훼0푀 푉퐶𝐶 퐵푉퐶�퐶 In writing the above equation we have assumed1− that�퐵푉 at퐶� breakdown,퐶� 1− the�퐵푉 collector퐶�퐶� to emitter voltage , is the same as the collector to base voltage . Rearranging terms, we obtain

퐵�퐶�퐶 �퐶𝐶 = (1 )1 (4.89) 휂 Since , we obtain 퐵�퐶�퐶 퐵�퐶푩� − 훼0 1 훽0 ≈ 1−훼0 = (4.90) 퐵푉퐶푩� 퐵푉퐶푩� 휂 퐶�퐶 1 휂 �훽0 For silicon, it is reported that has a value퐵� between≈ 훽2 0and 6.

휂

Example

Let us calculate by assuming that = 3 and = 125 for the transistor used in the previous example. 퐵�퐶�퐶 휂 훽0 = 퐶푩� 퐵휂� 퐵�퐶�퐶 From our last example, is calculated to be 143.8 V� 훽0 퐵�퐶푩� = 125 = 5 휂 3 Hence �훽0 √ 143.9 = = = 28.8 5퐶푩� 5 퐶�퐶 퐵� 퐵� �

Thus we see that is much smaller than . In the common-emitter mode imposes the limit on the collector voltage unless punch-through occurs earlier. 퐵�퐶�퐶 퐵�퐶푩� 퐵�퐶�퐶

Base Stored Charge

- 175 -

The minority carriers injected in the base are diffusing towards the collector base junction. At any instant of time, the minority carrier density distribution in the base region looks like what is shown in Figure (4.21A). This distribution is stationary and does not change with time, although individual electrons (minority carriers) are constantly moving. It is as though so many minority carriers are stored in the base region. The area under the plot in Figure (4.21A) multiplied by , where is the area of cross-section of the base region is called the stored minority carrier charge. The minority carrier charge stores in the base is given by 푞� 퐴

= ( ) (4.91) ′ 푊 ′ ′ To maintain charge neutrality, an 푄excess퐵 majority−푞 퐴 carrier∫0 ∆ density푛 푥 �( 푥 ( )) equal to ( ) also arises. The distribution of excess majority carriers is identical to that of excess ′minority carrier distribution,′ and is shown in Figure (4.21B). The stored majority carrier charge is equal∆푝 푥in magnitude ∆but푛 푥opposite in sign to . (Recall our discussion in chapter 3.)

푄퐵 In a well-designed transistor, most of the minority carriers emitted from the emitter reach the collector. Very few minority carriers are lost due to recombination in the base. Hence the distribution is almost linear. We can therefore define a term , called the Base-Transit time. The base-transit time denotes the average time taken by the injected minority carrier to traverse the neutral base region, and 퐵 reach the collector. The current in the base region휏 due to the injected electrons is given by

( ) = ( ) ( ) (4.92) ( ) ′ ′ ′ Where is the velocity of electrons퐼퐶� as푥 a function−푞 of퐴 푣. The푥 velocity∆푛 푥 of the minority carrier at some point in ′the neutral base region is ′ 푣′ 푥 푥 푥 ( ) ( ) = (4.93) ( ′ ) ′ − 퐼퐶� 푥 ′ The time taken to traverse a distance between푣 푥 푞and퐴 ∆ 푛 +푥 is ′ ′ ′ ′ �푥 푥 푥 ( �푥) = = (4.94) ( ′) ( )′ 푑푥 푞 퐴 ∆푛 푥 ′ ′ ′ The transit time is obtained by integrating𝑑 �, 푥and is equal− 퐼to퐶� 푥 �푥 𝑑 ( ) = = (4.95) ′ ( ) ′ 푊 − 푞 퐴 ∆푛 푥 ′ ′ Since we assume that there is negligible휏퐵 recombination∫ 𝑑 ∫0 in the퐼퐶 �base푥 region,�푥 the current is constant in the base, and equal to . Hence, we can take ( ) outside the integral, and obtain ′ 퐼퐶� 퐼퐶� 푥 = ( ) = (4.96) ′ 1 푊 ′ ′ 푄퐵 퐵 퐼퐶� 0 퐼퐶� 휏 =∫ − 푞 퐴=∆ 푛 푥 �푥 (4.97)

Since we assumed that there was negligible푄퐵 recombination휏퐵 퐼퐶� in휏퐵 theα 0base,퐼퐸 the excess carrier density can be approximated to vary linearly with , and is given by ′ 푥 - 176 -

( ) = 1 (4.98) ′ ′ 푥 ′ 0 푊 = ∆푛 푥 ∆(푛 ) � −= � (4.99) ′ ′ 푊 ′ ′ ∆푛0푊 But 푄퐵 − ∫0 푞 퐴 ∆푛 푥 �푥 −푞 퐴 2

= = (4.100)

훿∆푛 ∆푛0 ′ ′ 퐶� 푛 훿푥 푛 푊 퐼 푞 퐴 퐷= = −푞 퐴 퐷 (4.101) ′2 푄퐵 푊 휏퐵 퐼퐶� 2퐷푛 Example

Consider the base region in which the minority carrier diffusion constant is equal to 20 / . Assume = 2. Then the base transit time, is equal to 2 ′ 𝑐 푠𝑠 푊 (2휏퐵10 ) = = 10 2 20−4 2 푥 −9 휏퐵 푠𝑠 푥

Figure (4.21): Excess carrier distribution in the base region: A) minority carriers B) majority carriers.

Although we assumed no recombination in the base region for the purpose of determining the base transit time, in reality there is some recombination and the base transport factor , is less than unity. Recall that α� = (4.102) 1 � 푊′ α cosh퐿푛

- 177 -

If we now expand the denominator under the assumption , we obtain ′ 푊 ≪ 퐿푛 1 = 1 (4.103) ′2 ′2 1 푊 푊 푊′2 2 � 푛 푛 푛 α ≈ 1+ 2 ≈ − 2퐿 − 2퐷 휏 where is the minority carrier lifetime in2퐿 the푛 base.

휏푛 1 1 (4.104) ′2 푊 휏퐵 The above equation can be used toα give� ≈ a physical− 2 퐷interpretation푛휏푛 ≈ − to휏 the푛 base transport factor. We know that is the probability that the minority carrier in the base will recombine in the base in a time 푑� interval휏푛 . Hence if the base transit time is small compared with , then is the probability that 휏퐵 a minority carrier will recombine in the base퐵 region in its transit to the collector.푛 푛 Hence 1 is the �푡 휏 휏 휏 휏퐵 probability that an injected minority carrier will reach the collector without being lost by recombining. − 휏푛 By decreasing the base-transit time , in comparison with the lifetime , can be made closer to unity. When is very much smaller than , 1. 휏퐵 휏푛 α� 휏퐵 휏푛 α� ≈ Stored Charge Capacitance

The minority carriers stored in the base region give rise to capacitive effects. The Capacitor associated with minority carrier storage is called the diffusion capacitance, or storage capacitance. Consider the carrier density distributions in the base region shown in Figure (4.23A) for excess minority carriers and Figure (4.23B) for excess majority carriers when the emitter base junction is forward biased at some DC bias voltage, . Let us now apply a small incremental step voltage, in series with so that the net forward bias voltage is + as shown in Figure (4.22). Due to , the excess minority 퐵� 퐵� carrier density is increased� in the base to correspond to the new emitter-base voltage.𝑑 The stored � 퐵� negative charge7 in the base is increased� by𝑑 an amount, = , where 𝑑

𝑑퐵 −𝑑 = = ( ) (4.105) ′ 푊 ′ where ( ) is the incremental𝑑 change퐵 in− 𝑑the excess− minority푞� ∫0 carrier휕 ∆푛 densit𝑑y. The shaded region in Figure (4.23A) represents the increment in the negative charge, . This incremental negative charge entered휕 the∆푛 base region through the emitter from the step voltage source as shown in Figure (4.22). In order to preserve charge neutrality, the excess majority carrier in− the𝑑 base also increases by the same amount as the excess minority carriers. The incremental change in the positive charge due to majority carriers is equal to , and is shown in Figure (4.22B) by shaded lines. The incremental positive (majority carrier) charge ( ) flows into the base region through the base ohmic contact from the step voltage source as shown𝑑 in Figure (4.22). Thus we see that due to the application of the step voltage 𝑑 ( ), a charge + flows into the base through the base contact, and a charge – flows into the

7 𝑑Since electrons are𝑑 minority carriers, the increase in minority carriers corresponds to an increase𝑑 in the negative charge.

- 178 - base through the emitter, both from the step voltage source. When the step voltage is reduced to zero, the incremental charges flow back to the step voltage source from the base, the same way they entered the base region. Thus the transistor behaves as though there were a capacitor between the emitter and the base lead. This capacitor which is called the diffusion capacitor (or storage capacitor) has a capacitance, , equal to

푪풅𝒅� | | = = (4.106) 푑푄 푑 푄퐵 We can evaluate this capacitance by writing퐶푑𝑑� explicitly.푑� Recall푑푉퐵� that in a previous section, we expressed as 푄퐵 퐵 푄 = = | | | | | | = 푄퐵= 퐼 퐶� 휏 퐵 =α0 퐼퐸 휏퐵 = (4.107) 푑 푄퐵 푑 퐼퐸 푞 퐼퐸 α0 휏퐵 퐶푑𝑑� 푑푉퐵� α0 휏퐵 푑푉퐵� α0 휏퐵 푘푘 푟퐸

Example

Let us calculate the diffusion capacitance of a bipolar transistor carrying an emitter current of 1 with a transit time equal to 10 sec, and equal to 0.99. Let = 300 . −8 푚� α0 0.0259 푇 퐾 = = = 25.9 1 푘� � 푟퐸 훺 푞 퐼퐸0.99 × 10푚� = = = 3.82 × 10 25.9 −8 α0 휏퐵 −10 퐶푑𝑑� 퐹 푟퐸

- 179 -

Figure (4.22): Application of a step voltage , in series with so as to increase the injected minority carriers in the base. 𝑑 �퐵�

Figure (4.23): Incremental change in excess carrier densities A) minority carriers. The increase in stored minority carrier charge comes about due to + flowing from the emitter into the base. B) majority carries. The increase in stored majority carrier charge comes about due to flowing into the base 퐵 region from the𝑑 base ohmic contact. − 𝑑퐵

Frequency Response

We consider earlier the response of a bipolar transistor to a sinusoidally varying small signal input voltage. We had assumed that the small signal voltage varied sinusoidally at a very low frequency, so that the injected carrier density varied in step with the small signal input emitter base voltage. This is illustrated in Figure (4.14). When the sinusoidal input voltage is at its maximum value, the excess carrier density is increased everywhere in the base, as indicated by the top line in Figure (4.14). When is at its most negative value, the excess carrier density is decreased everywhere, as indicated by the bottom 푠 line in Figure (4.14). The middle line represents the excess carrier density due to the DC emitter푣�-base voltage. On the other hand, when the input signal varies at a very high frequency such that the period of the sinusoidal signal is small compared to the base-transit time, (i.e. < ) the excess carrier 1 density in the base has a spatial variation. The excess carrier density퐵 varies in phase in different regions, 휏 휔 휏퐵

- 180 - as illustrated in Figure (4.24). This variation arises because the excess carrier density injected at = 0 at some instant of time has not had enough time to travel to the collector before the excess carrier′ density at = 0 changes to a new value. Thus there is a phase difference and attenuation between푥 the (sinusoidal) ′current injected by the emitter and the (sinusoidal) current reaching the collector. Hence the current푥 gain decreases with frequency and also has a phase dependence.

We can now represent the high frequency response of the transistor in terms of an equivalent circuit. Let us initially consider the frequency response of the device, neglecting all parasitic capacitances and resistances. We consider the junction (depletion) capacitance as a parasitic element, and similarly the bulk resistances of the collector, base, and contact resistances8 are considered parasitic elements. On the other hand, the diffusion capacitance, , is not neglected and is included as part of the transistor. The diffusion capacitance arises due to the injection of minority carriers in the base, which is 퐶푑𝑑� vital to the transistor action. A transistor in which the parasitic elements are neglected, is called an intrinsic transistor. In the equivalent circuit for an intrinsic transistor, we included only those elements that are necessary for the transistor action.

Common Base Mode

Let us now consider the common-base mode. The equivalent circuit for the intrinsic transistor in the common-base mode is shown in Figure (4.25). In this circuit, the diffusion capacitance is shown to be connected parallel to (the emitter resistance). The reason for placing the diffusion capacitance in parallel with the input emitter resistance is that the input voltage supplies both the incremental 퐸 stored charge and the푟 injected current. The collector current is given as before by the current source 퐸 ( ) and is in phase with ; the collector푟 current is in phase with the current that flows through . The emitter current flows partially through and partially through . There is therefore a phase 푔푚푣�퐵� 푣�퐵� difference between the emitter current and the current through . Hence the emitter current and the 푟퐸 퐶푑𝑑� 푟퐸 collector current are not in phase. As the frequency of the input signal is increased, comparatively more 퐸 and more of the emitter current flows through , and hence the푟 collector current decreases. Thus the current gain decreases. 퐶푑𝑑� Let us assume that a sinusoidal input current of = sin is applied at the input terminals. Then the base-emitter voltage is given by 횤푒̃ �푒 휔� 퐵� �푣 × 1 = = (4.108) 푟퐸 �푗푗퐶 � 푑�푑푑 푖푒 푟퐸 1 퐵� 푒 푑�푑푑 퐸 �푣 � 푟퐸+푗푗퐶 1+ 푗�퐶 푟 The collector current us given by 푑�푑푑

= = (4.109)

푔푚푟퐸 � 푚 퐵� 푑�푑푑 퐸 푒 8 � 푔 푣� 1+ 푗� 퐶 푟 � The emitter bulk resistance is negligible because of the heavy doping

- 181 -

Hence the current gain , which is a function of frequency, is given by

α ( ) = = (4.110)

푖푐 푔푚푟퐸 At very low frequency ( 0), (α) 휔approaches푖푒 1, +the푗� low퐶푑 �frequency푑푑 푟퐸 gain. By putting = 0 in the equation for ( ) , we obtain 휔 → α 휔 α0 휔 α 휔 ( = 0) = = (4.111)

Hence α 휔 α0 푔푚푟퐸

( ) = (4.112)

α0 α 휔 1+ 푗� 퐶푑�푑푑 푟퐸

Figure (4.24): Spatial variation in the base of the excess minority carrier charge density due to a high frequency input signal.

Figure (4.25): High Frequency equivalent circuit of the intrinsic transistor in the common base mode

Let us now define a parameter, , as ∗ 𝑐 휔 (4.113)

∗ 1 𝑐 Then 휔 ≡ 퐶푑�푑푑 푟퐸

- 182 -

( ) = (4.114)

α0 � ∗ We have used a superscript to indicate thatα 휔 this and1+ other푗�푐푐 parameters that follow with a superscript refer to an intrinsic transistor. The same parameters without the superscript will refer to a real transistor in which the effect∗ of the parasitic elements are also included. As increases, ( ) ∗ decreases, and a phase difference arises between the output and input currents.∗ The frequency at which 휔 훼 휔 | ( )| falls to of its low frequency value is called the alpha-cutoff frequency. This happens when 1 = . Therefore, the alpha-cutoff frequency is given by 훼 휔 √2 ∗ 휔 휔𝑐 = = (4.115) ∗ ∗ 휔푐푐 1 𝑐 Recalling 푓 2휋 2휋 퐶푑�푑푑 푟퐸

= 0 퐵 푑𝑑� α 휏 We can write 퐶 푟퐸 = (4.116)

∗ 1 Thus we see that the maximum frequency is푓 limited𝑐 by2휋 theα0 휏base퐵 -transit time, . The phase of ( ) at = is 450. The variation of the magnitude of ( ) with is shown in Figure (4.26). In this figure, 퐵 | ( )| is plotted as a function of log . It is seen that | ( )| is equal to at휏 low frequenciesα and휔 falls 𝑐 푓off as푓 approaches . At high frequencies, α 휔 휔 0 훼 휔 ∗ 휔 훼 휔 α 휔 휔𝑐 ( ) = (4.117) ∗ α0 α0휔푐푐 � ∗ 푗휔 The gain falls off by a factor of 2 when αthe휔 frequency≈ 푗�푐푐 (or ) is doubled. In this region the gain is said to fall off at the rate of 6 db/octave. 휔

Figure (4.26): Plot of the magnitude of the current gain, ( ) , as a function of . is the value of at which | ( )| is equal to 0.707 of . α 휔 휔 휔𝑐 휔 훼 휔 α0

- 183 -

Example

Let us now determine the alpha cut-off frequency of an intrinsic n-p-n bipolar transistor, given: = 2 ′ 푊= 15 휇�/ 2 퐷푛 = 𝑐0.995푠 𝑠

0 × = α= = 1.33 x 10 ′2 × −4 2 푊 �1 10 � −9 휏퐵 2퐷푛 2 151 푠𝑠 = 2 ∗ 푓𝑐 Therefore, 휋 α0휏퐵 1 1 = = = 1.2 × 10 2 × 0.995 × 1.33 × 10 8.34 × 10 ∗ 8 푓𝑐 −9 −9 퐻� 휋

Common Emitter Mode

Let us now treat the frequency response of the device in the common emitter mode. The equivalent circuit of the intrinsic transistor is the same as what we derived earlier for the low frequency case, except that the input resistance, , which is in parallel with the diffusion capacitor ( ), is equal to . The equivalent circuit for the intrinsic transistor in the common emitter mode is shown 푟푖� 퐶푑𝑑� 푟퐸 in Figure (4.27). The input current (which is now the base current ) divides itself between two paths: 1− α0 one through , and the other through . The collector current ( ) is determined by the current 횤푏̃ that flows through , and therefore there is a phase difference between and . The base-emitter 푟푖� 퐶푑𝑑� 횤�̃ voltage ( ) is given by 푟푖� 횤�̃ 횤푏̃ 푣�퐵�

푟 1 = 퐸 = 푟퐸 (4.118) �1− α � 푗푗퐶 0 푑�푑푑 1− α0 푟퐸 1 푟퐸 퐵� 푏 푑�푑푑 푏 푣� 횤̃ 1− α + 푗푗퐶 1+ 푗� 퐶 1− α0 횤̃ The collector current is obtained as 0 푑�푑푑

= = 푟퐸 (4.119) 푚 푔 1− α 0 푟퐸 � 푚 퐵� 푑�푑푑 푏 The current gain is now a function횤̃ of frequency,푔 푣� and is1 +given푗� 퐶 by 1− α0 횤̃

- 184 -

( ) = = 푟퐸 (4.120) 푚 횤̃푐 푔 1− α 0 푟퐸 횤̃푏 푑�푑푑 At very low frequency, (i.e. 0),훽( 휔) is obtained 1by+ putting푗� 퐶 1=− 0α0, and is equal to .

0 휔 → 훽(휔) | = = 휔 훽 (4.121)

푟퐸 Therefore ( ) can be expressed in훽 terms휔 of휔 =0 as 훽0 푔푚 1− α0

0 훽 휔 ( 훽) = (4.122)

훽0 푟퐸 푑�푑푑 Let us define a parameter given by 훽 휔 1+ 푗�퐶 1− α0 ∗ 휔퐶� (4.123)

∗ 1 푟퐸 퐶� 푑�푑푑 Since 휔 ≡ 퐶 1− α0

= (4.124)

푑𝑑� 퐸 0 퐵 퐶 =푟 =α 휏 (4.125)

∗ 1 1 α0휏퐵 퐶� β0휏퐵 Then 휔 1− α0

( ) = (4.126)

훽0 � 훽 휔 1+ 푗�∗ The magnitude of the current gain (| ( )|) is plotted in Figure퐶퐶 (4.28) as a function of log . We define the beta cutoff frequency as that at which the magnitude of the gain, falls off to of its low frequency 훽 휔 휔 1 value. This happens when = . Hence beta cut-off frequency is given by √2 ∗ ∗ 휔 휔퐶� �푓퐶�� = = = (4.127) ∗ 휔퐶퐶 1 1 ∗ 푟 퐶� 퐸 2휋 푑�푑푑 2휋 β0휏퐵0 The phase of the collector current푓 differs from that2휋 of퐶 the 1base− α0 current by 45 when = . At very high frequencies, ∗ 푓 푓퐶�

( ) = (4.128) ∗ 훽0 훽0휔퐶퐶 � 훽 휔 ≈ 푗�∗ 푗 휔 In the range of frequencies over which the above equation퐶� is valid, the magnitude of | ( )| multiplied by is a constant, given by 훽 휔 휔 | ( )| = = = (4.129)

∗ 훽0 1 0 퐶� 휔 훽 휔 - 훽185휔 - β0휏퐵 휏퐵

As the frequency is increased, the gain falls by an amount such that the product of gain and frequency is a constant. This constant is called the gain-bandwidth product. The decrease in the gain with frequency at high frequencies, again demonstrates a 6db/octave characteristic, i.e. | | decreases by a factor of 2 when the frequency is doubled. The value of at which | ( )| becomes unity is denoted , and is found in Equation (4.129) to be 훽 ∗ 휔 훽 휔 휔� = (4.130) ∗ 1 According to Equation (4.129), the frequency at휔 which� 휏 |퐵 ( )| becomes unity is equal to 훽 휔 = = (4.131) ∗ ∗ 휔� 1 Using Equation (4.130) we can express 푓� as 2 휋 2 휋 휏퐵 ∗ 퐶� 휔 = = ∗ (4.132) ∗ 1 휔� To summarize, the alpha cut-off frequency휔 is퐶 � 훽0휏퐵 훽0

= (4.133)

∗ 1 1 1 The beta cut-off frequency is 푓𝑐 2휋 α0 휏퐵

= (4.134) ∗ 1 1 1 The gain bandwidth product is 푓퐶훽 2휋 훽0 휏퐵

= (4.135) ∗ 1 1 If any two of these three frequencies are known,푓� the2 휋third휏퐵 is readily obtained. Or alternately, if and either or are known, the three frequencies are readily obtained. Thus, of the five quantities, 퐵 , , , and either or , we can determine the other three if any two are known. 휏 0 0 ∗ ∗ α ∗ 훽 𝑐 퐶� � 퐵 0 0 푓 푓 푓 휏 α 훽 Example

Let us now determine the limiting frequencies and for the transistor discussed in the early example. We saw that = 1.33 × 10 . We∗ were∗ given as equal to 0.995. Hence 푓퐶� 푓� −9 휏퐵 1 푠𝑠1 α0 = = 1.194 × 10 2 1.33 × 10 ∗ 8 푓� −9 0.995 퐻� 휋 = = = 199 1 0.005 α0 훽0 = −× α0 × = 5.998 × 10 ∗ 1 1 1 5 푓𝑐 2휋 훽0 휏퐵 퐻�

- 186 -

Figure (4.27): The equivalent circuit of the intrinsic transistor in the common emitter mode.

Figure (4.28): Plot of the magnitude of the current gain in the common emitter mode as a function of log

ퟂ Inclusion of Junction Capacitances

Let us next consider the equivalent circuit for the bipolar transistor by including the emitter- base and collector-base junction capacitances. Previously, the α-cutoff frequency and the - cutoff frequency of an intrinsic transistor were discussed. The actual α - and - cutoff frequencies of the transistors will be different from what we obtained for the intrinsic transistor, due to the presence훽 of the emitter-base junction capacitances. The equivalent circuits that we훽 previously drew for the common-base mode and the common-emitter mode need to be modified to include the junction capacitances. Common Base Mode

The equivalent circuits for the common-base mode will have the emitter-base junction capacitance in parallel with the diffusion capacitance between the emitter and the base, and the collector-base capacitance will be in parallel with the current source in the output circuit between the collector and the base, as shown in Figure (4.29). If we go through the analysis exactly the same way as we did for the intrinsic transistor, we will find that the current gain is frequency dependent, with a time constant that is determined not only by the diffusion capacitance but also by the emitter-base junction

- 187 - capacitance. Without duplicating every step of the analysis, we can readily write the expression for α as a function of frequency:

( ) = (4.136) ( ) α0 훼 휔 1+푗� α0휏퐵+ 푟퐸퐶퐽𝐽

Figure (4.29): The equivalent circuit of the bipolar transistor in the common-base mode, including the emitter-base and the collector-base capacitances.

The expression differs from that for the intrinsic transistor due to the fact that the time constant of the input circuit is increased from to + . In terms of , ( ) can be written as

0 퐵 0 퐵 퐸 퐽𝐽 푑𝑑� α( 휏 ) =α 휏 푟 퐶 퐶 훼 휔 (4.137) ( ) α0 The expression for the alpha-cutoff frequency훼 휔 can1+ similarly푗�푟퐸 퐶푑 be�푑푑 written+퐶퐽𝐽 as 1 = 2 ( + ) 퐶� 푓 푑𝑑� 퐽𝐽 퐸 = 휋 퐶 퐶 푟 (4.138) ( ) 1 where is the emitter-base junction capacitance.2휋 α0 휏We퐵+ already퐶퐽𝐽푟퐸 know from previous discussion that the

emitter퐶 resistance,퐽𝐽 equals , and therefore decreases with the emitter current, . At very high 푘푘 emitter-currents, 퐸becomes small, making × negligible in comparison with 퐸 . 푟 푞퐼퐸 퐼 × At low currents,푟퐸 is less than 퐶, 퐽where𝐽 푟퐸 is the intrinsic α-cutoff frequency,α0 a휏nd퐵 at high currents, approaches . If we measure∗ the - cutoff∗ frequency at different values of the emitter 푓퐶� 푓퐶� 푓퐶� current, , and then plot the∗ reciprocal of versus , we will obtain a straight-line plot as show in 푓퐶� 푓퐶� 훼 1 퐸 퐶� Figure (4.30).퐼 If we now extrapolate the linear푓 plot, the퐼 퐸intercept on the vertical axis at = 0 will be 1 equal to . Thus we can obtain the intrinsic α - cutoff frequency by measuring the α -퐼 퐸cutoff 1 ∗ frequency푓 of퐶� a real transistor at different emitter-current.

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Common Emitter Mode

The equivalent circuit for the bipolar transistor in the common emitter mode including the junction capacitances is shown in Figure (4.31). The input capacitance is increased by having , the emitter-base junction capacitance in parallel with the diffusion capacitance, . In the common- 퐶퐽𝐽 emitter mode, the collector-base junction capacitance appears as a feedback capacitance between the 퐶푑𝑑� output and the input terminals. The feedback capacitance is usually referred to as a Miller Feedback Capacitance.

As we discussed in the common-base mode, the time constant of the input circuit now is increased from its intrinsic value to a larger value given by

= + ( ) 푟퐸 The expression for ( ) can퐼𝐼𝐼 be derived푡𝑡� as 𝑐𝑐𝑐𝑐before. We can write1−훼 0 �퐶푑𝑑� 퐶퐽𝐽�

훽 휔 ( ) = β 훽0( ) 푟퐸 퐵 퐽𝐽 훽 휔 1+푗� � 0휏 + 1−�0 퐶 � =

( ) α훽0 푗푗 0 퐵 퐸 퐽𝐽 1+ 1−�0 � 휏 + 푟 퐶 � =

( ) 훽0 푗푗 퐸 푑�푑푑 퐸 퐽𝐽 1+ 1−�0 �푟 퐶 + 푟 퐶 � = (4.139)

( )훽0 푟퐸 푑�푑푑 퐽𝐽 Thus we see that the effect of including the emitter1+-base푗� junction1−�0 �퐶 capacitance+ 퐶 � is as though the input circuit time constant is increased by . We can readily write the expression for the - cutoff ( ) 푟퐸 frequencies without any further analysis as 1−훼0 퐶퐽𝐽 훽 1 1 = 2 ( +0 ) 퐶� − 훼 푓 퐸 푑𝑑� 퐽𝐽 = 휋 푟 퐶 퐶 ( α ) 1−훼0 2휋 0휏퐵+푟퐸퐶퐽𝐽 = (4.140) ( ) 1 1 0 퐵 퐸 퐽𝐽 The - cutoff frequency is thus seen to be less than2휋훽 the휏 +intrinsic�0푟 퐶 - cutoff frequency. Again noticing that the emitter resistance, , is inversely proportional to the emitter-current. We can now determine the intrinsic훽 - cutoff frequency by measuring the actual - cutoff frequency훽 at several different emitter- 퐸 current values. This procedure푟 is to measure the - cutoff frequency at different values of the emitter- 훽 훽 훽- 189 -

current and plot the reciprocal of the measured - cutoff frequency, , as a reciprocal function of the 1 emitter current, as illustrated in Figure (4.32)훽. A linear plot will be푓 obtained퐶퐶 which when extrapolated 1 to intercept the vertical퐼퐸 axis at = 0 yields the reciprocal of the intrinsic - cutoff frequency, . 1 1 퐸 퐼 훽 푓퐶퐶

Example

Let us calculate the α - cutoff frequency and the - cutoff frequency of a transistor including junction capacitances. Let us consider the transistor which we used to calculate and in the previous 훽 examples. Assume = 1 , = 340 . ∗ ∗ 푓퐶� 푓퐶� is calculated to be퐼퐸 푚� 퐶퐽𝐽 푝�

퐸 . 푟 = = = 25.9

푘푘 25 9 푚� 퐸 Using the value for , and in the푟 previous푞퐼퐸 examples,1 푚� we calculate훺 the following. ∗ ∗ 0 퐶� 퐶� 훽 푓 푓 α = = 0.995 훽0 0 =훽0+11.33 10 −9 = 휏퐵 = 푥 푠𝑠 × ( α ) × ( . × . × . × × ) ∗ 1 1 −9 −12 퐶� 0 퐵 퐸 퐽𝐽 푓 2 휋 휏 + 푟 퐶 = 1.572 휋× 100 995 1 33 10 + 25 9 340 10 7 = 퐻� × ( β ) ∗ 1 α 푟퐸 퐶� 퐵 퐽𝐽 푓 2휋 0휏 + 1− 0퐶 = . = 78.56 × ( × . × × × ) 1 . −9 25 9 −12 Thus we see that both - cutoff and - cutoff2휋 199 frequencies1 33 10 are+ 0reduced.005 340 The10 junction capacitances퐾𝐾 decrease the frequency response of the transistor. 훼 훽

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Figure (4.30): Plot of the reciprocal α - cutoff frequency, , as a function of the reciprocal emitter- 1 current, . 푓퐶� 1 퐼퐸

Figure (4.31): Equivalent circuit in the common emitter mode, including junction capacitance.

Figure (4.32): Plot of as a function of . The intercept on the vertical axis at = 0 corresponds to 1 1 1

푓퐶퐶 퐼퐸 . 퐼퐸 1 ∗ 푓퐶퐶

Switching Transistors

A bipolar transistor can also be used as an inverter or as a switch. In digital applications, a bipolar transistor is used in either of these two functions. A bipolar transistor is typically usually used as what is called a saturated inverter, or a switch. The transistor is either off or on: when it is on, the transistor is in the saturation mode. When it is off, it is in the cutoff mode. Let us consider the circuit shown in Figure (4.33A). The transistor has a load resistance ( ) in series with the collector-supply voltage ( ) connected between the collector and the emitter. The base current is applied to the base 퐿 terminal as shown in this figure. Denoting the collector-emitter푅 voltage as the output voltage , �퐶퐶 = ��

� 퐶퐶 퐿 퐶 � - 191� - − 푅 퐼

As increases, decreases. Finally when

퐶 � 퐼 � = 푉퐶퐶 is zero. When the device is off, is nearly zero퐼 퐶and 푅=퐿 . Thus when the input current goes up from 0 to , the output voltage (which is the collector-emitter voltage) goes down from to 0. Hence � 퐶 � 퐶퐶 the� output voltage swing is opposite퐼 in phase to the input,� and� this circuit is called an inverter circuit. 퐵 퐶퐶 We assume퐼 that the transistor is initially off (i.e., the emitter-base voltage is reverse-biased).� At time = 0, a base current ( ) is applied. The base current is chosen to be large enough so that the device under steady-state condition will be driven into saturation. 푡 퐼퐵 Due to the base current, minority carriers are injected into the base from the emitter and the stored charge ( ) in the base increases. The collector current which is equal to , increases. As the 푄퐵 current increases퐵 in the collector circuit, the collector to the base voltage which was initially at a voltage 푄 휏퐵 nearly equal to the collector supply voltage starts to decrease. Ultimately, when sufficiently large collector current flows, the collector voltage falls below the base emitter voltage. The collector base junction becomes forward-biased, driving the transistor into saturation. When the collector to emitter voltage becomes 0, no further increase in collector current can arise. Thus the collector current reaches a maximum value equal to . However, the base current is still flowing, and the stored charge in the 푉퐶퐶 base continues to increase. The stored charge in the base increases even though the collector current 푅퐿 has reached its maximum value and stopped increasing.

Since we are discussing an - - transistor, the injected (and stored) minority carriers are electrons. The base current provides the majority carriers (holes) needed to neutralize the stored minority carrier charge, and to provide푛 푝 푛the hole current for: a) injecting holes into the emitter, b) providing majority carriers to recombine with the minority carriers in the base, and c) the flow of the collector-base junction leakage current, . We will first assume that the hole injection current into the emitter ( ) is negligible. This assumption is tantamount to letting the emitter injection efficiency 퐼CBO ( ) to be very nearly unity. Let us further assume that is negligibly small in comparison with the 퐼퐸� base current bias . The base current under these assumptions is given by 훾 퐼CBO 퐼퐵 | | = + (4.141) 푑 푄퐵 푄퐵 This equation is called the charge-control퐼 퐵equation푑�. The first휏푛 term on the right hand side denotes the base current needed for the buildup of majority carrier charge to neutralize the growing and the second term on the right hand side denotes the hole current needed for the recombination. 푄퐵 The boundary condition for solving the charge-control equation, when is turned on at time = 0, is 퐼퐵 푡 = 0 for < (4.142)

The solution is then easily seen to be 푄퐵 푡 푡0

- 192 -

| | = 1 푡 (4.143) − 휏푛 A plot of the growth of as a function푄 of퐵 time is퐼퐵 shown휏푛 � in −Figure푒 (4.34).� The stored charge is seen to grow exponentially with time. When > . 푄퐵 푡 휏 푛 | | = (4.144)

As long as the transistor is in the active mode (i.e.,푄퐵 the collector퐼퐵 휏푛 -base junction is reverse biased), the collector current ( ) and are related to each other as

퐼 퐶 푄 퐵 = (4.145)

Initially as grows, grows at the same rate,푄 and퐵 the퐼 퐶collec휏퐵 tor-base voltage, decreases. When becomes zero, the transistor is at the edge of saturation. When becomes negative, the device is in 퐵 퐶 퐶� 퐶� saturation 푄and the collector퐼 voltage is nearly equal to the emitter voltage. is at� its maximum value � �퐶� = since cannot increase beyond . Let us assume that at , the device퐼퐶 gets saturated and that 푉퐶퐶 퐶 1 becomes퐼 equal to . The plot� of푅퐿 the� growth of with time푡 is shown푡 in Figure (4.35). grows 푉퐶퐶 퐶 퐶 퐶 퐼exponentially with time푅퐿 as long as < . At = 퐼 , reaches the maximum value 퐼, and stays 푉퐶퐶 constant at this value. We can think of it 1as though 1 and퐶 becomes decoupled when > . While 푡 푡 푡 푡 퐼 � 푅퐿 � continues to grow exponentially with time even for > , remains constant at for > . 퐼퐶 푄퐵 푡 푡1 푉퐶퐶 퐵 1 퐶 1 푄 Let us define at = as . Therefore when푡 푡> 퐼 , the device is saturated.푅퐿 푡 푡

푄퐵 푡 푡1 푄푠 푄퐵 푄푆 = ( ) = 1 푡1 (4.146) − 휏푛 푠 퐵 1 퐵 푛 푄 푄 (푡 ) = 퐼 휏 � − 푒 � (4.147) 푉퐶퐶 But 퐼퐶 푡1 푅퐿

= ( ) = (4.148) 푉퐶퐶 Therefore 푄푆 퐼퐶 푡1 휏퐵 푅퐿 휏퐵

= 1 푡1 (4.149) 퐶퐶 푉 − 휏푛 From this equation, we can obtain 푅 퐿as 휏퐵 퐼퐵 휏푛 � − 푒 � 푡1 = ln (4.150)

1 1 푛 푉퐶퐶 휏퐵 represents the time taken by the device푡 to reach휏 steady�1−푅 퐿state퐼퐵 휏 푛collector� current, and is called the turn- on time. To keep small so as to obtain a high clock frequency in digital applications, should be kept 푡1 1 푛 푡 - 193 - 휏

small and should be kept small compared to . The latter can be achieved by keeping small

푉퐶퐶휏퐵 and increasing . Beyond = , continues to grow while remains constant. is determined by 푅퐿퐼퐵 휏푛 휏퐵 the gradient of ( ) in the base. In Figure (4.36), ( ) is plotted in the base region at several 퐼퐵 푡 푡1 푄퐵 퐼퐶 퐼퐶 instants of time. When′ < , the slope of ( ) plot′ increases with time indicates an increase of ∆푛 푥 ∆푛 푥 with time. When , stays constant which means′ the slope has to remain constant. However, the 1 퐶 growth of occurs now푡 in a푡 manner such that∆푛 푥 ( ) increase while the slope remains constant. 퐼 푡 ≥ 푡1 퐼퐶 ( = ) increases from zero when > , since′ the collector is also injecting minority carriers 푄퐵 ∆푛 푥 into the′ base.′ Ultimately ( = ) reaches a steady state value at = , determined by the ∆푛 푥 푊 푡 푡1 requirement that ′ ′ ∆푛 푥 푊 푡 ∞ ( ) = ( ) < The area under the plot of is described푄 by퐵 a∞ right-angled퐼퐵 휏푛 triangle for and by a trapezoid for > , is equal to times′ the area of the triangle at = . If ( ) is the excess carrier ∆푛 푥 푡 푡1 density at = for > , then the area of the trapezoid is equal ′to 1 푆 1 푡 푡 ′푄 ′ 푞 푡 푡 ∆푛 푊 푥 푊 푡 푡1 ( ) 퐴푡𝑡� = = ( ) + 푄퐵 ∞ ′ ′ 푄푆 What we have discussed so far푞� is called퐴 the푡𝑡� turn-on∆ behavior.푛 푊 푊Let us now푞� discuss the device behavior, when the transistor which has been on in the saturation mode for a long time with a large base current drive, is switched off by turning off to zero at time = 0. The transistor current will decrease to zero only after some time delay, and this behavior is called the turn-off transient. 퐼퐵 푡 Let us make the assumption that a base current has been flowing for a long time and that is sufficiently large to drive the transistor into saturation. The base current as a function of time is shown 퐵 퐵 in Figure (4.37). The charge control equation under this assumption퐼 becomes = 0. 퐼

| | | | 푡 = (4.151) 푑 푄퐵 푄퐵 The boundary condition is that at = 0, 푑� − 휏푛

푡 | ( = 0)| = (4.152)

The solution can be readily seen to be 푄퐵 푡 퐼퐵 휏푛

| | = 푡 (4.153) −휏푛 The stored charge decays exponentially with푄퐵 time,퐼 퐵as휏 illustrated푛푒 in Figure (4.38). The collector current continues to remain constant at the value it had before the base current was turned off. This situation continues until the device gets out of saturation. The device gets out of saturation when becomes equal to or less than . We define the time that it takes for the device to get out of saturation as the 푄퐵 storage time and denote it by the symbol . Therefore by our definition, at = , the stored charge is 푄푆 equal to . At = , 푡푠 푡 푡푠 푄푆 푡 푡푠 ( ) = = . 푡푠 (4.154) −휏푛 퐵 푠 푆 퐵 푛 푄 푡 - 194푄 - 퐼 휏 푒

Therefore, is obtained as

푠 푡 = ln = ln (4.155)

퐼퐵휏푛 퐼퐵휏푛 푉퐶퐶 휏퐵 푠 푛 푄푆 푛 When > , the device is in the 푡active mode휏 and� is� proportional휏 � to푅 퐿 .� Hence decays with time in step with . This is illustrated in Figure (4.39). The decay of is given by 푡 푡푠 퐼퐶 푄퐵 퐼퐶 푄퐵 퐼퐶 = 푡−푡푠 (4.156) 푆 푄 − 휏푛 퐶 휏퐵 To decay to of the initial value of the collector퐼 current,푒 , should be equal to . The total time 1 푠 푛 taken for the푒 collector current to decay to of its initial value푡 − 푡 from the instant of time휏 when is 1 reduced to zero is 퐵 푒 퐼 = + (4.157)

To speed up this process, should be decreased.푡 This푡푠 is equivalent휏푛 to saying that the device should not be operated in deep saturation. 푡푠

Example

Let us calculate the switching characteristics of an n-p-n transistor used as a saturated inverter, given = 2 , = 10 , = 150 , = 100, and = 10 . Let us further assume that the −9 emitter퐿 injection퐶퐶 efficiency퐵 is nearly unity,0 so that we can퐵 assume = . 푅 퐾� � � 퐼 휇� 훽 휏 푠𝑠 휏푛 0 Turn-on 훽 휏퐵 10 = = = 5 2 × 10 �퐶퐶 퐼𝑐𝑐 3 푚� = = 5푅 ×퐿 10 × 10 = 5 × 10 −3 −9 −12 푄푆 퐼𝑐𝑐= 휏 퐵 = 100 × 10 = 10 퐶 −9 −7 ( ) =휏푛 훽=0휏150퐵 × 10 × 10 = 1.5푠 𝑠× 10 −6 −7 −11 퐵 퐵 푛 푄 ∞ 퐼 휏 = ln 1 퐶 �퐶퐶휏퐵 푡1 −휏푛 � − � 5 × 10 푅퐿퐼퐵휏푛 = 10 × ln 1 = 10 × 0.405 1.5 × 10−12 −7 −7 − � − −11� - 195 -

= 4.05 × 10 −8 Turn-off 푠𝑠 5 10 = ln = 10 × ln 1.5 10−12 �퐶퐶휏퐵 −7 푥 푡푠 − 휏푛 � � − � −11� 푅퐿=퐼퐵1휏푛.1 × 10 푥 −7 Total time taken for to fall off to of its initial value is equal푠𝑠 to 1 퐶 퐼 + 푒= 10 + 1.1 10 = 2.1 × 10 sec −7 −7 −7 푡푠 휏푛 푥

Figure (4.33): A) A circuit configuration for using a bipolar transistor as a saturated inverter. B) The base current as a function of time.

Figure (4.34): The growth of the stored charge as a function of time, due to a constant base current drive, .

퐼퐵

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Figure (4.35): The increase in collector current as a function of time due to the base current drive . is assumed to be large enough to drive the device into saturation. The device gets saturated at = 퐵 퐵, ( ) 퐼 퐼 For < = . , 푡 푡1 푄퐵 푡 푡 푡1 퐼퐶 휏퐵

Figure (4.36): The growth of excess minority carrier density in the base in a saturated inverter.

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Figure (4.37): The base current during the turn-off phase.

Figure (4.38): The decay of stored charge in the base in a saturated inverter, when the base current drive, , is turned off at = 0.

퐼퐵 푡

- 198 -

Figure (4.39): The decay of collector current in a saturated inverter. The current starts to decay only after a time delay equal to the storage time.

Ebers-Moll Model

The model is a basic model for the bipolar transistor which relates dc currents to the junction voltages and is useful for analyzing the behavior of the devices under large signal input.

We saw earlier that in the active region of operation of the bipolar transistor, the current in the emitter, is given as

퐸 퐼 1 , 푞푉퐵� 푘푘 and the current in the collector is given as퐼 퐸� �푒 − �

= 1 + 1 푞푉퐵� 푞푉퐵� 푘푘 푘푘 The current in the퐼퐶 collector훼0퐼 퐸comprises� �푒 a collector− � base퐼퐶� junction�푒 diode− current� and a current due to minority carrier injection in the base from the emitter. If we want to describe the behavior of the device under saturation as well as in the inverse mode of operation, we must take into account the emitter current due to injection of minority carriers in the base from the collector also. This is what is done in the Ebers-Moll Model and is illustrated in Figure (1).

The emitter current, the base current and the collector current are all shown as going into the device. The emitter current comprises two components, one due to the junction diode current and the other due to minority carrier injection from the collector into the base.

= 1 + 1 (1) 푞푉퐵� 푞푉퐵� 푘푘 푘푘 퐸 퐸� 푟 퐶� where is the alpha of퐼 the device− 퐼 in the�푒 inverse− or reverse� 훼 mode퐼 �i.e.,푒 when− the collector� is used as the emitter and the emitter is used as the collector. The collector current is now written as 훼푟

- 199 -

= 1 1 (2) 푞푉퐵� 푞푉퐵� 푘푘 푘푘 where is the alpha of the퐼퐶 device훼푓 in퐼퐸 the� � 푒normal −mode.� The− ba퐼퐶se� � current푒 is− equal� to 훼 푓

= ( + ) = (1 ) 1 (1 ) 1 (3) 푞푉퐵� 푞푉퐵� 푘푘 푘푘 퐼 퐵 −The퐼 퐸transistor퐼퐶 is shown− as− two훼 푓diodes퐼퐸� connected�푒 − back� to− back with− 훼 a푟 current퐼퐶� � 푒source −connected� in parallel to each diode. The current is shown in the emitter and also in the collector as consisting of two components, one due to the diode current and the other from a current source due to minority carrier injection from the opposite region. The currents are controlled by the internal junction voltages. It is customary to write the equations as

= 1 + 1 (4) 푞푉퐵� 푞푉퐵� 푘푘 푘푘 and 퐼퐸 푎11 �푒 − � 푎12 �푒 − �

= 1 + 1 (5) 푞푉퐵� 푞푉퐵� 푘푘 푘푘 where 퐼퐶 푎21 �푒 − � 푎22 �푒 − � =

푎11 = − 퐼퐸� 푎12 = 훼푟 퐼퐶� 푎22 − 퐼퐶� and =

Using the reciprocity relationship of the푎21 two-port훼푓퐼 퐸device,� we can write = . Hence, = 푎12 훼21

Therefore only three out of the four parameters훼푟퐼퐶 viz.,� 훼,푓퐼퐸, � , are needed to model the device. 훼푓 훼푟 퐼퐸� 푎�푎 퐼퐶�

- 200 -

Figure 1: Equivalent circuit diagram for the bipolar transistor according to Ebers-Moll Model

Output Impedance

When we examined the DC characteristics of the transistor, we observed that the collector current increased slightly with collector voltage in the common-base mode, whereas in the common- emitter mode the collector current increased much more with collector voltage. The rate of increase of collector current with collector voltage is called the collector output impedance. The output impedance ( ) is defined as

푟� (4.158) 휕푉퐶 Theoretically, we would have expected the collector푟� ≡ 휕output퐼퐶 impedance to be infinite, since the reverse- biased collector-base junction acts as a constant current source. But the collector current depends on the width of the neutral region of the base. The width of the neutral region decreases with the collector voltage, due to the widening of the collector-base junction depletion region. The effect is called Early effect, and is illustrated in Figure (4.40). If we increase the collector-base voltage, the base width decreases and the slope of ( ) increases as shown in Figure (4.40). The collector current which is proportional to the slope of ( ′ ) at = increases, thus causing the output impedance to be ∆푛 푥 finite. ′ ′ ′ ∆푛 푥 푥 푊 If the emitter-base voltage is kept constant, the excess carrier density at = 0 is constant since it depends only on the impurity density in the base and the emitter-base voltage, i.e.′ 푥 ( = 0) = 1 푞푉퐵� ′ 푘푘 The output impedance under this∆ 푛condition푥 is given푛 by𝑃� �푒 − �

- 201 -

| = 휕푉퐶� | (4.159) 휕푉퐶� 휕푊′ 퐵�=푐𝑐𝑐𝑐� 휕퐼 퐵�=푐𝑐𝑐𝑐� � 퐶� 푉 퐶� 푉 푟 ≡ 휕퐼 휕푊′ can be obtained by differentiating Equation (4.79) with respect to ′ . 휕푊 퐶� 휕푉퐶� � Common-Base Mode Output Impedance

Let us now look at the output impedance when the transistor is used in the common-base mode. In Figure (4.9), the DC collector current-voltage characteristics were plotted in the common-base mode with the emitter current as a parameter with specific values. The output impedance is obtained from this plot under conditions of constant emitter current as

| (4.160) 휕푉퐶� 퐸=푐𝑐𝑐𝑐� The emitter current is determined both 푟by� ≡the emitter휕퐼퐶 -퐼base voltage and the gradient of the excess minority carrier density in the base. As the base width decreases, the slope increases and the emitter current will also increase. In order to keep the emitter current constant, the emitter-base forward voltage will have to decrease as illustrated in Figure (4.41). Thus the increase in the collector current with collector voltage is not as large as what would result if it were due to the decrease in the base width, with the emitter base voltage remaining constant.

Figure (4.40): Base width modulation due to Early Effect

- 202 -

Figure (4.41): The excess carrier density distribution in the base for two different collector voltages in the common-base mode. The base-emitter voltage has to decrease to keep the emitter current constant when the collector voltage is increased.

Common-Emitter Mode Output Impedance

Let us now consider the output impedance in the common-emitter mode. The DC characteristics were plotted in Figure (4.12) for the common emitter mode, and a family of curve was plotted for different values of base current. The output impedance under conditions of constant base output current is obtained from this plot and is equal to

= | (4.161) 휕푉퐶� 퐵=푐𝑐𝑐𝑐� When the collector to emitter voltage is푟 �increased휕퐼퐶 by a퐼 small amount, this increment is partially applied across the collector-base junction and partially across the base-emitter junction. This division between the two junctions occurs in a manner such that the base current remains constant.

The base current supplies a) the current for injection of minority carriers from the base into the emitter, and b) the majority carriers needed for recombination with minority carriers in the base. If the injection efficiency is assumed to be nearly unity, then the former is negligible, and the latter is the dominant component of the base current. The base current is then equal to | | (4.162) 푄퐵 where is the stored minority carrier charge,퐼 and퐵 ≈ is proportional휏푛 to the area under the plot of the excess minority carrier density in the base, as shown in Figure (4.42). is therefore constant when is 퐵 constant.푄 When is increased, the base width decreases due to the increase in the collector-base 퐵 퐵 voltage. In order to keep the stored charge constant, the excess carrier퐼 density at = 0 (height푄 of �퐶� the triangle) has to increase so that the area of the triangle is constant, as illustrated in′ Figure (4.42). 푄퐵 푥 - 203 -

The collector current increase is therefore not only due to the increase in slope arising from a decrease in base width, but also due to an increase in the injected minority carriers at = 0. Thus the output impedance is much smaller than what is obtained in the common base mode. ′ 푥

Figure (4.42): The excess carrier density in the base for two different values of the collector to emitter voltage in the common emitter mode under conditions of constant base current. The area under the plot has to remain constant if the base current is kept constant. Non-ideal base current

In our discussion of the bipolar transistor so far, we considered the emitter-base junction to be an ideal - junction. However, the emitter-base junction has non-ideal components, due to recombination in the depletion region. This is similar to our discussion of the forward-biased non-ideal - junction.푝 푛 We saw in our treatment of - junctions, that an additional component of current flows through the junction due to recombination in the depletion region. This current depends on the forward voltage푝 푛 as 푝 푛

= (4.163) 푞푉퐹 2푘푘 The non-ideal component gives rise to an 퐼extra푟𝑟 component퐼푆 푟𝑟 푒 of base current as shown in Figure (4.43). In this figure, represents the component corresponding to recombination in the emitter-base junction. It flows through the base lead, in addition to the other two components of base current that 퐵 푟𝑟 we discussed퐼 before. We saw earlier that the other two components ( and (1 ) ) have an exponential dependence on the emitter-base forward voltage. In other words, these two components 퐼퐸� − 훼� 퐼퐸� depend on the emitter-base voltage given by

+ (1 ) = + (1 ) (4.164) 푞푉퐵� 푞푉퐵� 푘푘 푘푘 where is the퐼 퐸saturation� current− 훼� of퐼퐸 the� emitter퐼퐸 푆�-base푒 junction due− to훼 �hole퐼퐸 injection푆� 푒 into the emitter, and is the saturation current of the emitter-base junction due to electron injection into the base. 퐼퐸 푆� The non-ideal component is added to these two to obtain to total base current. The plot of the 퐼퐸 푆� logarithm of ( ) is given in Figure (4.44). In this figure, at low values of emitter-base voltage, the base current is퐵 dominated퐵� by , and therefore the slope is equal to . However, as the emitter- 퐼 � 푞 base voltage is increased, the other퐵 푟𝑟 two components (the ideal components) increase faster with the 퐼 2푘푘

- 204 -

emitter-base voltage, and start to dominate. Hence the base-current has a slope given by . Thus we 푞 see that the base current can be approximated as comprising two linear regions: one for small emitter- 푘푘 base voltage with a slope , and another with a slope equal to for larger emitter-base voltage. The 푞 푞 collector-current on the other hand is determined only by the ideal components of base current, and 2푘푘 푘푘 hence the plot of the logarithm of is linear, with a slope equal to , and the ratio of to 푞 remains constant until we come퐶 to really low values of where the leakage current퐶 ( ) 퐼 푘푘 퐼 dominates. This plot of the logarithm of versus and the logarithm of versus in the same 퐵 푖𝑖𝑖 퐵� 퐶�퐶 figure퐼 is called the gummel plot. At low values of , the recombination� in the depletion region퐼 of the 퐼퐶 �퐵� 퐼퐵 �퐵� emitter-base junction dominates, and hence the base�퐵� current has the exponential dependence. At 푞푉퐵� higher values of , and (1 ) dominate, and hence the base-current has2푘푘 a 푞푉퐵� dependence. The 퐵collector� 퐸� -current depends� 퐸� exponentially on , since it is equal to ( ). is � 퐼 − 훼 퐼 푘푘 defined as the ratio of to , and therefore varies with . In Figure (4.45), is plotted as a 퐵� � 퐸� 0 function of , the emitter-base voltage ( or the emitter current� ). At low values of훼 퐼, decreases훽 퐶 퐵 퐵� 0 with a decrease in 퐼because퐼 in this region recombination� in the emitter-base훽 junction depletion 퐵� 퐵 퐵� 0 region dominates.� For moderate values of , is constant, since퐼 in this region both � and훽 have an �퐵� exponential voltage dependence with a slope퐵� to 0 . At very high values of , the current퐶 appears퐵 to � 훽 푞 퐼 퐼 decrease, and this is due to high-injection effects. We will discuss the high-injection퐵� effects in a later 푘푘 � section.

Figure (4.43): Various components of current in the bipolar transistor including the non-ideal current in the emitter-base junction.

- 205 -

Figure (4.44): Plot of the base and collector currents as a function of the emitter-base voltage.

Figure (4.45): Plot of as a function of or .

훽0 �퐵� 퐼퐸 Non-uniform doping in the base

We also have assumed earlier in our discussion of the transistor characteristics that the base had a uniform doping i.e., was constant. We further assumed that the minority-carrier flow in the base region was only due to diffusion, by assuming that the electric field was zero. We are going to see 퐴 now that in a real device, both푁 of these assumptions are invalid. Due to the method of fabrication of the bipolar transistor using an impurity diffusion process, the impurity profile in the base region typically has an exponential slope as shown in Figure (4.46). In this figure, the metallurgical junction is between where the logarithm of net goes to - (i.e., net is zero) on the emitter side and where it goes

푁퐴� ∞ 푁퐴� - 206 - to on the collector side. However, the actual neutral-base width is less than this, and equal to as indicated in the figure. In this neutral-base region therefore, it is possible to approximate the impurity′ profile∞ as 푊

( ) = −휂�′ (4.165) 푊′ ′= 0, where is the impurity concentration푁퐴� at 푥 푁퐴𝐴is푒 a parameter which depends on the variation of the impurity concentration with distance.′ Therefore, we find the hole density to be higher at 퐴𝐴 = 0 푁than at = . In order to prevent the푥 diffusion푎�푎 of휂 holes (majority carriers) in the direction of the′ collector, an ′electric′ field arises in the base region in a negative direction. A drift current due to the electric푥 filed is exactly푥 푊 balanced by the diffusion current. Thus there is no net hole current in spite of the concentration gradient of the acceptor impurity in the base. We can write the expression for the hole current density as

( ) = = ( ) (4.166) ′ 푑푝0 ′ 푑푁퐴� 푥 푝 푝0 푝 푝0 ′ 퐴� 푝 푝 ′ where is the퐽 majority푞 푝-carrier휇 ℰ density− 푞 퐷 in thermal푑푥 equilibrium,푞 푁 푥 and휇 thereforeℰ − 푞퐷 equal푑 to푥 , the acceptor density in the base-region. In order for the two currents to balance, has to equal 0. By 푝푝0 푁퐴� setting the equation for to 0, the electric field is expressed as 퐽푝 푝 퐽 ( ) ( ) = × × = × × (4.167) ′ ( ) ′ ( ) 퐷푝 푑푁퐴� 푥 1 푘푘 푑푁퐴� 푥 1 ′ ′ ′ ′ 푝 퐴� 퐴� We had assumedℰ earlier휇 that 푑 varies푥 exponentially푁 푥 with푞 , and substituting푑푥 this푁 exponential푥 function for we obtain the electric field as ′ 푁퐴 푥 퐴� 푁 = (4.168) 푘� 휂 ′ The electric field arises internally in the baseℰ region− to푞 prevent푊 the diffusion of holes from the highly doped side to the lightly doped side. For this reason the electric field is called the built-in electric field. The electric field influences the minority-carrier flow through the base region from the emitter to the collector. The direction of the built-in electric field is such as to aid the minority-carrier flow towards the collector. Hence the transit time ( ) is reduced. Since the transit time is reduced, the frequency response improves. There are some additional advantages due to the inhomogeneous doping of the 퐵 base. These are 1) increase in punch휏 -through voltage, 2) increase in output impedance, and 3) decrease in base resistance. Increase in punch-through voltage arises because as the depletion-region widens in the collector-base junction, a larger increase in the collector-voltage is needed to produce a given incremental charge in the depletion region, Hence it takes a larger amount of collector-voltage to punch through the base-region than what it would have taken if the entire base-region had a uniform doping concentration equal in value to what is obtained at = . Because of this, the Early effect also gets reduced, and therefore the output impedance increases.′ Due′ to the increase in the impurity concentration in the base, the base resistance decreases.푥 푊 However, there are also some disadvantages due to the non-uniform doping of the base region. First, the injection efficiency ( ) decreases, since the doping concentration on the base side of the emitter-base junction is larger than in a uniformly doped 훾

- 207 - base. Secondly, the emitter-base junction depletion-region width is smaller now, and hence the junction capacitance ( ) increases.

퐶퐽𝐽

Figure (4.46): Impurity Profile in the base region in a typical device.

- 208 -

Summary

• The bipolar transistor works on the principle of minority carrier injection into the lightly doped side of a one-sided abrupt - junction and the flow (or collection) of these minority carriers into a reverse-biased - junction which is placed nearby. 푝 푛

푝 푛 • The typical structure of a bipolar transistor comprises a heavily doped region and a lightly doped n-region separated by a narrow -region. The region is the emitter,+ the -region the 푛 base and the -region the collector. Such a transistor is+ called an - - transistor. A - - 푝 푛 푝 transistor is one in which the emitter is a region, and the collector a -region with the 푛 푛 푝 푛 푝 푛 푝 intervening base layer an -region. In normal+ operation, the emitter-base junction is forward- 푝 푝 biased and the collector-base junction reverse biased. 푛

• The total emitter current is the sum of minority carrier current injected from the emitter into the base and the minority carrier current injected from the base into the emitter. The emitter injection efficiency, , is defined as the ratio of minority carrier current injected into the base from the emitter to the total emitter current. In a well designed transistor, is made as close to 훾 1 as possible. 훾

• The minority carriers injected from the emitter diffuse through the base region to the collector- base junction. Before they reach the collector-base junction, some of the minority carrier current reaching the collector-base junction is less than the emitter injected minority carrier current. The ratio of the two is called the base transport factor, . In a well designed transistor, (which is less than 1) is made as close to unity as possible. 훼�

훼� • The minority carriers reaching the collector-base junction from the base are swiftly swept into the collector region by the electric field in the collector-base depletion region. This results in a collector current. In addition, a reverse leakage current due to the collector-base junction also flows and adds to the collector current.

• The current through the base is the difference between the current due to majority carriers recombining with minority carriers in the base and the reverse leakage current in the collector- base junction.

• The transistor can be normally operated in either of two modes viz., common-base or common- emitter. In the common-base mode the input signal is applied between the emitter and the base and the output signal is obtained between the collector and the base. In the common-emitter mode the input signal is applied between the base and the emitter and the output signal is obtained between the collector and the emitter.

• The short circuit current gain of the transistor is defined as the ratio of the incremental change in the output current and an incremental change in the input current with the output AC short

- 209 -

circuited. In the common-base mode, the current gain is called alpha of the device and is denoted by where the subscript 0 refers to the fact that the frequency is very low

0 훼 =

In the common-emitter mode, the short훼0 circuit훾 current훼� gain at low frequency is called beta of the device and is denoted by .

0 훽 = 1 0 0 훼 • 훽 In a well designed transistor, is made as close− to훼 0unity as possible. This is done by making the emitter heavily doped in comparison with base doping to get as close to unity as possible and 훼0 making the base region as narrow as possible in comparison with the minority carrier diffusion 훾 length in the base to get as close to unity as possible.

훼� • The transistor is said to be saturated or operating in the saturation regime when the collector- base junction also is forward biased. In saturation, the excess minority density in the base at the edge of the collector base space-charge region is increased due to injection from the collector. The transistor is said to be in the active regime of operation when the emitter-base junction is forward-biased and the collector-base junction is reverse biased.

• In the common-base mode of operation, the input resistance to a small signal input voltage is the small signal resistance of a forward biased p-n junction. The resistance is called the emitter resistance and is equal to = , where is the DC emitter current. The output circuit can 푘푘 be represented by having a 퐸dependent current퐸 source with a current value where is 푟 푞퐼퐸 퐼 the sinusoidal signal emitter-base voltage. 푔푚푣�퐵� 푣�퐵�

• In the common emitter mode the small signal input resistance is equal to and therefore

푟퐸 ( ) times larger than the input resistance in the common base mode. 1The− 훼 0output circuit can

1 be1− represented훼0 by the same equivalent circuit as for the common base mode.

• Due to injection of excess minority carriers in the base, at any given instant of time, there is an extra amount of charge in the base. This charge is called stored base charge,. In order to maintain charge neutrality majority carriers flow into the base from the base contact to keep an excess majority carrier density everywhere in the base exactly equal to the excess minority carrier density. A small increment in the emitter base voltage causes an increase in the stored charge by an incremental charge flowing into the base from the emitter while an incremental charge of same magnitude but of opposite polarity flows into the base from the base contact to maintain charge neutrality. Thus a capacitive effect arises due to the stored charge. This is called the diffusion capacitance.

- 210 -

• The time taken for the minority carriers to traverse the neutral base region from the emitter- base depletion region to the collector-base depletion junction when the device is in the active

mode is called the base transit time, . The collector current, times the base transit time, gives the stored base charge, 휏퐵 퐼퐶 = 푄퐵

퐵 퐶 퐵 The diffusion capacitance is related to 푄 as 퐼 휏

휏퐵 = 0 푑𝑑� 훼 퐵 퐶 퐸 휏 • Due to the diffusion capacitance the excess minority푟 carrier density does not change instantaneously with a change in the emitter base voltage. Hence the gain of the transistor falls off. The equivalent circuit for the transistor has the diffusion capacitor in parallel with the input resistance. The input circuit has a time constant due to and . When we ignore the effect of junction capacitances but include only the effect of the diffusion capacitance we call the 퐶푑𝑑� 푟푖� device an intrinsic transistor. In an intrinsic transistor, the alpha falls off at high frequencies and the frequency at which | | falls to of its low frequency is called the alpha-cut-off frequency, 1 . The gain | | falls of at a rate of 6 db/octave i.e., decreases by a factor of 2 for an increase in 훼 √2 frequency∗ by a factor of 2. In this range of frequencies the product of and | | is a constant. 푓𝑐 훼 The asterisk denotes that an intrinsic transistor is being considered. The beta also decreases 푓 훼 with frequency and the frequency at which | | falls to of its low frequency is called the beta 1 cut-off frequency, . Hence also, | | decreases by a factor of 2 when the frequency is 훽 √2 increased by a factor∗ of 2 at frequencies larger than . The frequency at which | | becomes 푓𝑐 훽 equal to 1, is called the gain-band width product and is∗ denoted by . The product of and | | 푓𝑐 훽 at frequencies larger than is a constant and equal to . ∗ 푓� 푓 훽 ∗ 푓𝑐 푓� • When the junction capacitances are included, the time constant of the input circuit is increased. In the output circuit capacitive shunting effects arise. Due to this, the frequency response is poorer than in the intrinsic device. The alpha cut-off frequency, the beta cut-off frequency and the gain band width product in an actual transistor are all reduced from the intrinsic transistor values.

• When the transistor is used as a saturated inverter, there is delay in the turn-on time. The delay in the turn-off time is longer than the turn-on time by the amount of storage time. The storage time is defined as the time needed to get off saturation regime and reach active regime.

- 211 -

• The punch-through voltage, , is the collector voltage at which the collector-base depletion region widens all the way until it touches the emitter-base junction depletion region, poses �푝� a limit on the maximum voltage that can be applied on the collector �푝�

• In the common base mode, the collector-base junction breakdown voltage represents the maximum voltage that can be applied on the collector unless punch-through occurs earlier. This voltage is denoted .

퐵�퐶�퐶 • In the common-emitter mode, the maximum collector voltage that can be applied is less than due to collector multiplication factor. This voltage is denoted .

퐵�퐶�퐶 퐵�퐶�퐶 • The output impedance of the transistor, , in the common base mode is not infinite due 휕퐼퐶 −1 to Early effect. Early effect is the reduction of the neutral base width with increasing collector �휕푉퐶�� voltage.

• The output impedance in the common emitter mode is smaller than in the common base mode due to the fact that the injection of minority carriers from the base has to increase in order to keep the base current constant when the collector voltage is increased.

• In an actual transistor, non-ideal components due to recombination in the emitter, base junction depletion region and surface recombination arise. This has the effect of lowering the gain at low emitter current whereas at higher emitter currents the gain remains at a high value.

• In a real device, the impurity profile is non-uniform in the base with the density decreasing from the emitter side to the collector side. Such a non-uniform impurity profile arises as a result of the diffusion process in the fabrication of the device. Due to this an internal electric field arises which helps the minority carriers to reach the collector in a shorter time interval. The base transit time is reduced. Hence the frequency response is improved.

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Glossary

= area of cross-section of the base region

퐴, = integration constants

퐴 퐵 = collector-base junction breakdown voltage

퐵�퐶�퐶 = collector breakdown voltage in the common emitter mode 퐵�퐶�퐶 = diffusion capacitance of the transistor 퐶푑𝑑� = emitter-base junction capacitance

퐶퐽𝐽 = electron diffusion constant

퐷푛 = hole diffusion constant 퐷푝 = hole diffusion constant in the collector

퐷푝� = incremental positive (majority carrier) charge

𝑑 = incremental time or time interval 𝑑 = small incremental step voltage 𝑑 = electric field ℰ = critical electric field ℰ𝑐𝑐 = - cut off frequency 푓Cα = α intrinsic - cut off frequency ∗ 푓Cα = - cut offα frequency 푓Cβ = 훽intrinsic - cut off frequency ∗ 푓Cβ = gain bandwidth훽 product

푓T = intrinsic gain bandwidth product ∗ 푓� = Farad. Unit of capacitance 퐹 = conductance 푔 = small signal input conductance 푔푚 = small signal sinusoidal base current 횤푏̃ = small signal sinusoidal collector current 횤�̃ = small signal sinusoidal emitter current 횤푒̃ = small signal sinusoidal input current 횤푖̃ � = small signal sinusoidal output current

횤𝑜�̃ = base current

퐵 퐼 - 213 -

= component of base current due to recombination in the emitter-base junction depletion region 퐼퐵 푟𝑟 = total base current

퐼퐵 � = collector current

퐼퐶 = collector-base junction leakage current 퐼퐶�0 = collector current with the base open

퐼퐶�퐶 = collector current due to the injection of electrons from the emitter into the base

퐼퐶� ( ) = current in the base region due to injected electrons

퐼퐶� 푥 = total collector current

퐼퐶� = DC current

퐼퐷 � = DC emitter current

퐼퐸 = electron current injected from the emitter into the base

퐼퐸� = DC emitter current due to hole injection from base into the emitter 퐸� 퐼 = total saturation current of the emitter-base junction

퐼퐸� = saturation current of the emitter-base junction due to electron injection into the base

퐼퐸𝐸 = saturation current of the emitter-base junction due to hole injection into the emitter

퐸𝐸 퐼 = saturation current due to recombination in the emitter-base depletion region

퐼푆 푟 𝑟 = total emitter current 퐼퐸 � = 1

푗 = √minority− carrier (electron) current density 퐽푛( ) = electron current density in the base ′ 퐽푛 푥 = hole current density 퐽푝 = Boltzmann constant

푘 = minority carrier (electron) diffusion length in the base 퐿푛 = minority carrier (hole) diffusion length in the collector 퐿푝� = minority carrier (hole) diffusion length in the emitter

퐿푝 � = avalanche multiplication factor

푀 = electron density

푛 = intrinsic carrier density

푛푖 = thermal equilibrium minority carrier (electron) density in the base 푛푝𝑝 - 214 -

= acceptor density

푁퐴 = acceptor impurity concentration in the base

푁퐴� = acceptor impurity concentration at = 0 ′ 푁퐴𝐴 = donor impurity concentration in the푥 collector

푁퐷� = donor impurity concentration in the emitter

푁퐷 � = majority carrier density in thermal equilibrium

푝0 = thermal equilibrium minority carrier (hole) density in the collector

푝푛�0 = thermal equilibrium minority carrier (hole) density in the emitter

푝 푛 � 0 = electron charge

푞 = minority carrier charge stored in the base

푄퐵 = stored charge, at t =

푄푠 = emitter resistance푄퐵 푡1

푟퐸 = output impedance

푟0 = load resistance

푅 퐿 = time

푡 = turn-on transient

푡1 = storage time, the time it takes for the device to get out of saturation

푡푠 = absolute temperature

푇 = base-emitter voltage

푣�퐵� = small signal sinusoidal input voltage

푣�푖� = small signal sinusoidal output voltage

푣�� � 푡 = small signal sinusoidal voltage

푣� 0( ) = velocity of electrons as a function of ′ ′ 푣 푥 = voltage applied across the junction 푥

� = voltage across the collector-base junction

�퐵� = built-in voltage of the collector-base junction

�푏𝑏 = built-in voltage of the emitter-base junction

�푏𝑏 = base-emitter voltage

�퐵� = breakdown voltage of the junction

�푏� = collector-base voltage �퐶� - 215 -

= DC collector supply voltage

�퐶퐶 = collector-emitter voltage

�퐶� = punch-through voltage �푝� = width of the neutral base region ′ 푊 = width of the base region between the metallurgical junction on the emitter side, and that on the collector side 푊퐵� = width of the neutral emitter region

푊 퐸 = coordinate

푥 = width of the depletion-region occurring in the base region due to the collector-base junction

푝� 푥 = width of the depletion-region occurring in the base region due to the emitter-base junction

푥푝� = coordinate system in the neutral base region where the origin is chosen at the edge of the neutral′ base region on the emitter side 푥 = current gain

훼 = low frequency current gain in the common-base mode

훼0 = base transport factor 훼| �( )| = alpha as a function of = 2 times frequency 훼 휔 = current gain in the common휔 휋 emitter mode

훽( ) = beta as a function of = 2 times frequency

훽 휔 = low frequency current휔 gain휋 in the common emitter mode 훽0 = differential 휕( ) = incremental change in the excess minority carrier density 휕 ∆ 푛 = short time interval 휕� = elementary distance ′ 휕푥 = increment ∆ = incremental change in base current ∆퐼퐵 = incremental change in collector current퐼퐵 ∆퐼퐶 = excess minority carrier in the base region퐼퐶 due to minority carrier injection from the emitter ∆푛 = incremental change in emitter current ∆퐼퐸( ) = excess carrier (electron) density as a function퐼퐸 of ′ ′ ∆푛 푥 = permittivity of semiconductors 푥 휖푠 = emitter injection efficiency

훾 - 216 -

= an empirical parameter in avalanche multiplication expression

휂 = an empirical parameter defining the impurity density variation in the base 휂 = electron mobility 휇푛 = hole mobility 휇푝 = base transit time

휏퐵 = minority carrier (electron) lifetime 휏푛 = 2 times frequency

휔 = 2휋 times alpha cut-off frequency ∗ 휔𝑐 = 2휋 times beta cut-off frequency ∗ 𝑐 휔 = value휋 of at which | ( )| becomes unity ∗ 휔 � = divergence휔 훽 휔 ∇

- 217 -

Problems

(Assume room temperature in all problems except when it is specifically states otherwise. Also assume silicon as the semiconductor material. Take , as 1 × 10 ) 10 −3 1. The expression for the excess minority푛푖 carrier density𝑐 in the neutral base region was derived in the text in terms of hyperbolic functions. Show under appropriate conditions that a) it becomes a linear function and b) it becomes an exponentially decaying function.

2. A n-p-n transistor has impurity concentrations of 5 × 10 in the emitter, 5 × 10 in the base and 2 × 10 in the collector.19 Assume−3 the distance between the 𝑐 metallurgical16 −3 junction on the emitter15 side −and3 that on the collector side called the metallurgical 𝑐 𝑐 base width is equal to 2 . Assume the diffusion constant and the lifetime of the minority carriers in the base region to be 30 / ane 3 × 10 respectively. Assume that the 휇� emitter-base voltage is 0.5 and the 2collector-base voltage−6 is 5 . Plot on a graph sheet the 𝑐 푠𝑠 푠𝑠 excess carrier density in the base region. − � �

3. We derived in the class the expression for the base-transport factor, the emitter injection efficiency and the current gain for a n-p-n transistor. Do the same for a p-n-p transistor.

4. For the device in problem 2, the area of the emitter-base junction and that of the collector-base junction are the same and equal to 10 . Assume the minority carrier mobility and lifetime in the emitter to be 500 / −and4 0.225 10 sec respectively. What is the injection 𝑐 efficiency at the voltage biases2 given in problem 2. −6 𝑐 � − 푠𝑠 푥

5. For the device in problem 2, determine the base transport factor at the specified voltage.

6. A p-n-p transistor has impurity concentrations of 5 × 10 , 2 × 10 and 5 × 10 in the emitter, base and collector regions19 respectively.−3 Assume16 that−3 the 𝑐 𝑐 distance15 between−3 the two metallurgical junctions, is 2.5 × 10 . Assume the emitter- 𝑐 base junction is forward biased by 0.5 and that the collector-base− junction4 is reverse biased by 푊퐵� 𝑐 10 V. The minority carrier parameters are given below in each of the three regions. � Emitter Base Collector

Lifetime (sec) 10 10 2x10 −6 −4 −4 Diffusion Constant 10 15 16 ( / ) 2 Calculate , , . 𝑐 푠𝑠

훾 훼� 훼0 푎�푎 훽0

- 218 -

7. For the device in problem 6, calculate and . Take the area of both the collecetor and the emitter junctions to be 10 . 퐼퐶�퐶 퐼퐶�퐶 8. For the device in problem 6, draw−3 the2 equivalent circuit for a) common base mode and b) 𝑐 common emitter base mode.

9. For the device in problem 6, calculate the punch-through voltage, and . Assume that the critical electric filed for breakdown in the collector-base junction is 200,000 V/cm. Take 퐵�퐶�퐶 퐵�퐶�퐶 as = 3

휂

10. An n-p-n transistor has impurity concentrations of 10 , 10 and 10 in the emitter, base and collector regions respectively. Assume19 that−3 the 16distance−3 between15 the −two3 𝑐 𝑐 𝑐 metallurgical junctions, is 2.5 × 10 . Assume that the emitter-base junction is forward biased to 0.5 V. Find the punch-through−4 voltage. 푊퐵� 𝑐

11. Assume that for the device in problem 10, = 5 x10 sec and the minority carrier lifetime in the base is 10 sec. Assume the transistor to be intrinsic.−8 What is , assuming that is 휏퐵 nearly unity. Take−5 the value of the parameter as 3 and to be 250V. 퐵�퐶�퐶 훾

휂 퐵�퐶�퐶

12. For the device in problem 11, determine , and . Assume the device is intrinsic i.e., you can neglect the parasitic elements and the junction capacitances. 푓Cα 푓Cβ 푓T

13. Assume an area of 10 for emitter-base junction for the device in problem 10. Assume is not the value given −in3 problem2 11. Assume the width of the neutral base region to be 𝑐 휏퐵 2 × 10 cm. Let the minority carrier diffusion constant in the base be 16 ′ . 푊 Calculate −4 and at an emitter-base bias voltage of 0.6 V and a collector-base2 reverse−1 bias √ �푚 푠𝑠 of 5 V. (Note: You are still assuming that is nearly unity.) 푓Cα 푓Cβ

훾

14. For the operating voltage given in the last problem determine the equivalent circuit. Assuming the area of the collector-base junction to be the same as that for the emitter-base junction.

15. Assume that is 10 for the device in the last problem. Calculate the storage time when the device is used as an inverter with a load resistor of resistance equal to 10 . Assume that the �퐶퐶 � base current drive is 2 and that the current has been on for a long time for the purpose of 퐾� calculating the storage time. 푚�

16. A switching transistor has the following characteristics = 10 and = 10 . The device is used as a saturated inverter and assume (a) , the∗ emitter6 injection∗ efficiency8 is nearly 푓Cβ 퐻� 푓T 퐻� 훾 - 219 -

unity, b) the load resistor is 5 and (c) the collector supply voltage is 10 . A base drive current of 50 is applied at = 0. Calculate the time required to turn the device on. 퐾훺 � 17. In the device in problem 16, assume that the base current of 50 is flowing for sufficiently 휇� 푡 long time to establish steady state condition. Assume that at = 0, the base current is switched 휇� off. Calculate the turn-off time which is defined as the time required for the collector current to 푡 decay to a predetermined fraction such as or 10% of the initial value. 1

푒 18. For the condition given in problem 17, determine the emitter-base voltage at time = 0, assuming that the device is a - - transistor and the base doping is 10 . 푡 16 −3 푝 푛 푝 𝑐

19. The impurity concentration in the base of a n-p-n transistor is given approximately by the function ( ) = exp( ) ′ ′ Where = impurity concentration at = 0 and is− a parameter.휂푥 Take as equal to 2 푁퐵 푥 푁퐵� ′ μm. ′ 푊 ′ 푁퐵� 푥 휂 푊 (a) What is the value of if = ( = ) = 10 . Do not confuse this with the

one that we used to calculate . ′ ′ −3 휂 퐵� 퐵 퐵� 휂 (b) What is the value of the electric푁 field푁 푥in the푊 base? 푁 퐵�퐶�퐶

20. Design a n-p-n transistor with the following specifications. It should have a minimum of 40 and it should be capable of withstanding a collector voltage of 100 . It should have an intrinsic 훽0 beta cut-off frequency of 150 or larger. Assume you can alter the minority carrier lifetime � to any specific design value by adding g-r centers with a capture cross-section of 10 . 퐾𝐾 Assume that the starting material has a lifetime of 100 . (Note: There is no unique−16 design.)2 𝑐 휇𝜇�

- 220 -

Chapter 5 MOS Devices

MOS devices refer to a class of devices in which a thin insulating layer such as silicon-di-oxide (usually a thermally grown oxide) is sandwiched between a metallic plate and a semiconductor. The term MOS stands for Metal-Oxide-Semiconductor. These types of devices are called field-effect devices, since a voltage applied between the metallic plate and the semiconductor gives rise to an electric field perpendicular to the interface between the semiconductor and the insulating layer. The electric field alters the electrical properties of the semiconductor in a region close to the interface by inducing a space charge. The term MIS is used to describe in which, instead of an oxide, any insulting layer is used.

MOS Capacitor

We will first consider a simple structure called an MOS capacitor, which is also sometimes called an MOS diode. The device is just a parallel-plate capacitor with the metallic layer as one plate, and the semiconductor as the other parallel plate, with the insulating layer as the dielectric region between the parallel plates. The device is illustrated in Figure (5.1). This type of structure is fabricated by oxidizing the surface of silicon in a furnace which is heated to a very high temperature, such that a thin oxide layer of thickness typically less than 1000 is grown on the surface. Silicon has a great affinity for oxygen, and therefore the thermal oxidation of silicon is easy to accomplish. The interface between silicon and the oxide has a low interfacial charge.Å A metallic film is then deposited on top of the oxide layer to form the MOS sandwich.

The metal is called the gate. In modern devices the gate is made of polysilicon instead of a metallic layer. The semiconductor is called the substrate. A voltage applied on the gate with respect to the substrate give rise to a perpendicular electric field in the oxide layer. The electric field induces charge in the semiconductor. The induced charge generally exists over a certain spatial region of the semiconductor, and hence it is called the spacer charge. The space charge is induced in a region very close to the semiconductor-oxide interface. This field-effect is illustrated in Figure (5.2). is the electric field in the oxide layer. The electric field lines terminate on the charges in the space charge 𝑜 region. Choosing the interface between the oxide and the substrate as the origin of the ℰ-axis in Figure (5.2), the space charge region extends up to a depth of and the neutral region is below the space charge region. 푥 푥 푑 In the neutral region of the semiconductor, the net charge density is zero, and in the space charge region of the semiconductor, the charge density is not zero. The total charge of the free carriers in a neutral semiconductor, which is mostly due to majority carriers, is equal and opposite to that of the ionized impurities. Hence, in the neutral region the charge density is zero, as shown in the following equation. 휌 = [ + ] = 0 (5.1) + − As done in earlier chapters, we휌 use a푞 subscript푝0 − 푛 0 0to denote푁퐷 − the푁 thermal퐴 equilibrium carrier density. On the other hand, in the space-charge region, the free-carrier densities are less than in the neutral region,

- 221 - and hence a net charge density results. The electrostatic potential, , varies in the space charge region as given by Poisson equation. ∅ = (5.2) 2 휌 The potential energy of the electron, which is∇ equal∅ to− 휖, 푠varies in the space-charge region with distance, and therefore and also vary with distance as we saw in our discussion of the depletion � region in junctions. This is usually called the bending퐸 of the bands in the space-charge region. In the 푖 � neutral region of the semiconductor,퐸 퐸 where the electric field is zero, the potential and therefore the potential 푝energy� are constant with distance, and hence the band is said to be flat. This is illustrated in Figure (5.3). Since there is no electric current in the direction of , the Fermi energy, , does not vary with . In the example shown in Figure (5.3) we have assumed the substrate be -type. 푥 퐸퐹 푥 푝

Figure (5.1): Cross-section of an MOS capacitor

- 222 -

Figure (5.2): Illustration of the field-effect in the MOS structure. A sheet charge is induced in the gate and a space-charge is induced in the semiconductor

Figure (5.3): The space-charge region in the substrate of an MOS capacitor. (a) Four distinct regions of the MOS capacitor (b) The band-bending in the space-charge region of the semiconductor

The electrostatic potential, which is equal to plus an additive constant, can therefore be 퐸푐 written as − 푞 = (5.3) 퐸퐹−퐸푖 ∅ - 223 -푞

Recalling our previous discussions on carrier density, we can write

= = (5.4) 퐸퐹−퐸푖 푞∅ � 푘푘 � �푘푘� and 푛 푛푖푒 푛푖푒

= = (5.5) 퐸퐹−퐸푖 푞∅ �− 푘푘 � �− 푘푘� The neutral region beneath the space푝-charge푛 region푖푒 is referred푛 to푖 푒as the bulk, where the carrier densities are at their thermal equilibrium values. The carrier densities in the bulk are denoted and , and are given by 푛푏 푝푏

= (5.6) 푞∅푏 � 푘푘 � and 푛푏 푛푖푒

= (5.7) 푞∅푏 �− 푘푘 � Here we have defined a bulk electrostatic potential푝푏 푛 as푖푒 equal to = (5.8) 퐸퐹−퐸푖� where is the intrinsic Fermi energy level,∅ and푏 푞 represents the separation between the Fermi energy and the intrinsic Fermi energy in the bulk. We can now write the expression for the electrostatic 푖� 퐹 푖� potential퐸 in the bulk in terms of the thermal equilibrium퐸 − 퐸 carrier densities.

∅푏 = = (5.9) 푘푘 푛푏 푘푘 푝푏 In an -type substrate, is larger∅푏 than 푞 , 𝑙and� hence푛푖 � −is positive푞 𝑙 �,푛 while푖 � in a -type substrate is larger than , and therefore it is negative. Again we can express this in terms of the impurity 푏 푖 b 푏 concentrations푛 in the bulk푛 by assuming that푛 the majority∅ carrier density is equal푝 to the donor density푝 in 푖 the -type substrate,푛 and equal to the acceptor density in the -type substrate. For example, in an - type substrate, is equal to which is equal to . Hence 푛 + 푝 푛 푏 푛0 퐷 푛 푛 = 푁 + (5.10) 푘푘 푁퐷 푘푘 푁퐷 Similarly, in a -type substrate, ∅is푏 equal푞 to𝑙 � ,푛 which푖 � ≈ is 푞equal𝑙 �to푛 푖 �. Hence − 푏 푝0 퐴 푝 푝 = 푝 푁 − (5.11) 푘푘 푁퐴 푘푘 푁퐴 The reader should easily recall that∅푏 is− the푞 same𝑙 � electrostatic푛푖 � ≈ − 푞potential𝑙 � 푛푖 � or we defined in Chapter 2. In the above two equations we have made the assumption that at room temperature all the ∅푏 ∅푛 ∅푝 impurity atoms are ionized, i.e., = and = . In the space-charge region, the electrostatic potential varies due to band bending,+ and hence the− carrier densities vary. 푁퐷 푁퐷 푁퐴 푁퐴

- 224 -

( ) ( ) = 푞 ∅ 푥 (5.12) � 푘푘 � and 푛 푥 푛푖 푒 ( ) ( ) = 푞 ∅ 푥 (5.13) �− 푘푘 � 푖 Multiplying both sides of the equation by푝 푥 ×푛 푒 , which is equal to 1, we get 푞∅푏 푞∅푏 � 푘푘 � �− 푘푘 � 푒 푒 ( ( ) ) ( ) = 푞∅푏 푞 ∅ 푥 −∅푏 � � � � 푘푘 푘푘 푖 ( ( ) ) 푛 푥 = 푛 푒 푒 (5.14) 푞 ∅ 푥 − ∅푏 � 푘푘 � and similarly, 푛푏 푒 ( ( ) )

( ) = (5.15) 푞 ∅ 푥 −∅푏 �− 푘푘 � ( ) We will now define as electrostatic푝 potential푥 푝difference푏푒 , between and ,as ( ) = ( ) 흍 풙 푥 푥 푑 (5.16) ( ) We can think of as representing휓 the푥 electrostatic∅ 푥 − potential∅푏 at some point , measured with respect to the bulk. The carrier densities at in the space-charge region can be now expressed in terms of this electrostatic휓 푥 potential difference as 푥 푥 ( ) ( 휓) = 푞 휓 푥 (5.17) � 푘푘 � and 푛 푥 푛푏 푒 ( ) ( ) = 푞 휓 푥 (5.18) �− 푘푘 � We notice that 푝 푥 푝푏 푒 ( ) ( ) = = 2 The law of mass action is valid in the 푛space푥 -푝charge푥 region,푛푏 푝 since푏 the푛푖 space-charge region is in thermal equilibrium. The value of at x = 0 is denoted , and is defined as the surface potential (or more correctly, the interface potential). ψ 휓푠 = ( = 0) (5.19)

The electron and hole concentrations at 휓the푠 interface휓 푥 are obtained by substituting in Equations (5.17) and (5.18) 휓푠

= ( = 0) = (5.20) 푞휓푠 � 푘푘 � 푛푠 푛 푥 푛푏푒 - 225 -

= ( = 0) = (5.21) 푞휓푠 �− 푘푘 � q ( ) E denotes the value by which푝푠 the 푝potential푥 energy푝 ( 푏and푒 therefore ) is different at some value x, from the value in the bulk. In other words, ( ) denotes the amount of band bending. If ( ) is c positive,− 휓 푥 the potential energy at x is less than in the bulk, and the bands bend downward. The electron concentration increases, and the hole concentration푞 휓 푥 decreases from their respective bulk values휓 푥 (i.e., the values in the neutral region) as the interface is approached. If ( ) is negative, the band bends upward as the surface is approached. The hole concentration increases and the electron concentration decreases. 휓 푥

The amount of space charge induced in the semiconductor under a unit area of cross-section at the interface is defined as the space-charge density . If is the width of the space-charge region, then is equal to 푄푠� 푥푠� 푠� 푄 = ( ) (5.22) 푥푠� can be visualized as the amount of 푄charge푠� ∫in0 the휌 space푥 -charge𝑑 region contained in a cylinder of unit area of cross-section and length equal to the width of the space-charge region, as shown in Figure (5.4). 푄푠� We will now consider the behavior of the space-charge region under different bias conditions. For the purposes of discussion of the principles involved, let us consider a -type substrate, although we could have (equally well) chosen an -type semiconductor. 푝 푛

Figure (5.4): The space-charge density , is defined as the amount of charge in the space-charge region under a unit area of cross-section at the interface. Therefore is equal to the charge contained 푠� in a cylinder in the space-charge region of푄 length equal to the width of the space-charge region and unit 푠� area of cross-section. 푄

Accumulation

Let us first assume a negative charge voltage on the gate. The voltage is divided between the oxide layer and the space-charge region. The potential across space-charge region is the surface

- 226 - potential , and is therefore negative. Another way of looking at this is that the electric field at the interface is directed away from the semiconductor towards the gate. Potential increases from the 푠 푠 surface as휓 the bulk휓 is approached and hence is negative. The induced space-charge is therefore positive. The hole concentration increases from its bulk value, so as to give rise to a net positive charge 푠 in the space-charge region. The hole concentration휓 at the surface is given by

= (5.23) 푞휓푠 �− 푘푘 � | | For a small increase in , increases푝 푠enormously푝푏 푒 because of the exponential dependence on , and also because (which is the thermal equilibrium majority carrier density) is large. The space-charge 푠 푠 푠 region occurs in a very thin휓 region푝 near the surface, so that the space-charge can be considered as휓 a 푏 sheet charge.푝 Since the majority carrier charge is increased at the interface by accumulating majority carriers, the space-charge region is called the accumulation region. Under this condition, the semiconductor is said to be accumulated.

The band-bending under conditions of accumulation is illustrated in Figure (5.5). The space charge region extends to a width = . Since the space charge due to accumulation extends over a small region of very narrow width, the accumulation charge can be considered to be a sheet charge for 푠� 푎𝑎 all practical purposes. 푥 푥

Figure (5.5): Band-bending under conditions of accumulation at the surface of a -type substrate in an MOS capacitor. In this figure, the conduction band and the valence band in the oxide are also shown. 푝

- 227 -

Example

Let us calculate the carrier densities at the surface, given the surface is accumulated with = 0.1 V and the bulk impurity concentration = 5 × 10 . Let us consider room temperature and 휓푠 − therefore we can assume that all the acceptor atoms15 are− ionized3 at room temperature. 푁퐴 𝑐 = 5 × 10 15 −3 푏 푝 = 𝑐 푞휓푠 �− 푘푘 � . 푠 푏 푝 푝 푒 . = 5 × 10 0 1 15 � 0 0259� −3 = 2.4 × 10푒 𝑐 17 −3 = = 20,000𝑐 2 푛푖 −3 푏 푏 푛 푝 𝑐 = 푞�푠 � � 푘푘 . 푠 푏 푛 푛 푒 . = 20000 × 0 1 �− 0 0259� −3 = 421 푒 𝑐 −3 The hole concentration at the surface is much larger than𝑐 the hole (majority carrier) density in the bulk and hence the surface is accumulated. Alternately, we can calculate , by using the law of mass action. 10 푛푠 = = = 421 2 2.4 ×2010 푛푖 −3 푛푠 17 𝑐 푝푠 Depletion

When the gate voltage is positive, is positive and the induced space-charge is negative. The hole concentration in the space charge region is decreased from its thermal equilibrium value, and 푠 hence the net charge is negative in the space휓 -charge region. The space-charge is assumed to be only due to the ionized impurity atoms, and not due to free carriers. It is therefore like the depletion region that we considered earlier in the - junction. To calculate the potential variation in the space-charge region, we make the approximation that, throughout the space-charge region, the free carrier density is small compared with the impurity 푝density.푛 Hence we will refer to the space-charge region under these conditions as the depletion region. The free carrier density varies in the depletion region due to the potential variation as given by Equation (5.17) and (5.18). The hole density decreases and the electron density increases as they move towards the surface. The assumption that the free carrier density in the space-charge region is negligibly small in comparison with the impurity concentration is valid everywhere in the depletion region except near the edge of the depletion region. However, we will assume that even at the edge of the depletion region the free carrier density is negligible, by taking the

- 228 - majority carrier density to change abruptly from its thermal equilibrium value in the bulk to zero in the depletion region.

We can express in terms of , the thermal equilibrium majority carrier density in the neutral region as 푛푠 푝푏

= = (5.24) 푞휓푠 2 푞휓푠 � 푘푘 � 푛푖 � 푘푘 � But is given by Equation (5.7). Rearranging푛푠 푛푏 Equation푒 (5.7)푝 as푏 푒

푏 푝 = (5.25) 푞∅푏 � 푘푘 � and substituting, we obtain 푛푖 푝푏푒

= = 푞 2 �∅ � (5.26) 2 푞 ∅푏 푏 푛푖 � 2 푘푘 � �− 푘푘 � Therefore 푝푏 푝푏푒 푝푏푒

( | | ) = 푞 (5.27) �푘푘 �푠−2 ∅푏 � is < 2 | | From this we see that as long푛푠 as 푝푏 푒 , the electron density at the surface is less than , and therefore less than . If the electron density at the surface is less than (= ), it is definitely 푠 푏 푏 also less than throughout the space-charge휓 region∅ since ( ) is less than when is positive. For푝 퐴 푏 퐴 the entire range of values푁 between 0 and 2 | |, the assumption that the푝 space-푁charge region is just 퐴 푠 푠 a depletion region푁 is valid, and the semiconductor surface is푛 said푥 to be depleted.푛 휓 휓푠 ∅푏 When = | |, = as can be readily seen. Therefore, when > | |, > and < , In other words, at the surface, the electron concentration is larger than the hole concentration, 푠 푏 푠 푖 풔 풃 푠 푖 and hence the 휓surface∅ region푛 is 푛more like an -type material as far as the carrier흍 densities∅ 푛 are푛 푠 푖 푝concerned,푛 and hence the surface is said to be weakly inverted. We say that the surface is weakly inverted because at the surface the electron 푛concentration, is still less than the bulk majority carrier density, . It is therefore possible to subdivide the range of values of 0 to 2 | | as depletion when 푠 lies between 0 and | | and weak inversion, when lies푛 between | | and 2| |. 푝푏 휓푠 ∅푏 휓 푠 The band bending∅푏 and the charge density in the휓푠 space-charge region∅푏 are shown∅푏 in Figure (5.6), when the surface is depleted. The band-bending under the condition that satisfies 0 < < | |, is shown in Figure (5.6 A), while the band-bending under the condition that satisfies | | < < 2| |, 푠 푏 corresponding to weak inversion is shown in Figure (5.6 B). In both cases휓 the charge density휓 in ∅the 푠 푏 푠 푏 space-charge region is still equal to – . For our purposes, the entire range휓 of value∅ s between∅ ∅ 0 and 2| | is said to correspond to depletion condition, for in this range the charge density in the space- 퐴 푠 charge region is equal to – . When푞 �the gate voltage, , is of such a magnitude휓 and polarity as to 푏 make ∅ lie between 0 and 2| |, the MOS device is said in the depletion region of operation. 푞�퐴 �퐺 휓푠 ∅푏

- 229 -

Figure (5.6): Band-bending under conditions of depletion at the surface: A) Depletion B) Weak inversion

Strong Inversion

When becomes larger than 2| |, according to Equation (5.27) the surface electron density becomes larger than , and hence . The surface is said to be strongly inverted under this conditions. 푠 푏 The band-bending휓 and the charge density∅ under strong inversion conditions are shown in Figure (5.7). 푝푏 푁퐴 The space-charge region under these conditions can be viewed as comprising two regions: one in which the electron density is > , and the other in which the electron density is < . The former, which lies close to the surface is called the inversion region and the latter, that lies between the 퐴 퐴 inversion region and the bulk neutral푁 region, is called the depletion region. The charge푁 density in the inversion region is determined by the electron density (since ( ) > ) while that in the depletion region is just equal to since ( ) < . 푛 푥 푁퐴 Referring to Figure푞�퐴 (5.7),푛 the푥 region푁퐴 lying between the surface ( = 0) and , where ( ) = 2| |, represents the inversion region. The region to the right of is the depletion region. 퐼 Thus when = 2| |, occurs at = 0. The inversion region has zero푥 width. Hence,푥 the condition 퐼 푏 퐼 휓 푥= 2| |∅ is called the on-set of inversion. As increases beyond 2| |푥, the inversion region grows. 푠 푏 퐼 The surface 휓is said to∅ be strongly푥 inverted.푥 We denote the boundary between the space charge region 휓푠 ∅푏 휓푠 ∅푏 and the neutral bulk region . Hence the depletion region occurs between and .

In reality, the inversion푥푑 푚layer𝑚 is a thin layer, the charge in this region can be푥퐼 approximated푥푑 푚𝑚 as a sheet charge, and can be assumed to be negligibly small.

푥퐼 - 230 -

Figure (5.7): Band-bending under inversion conditions

Relation Between V and

In this section, letG us determineψs the relation between the gate voltage and the surface potential or the space-charge density. Referring to Figure (5.8), when a voltage is applied between the gate and the substrate, the electric field, , in the oxide induces a space-charge of density in the 퐺 semiconductor. If is positive, i.e., (directed into the semiconductor,)� is negative. Using Gauss’ 𝑜 푆� law, is given by ℰ 푄 ℰ𝑜 푄푆� 푄 푆 � = (5.28) where is the permittivity of the oxide,푄푆 which� is− equal휖𝑜 ℰto𝑜 휖 𝑜 = (5.29) where is the dielectric constant of the휖𝑜 oxide,퐾 and𝑜휖 � is the permittivity of free space. For silicon-di- oxide, 3.9. The voltage across the oxide layer is denoted , and equal to 퐾𝑜 휖� 퐾 𝑜 ≈ = � 𝑜 (5.30)

where is the thickness of the oxide layer.�𝑜 Usingℰ𝑜 Equation푡𝑜 (5.28), we can write 𝑜 푡 = (5.31) 푄푆� A parallel plate capacitor using the oxide�𝑜 layer as− the휖𝑜 dielectric푡𝑜 and with as the thickness of the dielectric layer will have a capacitance per unit area (denoted ) equal to 푡𝑜 퐶𝑜 - 231 -

= (5.32) 휖𝑜 We call the oxide layer capacitance.퐶 Equation𝑜 푡𝑜 (5.31) can be rewritten as

𝑜 퐶 = (5.33) 푄푆� Since the gate voltage is applied across �the�푥 oxide− layer퐶𝑜 and the space-charge region, is divided into two components, one, , across the oxide layer, and the other, , across the space-charge region. �퐺 � 𝑜 = + 휓 푠 (5.34) ( ) Since depends on the surface , we�퐺 will denote�𝑜 휓 as푠 . Combining the equation for with the above equation, we can write as 푄푆� 휓푠 푄푆� 푄푆� 휓푠 �𝑜 �퐺 ( ) = + (5.35) 푄푆� �푠 We can now determine the relation between�퐺 − and퐶 𝑜 for accumulation,휓푠 depletion and inversion conditions. 휓푠 �퐺

Figure 5.8: The MOS diode under the application of a DC bias

- 232 -

Figure (5.9): Sheet charge of majority carriers under accumulation condition

Accumulation

As stated before, we will assume that the space-charge is a sheet charge under accumulation conditions, and we take 0. By putting = 0 in Equation (5.35), we get

휓푠 ≈ 휓푠 = 푄푆� �퐺 − Denoting the space-charge density in accumulation as 퐶𝑜, we get = 푄푎𝑎 (5.36)

In -type substrates, the accumulation charge푄푎𝑎 is due− to퐶 holes𝑜 � 퐺and it is also customary to write 푝 =

Hence 푄푎𝑎 푄푝 = = (5.37)

Since is negative in the accumulation푄푝 condition푄푎𝑎 for a −-type퐶𝑜 substrate,�퐺 we see that � 퐺 = | 푝 | (5.38)

The accumulation sheet charge is illustrated푄푝 in Figure퐶𝑜 (5.9).�퐺 The origin of the -axis is chosen at the oxide silicon interface, and the gate electrode is located at = in this figure. We see that the charge on the gate is a sheet charge and the space-charge is also a sheet charge푥 at the surface of the 𝑜 semiconductor. 푥 −푡

- 233 -

Example

Consider an MOS capacitor with a p-type substrate. Let the oxide thickness be 500 . Calculate the accumulation charge density at a gate voltage of – 5V. Å . × . × = = = = 6.9 × 10 / × −14 휖𝑜 퐾𝑜휖𝑜 3 9 8 85 10 −8 2 𝑜 −8 ( A useful number퐶 to remember푡𝑜 for 푡𝑜 is 3.45 ×500 10 10 / for 1000 of oxide퐹 thickness.)𝑐 −8 2 = = 퐶𝑜= 6.9 × 10퐹 𝑐× ( 5) = Å 3.45 × 10 / −8 −7 2 푄푎𝑎 푄푝 − 퐶𝑜 �퐺 − − 퐶 𝑐 Depletion

We now apply a positive gate voltage of such a magnitude as to keep less than 2| |. is positive. As stated before, the space-charge is due to ionized impurities, and hence the space charge 푠 푏 푠 region is a depletion region. As discussed in chapter 3, all the impurity atoms (acceptors휓 for a∅ -type휓 substrate) are ionized in the depletion even at low temperatures, in contrast to the neutral region where the impurity atoms are only ionized partially at low temperatures. 푝

The charge density in the depletion region is

= (5.39)

and the potential difference across the depletion휌 − region푞 푁퐴 is . Although the depletion region can be treated as similar to that arising in a one-sided abrupt junction, and the expression for the space-charge 푠 density written straight away, we derive the expression for휓 ( ) again using Gauss theorem approach. Consider the band-bending under depletion condition shown in Figure (5.10 A), and the corresponding 푠 space-charge region shown in Figure (5.10 B). The electric field휓 푥(x) at some point in the depletion region is due to negatively ionized acceptor atoms lying between and in the depletion region and the direction of the electric field is positive since the charge is negative.ℰ 푥 푥 푥푑 ( ) (x) = (5.40) 푞푁퐴 푥푑 − 푥 Integrating, we obtain ℰ 휖푠

( ) = +

휓푠 푥 − ∫ ℰ 𝑑( 퐶 ) = 푥2 + C (5.41) 푞 푁퐴 푥푑 푥 − 2 − 휖푠

- 234 -

Figure (5.10): A) Band-bending under depletion conditions. B) The space-charge is due to depletion charge. where C is an integration constant. At = , ( ) = 0. Hence

푑 푠 푑 푥 푥= 휓 푥 2 (5.42) 푞 푁퐴 푥푑 Substituting this expression for C, 퐶 2휖푠

( ) 2 2 ( ) = 푥 푥푑 푞 푁퐴 푥푑푥− 2 − 2 푠 휖푠 휓 푥= − ( ) (5.43) 푁퐴 2 at = 0, = . Hence 푞 2휖푠 푥푑 − 푥 푠 푥 휓 휓 = 2 (5.44) 푞 푁퐴 푥푑 Hence the expression for ( ) can be expressed휓푠 as 2휖푠

휓 푥 ( ) = ( ) 푁퐴 2 휓 푥 푞 2휖푠 푥푑 − 푥 = 1 2 푁퐴 2 푥 푑 푞 -2 235휖푠 푥 - � − 푥푑�

= 1 (5.45) 푥 2 The space-charge density under depletion conditions휓푠 � is denoted− 푥푑� , since the charge is only due to depletion charge. 푄푑 = = (5.46)

We can now express as 푄푆� 푄푑 − 푞 푁퐴 푥푑 퐺 � = + = + (5.47) 푄푑 푞 푁퐴 푥푑 From Equation (5.44), we can write�퐺 as− 퐶𝑜 휓푠 퐶𝑜 휓푠 푥푑

= (5.48)

2 휖푠 �푠 푑 Using this expression for , the depletion푥 charge� density푞 푁퐴 can be expressed as

푑 푑 푥 푄 = = = 2 (5.49)

2 휖푠 �푠 푑 퐴 푑 퐴 푠 퐴 푠 Substituting the푄 expression− 푞 for푁 푥 in that− for푞 푁, we� get푞 푁 퐴 − � 휖 푞 푁 휓

푄푑 �퐺 = + (5.50) �2휖푠푞 푁퐴 �푠 �퐺 퐶𝑜 휓푠 Example

Let us calculate the depletion region width in a -type substrate when the band-bending is 0.5 . Assume the net acceptor density in the substrate is 5 × 10 . = 0.5 푝 15 −3 � 푠 𝑐 휓 � = =

2휖푠 �푠 2 퐾푠휖0 �푠 푑 푞 푁퐴 푞 푁퐴 푥 × �. × . × � × . = = 1.316 × 10 . × × ×−14 2 11 9 8 85 10 0 5 −19 15 −9 =� 3.1636 ×1010 5 cm10 = 0.363√ −5 μm

- 236 -

This result can be scaled for other values of by multiplying the result by and for other values of . �푠 푠 × 휓 0 5 by multiplying by . A useful number to remember is 1.15 μm at 1� volt drop across the 15 5 10 퐴 10 depletion푁 region for an� impurity푁퐴 concentration of . 15 −3 𝑐 Example

Let us calculate the gate voltage needed to bend the band by 0.5 under depletion in an MOS capacitor with an oxide thickness of 500 and a net acceptor density, = 5 × 10 . = 6.9 × � 10 / 15 −3 from one of the previous examples. 퐴 𝑜 −8 2 Å 푁 𝑐 퐶 퐹 𝑐 = + �2휖푠푞 푁퐴 �푠 퐺 푠 � × 퐶𝑜. × . × 휓 × . × × × × . = + 0.5 −14. × −19 15 √2 11 9 8 85 10 1 6 10 5 10 0 5 −8 = 0.92 6 9 10

�

To obtain an expression for in terms of , we rewrite equation (5.50) as a quadratic equation in 1 as 2 휓푠 �퐺 휓푠 + = 0 1 2 where = . Solving this quadratic휓푠 equation퐾 휓푠 − �퐺 �2휖푠푞 푁퐴 𝑜 퐾 퐶 ± = (5.51) 1 2 퐺 2 −퐾 �퐾 +4 푉 푠 2 Substituting this expression for 1 in the휓 quadratic equation and rearranging, we get 2 휓푠 ± = + 2 2 −퐾 퐾�퐾 +4푉퐺 퐺 푠 � 휓 ± 2 = + 2 2 4푉퐺 (5.52) −퐾 퐾 �1+ 퐾2 Since = 0 when = 0, we take only휓푠 the positive sign2 in front of the radical 푠 퐺 휓 � = + 1 + 1 (5.53) 2 퐾 4푉퐺 2 �퐺 휓푠 2 �� 퐾 − �

- 237 -

Or

= 1 + 1 (5.54) 2 퐾 4푉퐺 푠 퐺 2 휓 � − 2 �� 퐾 − �

Example

Let us calculate the surface potential of the MOS capacitor discussed in the previous example for a gate voltage of 0.7 V.

= 1 + 1 2 퐾 4푉퐺 푠 퐺 2 휓 � − 2 �� 퐾 − �

× . × . × × . × × × = = .−14× −19 15 �2휖푠푞 푁퐴 √2 11 9 8 85 10 1 6 10 5 10 −8 퐾 퐶 𝑜 = 0.59 6 9 10

. . = 0.7 √𝑉𝑉 1 + 1 2 . 0 59 2 8 2 푠 2 0 59 휓 − = 0�.35� − �

� Onset of Strong Inversion

We can now derive an expression for the gate voltage which corresponds to the onset of strong inversion. Onset of inversion occurs when = 2| |. We denote this value . When the gate voltage exceeds the value corresponding to the onset of inversion, strong inversion occurs. For this 푠 푏 푠 푖𝑖 reason, the gate voltage is called the turn-on휓 voltage∅, since the gate voltage has휓 to exceed this voltage to turn the device on. Denoting the threshold voltage as , we find

푻 푽 = + �2휖푠푞 푁퐴 �푠 푖𝑖 � 푠 푖𝑖 � 퐶𝑜 | | 휓 = + 2| | (5.55) �2휖푠푞 푁퐴2 ∅푏 We are assuming that at the onset of inversion, there퐶𝑜 is no contribution∅푏 to the space charge from electrons, and that the space-charge is essentially depletion charge.

- 238 -

Example

Let us determine the threshold voltage, , for the MOS capacitor discussed in the previous examples. �� = 5 × 10 15 −3 Hence 푁퐴 𝑐 = ln ln = 0.34 − 퐾� 푁퐴 퐾� 푁퐴 푏 푞 푛푖 푞 푛푖 ∅ − � =� 2≈| −| = 0.68� � − �

푠 푖𝑖 푏 휓= = ∅ + � �2휖푠푞 푁퐴 �푠 푖𝑖 � 푠 푖𝑖 × �. × . × ×퐶𝑜. × ×휓× × . = + 0.68 −14. × −19 15 √2 11 9 8 85 10 1 6 10 5 10 0 68 −8 =6 91.1710

� Strong Inversion

The space charge, under strong inversion, arise due to both the inversion charge (the minority carrier charge at the surface) and the depletion charge. Denoting the inversion charge per unit area of the surface as .

푄 푖 𝑖 = + (5.56) 2| | For a small increase, Δ , in the surface푄 potential푆� 푄 beyond푖𝑖 푄푑 (equal to ), the inversion charge density, , increase enormously since the minority carrier density, , increase exponentially with 푠 푠 푖𝑖 푏 surface potential. On the휓 other hand, the depletion charge density,휓 , increase∅ only slightly since it is 푄푖𝑖 푛푠 proportional to . The exact calculation of ( ) in strong inversion푑 is treated in advanced text 1 푄 books, and is beyond2 the scope of this book. Hence we will calculate ( ) using an approximation 푠 푆� 푠 called Depletion 휓Approximation. According to푄 this 휓approximation, in strong inversion the depletion 푆� 푠 region width, and hence the depletion charge density do not increase푄 once휓 inversion sets in i.e., they remain constant at a value they had at the on-set of inversion. i.e., corresponding to = = 2| |. This means that beyond the on-set of inversion the space-charge density, , increases only , 푠 푠 푖𝑖 due to an increase in the inversion charge density. Since the depletion region width does휓 not휓 increase 푏 푄푆� once∅ inversion sets in, the depletion region width is presumed to have attained a maximum value at the on-set of strong inversion and is denoted where

푑 푚𝑚 푥 | | = = (5.57)

2휖푠 �푠 푖𝑖 2휖푠2 ∅푏 푑 푚𝑚 The depletion layer charge density,푥 which also� does푞 푁not퐴 increase� once푞 푁 퐴inversion sets in, is denoted where

푄푑 푚𝑚 - 239 -

= = 2 2| | (5.58)

Therefore 푄푑 푚𝑚 −푞푁퐴푥푑 푚𝑚 − � 휖푠푞 푁퐴 ∅푏

= + = 2 2| | (5.59)

Under strong inversion,푄 푆� can be푄 푖written𝑖 푄as푑 푚𝑚 푄푖𝑖 − � 휖푠푞 푁퐴 ∅푏 푠 휓 = + = 2| | + (5.60)

As already stated, for a small 휓 푠 , 휓 increases푠 푖𝑖 enormously.훥 휓푠 As∅ 푏 is increased,훥 휓푠 the term - will 푄푖𝑖 푠 푖𝑖 퐺 increase enormously, while 훥 휓 will푄 change only slightly. Hence in �the depletion approximation,퐶𝑜 is neglected and at inversion is taken as 2| |. 훥 휓푠 훥 휓푠 푠 푏 휓 = + = ∅ + 2| | + 푄푆� 푄 푖𝑖 푄푑 푚𝑚 퐺 푠 푏 푠 � − 퐶𝑜 휓 − 퐶𝑜| −| 퐶𝑜 ∅ 훥 휓 = + + 2| | + (5.61) 푄 푖𝑖 �2휖푠푞 푁퐴2 ∅푏 Using Equation (5.55), we can− write퐶𝑜 퐶𝑜 ∅푏 훥 휓푠

= + + (5.62) 푄 푖𝑖 Hence we can write �퐺 − 퐶𝑜 �� 훥 휓푠

= + (5.63) 푄 푖𝑖 Therefore �퐺 − 퐶𝑜 ��

= ( ) (5.64)

This equation implies that the푄 푖inversion𝑖 − charge퐶𝑜 � density퐺 − � �is zero when . However, in reality it is not true, and there is some inversion charge density even when is equal to or slightly 퐺 � less than due to weak inversion. However, in the region of strong inversion,� i.e.,≤ � when > , the 퐺 � depletion approximation is reasonable and Equation (5.64) is valid. The inversion� charge density� � 퐺 � increases �linearly with the gate voltage. Since the inversion charge density is due to electrons� in �a -type substrate, the inversion charge density, , is also denoted . 푝 =푄 푖𝑖 = ( 푄푛 ) (5.65)

So far we discussed accumulation,푄푖𝑖 푄 depletion푛 − and퐶𝑜 inversion�퐺 − � using� a -type substrate as a reference. If we use an -type substrate, we will get similar expressions for the various parameters, excepting in accumulation we will have a sheet charge of electrons, and in 푝inversion a sheet charge of holes. Also the signs will푛 be opposite for different parameters. Table 1 summarizes the expressions for different parameters in both -type and -type substrates when the MOS capacitor is biased into

푝 푛

- 240 - accumulation, depletion or inversion. Referring to Table 1, it can be seen that the surface potential is of opposite sign in an -type substrate in comparison to the -type substrate.

푛 푝 Table 5.1 Difference between substrate types

Bias Parameter -type -type

e.s. potential, 푝 푛 = ln = ln − − 푏 퐾� 푁퐴 퐾� 푁퐷 ∅ − � 푖 � � 푖 � Surface Potential, 푞< 0 푛 푞 > 0 푛 Accumulation 푠 Space charge density,휓 = = = = 푝 𝑜 퐺 푛 𝑜 퐺 푄 ≈ − 퐶 � 푄 ≈ − 퐶 � 푆� 푎𝑎 푄 푄 Surface Potential, 2 > > 0 2 < < 0 Depletion 휓푠 − ∅푏 휓푠 − ∅푏 휓푠 Space charge density, = = 2 = = 2 | |

푆� 푑 푠 퐴 푠 푑 푠 퐷 푠 푄 푄 −� 휖 푞 푁 휓 푄 � 휖 푞 푁 휓 Surface Potential, = 2 = 2| | = 2 On-set of 휓푠 − ∅푏 ∅푏 − ∅푏 Inversion Space charge density, = = 2 2| | = = 2 2

푆� 푑 푚𝑚 푠 퐴 푏 푑 푚𝑚 푠 퐷 푏 푄 푄 − � 휖 푞 푁 ∅ 푄 � 휖 푞 푁 ∅ Surface Potential, > 2 but 2 < 2 but 2 1 1 푠 푏 푏 푏 푏 휓 − ∅ ≈ − ∅ − ∅ ≈ − ∅ Space charge density, = + = + Strong 푆� 푛 푑 푚𝑚 푝 푑 푚𝑚 Inversion 푄 푄 푄 푄 푄 Threshold Voltage, = + 2| | = + 2

� 푄푑 푚𝑚 푑 푚𝑚 � 푏 푄 푏 − 𝑜 ∅ − 𝑜 ∅ Inversion charge density, = = 퐶 ( ) = = 퐶 ( )

1- Depletion Approximation 푄푖𝑖 푄푛 − 퐶𝑜 �퐺 − �� 푄푝 − 퐶𝑜 �퐺 − ��

Non-Ideal MOS Device

Till now we discussed as ideal MOS Device. What do we mean by an ideal MOS device? An ideal MOS device is one in which there is no induced space charge in the semiconductor in the absence of the gate voltage. An non-ideal MOS device, on the other hand, is one in which there is a space-charge in the semiconductor even in the absence of a gate voltage. Real MOS device have a space charge in the semiconductor even when the gate voltage V is equal to zero. There are three causes for this non-ideal behavior, and they are: G - 241 -

1. Charges in the oxide 2. Difference in work function between the gate and the substrate 3. Surface state charges

We can apply a voltage on the gate that will reduce the induced space charge to zero, and we define that voltage as the flat-band voltage, . The reason for using the concept of the flat-band voltage is to describe the behavior of a non-ideal MOS device as though it were an ideal device with a bias voltage 푭� applied on the gate equal to the flat-band푽 voltage . For example the threshold voltage of a non-ideal MOS device can be written as �퐹� = + (5.66)

Actually, this relationship is not exactly�� 푛valid�푛− due푖𝑖𝑖 to the� presence� 푖𝑖𝑖 of surface�퐹� states, but for all practical purposes this can be taken as valid. Let us now determine the flat-band voltage arising from the three different sources of non-ideal behavior.

Charges in the Oxide

Due to the manner in which the silicon atoms are bonded to the oxygen atoms in the growth of the oxide layer in a thermal oxidation process, the oxide layer contains some positive charge. The positive charges in the oxide induced a negative space charge in the semiconductor, and the magnitude of the induced charge will depend on the relative location of the charge in the oxide between the metal and the substrate. Let the positive charge in the oxide have a density (x), and let it vary in some arbitrary fashion as shown in the Figure (5.11). In this figure we have chosen the origin, ( = 0), at the oxide-silicon interface, and the oxide-gate interface at = where휌 is the thickness of the oxide layer. Consider the charge in the oxide in an elementary region between and + . This푥 charge can 𝑜 𝑜 be thought of as a sheet charge of density ( ) . According푥 −푡 to the theory푡 of images, the sheet charge ( ) , located at a distance on the -axis from the oxide-silicon interface,푥 will푥 induce𝑑 charges both on the gate and in the semiconductor. The 휌charge푥 𝑑 induced in the semiconductor is proportional to the distance휌 푥 𝑑 between the sheet charge푥 and푥 the gate and similarly the charge induced in the gate is proportional to the distance between the sheet charge and the semiconductor. Hence it is possible to write the space charge induced in the semiconductor per unit area of the interface as

( ) ( ) ( ) = ( ) = (5.67) 푥− −푡𝑜 휌 푥 푡𝑜+푥 푑� In writing this equation,� we푄푆 must� note− 휌 that푥 𝑑 is negative푡𝑜 due to− our choice푡𝑜 of the origin of the -axis and can be shown to be the proportionality constant. The total induced space charge per unit area of the 푥 푥 1 interface due to the entire positive charge distribution in the oxide is then given by 푡𝑜 ( ) ( ) = = (5.68) 0 휌 푥 푡𝑜+푥 푑� 푆� 푆� 𝑜 𝑜 Let us now apply on 푄the gate a∫ voltage�푄 equal− to ∫−푡 . This voltage푡 will give rise to a charge, 푄푆� on the gate. According to Gauss’ theorem, all the electric field lines emanating from the positive charge푆� 퐶𝑜 푄

- 242 - in the oxide now terminate on the charges on the gate. The space-charge in the semiconductor becomes zero. Hence, the flat-band voltage can be expressed as the voltage necessary to apply on the gate such that the charge on the gate is . �퐹� 푆� =푄 푄푆� 퐹� 퐶𝑜 � = ( ) ( + ) (5.69) 1 0 퐶𝑜푡𝑜 −푡𝑜 𝑜 = − ∫ ( 휌) (푥 푡+ ) 푥 𝑑 (5.70) 1 0 We define an effective sheet charge− 퐶𝑜 density∫−푡𝑜 휌 in 푥the oxide푡𝑜 as 푥 𝑑, which is equal to 𝑜 = = ( ) ( 푄 + ) (5.71) 1 0 Hence the flat-band voltage푄 𝑜is then− given푄푆� by 푡𝑜 ∫−푡𝑜 휌 푥 푡𝑜 푥 �푥

= (5.71) 푄푆� �퐹� − 퐶𝑜

Figure (5.11): Am arbitrary charge density in the oxide.

Example

Let us now consider an example in which we assume that the oxide layer has a thickness of 1000 , and has a uniform charge density of 10 ions/ . Since ( ) is a constant, say equal to , (or ) can be evaluated taking outside15 the integral.3 Å 𝑐 휌 푥 휌0 푄푆� 푄𝑜 휌 - 243 -

( ) ( ) = = 0 휌 푥 푡𝑜+푥 푆� 𝑜 −푡𝑜 𝑜 푄 = − 푄 (− ∫+ ) 푡 𝑑 − 휌0 0 푡𝑜 ∫−푡𝑜 푡𝑜 푥 𝑑 = + = 2 0 휌0 푥 휌0푡𝑜 𝑜 In this example, becomes equal to − 푡𝑜 �푡 푥 2 �−푡𝑜 − 2

푄푆� × . × × × = = 0.8 × 10 / (5.73) 15 −19 −8 10 1 6 10 1000 10 −9 2 푆� 2 . × . × 푄 = − = = −3.45 × 10 � 𝑐/ (5.74) × −14 휖𝑜 퐾𝑜휖� 3 9 8 84 10 −8 2 −8 Hence 퐶𝑜 퐶𝑜 푡𝑜 1000 10 ≈ 퐹 𝑐 . × = = 0.023 (5.75) . × −9 −0 8 10 −8 Note that the flat-band voltage �due퐹� to the3 positive4 10 charge− in the oxide� is the same for - and - type substrates. It is negative since the charge in the oxide is positive. 푛 푝

Work Function Difference

Every material has a characteristic work function. Work function is defined as the difference between the Fermi energy and the surface barrier energy, where the surface barrier energy is the energy that an electron inside the solid must have in order to leave the solid and escape into vacuum. The surface barrier energy therefore, is also called the vacuum level, . This is illustrated in Figure (5.12). We denote the work-function of the solid as and is given by 퐸vac = 푞푞 (5.76)

The student should note that Φ, although푞 itΦ is in units퐸vac of− volt,퐸F is not the electrostatic potential.

Figure (5.12): Work-function of a solid

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Let us now consider two solids A and B with work functions and respectively. Let be less than . Let the two solids be not connected to each other as shown in Figure (5.13 A). Since the 퐴 퐵 퐵 two solids are not connected to each other, the vacuum levels for the푞Φ two solids푞Φ are at the same levelΦ 퐴 since an electronΦ that is excited with enough energy to overcome the surface barrier energy will have zero kinetic energy in vacuum in both solids. The Fermi energy is at a higher level in B than in A since it has a smaller work-function. If the materials A and B are connected to form a junction as shown in Figure (5.13 B), electrons will move from material B into material A so that the Fermi levels are lined up. Hence B will be positive with respect to A. A potential difference, = will arise between B and A. This potential difference is called the contact potential. ΦAB ΦA − ΦB

Figure (5.13): Two solids with different work-function: A): The solids are not connected to each other, and B): The solids are connected to each other.

When we consider the work function of a semiconductor, a similar definition follows. The vacuum level and the conduction and valence bands are shown in Figure (5.14). We notice that the location of the Fermi energy in the bandgap is dependent on the type and density of dopant (impurity) atoms. The work-function is therefore not a constant for a given semiconductor material and depends on whether it is or type material and also on the resistivity of the material. On the other hand, the difference in energy between the conduction band minimum, , and the vacuum level, , is a constant for a given푛 semiconductor푝 material. This difference is called electron affinity, and denoting � 𝑣� affinity as χ we have 퐸 퐸

= (5.77) 퐸푣𝑣−퐸푐 In a junction, - and -type semiconductor휒 materials푞 form a junction. This is equivalent to connecting a -type and an -type semiconductor. We saw that a potential difference, called the built-in voltage, , 푝exists� between푛 the 푝-region and the -region in a junction. The built-in voltage is just the contact potential푝 as shown푛 below: �푏� 푛 푝 푝� Let us denote the work-function of the -type semiconductor , and that of the -type semiconductor as . 푝 Φ1 푛 Φ 2 = = + (5.78)

1 𝑣� 퐹1 � 퐹1 푞Φ 퐸 − 퐸 - 245푞� - 퐸 − 퐸

Where is the Fermi level in the -type semiconductor. But

퐸퐹1 푝 = + = + (5.79) 퐸푔 퐸푔 Where is the bulk electrostatic퐸� − 퐸 potential퐹1 2 of the퐸 푖-type− 퐸 semiconductor.퐹1 2 푞� ∅Therefore푝� ∅푝 푝 = + + (5.80) 퐸푔 Similarly, it can be shown that 푞∅1 푞� 2 푞�∅푝�

= + (5.81) 퐸푔 where is the bulk electrostatic potential푞∅2 of푞 �the -2region.− 푞 ∅푛 ∅Hence푛 the work-function difference, =푛 is equal to

12 1 2 = Φ =Φ − Φ + = (5.82)

In an MOS device whenΦ 12there isΦ a1 difference− Φ2 between�∅푝� the∅푛 gate and�푏� the semiconductor work function, a potential difference arises. Let be the work function of the gate and be the work function of the substrate semiconductor. Therefore the substrate is positive with respect to the gate by 푀 푆 an amount equal to the contact potential Φ where = . This will thereforeΦ give rise to a space charge in the semiconductor. If we now apply a voltage , it will exactly balance out the 푀� 푀푆 푀 푆 contact potential difference, and hence noΦ space chargeΦ will beΦ induced− Φ in the substrate. We can 푀� therefore conclude that the flat-band voltage V required toΦ account for the work function difference is given by 퐹� V = = (5.83)

퐹� Φ푀� Φ푀 − Φ푆

Figure (5.14): Electron affinity and work function in a semiconductor

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Surface States (Interface States)

At the interface between the oxide and the substrate, there are a large number of states due to unfilled covalent bonds which are called dangling bonds. These states can be charged or neutral. These are called interface states or surface states. Usually these states are distributed continuously in energy in the band gap. Let us define as the number of states per unit area per unit energy interval at some energy level . Then in an energy level between and + , the number of states per unit area is 푖� . A surface state will be more퐷 positively charged if the Fermi energy is below it. On the other hand, it will be less퐸 positively charged if the Fermi𝑑 energy is퐸 above퐸 it. Thus,𝑑 the location of the Fermi energy in 푖� 퐷the𝑑 band gap at the surface determines the amount of charge in the surface states. Although the Fermi energy remains constant with distance, its location in the band gap at the surface changes as the band bending changes (i.e., as changes). Hence the amount of charge in the surface states changes. The charge in the surface states is located at the interface, and therefore is a sheet charge. If is the sheet 푠 charge density at the surface,휓 then is a function of . The sheet charge at the interface between 푠� the semiconductor and the oxide induces a space charge of density in the semiconductor.푄 Hence if 푠� 푠 we apply a charge on the gate equal푄 to , then the space휓 charge will be completely removed from 푠� the semiconductor. This is again based on the principle of Gauss’ theorem−푄 that the electric lines of force 푠� emanating from the surface state sheet− charge푄 now will terminate on the sheet charge on the gate. Hence we can write the flat band voltage as equal to 푠� 푠� 푄 −푄 �퐹� ( ) = (5.84) −푄푠� �푠=0 The total flat band voltage due to all the �three퐹� sources퐶 is𝑜 ( ) = + (5.85) 푄𝑜 푄푠� �푠= 0 On the other hand, the threshold� voltage퐹� − V 퐶of𝑜 a nonΦ-ideal푀� MOS− structure퐶𝑜 is � ( | |) V = V + 푄𝑜 푄푠� �푠 = 2 ∅푏 � 푛�푛−푖𝑖𝑖 � 푖𝑖𝑖 − 퐶𝑜 Φ푀� − 퐶𝑜 = V + (5.86) ∆푄푠� Where � 푖𝑖𝑖 �퐹� − 퐶𝑜

= (| | = 2| |) ( = 0) (5.87)

In most modern devices, ∆푄 is푠 �very small,푄푠� and휓푠 hence it∅ is푏 reasonable− 푄푠� to휓 express푠 the threshold voltage as ∆푄푠� V = V + V (5.88)

and neglect the term. � � 푖𝑖𝑖 퐹� ∆푄푠� − 퐶𝑜 Example

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Let us assume that the surface state density, , is a constant with energy (i.e., it has the same value throughout the band-gap) with a value equal to 5 × 10 . Let the oxide thickness be 1000 . 퐷푖� Let us assume a -type substrate with = 5 × 10 / 10 . Let−1 us −calculate2 the value for . The 푒푒 𝑐 Å 15 3 ∆푄푠� change in between flat-band and on-퐴set of inversion is equal to = 2| |. 푝 푁 𝑐 − 퐶𝑜 푠 × 푠 푏 휓 = = 0.0259 ln 훥 휓 =∅ 0.298 × 15 푘푘 푁퐴 5 10 10 ∅푏 − 푞 𝑙 �=푛푖 �2 × 0−.298 = 0.596�5 10 � −

is negative, since the bands bend훥 휓푠 downward at inversion (Fermi� energy rises in the band gap at the surface). ∆푄푠� | | = = �푠=2 ∅푏 ∆ 푄 푠 � = −1푞.6∫ �×푠= 010 푁×푖� 𝑑5 × 10−푞×푁0푖�.596훥 휓푠 −19 10 = − 4.77 × 10 −9 −2 Therefore − � 𝑐 4.77 × 10 = = 0.14 3.45 × 10−9 ∆푄푠� − −8 � In modern semiconductor devices, 퐶the𝑜 interface state density is lower by a factor of 10, and the oxide thickness is of the order of 200 . Hence is a couple of millivollts, and hence negligible. ∆푄푠� Å − 퐶𝑜 Capacitance Effect

In a parallel plate capacitor, which is formed by having a dielectric medium between two parallel metal plates, equal sheet charges but of opposite sign are induced on the two plates when a voltage is applied across this capacitor. The charge on the metal plate which is connected to the positive terminal of the voltage source has a positive sheet charge and that which is connected to the negative terminal has a negative sheet charge. The capacitance is determined only by the geometrical structure of the parallel plate capacitor. The capacitance is inversely proportional to the thickness of the dielectric material between the plates and proportional to the area of the plate. On the other hand, when an MOS capacitor is fabricated with silicon (the substrate) as one of the two parallel plates, the induced charge in the semiconductor is not a sheet charge but a space charge which under certain specific conditions can be approximated as a sheet charge as discussed earlier. The magnitude of the space charge is not directly proportional to the potential drop across the space charge region, (the surface potential). Hence, the capacitance in an MOS capacitor is voltage dependent. In this section we will examine the voltage dependence of the capacitance of an MOS capacitor. Let us first consider an ideal MOS capacitor.

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Ideal MOS Capacitor

Let us now assume that a DC voltage is applied on the gate of an MOS capacitor so as to bias the device in one of the three regimes of operation, i.e., accumulation, depletion or inversion. The 퐺 applied voltage is divided into a voltage, , across� the oxide layer and a potential drop, , across the space charge region. Let us assume that additionally a step voltage of a very small amplitude , is 𝑜 푠 applied in series with the DC voltage , �as shown in Figure (5.15). The applied voltage 휓 is also 퐺 distributed partially across the oxide and partially across the space charge region, giving us the�� 퐺 퐺 relationship � 𝑑

= + (5.89)

where and , are the incremental𝑑퐺 changes𝑑 𝑜in and𝑑 푠 respectively. gives rise to a small sheet charge on the gate electrode in addition to the already existing sheet charge due to the bias 𝑜 푠 𝑜 푠 퐺 voltage 𝑑. Similarly,𝑑 the space charge density in the semiconductor� 휓 increases 𝑑by 𝑑 where �퐺 𝑑푆�

= (5.90) | | We divide the above equation for by 𝑑푆� to obtain−𝑑 𝑑퐺 𝑑푆� = + (5.91) | | | | dV퐺 dV𝑜 푑푑푠 푆� 푆� The differential of the gate (sheet) charge푑� , with푑� respect to푑� , given by , denotes the ratio of the

푑� 퐺 incremental change in the gate charge 푄 to the incremental� change dV 퐺in the gate voltage, and is defined as the small signal capacitance C of the MOS capacitor of gate area equal to unity. 𝑑 푄 𝑑퐺

= = (5.92) dV퐺 1 1 푑푑 푑� 퐶 dV퐺

Figure (5.15): A small step voltage in series with a DC bias voltage V applied across an MOS capacitor

퐺

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The differential of with respect to similarly is equal to the capacitance of a parallel plate capacitor with the oxide layer between two metallic plates. Denoting this as the oxide capacitance per unit area, 𝑜 C , we obtain 푄 �

ox = = (5.93) | | dV𝑜 dV𝑜 1 where 푑�푆� 푑� 퐶𝑜

= (5.94) 휖𝑜 The differential of | | with respect to is defined퐶𝑜 as th푡e𝑜 space charge capacitance and is given by 𝑑푆� 휓푠 퐶푆� = (5.95) | | 푑푑푠 1 Therefore 푑�푆� 퐶푆�

= + (5.96) 1 1 1 The small signal equivalent capacitance of an MOS퐶 capacitor퐶𝑜 퐶푆 of� unit area can therefore be described as a series combination of two capacitors, one , and the other , and this is illustrated in Figure (5.16). Instead of an incremental voltage , let us now apply a small sinusoidal voltage in series with the DC 𝑜 푆� voltage across the MOS capacitor, as shown퐶 in Figure (5.17). Due퐶 to the small signal voltage , a small 퐺 signal current flows. By measuring𝑑 the ratio of and , the capacitance can be 푣determined� from the relation 푣� 횤̃ 푣� 횤̃ 퐶 = (5.97)

1 where is 2 time the frequency of the small퐶 signal 푗voltage� �� . If is the component of space charge that varies sinusoidally in step with , then 휔 휋 푣� 푄� 푣� = (5.98)

푄� The measured capacitance is 퐶 ��

퐶 = (5.99)

퐶𝑜 퐶푆� where the area of the gate is assumed to be unity,퐶 퐶𝑜 is+ the퐶푆� oxide capacitance per unit area, and is equal to space charge capacitance per unit area. If we were to consider a MOS capacitor of area , then 𝑜 푆� the measured capacitance, is equal to × where퐶 is the capacitance per unit area. 퐶 퐴 푚 퐶 퐴 퐶= 퐶 (5.100)

푄� 푄�푎𝑎

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Figure (5.16) Equivalent circuit of an MOS capacitor

Figure (5.17): A sinusoidal voltage in series with a DC bias voltage applied across an MOS capacitor

Let us now evaluate the small signal capacitance for different regimes. Accumulation

The charge distribution in the MOS device under accumulation conditions can be visualized as shown in Figure (5.18). Let us assume that = 0 represents the interface between the oxide and silicon. Let the interface between the gate and the oxide be located a = . 푥 𝑜 The accumulation charge is located at = 0 as shown in Figure푥 (5.18).− 푡 is a small component of the total sheet charge which varies in step with the applied small signal voltage . 푥 푄�푎𝑎 | | 푣� = = (5.101) 푑�푆� 푑�푎𝑎 Since changes enormously for a small퐶푆� change푑푑 in푠 , 푑푑 is푠 very large. Hence

푎𝑎 푠 푆� 푄 = + 휓 =퐶 = (5.102) 1 1 1 1 1 푡𝑜 퐶 퐶𝑜 퐶푆� ≈ 퐶 퐶𝑜 휖𝑜 - 251 -

The above equation shows that the measured capacitance of the MOS capacitor in the accumulation regime of operation is essentially equal to the oxide capacitance. The equivalent circuit is just a simple capacitor as shown in Figure (5.19)

퐶𝑜

Figure (5.18): Charge distribution in the MOS capacitor under accumulation conditions.

Figure (5.19): Equivalent circuit of an MOS capacitor when the surface is accumulated. The MOS capacitor has a capacitance equal to in accumulation. Depletion 퐶𝑜

The space charge under depletion conditions is the charge in the depletion region in the semiconductor. The applied elementary sinusoidal voltage induces a sinusoidally varying depletion charge in the semiconductor. The sinusoidal depletion charge variation occurs at the edge of the depletion region, as shown in Figure (5.20) The space charge capacitance is the depletion region capacitance which is equal to

= = (5.103) 휖푠 where is the depletion region width. The 퐶equivalent푆� 퐶푑 circuit푥 푑is shown in Figure (5.21). Since the depletion region capacitance is comparable to or less than the oxide capacitance, the net capacitance of 푑 the series푥 combination of and is less tha . As the gate voltage is increased so as to induce a larger space charge region, the width of the depletion region increases, and therefore the space 푆� 𝑜 𝑜 charge capacitance (=퐶 ) decreases.퐶 Hence퐶 the capacitance of the MOS capacitor decreases as 푑 the gate voltage is increased. 푥 퐶푆� 퐶푑

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Figure (5.20): Charge distribution in the MOS capacitor under depletion conditions.

Figure (5.21): Equivalent circuit of an MOS capacitor when the surface is depleted. Inversion

When the gate voltage is increased to a value larger than the threshold voltage, we saw earlier that the space charge is due to both the inversion charge and the depletion charge. Due to the applied voltage , an increase in the inversion charge occurs if minority carriers can be generated fast enough in the depletion region. Then the inversion charge has a component that varies sinusoidally in step with 퐺 the small� �signal voltage . For the inversion charge to change in step with applied voltage, the sinusoidal voltage should change slowly with time. The period of oscillation of the sinusoidal voltage has to be 푣� much larger than the minority carrier generation lifetime (i.e., ). The sinusoidal change in the < 푣� 1 푔 space charge density therefore is equal to 휏 휔 휏푔 =

and occurs at the silicon surface as shown in 푄Figure�푆� (5.22).푄�푖𝑖 The range of frequencies for which < is 1 called the low frequency region. 휔 휏푔

= = 푆� 푖𝑖 푆� �푄 �푄 퐶 푠 푠 𝑑- 253 - 𝑑

Figure (5.22): Low frequency charger distribution in the MOS capacitor under inversion conditions.

Hence it is large since the inversion charge increases enormously for a small change in . At low frequencies the MOS device behaves like a parallel plate capacitor with an oxide dielectric between the 푠 two parallel plates since the sinusoidal change occur on the gate and in the semiconductor휓 at the oxide interface. The capacitance of the MOS capacitor under these conditions is the same as the oxide capacitance. The equivalent circuit under inversion conditions is shown in Figure (5.23).

Figure (5.23): Low frequency equivalent circuit of an MOS capacitor under inversion conditions

If sufficient time is not allowed for minority carriers to be generated, the inversion charge is not able to change in step with applied voltage and hence the incremental change in the space charge density occurs at the edge of the depletion region adjacent to the neutral region of the semiconductor as shown in Figure (5.24). In other words, if the period of the sinusoidal voltage is much less than the minority carrier generation lifetime , then the elementary sinusoidal change in the space charge 푣� occurs at the edge of the depletion region. The range of frequencies for which > is called the high 휏푔 𝑑 1 frequency region. The measured capacitance at high frequencies therefore will휔 be the휏푔 oxide capacitance in series with the depletion capacitance corresponding to a depletion region with a maximum width, . Figure (5.25) shows the equivalent circuit of the MOS capacitor under high frequency conditions. Due to the different responses of the space charge density at low and high frequencies, the space 푑 푚푎푥 charge푥 capacitance will have different values at low and high frequencies.

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Figure 5.24 High frequency charge distribution in an MOS capacitor under inversion conditions.

Figure 5.25 High frequency equivalent circuit of an MOS capacitor under inversion conditions.

C-V Curve

The plot of a small signal capacitance of an MOS capacitor as a function of the DC bias (gate) voltage is called a capacitance-voltage (C-V) curve. While the C-V curve will be the same for both high and low frequencies in the accumulation and depletion regions, there will be a difference in the inversion region behavior. At intermediate frequencies, the C-V curve will lie between the high frequency and the low frequency curves, under inversion conditions. This is illustrated in Figure (5.26). is the value of the capacitance measured in inversion under high frequency conditions, the oxide capacitance. At intermediate frequencies, the capacitance measured in inversion region has a 푚𝑚 𝑜 퐶value intermediate between and . 퐶

퐶푚𝑚 퐶𝑜

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Figure (5.26): The capacitance-voltage (C-V) characteristics of an MOS capacitor at low, intermediate and high measurement frequencies.

Deep Depletion

Let us now assume that a step voltage of amplitude (> ) is applied on the gate at time = 0. Let us also apply a small sinusoidal voltage in series with the step voltage as shown in Figure 퐺 � (5.27). The purpose of is to measure the small signal capacitance,� 𝑑 and it is therefore called the probing 퐺 푡signal. Immediately after the application of the step푣� voltage, that is at = 0 , the �induced space charge 푣� will essentially be due to the depletion charge. This is because there is no time+ for the inversion charge to be generated by thermal generation of minority carriers. The depletion푡 charge is increased beyond the maximum depletion charge obtained under steady state conditions. The space charge region under this condition represents the non-equilibrium condition. At = 0 , the depletion region widens to a large value which is much larger than . Hence, the condition+ is called “deep depletion.” The 푡 generation and+ recombination rates in the space charge are not what wll be obtained in thermal 푑0 푑 푚푎푥 equilibrium,푥 and do not balance each other 푥out as required by the law of mass action. Minority carriers are generated in the space charge region in excess of what are lost due to recombination. These minority carriers will be driven towards the interface, where they will form the inversion charge, while the majority carriers proceed in the opposite direction. Part of the majority carriers generated will neutralize the ionized impurities at the edge of the depletion region, thereby reducing the depletion region width while the rest of the majority carriers will be flown out of the ohmic contact. It must be pointed out that the increase in the inversion charge is more than the reduction in the depletion charge. As time progresses, the inversion charge increases and the depletion charge decreases correspondingly. Ultimately at = , a depletion width of is obtained, and the inversion charge and the depletion charge attain steady state values. The space charge region now is in thermal equilibrium. The 푑 푚푎� space-charge 푡region∞ thus exhibits a transient푥 behavior before steady state conditions are reached. Therefore the capacitance also exhibits a transient behavior. The transient response of the capacitance is shown in Figure (5.28). In deep depletion, (i.e., at = 0 ), there is no inversion charge and there is + 푡 - 256 - only depletion charge. Hence, the surface potential (0 ) is related to the step voltage by the relationship + 휓푠 �퐺 ( ) ( ) = + (0 ) = + (0 ) (5.104) + + 푄푑 0 + �2휖푠푞푁퐴�푠 0 + This can be solved to� yield,퐺 as− was퐶 𝑜done earlier,휓푠 퐶𝑜 휓푠

(0 ) = + 1 1 (5.105) 2 + 퐾 4푉퐺 푠 퐺 2 where K is defined as equal to 휓 � − 2 �� 퐾 − �

= (5.106) �2휖푠푞푁퐴 The band-bending at = 0 is shown in Figure퐾 (5.29).퐶𝑜 The surface potential represents a potential well at = 0 for an electron+ (remember we are assuming a -type substrate), of depth equal to (0 ). 푡 휓푠 This potential well can be used to store electrons. This storage can be done only momentarily due to the+ 푠 transient푥 nature of deep depletion. By placing a large number푝 of MOS capacitors at close spacing휓 and by pulsing these in a suitable sequence, the charge stored in the potential well under these capacitors can be successfully transferred from one to the next all the way to the end of the string of capacitors. The string of closely spaced MOS capacitors is called the charge couple device (CCD). The charge coupled device has many applications in digital, analog and imaging circuits.

The capacitance at = 0 is very small since the depletion region width is large and hence at = 0 is small. Hence is also small.+ As time increases from = 0 , decreases and the depletion 푡 퐶푑 region+ width decreases and increases. Therefore the capacitance+ of the MOS capacitor also increases 푠 푡and finally reaches a steady퐶 state value corresponding to =푡 휓 after a long time. This is called 푑 the transient behavior of the퐶 MOS capacitor. 푥푑 푥푑 푚𝑚

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Figure 5.27: Application of a large step voltage on the gate of an MOS capacitor to produce deep depletion.

Figure 5.28 Transient behavior of the capacitance of an MOS capacitor.

Figure 5.29 Band-bending in deep depletion.

MOS Transistors

The MOS transistor is a device which is based on the principle that a conducting channel is induced in the surface of a semiconductor by an electric field of suitable polarity and magnitude. In our study of MOS capacitors, we saw that when we apply a gate voltage larger than the threshold voltage, the surface gets inverted. The inversion layer at the surface can provide a conducting path, and therefore can carry electric current between two regions which are doped opposite to the substrate. As an example, consider a MOS structure on a -type substrate as shown in Figure (5.30). Two regions, one called the source and ther other called the drain are formed by diffusing (or implanting) donor+ atoms. A thin oxide layer called the gate oxide푝 is grown on the surface of the substrate between푛 the source and the drain. A gate electrode extending from the edge of the source region to the edge of the drain region is deposited on top of the gate oxide layer. When the voltage applied on the gate is less than the threshold voltage , no conducting channel exists between the source and the drain. The path between the source and the drain comprises two - junctions connected back-to-back. When the � � + 푛 푝 - 258 - gate voltage is made larger than , an inversion layer of electrons provides a conducting channel between the source and the drain. The conductance is dependent on the inversion charge density, and 퐺 � hence on the� gate voltage . Thus we� are able to control the conductance (and therefore the resistance) between the source and the drain through the application of a voltage on the gate. The 퐺 device is therefore called a� transistor.

The particular structure which we discussed is called an -channel device since the charge carriers in the conducting channel are electrons. If we had started with an -type substrate and 푛 diffusions for the source and the drain, the resulting device would have been called a -channel device.+ The drain current, I , versus the gate voltage, V , characteristic is called the푛 transfer characteristic푝 , and it is shown in Figure (5.31) for - and - channel devices. For the -channel device when푝 the gate D G voltage is less than no current flows, and when is larger than the current increases with . We will later on see that this increase푛 푝is linear for small drain voltages.푛 In the case of the - channel 퐺 � 퐺 � 퐺 device, the� threshold voltage� is negative, and hence when� the magnitude� of the negative gate voltage� is larger than the magnitude of the threshold voltage, conduction results in the -channel device.푝 The symbols for the two types of devices are also shown in Figure (5.31). The direction of the arrow in the substrate lead is to indicate whether it is a - channel or an - channel device,푝 and the convention is the same as in a junction. The direction of the arrow is from the to the -region. In the case of the - channel device, the substrate is and the channel푝 is . Therefore푛 the arrow is inward. In the case of the -channel, the푝 푛substrate is -type and the channel is -type, and푝 therefore푛 the arrow is outward. 푛 푝 푛 푝 Let us now calculate푛 the drain current. Assume푝 that the length of the channel between the source and the drain lies along the axis with the source at = 0, and the drain at = . Let be the width of the channel. Then the current, , in the channel is given by 푦 푦 푦 퐿 푊 퐽 = (5.107) 푑퐸퐹 where is the mobility of the electrons, 퐽 is the휇 electron푛푛 푑� density, and is the Fermi energy. The current density is in units of amperes per square centimeter. Let us assume that a voltage between 푛 퐹 the source휇 and the drain. At any point, the푛 voltage drop along the channel퐸 measured with respect to the 퐽 �퐷 source will be denoted ( ). The gradient of the voltage along the channel is . The gradient of Fermi 푑� energy is related to the gradient of the voltage drop along the channel through the relationship � 푦 푑�

= (5.108) 푑퐸퐹 푑� We can therefore write the current density as푑� equal− to푞 푑�

= (5.109) 푑� Let be the thickness of the inversion layer퐽 of −electrons.푞휇푛 푛 Furthermore,푑푦 let us assume that the electron density, , is constant in the inversion layer perpendicular to the channel. Then the drain current is 퐼 equal푥 to the current density, , multiplied by W , and is given by 푛 퐼퐷 퐼 퐽 = 푥 (5.110) 푑� 퐼퐷 − 푞휇푛 푛 푊푥퐼 푑� - 259 -

Consider a small section of the inversion layer a shown in Figure (5.32). Let us now describe a rectangular strip of length , contained between and + . The charge contained in this rectangular strip is equal to since is the electron (inversion) charge per unit area of the surface. But this is also equal𝑑 to 푦 푦 𝑑 푄푛푊 𝑑 푄푛 = (5.111)

Hence 푄푛푊 𝑑 − 푞 푛 푊푥퐼 𝑑 = (5.112)

Using this expression for we can express푄 푛the drain− 푞 current푛 푥퐼 as 푛 푄 = (5.113) 푑� 퐼퐷 휇푛 푄푛 푊 푑�

Figure (5.30) A schematic representation of the n-channel MOS transistor.

Figure (5-31) The transfer characteristics and the symbol for a) n-channel device and b) p-channel device.

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Figure (5-32) A section of the inversion layer between and +

We will now distinguish three cases: 푦 푦 𝑑 Case 1: Small

Let the source�퐷 and the substrate be connected together to the ground. We will assume that the drain voltage is small. Therefore it is reasonable to assume that the inversion charge density is the same everywhere in the channel i.e., does not vary with . The inversion charge density is given by �퐷 = = (푦 ) 푄 푖 𝑖 (5.114)

where is the oxide capacitance푄푖𝑖 per unit푄 푛area. −The퐶𝑜 inversion�퐺 − layer�� now looks like a rectangular sheet of electron charge of dimensions , and as shown in Figure (5.33). Therefore the drain current is 𝑜 given by퐶 푊 퐿 푥퐼 = ( ) (5.115) 푑� 퐷 푛 𝑜 퐺 � We can now approximate the gradient퐼 − 휇as equal푊퐶 to � since− � the 푑drain� voltage is very small. 푑� 푉퐷 퐷 Hence 푑� 퐿 �

= ( ) (5.116) 푊 We see therefore that the drain current퐼퐷 −varies휇푛 linearly퐿 퐶𝑜 with� 퐺the− drain�� voltage�퐷 and also linearly with the gate voltage.

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Figure (5-33) Approximation of the inversion layer as a sheet charge.

Case 2: Large (Simplified Treatment)

We will now�퐷 consider the case when the drain voltage, is large. We present a simplified treatment. Since the drain voltage is large, it is no longer possible�퐷 to assume that the gradient is 푑� constant everywhere between the source퐷 and the drain. Due to ( ), the inversion charge density will � 푑� be different for different values of . In the early days, it was assumed that a threshold voltage ( ) at some value of was related to at = 0 at the source by the �relation푦 푦 �� 푦 푦 � � 푦( ) = ( = 0) + ( ) (5.117)

At the drain therefore, the threshold�� voltage푦 will�� be푦 given by � 푦 ( = ) = ( = 0) + (5.118)

Hence the inversion charge density,�� 푦, according퐿 to� Equation� 푦 (5.114)�퐷 is at a maximum value at the source, decreases along the channel, and is at minimum value at the drain. Another way of looking at 푛 this is that the electric field between 푄the gate and the substrate is large at the source and decreases as one proceeds towards the drain. Hence the induced inversion charge gets smaller as you approach the drain. The inversion charge density ( ), at some point in the channel is therefore given by

( ) = [ 푄푛 푦( )] = [푦 ( = 0) ( )] (5.119)

Substituting푄푛 this푦 value− of퐶 𝑜 in� 퐺the− expression�� 푦 for −drain퐶𝑜 current�퐺 − given�� 푦before, we− get� 푦 푛 =푄 [ ( = 0) ( )] (5.120) 푑� Multiplying both sides퐼퐷 of the− above휇푛푊 equation퐶𝑜 �퐺 by− � and� 푦 integrating,− � we푦 obtain푑� 𝑑 = [ ( = 0) ( )] 푉퐷 퐼퐷퐿 − � 휇푛푊퐶𝑜 �퐺 − �� 푦 − � 푦 𝑑 = 0 ( = 0) (5.121) 푉퐷 Dividing both sides of the equation− 휇푛푊 by퐶 𝑜, the�� drain퐺 − current�� 푦 is obtained− 2 as� � 퐷 퐿 = ( = 0) (5.122) 휇푛 푊퐶𝑜 푉퐷 Writing the threshold퐼 퐷voltage− at the 퐿source� �퐺=−0 as�� 푦 , we obtain− 2 �as� equal퐷 to 푦 �� 퐼퐷 = (5.123) 휇푛 푊퐶𝑜 푉퐷 At the drain, is equal to ,퐼 퐷and therefore− 퐿 at ��=퐺 − is �equal� − to2 � �퐷 푦 퐿 푄푛 푦 퐿 - 262 -

( = ) = [ ] (5.124)

The inversion charge density푄 at푛 the푦 drain퐿 decreases,− 퐶𝑜 as �퐺is− increased,�� − �퐷 and at some value of denoted , the inversion charge density ( = ), becomes zero. at the drain becomes zero when �퐷 �퐷 퐷 푠𝑠 푛 푛 � 푄 =푦 퐿 = 푄 (5.125)

At this value of the drain voltage, the�퐷 channel�퐷 is푠𝑠 said to� be퐺 −“pinched�� -off” at the drain, and hence is also denoted . We denote the drain current at equal to , as . Substituting for in �퐷 푠𝑠 Equation (5.122), we get �푝 �퐷 �퐷 푠𝑠 퐼퐷 푠𝑠 �퐷 푠𝑠 �퐷

= = ( ) (5.126)

휇푛 푊퐶𝑜 푉퐷 푠𝑠 휇푛 푊퐶𝑜 2 When퐼퐷 푠 the𝑠 channel− is pinched퐿 �� off퐺 −at the�� drain,− 2 the� drain�퐷 푠 𝑠current− is proportional2 퐿 � to퐺 the− �square� of the gate voltage as shown in Equation (5.126). If is increased beyond , the channel is pinched off at some other value of < . In other words, the pinch off point moves towards the source. The voltage �퐷 �퐷 푠𝑠 at the pinch-off point is now equal to . Another way of looking at this is at the pinch off point, the gate voltage is equal푦 to퐿 the local threshold voltage. Hence the inversion charge density is zero. 푦 �퐷 푠𝑠 When > , let be the distance of the pinch-off point from the source as shown in Figure (5.34). The region between′ = and = , has a voltage drop . This region is a �퐷 �퐷 푠𝑠 퐿 depletion region. Although the channel is′ pinched off at = , the current through the device is what 푦 퐿 푦 퐿 �퐷 − �퐷 푠𝑠 would flow in a device with = and of length , since′ the integration carried out in Equation 푦 퐿 (5.121) to obtain is now carried out between the source′ and the pinch off point. In other words, the 퐷 퐷 푠𝑠 current is given by � � 퐿 퐼퐷

( > ) = ( ) (5.127)

휇푛 푊퐶𝑜 2 ′ In the olden days, devices퐼퐷 were�퐷 fabricated�퐷 푠𝑠 with very− large2 퐿 , and� hence퐺 − �� was negligible in comparison with . Hence at voltages greater than was essentially the′ same as what was 퐿 퐿 − 퐿 obtained at = . It is though the drain current saturates to a constant value . 퐿 퐼퐷 �퐷 푠𝑠 퐷 퐷 푠𝑠 퐷 푠𝑠 � � ( > ) = = ( ) 퐼 (5.128)

휇푛 푊퐶𝑜 2 ′ However, in modern퐼퐷 short�퐷 channel�퐷 푠devices𝑠 the퐼퐷 distance푠𝑠 −( 2 )퐿 between�퐺 the− � pinch� off point and the drain is not negligible in comparison with . Hence the drain current′ does not remain constant with the drain voltage, and increases with the drain voltage in saturation.퐿 − 퐿 The drain voltage drain current characteristic is shown in Figure (5.35). The퐿 drain conductance g is defined as

D = | = (5.129) 휕�퐷 and the transconductance is define푔퐷 d as 휕�퐷 �퐺 𝑐𝑐𝑐𝑐 푔푚 = | = (5.130) 휕�퐷 푔푚 휕�퐺 �퐷 𝑐𝑐𝑐𝑐

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Figure (5.34) Schematics representation of the channel pinch-off.

Figure (5.35) The drain current/drain voltage characteristics, called characteristics, of an n- channel MOS transistor. 퐼퐷 − �퐷 We will call the region of operation when < as linear, and the region when > as the saturation region. This is shown in the table given below. Thus we see that as the �퐷 �퐷 푠𝑠 drain voltage is increased, the current deviates from a linear behavior and ultimately saturates to . �퐷 �퐷 푠𝑠 Linear Saturation 퐼퐷 푠𝑠

0 ( )

푛 𝑜 푔퐷 휇 푊퐶 ≈ �퐺 − �� 퐿

( )

푛 𝑜 푛 𝑜 푔푚 휇 푊퐶 휇 푊퐶 �퐷 �퐺 − �� 퐿 퐿 Case 3: Large ( A More Exact Treatment)

�퐷 - 264 -

The assumption made in Case 2 that the threshold voltage at any point in the channel increases from its value at the source by the amount ( ) presupposes that the depletion region width does not change with the voltage drop ( ). In other words, it is implicit in this푦 assumption that the depletion region width is the same from the source to� the푦 drain independent of ( ). This is not correct. If we refer to Figure (5.36) we will notice� that푦 due to the voltage drop ( ) the Fermi level at point is decreased from the Fermi level in the bulk by an amount equal to ( ) and therefore� 푦 the band has to be bent by ( ( ) + 2| |) to produce the onset of inversion. Hence� the푦 condition for the inversion푦 to set in at is dictated by ( ). The band-bending for the onset of inversion푞� 푦 at is given by 푞 � 푦 ∅푏 푦 � 푦 ( ) = ( ) + 2| | 푦 (5.131)

Under this condition, the depletion휓 region푠 푖𝑖 width푦 at� is푦 given by ∅푏 푦 ( ) = [ ( ) + 2| |] (5.132)

2 휖푠 푑 푏 And the depletion charge density푥 at 푦 is given� by푞 푁 퐴 � 푦 ∅

( ) = 푦 ( ) = 2 [ ( ) + 2| |] (5.133)

Now if we were to푄 write푑 푦 the expression−푞 푁퐴 푥for푑 gate푦 voltage−� 푞in 푁terms퐴 휖of푠 the� space푦 charge∅ 푏density and , we get 휓푠 ( ) = + ( ) = + ( ) (5.134) 푄푠� 푦 �퐺 �𝑜 휓푠 푦 − 퐶𝑜 휓푠 푦

Figure (5.36): Band-bending at when the voltage drop is ( ).

We will now use the depletion approximation푦 according to which the band� 푦 bending in strong inversion is the same as the band bending at the onset of inversion. That is, the surface potential continues to remain at

( ) = ( ) + 2| | (5.135)

even in strong inversion. Using the above휓푠 two푦 equations,� 푦 we find∅ that푏 the total space-charge density is given by

( ) = [ ( )] = [ ( ) 2| |] (5.136)

푄푠� 푦 − 퐶𝑜 �퐺 − 휓푠 푦 - 265 -− 퐶𝑜 �퐺 − � 푦 − ∅푏

The inversion charge density is the difference between the total space charge density and the depletion charge density. Hence, the inversion charge density is given by

( ) = ( ) ( )

푄푛 푦 푄푠� 푦 − 푄푑2 푦 = ( ) 2| | ( ( ) + 2| |) 퐴 푠 1 �푞 푁 휖 2 − 퐶𝑜 ��퐺 − � 푦 − ∅푏 − � 푦 ∅푏 � If we substitute this in the expression for the drain current which퐶𝑜 is

= ( ) 𝑑 퐼퐷 휇푛 푊 푄푛 푦 and then integrate after multiplying both sides by , we obtain𝑑 푑� 퐿 = 2| | ( + 2| |) 3 푛 𝑜 퐷 퐴 푠 휇 퐶 푊 푉 2 �푞 푁 2 휖 2 (2퐷| |) 퐿 퐺 2 푏 퐷 3 퐶𝑜 퐷 푏 퐼 −3 � � � − − ∅ � � − � � ∅ − (5.137) 2 ∅푏 �� We could have arrived at the same result by determining the local threshold voltage ( ) as

[ ( ) | |] �� 푦 ( ) = + ( ) + 2| | (5.138) �푞 푁퐴2 휖푠 푉 푦 + 2 ∅푏 and obtaining the inversion�� 푦 charge density퐶 as𝑜 � 푦 ∅푏

( ) = [ ( )] (5.139)

This expression for the drain current푄푛 is푦 valid as− long퐶𝑜 as the�퐺 channel− �� is푦 not pinched off due to a high drain voltage. The channel is pinched off at = if the gate voltage is equal to the threshold voltage at = . therefore can be obtain from Equation (5.138) by putting ( = ) as and ( = ) as 푦 퐿 푦 . Hence solving for (left as an exercise for the student), we get 퐿 �퐷 푠𝑠 �� 푦 퐿 �퐺 � 푦 퐿 �퐷 푠𝑠 �퐷 푠𝑠 = + 1 1 2| | (5.140) 2 퐾 4푉퐺 퐷 푠𝑠 퐷 2 푏 where � � − 2 �� 퐾 − � − ∅

= (5.141) �2휖푠푞푁퐴 K is called the body factor, and is also sometimes퐾 denoted퐶𝑜 .

The expression for the drain current given in Equation휸 (5.137) is valid in strong inversion regions ( > ). This expression is modified in the presence of a substrate bias voltage. If a reverse 퐼퐷 �퐺 �� - 266 - bias voltage is applied between the source and the body, then the amount of band bending needed at the source ( = 0) for the onset of inversion is �푆� 푦 ( = 0) = + 2| | (5.142)

The band-bending required at the onset휓푠 푦of inversion at�푆 some� point∅ 푏 in the channel is therefore equal to 푦 ( ) = + ( ) + 2| | (5.143)

Hence the expression for the drain휓 current푠 푦 in the�푆 presence� � of푦 a substrate∅푏 bias is modified such that 2| | is replaced by + 2| | and is therefore given by

∅푏 �푆퐵 ∅푏 = 2| | ( + +

휇푛 퐶𝑜 푊 푉퐷� 2 2퐷| |) ( + 2퐺| |) 푆� 푏 퐷� 퐷� 푆� 퐼 −3 퐿 �� � − 32 − � − ∅ � � − 3 훾 � � � (5.144) 2 2 ∅푏 − �푆� ∅푏 � � where is the drain to source voltage, is measured with respect to the substrate, and is equal to given in Equation (5.141) �퐷� �퐺 훾 퐾

Measurement of

The threshold �voltage,� can be obtained by measuring the drain current as a function of the gate voltage for a small value of drain voltage. When the drain current is plotted as a function of , the characteristics shown in Figure (5.37) are obtained. If we now extrapolate the linear portion to intercept 퐺 the voltage axis, the intercept on the voltage axis is the threshold voltage. �

We note that does not go to zero when the gate voltage, is equal to . This is because the inversion charge density does not actually go to zero at this voltage. In fact, a small drain current 퐷 퐺 � flows through the device,퐼 even when the gate voltage is less than the� threshold voltage� under weak inversion conditions. This region of operation is called the subthreshold region. The calculation of the drain current in the subthreshold region is beyond the scope of this course. Hence we will assume that the drain current is zero at = .

Equivalent Circuit �퐺 ��

We can draw the equivalent circuit for the MOS transistor. This is illustrated in Figure (5.38). The input voltage is applied between the gate and the source, and the output current is taken between the drain and the source. and are the capacitance and the conductance that the input voltage source looks into. In our case, is given by 퐶푖� 퐺푖� 푖� 퐶 = (5.145)

푑 푄퐺 퐶푖� 푑 푉퐺 ≈ 푊 퐿 퐶𝑜

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The value of is obtained in strong inversion. The input conductance, , is essentially due to leakage current in the oxide. The feedback capacitance between the drain and the gate, , is the sum 퐶푖� 퐺푖� of the gate overlap capacitance and the fringing field capacitance. The overlap capacitance arises 퐶푓� because the gate electrode extends beyond the length of the channel over the source and the drain region. This fringing field capacitance arises due to edge effects. The output conductance is given by

푔퐷 = (5.146)

훿 퐼퐷 퐷 퐷 The transconductance is given by 푔 훿 푉 푚 푔 =

훿 퐼퐷 The output capacitance is the capacitance between푔푚 the훿 drain푉퐺 and the source region, and that is essentially equal to the capacitances of the source-to-substrate -junction and the drain to substrate -junction. 푝� 푝� We can now calculate the maximum frequency up to which the MOS transistor can be used. The constant current source in the output circuit is equal to . This constant current source therefore, is directly related to the voltage across the conductance , that is in other words, the current through 푔푚 푣�푔 the conductance ,. Therefore as the frequency is increased, the input current flows more and more 퐺푖� through the input capacitance ,, and less through the conductance . We can now calculate the 퐺푖� gain of the device at any given frequency, just as we did for the bipolar transistor in the last chapter. The 퐶푖� 퐺푖� maximum frequency is defined as that at which the current through the input capacitance, , is equal to . The current through is given by . Equating this to we get 퐶푖� 푚 퐺 푖� 푔 푖� 푚 푔 푔 � 퐶 푣� =푗� 퐶 푔 푣 � (5.147)

푔푚 This now leads to the maximum frequency휔 as푚 𝑚equal to퐶 푖�

= (5.148)

푔푚 푚𝑚 The ratio is called the figure of merit푓. In the linear2 휋 region퐶푖� we saw that the transconductance is

품풎 given by 푚 푪𝒊 푔

=

휇푛 퐶𝑜 푊 and 푔푚 �퐷 퐿 =

Substituting these two values to calculate the퐶푖� figure푊 of merit,퐿 퐶𝑜 we find the maximum frequency to be 푓푚𝑚 = (5.149)

1 휇푛푉퐷 2 푓푚𝑚 2 휋 퐿 - 268 -

Assuming is very small because we are considering the linear region, is approximately the 푉퐷 electric field,퐷 and multiplying it by gives us the velocity, and therefore, the maximum frequency � 퐿 is rewritten as 휇푛 푓푚𝑚 = = (5.150)

� 1 푚𝑚 where is the velocity of the carriers in푓 the channel,2 휋 and퐿 2is 휋the휏푡 �reciprocal of the transit time . Thus � we find that the maximum frequency at which the device can be operated is related to the time 푡of� 푣 퐿 휏 transit of carriers from the source to the drain. The intrinsic limit on the maximum frequency of operation is imposed by the transit time of the carriers. This is true not only in MOS transistors and bipolar transistors but also in all other devices.

Figure (5.37) Plot of the drain current as a function of the gate voltage to obtain the threshold voltage

Figure (5.38) Equivalent circuit for the MOS transistor.

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Summary

• An MOS device uses the electric field applied perpendicular to the surface of a semiconductor to induce space-charge in the semiconductor. The MOS structure consists of a semiconductor substrate on the surface of which a thin oxide layer is grown. A conducting layer (metal), deposited on top of the oxide, is called the gate and a voltage applied on the gate generates the electric field perpendicular to the semiconductor surface (also called the interface). • The space charge induced by the voltage on the gate bends the bands in the semiconductor up or down towards the interface according to whether the gate voltage is negative or positive, respectively. • When the induced charge is of the same polarity as that of the majority carriers in the substrate (bulk), excess majority carriers accumulate at the interface. The surface is said to be accumulated in this case and the device is in accumulation region of operation. The space charge is due to the excess majority carriers. • When the induced charge is of a polarity opposite to that of the majority carriers in the semiconductor, the induced space charge is initially due to ionized impurity atoms in the substrate and the surface is said to be depleted. The device is said to be in the depletion region of operation. For larger values of gate voltage the minority carrier density at the interface becomes comparable to or large than the majority carrier density in the bulk. Under these conditions, the surface is said to be inverted. The charge due to the excess minority carriers is called the inversion charge and the layer of excess minority carriers is called the inversion layer. The inversion layer is very thin and therefore the inversion charge can be considered as a sheet charge. The space charge is due to both depletion charge and inversion charge. • The space charge density, which is defined as the charge contained in the space charge under unit area of the interface, determines the electric field in the oxide and hence the potential difference across the oxide. • The potential difference across the space charge region is the amount of band bending in the space charge region and is called the surface potential. The surface potential determines whether the substrate is accumulated, depleted or inverted. • The charge in the accumulation layer or in the inversion layer is approximated as a sheet charge and hence the potential difference across these layers is taken as negligible. On the other hand, the depletion region is of non-zero width. The width of the space charge region is therefore approximated as the width of the depletion region. • Inversion sets in when the bending of the band is such that the difference between the Fermi energy and the intrinsic Fermi energy at the interface is the same in magnitude as the difference between these two energies in the bulk but of opposite polarity. This condition is called “on-set of inversion”. The gate voltage at which the surface starts to be inverted (on-set of inversion) is called the threshold voltage.

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• According to the depletion approximation, the depletion region width remains constant in strong inversion at a value it had at the on-set of inversion. The width of the depletion region remains at a maximum value under inversion condition. • An ideal MOS device is one in which there is no space charge in the semiconductor when the 푥푑 푚𝑚 gate voltage is zero. A voltage is required to be applied on the gate to induce a space charge. • A non-ideal MOS device is one in which there is a space charge even when there is no applied voltage on the gate, Charge in the gate oxide, interface states and work function difference between the gate and the substrate give rise to non-ideal behavior. • Flat band voltage is the voltage needed to be applied on the gate to reduce the space charge to zero so that the device with a voltage on the gate equal to the flat –band voltage looks like an ideal device with zero gate voltage. • The space charge induced by the oxide charge and the work function difference is independent of band-bending while that due to the interface states depends on band bending. • Interface states can be acceptor type or donor type. It is conventional to assume that the interface states in the upper half of the band gap are acceptor type while those in the lower half are donor type. • The space charge region gives rise to small signal capacitance effect. This capacitance is called space charge capacitance and is equal to the rate of increase of space charge with surface potential. • Since the inversion and accumulation charge increases exponentially with surface potential, the space charge capacitance of accumulation and inversion charges is much larger than the capacitance due to the oxide layer. The capacitance due to the oxide layer called the oxide capacitance is the same as a parallel plate capacitor with the gate oxide as the dielectric. The MOS capacitor behaves equivalent to a series combination of the oxide capacitance and the space charge capacitance. • Due to the dependence of the space charge capacitance on the surface potential which in turn depends on the gate voltage, the capacitance of a MOS capacitor varies with the gate voltage. The capacitance-voltage characteristic is called the C-V curve. • In accumulation, the MOS capacitance is equal to the oxide capacitance itself since the space charge capacitance is very much larger. • As the device is biased into depletion, the space charge capacitance (equal to the depletion region capacitance ) decreases (due to the increase in depletion region width) and hence 휖푠 the capacitance of the MOS device decreases. ≈ 푥푑 • At strong inversion, the capacitance behavior is different at high and low frequency of the measuring small signal. At high frequencies, the inversion charge does not vary in step with the small measuring signal and hence the space charge capacitance remains at a minimum value and the MOS capacitance remains constant at a minimum value. 휖 푠 • ≈At low푥푑 푚 frequencies,𝑚 the inversion charge has time to be generated thermally so that it can vary in step with the measuring signal; hence the space charge capacitance is determined by the

- 271 -

variation with time of the inversion charge and is therefore very large. The MOS capacitance in strong inversion becomes equal to the oxide capacitance as in the accumulation case. • In the MOS transistor, the conductance between two heavily doped regions, in the substrate, (called the source and the drain) is controlled by the surface inversion of the region between the source and the drain. • In the simplest model, the drain voltage is assumed to be small and the inversion charge density is assumed to be constant between the source and the drain. • If the inversion charge is due to electrons providing a conductance between the source and drain, the device is called an -channel MOS transistor. + 푛 • If the source and drain are heavily doped regions and the inversion charge is due to holes, the 푛 device is called a -channel MOS transistor. 푝 • When the drain voltage is high, the inversion charge density decreases towards the drain due to 푝 the voltage drop along the channel. This is equivalent to the local threshold voltage increasing from its lowest value near the source to the highest value near the drain. • In one approximate model, the local threshold voltage at some point in the channel is taken as sum of the voltage drop at measured with respect to the source and the threshold voltage at 푦 the source. 푦 • In a more exact model, the band bending required to attain inversion at some point in the channel is taken as the sum of voltage drop at and the band bending needed at the source to 푦 produce inversion. 푦 • As the drain voltage is increased, ultimately the inversion charge density at the drain decreases and becomes zero. The channel is said to be pinched off at the drain. When the drain voltage is increased further, the pinch – off point moves toward the source and the drain current increases very slowly with the drain voltage as though the current has saturated. Hence the transistor is said to be operating in the saturation region. •

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Glossary

= Angstrom (s) = an integration constant Å C = capacitance per unit area of the MOS capacitor 퐶 = depletion layer capacitance = feedback capacitance between the drain and the gate 퐶푑 = input capacitance 퐶푓� = measured capacitance 퐶푖� = minimum capacitance in the high frequency curve 퐶푚 = oxide layer capacitance per unit area 퐶푚𝑚 퐶 − � = space charge capacitance per unit area 퐶𝑜 . . = electrostatic potential 퐶푆� = energy at the bottom of the conduction band (also potential energy of electrons) 푒 푠 = Fermi energy 퐸c = Fermi level in a p-type semiconductor 퐸퐹 = energy gap 퐸퐹1 = intrinsic Fermi energy 퐸푔 = intrinsic Fermi energy level in the bulk 퐸푖 = vacuum level 퐸푖� = energy at the top of the valence band (also potential energy of holes) 퐸𝑣� = electric field 퐸푉 = electric field in the oxide ℇ ( ) = electric field at ℇ𝑜 = maximum frequency ℇ 푥 푥 = drain conductance or output conductance 푓푚𝑚 = transconductance 푔퐷 G = input conductance 푔푚 푖 � = small signal current = drain current 횤̃ = drain current at saturation 퐼퐷 = current density 퐼퐷 푠𝑠 = Boltzmann constant 퐽 = body factor, also 푘 = dielectric constant of the oxide 퐾 훾 = length of the channel 퐾𝑜 = distance of the pinch-off point from the source 퐿 ′ = electron density 퐿 = thermal equilibrium electron density 푛 = electron density in the bulk 푛0 = intrinsic carrier density 푛푏 푖 푛 - 273 -

= thermal equilibrium electron density in n-type material = electron concentration at the interface or surface 푛푛0 ( ) = electron density at 푛푠 = acceptor impurity density 푛 푥 푥 = ionized acceptor impurity density 푁퐴 − = donor impurity density 푁퐴 = ionized donor impurity density 푁퐷 + = number of interface states per unit area per unit energy interval at some energy level 퐷 푁 = hole density 푖� 푁 = thermal equilibrium hole density 퐸 푝 = hole density in the bulk 0 푝 = hole concentration at the interface or surface 푝푏 ( ) = hole density at 푝푠 = electron charge (magnitude) 푝 푥 푥 = work function of the solid 푞 = work function of solid 푞푞 = work function of solid 푞푞퐴 퐴 = sinusoidally varying small signal charge 푞푞퐵 퐴 = space charge density in accumulation 푄� = depletion charge density 푄푎𝑎 = maximum depletion charge density 푄푑 = charge on the gate 푄푑 푚𝑚 = inversion charge density 푄퐺 = electron density at the surface or interface 푄푖𝑖 = effective sheet charge density 푄푛 = hole density in the surface or interface 푄𝑜 = space charge density 푄푝 = sheet charge density at the interface or surface 푄푆� t = time 푄푠� = thickness of the oxide layer = velocity of the carriers in the channel 푡𝑜 = small signal sinusoidal voltage 푣 = small signal sinusoidal gate voltage 푣� = built-in voltage, contact potential 푣�푔 = drain voltage �푏� = saturation drain voltage �퐷 = drain to source voltage �퐷 푠𝑠 = flat-band voltage �퐷� = thermal equilibrium density of holes �퐹� = gate voltage 푝0 = voltage across the oxide layer �퐺 = pinch-off voltage �𝑜 �푝 - 274 -

= reverse substrate bias voltage = threshold voltage �푆� = threshold voltage of an ideal MOS device �� = threshold voltage of a non-ideal MOS device �� 푖𝑖𝑖 ( ) = voltage at �� 푛�푛−푖𝑖𝑖 = width of the channel � 푦 푦 = position value 푊 = width of the space-charge or depletion region 푥 = depletion region width at the on-set of strong inversion 푥푑 = thickness of the inversion layer of electrons 푥푑 푚𝑚 y = position value 푥퐼 = change in surface potential = permittivity of free space ∆휓푠 = permittivity of the oxide 휖0 = permittivity of the semiconductor 휖𝑜 = body factor, same as 휖푠 = electrostatic potential 훾 퐾 = bulk potential ∅ = bulk electrostatic potential of the -region ∅푏 = bulk electrostatic potential of a -type semiconductor ∅푛 푛 ( ) = electrostatic potential at ∅푝 푝 = work function ∅ 푥 푥 = work function of a -type semiconductor Φ = work function of an -type semiconductor Φ1 푝 = contact potential Φ2 푛 = work function of the gate Φ퐴� = work function difference between the gate and the substrate Φ푀 = work function of the substrate semiconductor Φ푀� = charge density Φ푆 = minority carrier generation lifetime 휌 = transit time 휏푔 = electrostatic potential difference 휏푡� = surface potential or interface potential 흍 = band bending at the onset of inversion or surface potential at onset of inversion 휓푠

휓푖𝑖 (x) = electrostatic potential difference at x i.e., ( ) = electron affinity 흍 ∅ 푥 − ∅푏 = 2 times the frequency of the small signal voltage 휒 = 2 times the maximum frequency 휔 휋 푣� 휔 푚𝑚 휋

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Problems

1. Show that = when = for a MOS device with a -type substrate. 2. For an ideal MOS diode (capacitor) with = 500 and = 10 , Find the value of the 푛푠 푛푖 휓푠 − ∅푏 푝 surface potential and the voltage across the oxide layer required15 to produce−3 a) intrinsic 푡𝑜 Å 푁퐴 𝑐 condition at the interface and b) on-set of strong inversion at the interface. 휓푠 3. In a MOS capacitor with a -type substrate determine the space-charge density for a gate voltage of 0.75 assuming the oxide layer thickness to be 1000 and the substrate impurity 푝 concentration to be 5 × 10 . � Å 4. Consider a MOS capacitor with15 a −substrate3 doping of 2 × 10 of donor atoms and an 𝑐 oxide thickness of 600 , Determine the gate voltage at which15 the− surface3 becomes intrinsic. 𝑐 5. For the device in the previous problem, find the value of the space-charge density at a gate Å voltage equal to a. The threshold voltage b. Twice the threshold voltage 6. For the device in the previous problem, find the inversion charge density and the depletion region charge density when the gate voltage is three times the threshold voltage. 7. For the device in the previous problem, find the high and low frequency capacitance at a gate voltage of 4 . Assume an area of 10 for the capacitor. 8. Show that the potential (x) in the depletion−4 2 region of a MOS capacitor varies as − � 𝑐 (x) = 1 . 흍 푥 2 9. 흍Show that휓푠 �in deep− 푥푑� depletion, the relation between the gate voltage and the surface potential is given by

휓푠 4 = + 1 1 + 22 퐾 �퐺 �퐺 �� 2 − � 휓푠 Where is defined as 퐾 2 퐾 = � 휖푠푞푁퐴 10. Consider a MOS device in which the oxide퐾 layer contains positive ions with a density of 퐶𝑜 10 . If the oxide thickness is 400 calculate the flat-band voltage . 11. Assume16 −the3 gate is -type poly-silicon with an impurity concentration of 10 and the 𝑐 Å �퐹� substrate to be -type with a resistivity equal to 5 -cm. Assuming the electron17 affinity−3 of silicon 푝 𝑐 to be 4.1 , determine the flat-band voltage. 푛 훺 12. Consider a MOS capacitor with a -type substrate and an oxide thickness of 500 . Let the � substrate impurity density be 5 × 10 . Assume the device is ideal. 푝 Å a. Calculate the high frequency and low 15frequency−3 capacitances in strong inversion. 𝑐 b. Assuming that a step voltage of 5 is applied on the gate at = 0. Calculate the small signal capacitance at = 0 . + � 푡 푡

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13. Consider a MOS transistor with a -type substrate containing 10 acceptor impurities and an oxide thickness of 500 . The flat-band voltage is 0.5 . Determine15 − the3 drain conductance for 푝 𝑐 a gate voltage of 5 in the linear region. Take the mobility as 900 , as Å � 20 and as 10μm. 2 −1 −1 � 휇푛 𝑐 � 푠𝑠 푊 14. For the device in the previous problem, determine the transconductance in the saturation 휇� 퐿 region. 15. Consider an -channel MOS transistor with = 150 . The substrate impurity concentration is 10 . Let the channel length, , be 2 , the channel width, be 20 and the 푛 푡𝑜 Å mobility16 of− the3 channel carriers be 750 . 𝑐 퐿 휇� 푊 휇� a. What is the value of the transconductance2 −1 −1 for a drain voltage of 20 ? 𝑐 � 푠𝑠 b. Assume that the oxide has positive ions of density equal to 10 . Assume that 푚� the gate is -type poly-silicon with an impurity density of 10 17 −. 3Neglect the 𝑐 contribution of surface states to the flat-band voltage. Using18 the simplified−3 푝 𝑐 treatment given in case 2, determine the saturation drain voltage, for a gate voltage 5 . �퐷𝐷� �

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