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Generation of Free and Holes In an intrinsic , the number of free electrons equals the number of holes. Thermal : The concentration of free electrons and holes increases with increasing temperature. Thermal : At a fixed temperature, an with a large energy gap has smaller free and concentrations than a semiconductor with a small energy gap. Optical: Light can also generate free electrons and holes in a semiconductor. Optical: The energy of the (hν) must equal or exceed the energy gap of the semiconductor (Eg) . If hν > Eg , a can be absorbed, creating a free electron and a free hole. This absorption process underlies the operation of photoconductive light detectors, , photovoltaic (solar) cells, and state camera “chips”. Electrons and Holes in Intrinsic Semiconductor Electron energy

Ec+χ CB

Ec FREE e– hυ > E hυ g E hole g – HOLE e Ev

VB

0

(a) (b)

Fig. 5.3: (a) A photon with an energy greater than Eg can excite an electron from the VB to the CB. (b) When a photon breaks a Si-Si bond, a free electron and a hole in the Si-Si bond is created. Electrons and Holes in under

Conduction band — free electron Valence band - holes

Energy in the presence of a uniform electric field. Shown are the upper almost-empty band and the lower almost-filled band. The tilt of the bands is caused by an externally applied electric field Intrinsic Semiconductor • Semiconductor contain no impurities. •Electron density equals to hole density. • Resulting from thermal activation or photon excitation. • Intrinsic carrier density Schematic representation of Carrier densities (a) (b) (c) (d) 1/2 g(E) ∝ (E-Ec) E E E Ec+χ [1-f(E)]

CB For Area =∫ nE ( ) = ndEE electrons n (E) Ec Ec E

E EF F

E v Ev pE(E)

For holes Area = p VB

0 g(E) f(E) nE(E) or pE(E) Fig. 5.7: (a) Energy band diagram. (b) Density of states (number of states per unit energy per unit volume). (c) Fermi-Dirac probability function (probability of occupancy of a state). (d) The product of g(E) and f(E) is the energy density of electrons in the CB (number of electrons per unit energy per unit volume). The area under nE(E) vs. E is the electron concentration in the conduction band.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Non-degenerate and Degenerate semiconductors

E CB n > Nc Impurities forming CB E a band Fn Ec Ec g(E)

Ev Ev EFp VB

(a) (b)

Fig. 5.21: (a) Degenerate n-type semiconductor. Large number of donors form a band that overlaps the CB. (b) Degenerate p-type semiconductor.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Extrinsic Semiconductors Extrinsic semiconductors : impurity dictate the properties Almost all commercial semiconductors are extrinsic Impurity concentrations of 1 in 1012 is enough to make extrinsic at room T! Impurity atoms can create states that are in the . In most cases, the of a semiconductor leads either to the creation of donor or acceptor levels n-type semiconductors n-Type In these, the charge carriers are negative p-type semiconductors. In these, the charge carriers are positive p-Type Band Diagram: Acceptor Dopant in Semiconductor

For Si, add a group III element to “accept” an electron and make p-type Si (more positive “holes”). EC “Missing” electron results in an extra “hole”, with an acceptor energy level E just above E A F E the valence band E . A V EV p-type Si Holes easily formed in valence band, greatly increasing the electrical conductivity.

Fermi level EF moves down towards EV. Typical acceptor elements are , Aluminum, Gallium, Indium. Band Diagram: Donor Dopant in Semiconductor Increase the conductivity of a semiconductor by adding a small amount of another material called a dopant (instead of heating it!) For group IV Si, add a group V element to “donate” an electron and make n-type Si n-type Si (more negative electrons!). EC ED “Extra” electron is weakly bound, with EF donor energy level ED just below conduction band EC. Egap~ 1 eV EV Dopant electrons easily promoted to conduction band, increasing electrical conductivity by increasing carrier density n.

Fermi level EF moves up towards EC. Typical donor elements that are added to Si or Ge are phosphorus, arsenic, antimonium. Silicon n-type semiconductors: Si Si Si Si Bonding model description: Element with 5 bonding electrons. Only 4 electrons Si Si Si Si - participate in bonding the extra e can easily Si P Si Si become a conduction electron Si Si Si Si p-type semiconductors: Si Si Si Si Bonding model description: Element with 3 bonding electrons. Since 4 Si Si Si Si electrons participate in bonding and only 3 are Si Si B Si available the left over “hole” can carry charge Si Si Si Si

Question: How many electrons and holes are there in an intrinsic semiconductor in thermal equilibrium? Define: -3 no equilibrium (free) electron concentration in conduction band [cm ] -3 po equilibrium hole concentration in valence band [cm ] Certainly in intrinsic semiconductor: no = po = ni -3 ni intrinsic carrier concentration [cm ] As T ↑ then ni ↑ = = npn iOO As Eg ↑ then ni ↓ What is the detailed form of these dependencies? We will use analogies to chemical reactions. The electron-hole formation can be though of as a chemical reaction…….. Similar to the chemical reaction………

+− +⇔ hebond + − 2 +⇔ OHHOH )( The Law-of-Mass-Action relates concentration of reactants and reaction products. For water…… Where E is the energy released or consumed during the reaction…………. OHH −+ ]][[ ⎛ E ⎞ K = exp⎜−≈ ⎟ 2OH ][ ⎝ kT ⎠ This is a thermally activated process, where the rate of the reaction is limited by the need to overcome an energy barrier (activation energy).

By analogy, for electron-hole formation:

pn oo ]][[ ⎛ Eg ⎞ K exp⎜−≈= ⎟ bonds][ ⎝ kT ⎠ Where [bonds] is the concentration of unbroken bonds and Eg is the activation energy >> ,pn[bonds] In general, relative few bonds are broken to oo form an electron-hole and therefore the [bonds] = tantcons number of bonds are approximately constant. ⎛ E ⎞ pn exp⎜−≈ g ⎟ Two important results: oo ⎜ kT ⎟ ⎛ E ⎞ ⎝ ⎠ 1)…………. ⎜ g ⎟ ni exp⎜−≈ ⎟ ⎝ 2kT ⎠

2)……………. 2 =× npn iOO

The equilibrium np product in a semiconductor at a certain temperature is a constant, specific to the semiconductor. Effect of Temperature on Intrinsic Semiconductivity The concentration of electrons with sufficient thermal energy to enter the conduction band (and thus creating the same concentration of

holes in the valence band) ni is given by ⎛ Δ− E ⎞ ⎜ ⎟ n i ≈ exp ⎜ ⎟ ⎝ BTk ⎠ For intrinsic semiconductor, the energy is half way across the gap, so that ⎛ − E ⎞ ⎜ g ⎟ ni ≈ exp⎜ ⎟ ⎝ 2 BTk ⎠ Since the electrical conductivity σ is proportional to the concentration of electrical charge carriers, then ⎛ − E ⎞ ⎜ g ⎟ = σσO exp⎜ ⎟ ⎝ 2 BTk ⎠ Thermal Stimulation

P = Ratio of the number of electrons ⎛ Δ− E ⎞ P = exp⎜ ⎟ promoted to conduction band and the ⎜ Tk ⎟ ⎝ B ⎠ number of electrons in the system

Suppose the band gap is Eg = 1.0 eV

T(°K) kBT(eV) E/kBT ⎛ ΔE ⎞ Δ exp ⎜ − ⎟ ⎝ k BT⎠ 00∞ 0 100 0.0086 58 0.06x10-24 200 0.0172 29 0.25x10-12 300 0.0258 19.4 3.7 x10-9 400 0.0344 14.5 0.5x10-6 Example Calculate the number of Si atoms per cubic meter. The density of silicon is 2.33g.cm-3 and its atomic mass is 28.03g.mol-1. Then, calculate the electrical resistivity of intrinsic silicon at 300K. For Si at 300K 16 -3 -19 2 -1 2 -1 ni=1.5x10 carriers.m , q=1.60x10 C, μe=0.135m (V.s) and μh=0.048m .(V.s)

Solution

N × ρSiA 28 −3 22 −3 nSi = 1000.5 −×= matomsSi 1000.5. −×= .cmatomsSi ASi

σ i qn ()+××= μμhe σ ()×= 105.1 16 mcarriers 3 ()× −19 ()2 ()sVmC −1 + 2 ..048.0..135.0106.1/ sVm −1 ()= σ −3 −Ω×= m)(104385.0 −1 1 ρ = yresistivit 1028.2 3 −Ω×== m σ Example The electrical resistivity of pure silicon is 2.3x103Ω-m at room temperature (27oC ~ 300K). Calculate its electrical conductivity at 200oC (473K). -5 Assume that the Eg of Si is 1.1eV ; kB =8.62x10 eV/K ⎛ − E ⎞ ⎛ − E ⎞ g g ⎛ − Eg ⎞ σ = C exp. ⎜ ⎟ σ 473 = C exp. ⎜ ⎟ σ = C exp. ⎜ ⎟ ⎜ 2 Tk ⎟ ⎜ k 4732 )( ⎟ 300 ⎜ ⎟ ⎝ B ⎠ ⎝ B ⎠ ⎝ kB 3002 )( ⎠

σ 473 ⎛ − Eg − Eg ⎞ = exp⎜ − ⎟ σ 300 ⎝ kB kB 30024732 )()( ⎠

⎛σ ⎞ − Eg Eg Eg ⎛ 1 1 ⎞ .11 eV ⎛ 1 1 ⎞ ln⎜ 473 ⎟ = + = − = − ⎜ ⎟ ⎜ ⎟ −5 ⎜ ⎟ ⎝σ 300 ⎠ kB kB )()( 230024732 kB ⎝ 300 473 ⎠ ×106282 ).( ⎝ 300 473 ⎠

⎛σ 473 ⎞ ln⎜ ⎟ = .7777 ⎝σ 300 ⎠ 1 = σσ2385)( = 0412385 Ω= m).(.)( −1 473 300 . ×1032 3 Example: For germanium at 25oC estimate (a) the number of charge carriers, (b) the fraction of total electrons in the valence band that are excited into the conduction band and ⎛ − E ⎞ (c) the constant A in the expression g when E=E /2 = An exp ⎜ ⎟ g ⎝ 2 BTk ⎠ Data: Ge has a diamond cubic structure with 8 atoms per cell and valence of 4 ; 2 2 a=0.56575nm ; Eg for Ge = 0.67eV ; μe = 3900cm /V.s ; μh = 1900cm /V.s ; -5 ρ = 43Ω-cm ; kB=8.63x10 eV/K (a) Number of carriers = 25o CT −5 BTk .)(( ×= 1063822 KeV =+ .))(/ 0514025273 eV

σ .0230 13 electrons n = = −19 . ×= 1052 3 q + μμhe )( × (. +190039001061 ) cm There are 2.5x1013 electrons/cm3 and 2.5x1013 holes/cm3 helping to conduct a charge in germanium at room temperature. b) the fraction of total electrons in the valence band that are excited into the conduction band

The total number of electrons in the valence band of germanium is : 8 cellatoms )(/( 4valence−electrons atoms)/ Valence − electrons = .( 10565750 − cmx )37

Total valence electrons . ×=−− 10771 23 /cmelectrons 3 number excited −− /cmelectrons 3 . ×1052 13 Fraction excited =− = . ×= 10411 −10 Total valence −− /cmelectrons 3 . ×10771 23

(c) the constant A n . ×1052 13 A == . ×= 10141 −19 /cmcarriers 3 ⎛ −E ⎞ ⎛ − .670 ⎞ ⎜ g ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ .05140 ⎠ e⎝ 2 BTk ⎠ e The Mass Action Law This relationship is valid for both intrinsic and extrinsic semiconductors. In an the increase in one type of carrier (n or p) reduces the concentration of the other through recombination so that the product of the two (n and p) is a constant at a any given temperature. The carriers whose concentration in extrinsic semiconductors is the larger are designated the majority carriers, and those whose concentration is the smaller the minority carriers. At equilibrium, with no external influences such as light sources or applied voltages, the concentration of electrons,n0, and the concentration of holes, p0, are related by 2 ni denotes the carrier concentration in =× npn ioo intrinsic silicon A material is defined as intrinsic when it consists purely of one element and no outside force (like light energy) affects the number of free carrier other than heat energy. In intrinsic Si, the heat energy available at room temperature generates approximately 1.5x1010 carriers per cm3 of each type (holes and electrons) . The number of free carriers doubles for approximately every 11°C increase in temperature. This number represents a very important constant (at room temperature), and we define 10 -3 ni = 1.5x10 cm where ni denotes the carrier concentration in intrinsic silicon at room temperature (constant for a given temperature). Based on charge neutrality, for a sample doped with ND donor -3 -3 atoms per cm and NA acceptor atoms per cm we can write

no + NA = po + ND which shows that the sum of the electron concentration plus the ionized acceptor atoms is equal to the sum of the hole concentration plus the ionized donor atoms. The equation assumes that all donors and acceptors are fully ionized, which is generally true at or above room temperature. Given the impurity concentration, the above equations can be solved simultaneously to determine electron and hole concentrations. In electronic devices, we typically add only one type of impurity within a given area to form either n-type or p-type regions. In n-type regions there are typically only donor impurities and the donor concentration is much greater than the intrinsic carrier concentration, NA=0 and ND>>ni.

Under these conditions we can write nn = ND where nn is the free electron concentration in the n-type material and 3 ND is the donor concentration (number of added impurity atoms/cm ). Since there are many extra electrons in n-type material due to donor impurities, the number of holes will be much less than in intrinsic silicon and is given by, 2 pn = ni / ND where pn is the hole concentration in an n-type material and ni is the intrinsic carrier concentration in silicon. Similarly, in p-type regions we can generally assume that ND=0 and NA>>ni. In p-type regions, the concentration of positive carriers (holes), pp, will be approximately equal to the acceptor concentration, NA.

pp = NA and the number of negative carriers in the p-type material, np, is given by 2 np = ni / NA Notice the use of notation, where negative charged carriers are n, positive charged carriers are p, and the subscripts denote the material, either n-type or p-type. This notation will be used throughout our discussion of p-n junctions and bipolar . The above relationships are only valid when ND or NA is >> ni, which will always be the case in the problems related to design. Example : Calculate the conductivity and the resistivity of an n-type silicon wafer which contains 1016 electrons per cubic centimeter with an of 1400 cm2/Vs and a hole mobility of 480 cm2/Vs

Solution: The conductivity is obtained by adding the product of the electronic charge, q, the carrier mobility, and the density of carriers of each carrier type, or:

σ = (μ + μpn pnq ) As n-type material contains almost no holes, the conductivity equals:

-19 16 σ= q μn n = 1.6 x 10 x 1400 x 10 = 2.24 1/Ωcm. The resistivity equals the inverse of the conductivity or: 11 ρ == ()+ μμσpn pnq and equals ρ = 1/σ = 1/2.24 = 0.446 Ωcm. A more precise solution: 2 =× npn i As the number of holes (p) is small with respect to the number of +=+ NNpNn = 0...... A D A donors (ND) then n~ND

+= Npn D 2 10 2 ni ()×105.1 4 p == 16 ×= 1025.2 holes ND 10

σ = (μ + μpn pnq ) σ ()1619 4 ××+××= 4801025.2140010106.1 σ 241.2 ()−Ω= cm −1 Example An n-type piece of silicon of length L = 10 micron has a cross sectional area A = 0.001 cm2. A voltage V = 10 Volt is applied across the sample yielding a current I = 100 mA. What is the resistance, R of the silicon sample, its conductivity, σ, and 2 electron density, n ? μn= 1400 cm /Vs

Solution The resistance of the sample equals R = V/I = 10/0.1 = 100 Ω. Since R = L /(σA) the conductivity is obtained from: σ = L/(R A) = 0.001/(100 x 0.001) = 0.01 1/Ωcm. The required electron density is related to the conductivity by:

σ = q n μ n so that the density equals: -19 13 -3 n = σ/(q μ n) = 0.01/(1.6 x 10 x 1400) = 4.46 x 10 cm . Example A Si sample at room temperature is doped with 1011 As atoms/cm3. What are the equilibrium electron and hole concentrations at 300 K? Solution 2 Since the NA is zero we can write, n p = ni

And n + NA = p + ND 2 2 → n –ND n – ni = 0 Solving this quadratic equations results in n = 1.02x1011 [cm-3] 2 20 11 and thus, p = ni / n = 2.25x10 / 1.02x10 p = 2.2x109 [cm-3]

Notice that, since ND>ni, the results would be very similar if we assumed 11 -3 nn=ND=10 cm , although there would be a slight error since ND is not much greater than ni. SemiconductorsSemiconductors ThermalThermal excitationexcitation ofof electronselectrons Fermi level now lies in the gap

EF VB CB pure solid

Fermi-Dirac EF distribution n-type

E F p-type Conductivity Intrinsic semiconductor (Germanium, Silicon). For every electron, “e”, promoted to the conduction band, a hole, “h”, is left in the valence band (+ charge). The conductivity is determined by the number of electron-hole pairs.

Total conductivity σ = σe + σh = nqμe + pqμh

For intrinsic semiconductors: n = p & σ = nq(μe + μh)

Extrinsic semiconductor (doping). n-type. The number of electrons in the conduction band far exceeds the number of holes in the valence band (or n>>p).

σ = σe = nqμe p-type. The number of holes in the valence band far exceeds the number of electrons in the conduction band (or p>>n)

σ = σh = pqμh Temperature dependent conductivity

600 °C 400 °C 200 °C 27 °C 0LLLL °C 18 ) 10 3 -

2.4×1013 cm-3 1015

Ge 1012

1.45 1010 cm-3 109 × Si

6 Intrinsic Concentration (cm Concentration Intrinsic 10 2.1×106 cm-3

GaAs 103 1 1.5 2 2.5 3 3.5 4 1000/T (1/K)

Fig. 5.16: The temperature dependence of the intrinsic concentration.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Temperature dependence of Conductivity for Semiconductor ln(n)

Intrinsic slope = -Eg/2k

T Ionization Extrinsic s slope = -ΔE/2k ln(Nd) Ti

ni(T) 1/T

Fig. 5.15: The temperature dependence of the electron concentration in an n-type semiconductor.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Temperature variation of conductivity – Intrinsic Semiconductors

σ = n|q|μe + p|q|μh Strong exponential dependence of carrier concentration in intrinsic semiconductors Temperature dependence of carrier mobility is weaker.

⎛ − Eg ⎞ Apn ×≅= exp⎜ ⎟ ⎝ 2 BTk ⎠

⎛ − Eg ⎞ σ C ×≅ exp⎜ ⎟ ⎝ 2 BTk ⎠ Temperature variation of conductivity - Intrinsic Semicoductor

⎛ Eg ⎞ Apn exp⎜−×≅= ⎟ ⎝ 2 BTk ⎠

⎛ Eg ⎞ σ C exp⎜−×≅ ⎟ ⎝ 2 BTk ⎠

The constant A is related to the density of states and the effective ln(n) = ln(p) ≅ ln(A) - Eg /2 kT masses of electrons and holes.

Plotting log of σ , p, or n vs. 1/T produces a straight line.

Slope is Eg/2kB; gives band gap energy. Δlnp − E = g Δ()T 21 kB Temperature variation of conductivity – Extrinsic Semiconductor

Extrinsic semiconductors low T: all carriers due to extrinsic excitations mid T: most dopants ionized (saturation region) high T: intrinsic generation of carriers dominates provided ND >> ni the number of carriers is ln(n) dominated by nd At very high Intrinsic temperatures, ni increases beyond N slope = -Eg/2k dopants are activated as T D > 50 - 100K so carrier concentration increases

T Ionization Extrinsic s slope = -ΔE/2k ln(Nd) Ti

ni(T) 1/T

Fig. 5.15: The temperature dependence of the electron concentration in an n-type semiconductor.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

Semiconductor: Dopant Density via

• Why Useful? Determines carrier type (electron vs. hole) and carrier density n for a semiconductor. •How? Place semiconductor into external B field, push current along one axis, and

measure induced Hall voltage VH along perpendicular axis. MagneticICurrent BField )__)(_( Carrier __ nDensity = VVoltageHalltThicknessqeChCarrier H )__)(_)(_arg_(

• Derived from Lorentz equation FE (qE) = FB (qvB). r r r Hole FqvBB = × Electron + charge – charge The Hall Effect and the Lorentz Force The basic physical principle is the Lorentz force. When an electron (e-) moves along a direction perpendicular to an applied magnetic field (B), it experiences a force acting normal to both directions and moves in response to this force and the force effected by the internal electric field. For an n-type, bar-shaped semiconductor shown in Fig.1, the carriers are predominately electrons of bulk density n.

We assume that a constant current I flows along the x-axis in the presence of a z-directed magnetic field. Electrons subject to the Lorentz force drift away from the current line toward the negative y-axis, resulting in an excess surface electrical charge on the side of the sample. This charge results in the Hall voltage, a potential drop across the two sides of the sample.

This transverse voltage is the Hall voltage VH and its magnitude is equal to IB/qnd, where I is the current, B is the magnetic field, d is the sample thickness, and q (1.602 x 10-19 C) is the elementary

charge. In some cases, it is convenient to use layer or sheet density (ns = nd) instead of bulk density. Semiconductors Devices

Impurities Put Allowed Levels in the Band Gap of Silicon

“p Type” “n Type”

Many ELECTRONS! Conduction Band Conduction Band

Donor Level = where Acceptor Level thermal electrons can easily go Many HOLES!

Valence Band Valence Band

Boron Doped Phosphorous Doped “Majority Carrier” and Current Flow in p-type Silicon + p-type Silicon - Hole Flow

Current Flow

“Majority Carrier” and Current Flow in n-type Silicon

+ n-type Silicon - Electron Flow Current Flow The p-n Junction

p n 0 Volts Hole Diffusion Electron Diffusion

Holes and Electrons “Recombine” at the Junction

A Depletion Zone (D) and a Barrier Field Forms at the p-n Junction

Barrier Field p -- D ++ n 0 Volts

Acceptor Ions Donor Ions Hole (+) Diffusion Electron (-) Diffusion The Barrier Field Opposes Further Diffusion (Equilibrium Condition) When a p-n junction is formed, some of the free electrons in the n-region diffuse across the junction and combine with holes to form negative ions. In so doing they leave behind positive ions at the donor impurity sites. In the p-type region there are holes from the acceptor impurities and in the n-type region there are extra electrons. When a p-n junction is formed, some of the electrons from the n-region which have reached the conduction band are free to diffuse across the junction and combine with holes. Filling a hole makes a negative ion and leaves behind a positive ion on the n-side. A space charge builds up, creating a depletion region which inhibits any further electron transfer unless it is helped by putting a forward bias on the junction. Equilibrium of junction Coulomb force from ions prevents further migration across the p-n junction. The electrons which had migrated across from the N to the P region in the forming of the depletion layer have now reached equilibrium. Other electrons from the N region cannot migrate because they are repelled by the negative ions in the N region and attracted by the positive ions in the N region. Reverse bias An applied voltage with the indicated polarity further impedes the flow of electrons across the junction. For conduction in the device, electrons from the N region must move to the junction and combine with holes in the P region. A reverse voltage drives the electrons away from the junction, preventing conduction. Forward bias An applied voltage in the forward direction as indicated assists electrons in overcoming the coulomb barrier of the space charge in depletion region. Electrons will flow with very small resistance in the forward direction. “Forward Bias” of a p-n Junction p - + n + Volts - Volts Current •Applied voltage reduces the barrier field •Holes and electrons are “pushed” toward the junction and the depletion zone shrinks in size •Carriers are swept across the junction and the depletion zone •There is a net carrier flow in both the P and N sides = current flow!

“Reverse Bias” of a p-n Junction

p --- D +++ n -Volts Current + Volts •Applied voltage adds to the barrier field •Holes and electrons are “pulled” toward the terminals, increasing the size of the depletion zone. •The depletion zone becomes, in effect, an insulator for majority carriers. •Only a very small current can flow, due to a small number of minority carriers randomly crossing D (= reverse saturation current) p-n Junction: Band Diagram p-n regions “touch” & free carriers move • Due to diffusion, electrons n-type electrons move from n to p-side and EC holes from p to n-side. EF

• Causes depletion zone at EF junction where immobile E V holes charged ion cores remain. p-type • Results in a built-in electric field or potential, which p-n regions in equilibrium opposes further diffusion. –– E +––– C ++ – – E + – F +++––– +++– ++ E V Depletion Zone Built-in potential

•For example: pn Junction: IV Characteristics

• Current-Voltage Relationship eV/ kT IIe=−o[1] Forward •Forward Bias:current Bias exponentially increases. • Reverse Bias: low leakage

current equal to ~Io. • Ability of p-n junction to pass Reverse Bias current in only one direction is known as “rectifying” behavior. The • The diode is a two terminal that allows current to flow in only one direction. • It is constructed of a P and an N junction connected together. Diode Operation • No current flows because the holes and electrons are moving in the wrong direction. • If you flip the battery around, the electrons are repelled by the negative terminal and the holes are repelled by the positive terminal allowing current to flow. Diode Characteristic Curve Diode Characteristics • An ideal diode would block all current when reverse biased. From the graph we see that this is not the case. • A small current (≈10μAmps) will flow when reverse biased and if the reverse voltage is increased enough the junction breaks down and current will begin to flow (Avalanche and Zener Breakdown).

When forward biased, a small voltage is required to get the diode current flowing. For a silicon diode this voltage is approximately 0.7V For a germanium diode, this voltage is approximately 0.3V Current Ge Si GaAs

~0.1 mA Voltage 0 0.2 0.4 0.6 0.8 1.0

Fig.6.4: Schematic sketch of the I-V characteristics of Ge, Si and GaAs pn Junctions

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

Diode Symbols

BAAl SiO2 A Al A p p n n B B Cross-section of p-n junction in an IC One-dimensional representation diode symbol

Current flow through a reverse biased p-n junction Diode: tiny reverse current under reverse bias Illuminated : large reverse current under forward bias Base-collector junction: large reverse current under reverse bias

Reverse current: A large reverse current requires a source of minority carriers • light (solar cell) OR • a nearby forward-biased junction (Bipolar Junction BJT) Quiz A. Design a p-type semiconductor based on silicon, which provides a constant conductivity of 100 ohm-1.cm-1 over a range of temperatures. Comment on the level of purity needed. (for 2 -1 -1 -19 silicon μh = 480 cm .V .s ; q = 1.6x10 Coulomb or Ampere.seconds ; silicon lattice parameter = 5.4307x10-8 cm ; 8 atoms per unit cell) B. A light-emitting diode display made using GaAs-GaP solid solution of composition 0.4GaP- 0.6GaAs has a direct bandgap of 1.9eV. What will be the color this LED display?

Solution A: In order to obtain the desired conductivity, we must dope the silicon with atoms having a valence +3.

If we assume that the number of intrinsic carriers is small: σ = σh = pqμh. -1 -1 2 -1 -1 -19 Where σ = 100 ohm .cm and μh = 480 cm .V .s . q = 1.6x10 Coulomb or Ampere.seconds

Then σ 100 18 3 p == −19 = . 10301 donorx − /cmatoms μh xe 4801061 ))(.( 1 electron dopant dopantXatom atomsiliconperatom 8 silicon cellunitperatoms )_____)(_____)(___( p = volumecellunit )__( x 18 x10430751031 − ).)(.( 38 x = = x1026 −6 -dopant -atoms -per -silicon atoms 8 Solution B

−15 −18 hc hc 1041.4( − smxseVx ).10998.2)( −6 g hE ν λ ==⇒== = 10652.0 mx λ Eg 9.1 eV

Or 652nm, the wavelength of red light. A Deeper Knowledge of Energy Gap A semiconductor establishes a periodic arrangement of atoms, leading to a periodic spatial variation of the potential energy throughout the crystal. must be used as the basis for allowed energy levels and other properties related to the semiconductor. A coherent discussion of these quantum mechanical results is beyond the scope of this course. Different semiconductor (with their different atomic elements and different inter- atomic spacings) lead to different characteristics. In semiconductors, a central result is the energy momentum functions determining the state of the electronic charge carriers. In addition to electrons, semiconductors also provide holes (i.e. positively charged ) which behave similarly to the electrons. Two energy levels are important: one is the energy level (conduction band) corresponding to electrons which are not bound to crystal atoms and which can move through the crystal and the other energy level (valence band) corresponds to holes which can move through the crystal. Between these two energy levels, there is a region of “forbidden" energies (i.e., energies for which a free carrier can not exist). The separation between the conduction and valence band minima is called the energy gap or band gap. The energy bands and the energy gap are fundamentally important features of the semiconductor material. Quantum Mechanics in Free Space In quantum mechanics, a “" is represented by a collection of plane waves r r where the frequency ω is related to the energy E according to ω ⋅− xkt )( and the momentum e p is related to the by ~p = ¹h~k. In the case of a classical particle with mass m moving in free space, the energy and momentum are related by E = p2=(2m) which, using the relationship between momentum and wave vector, can be expressed as E = (¹hk)2=(2m). Due to the arrangement of atoms in the crystal there is a different surface density of atoms on different crystallographic planes. For example, in Si the (100), (110), and (111) planes have surface atom densities (atoms per cm2) of 6.78x1014, 9.59x1014, and 7.83x1014, respectively. Direct and Indirect Semiconductors The real band structure in 3D is calculated with various numerical methods, plotted as E vs k. k is called wave vector = hkp p is momentum For electron transition, both E and p (k) must be conserved.

A semiconductor is direct if the maximum of the conduction band and the minimum of the valence band has the same k value

A semiconductor is indirect if the …do not have the same k value Direct semiconductors are suitable for making light-emitting devices, whereas the indirect semiconductors are not. Direct and indirect bandgap semiconductors E E

CB Indirect Band Gap, Ec Eg Direct Band Gap Eg Photon CB Ec Ev kcb Ev VB VB kvb -k k -k k

(a) GaAs (b) Si E

CB

Er Ec Ev VB -k k (c) Si with a recombination center

Fig. 5.50: (a) In GaAs the minimum of the CB is directly above the maximum of the VB. GaAs is therefore a direct band gap semiconductor. (b) In Si, the minimum of the CB is displaced from the maximum of the VB and Si is an indirect band gap semiconductor. (c) Recombination of an electron and a hole in Si involves a recombination center.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca What happens with the absorbed photons ? Part of it is re-emitted as light called photoluminescence CB electron

Radiative transition: emission of photon absorption non-radiative transition

hole VB

Luminescence = emission of optical radiation as a result of an electronic excitation Photoluminescence: optical excitation Catholuminescence: excitation by electron irradiation Electroluminescence: excitation by current Radiative and Non-radiative recombination

CB E CB c Energy Er Er Er Phonons Recombination E center Ec v VB ψcb(kcb) hυ = Eg (a) Recombination ψvb(kvb) Ev CB

Ec E VB t Et Et Distance Trapping center Ev VB

Fig.5.22: Direct recombination in GaAs. kcb = kvb so that momentum conservation is satisfied (b) Trapping Fig. 5.23: Recombination and trapping. (a) Recombination in Si via a generated by a recombination recombination center which has a localized energy level at Er in the center (close to the mid gap) bandgap, usually near the middle. (b) Trapping and detrapping of electrons by trapping centers. A trapping center has a localized In recombination, a free electron encounters energy level in the band gap. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) an incomplete bond with a messing electron http://Materials.Usask.Ca and the free electron in the CB and the free Trapping center: a localized state (close hole in the VB are annihilated. to the edge, e can detrap again)

E α (1/micron) 1000 Vacuum B Optical absorption 100 GaAs CB hυB B 10 Photon A 1 hυA Si 0.1 VB 0.01 A 0.001 g(E) 0123456 Photon energy (eV) E +χ CB c Fig. 5.37: The absorption coefficient α depends on the photon energy hυ and hence on the wavelength. Density of states increases from band edges Thermalization and usually exhibits peaks and troughs. Generally α increases with the photon energy greater than Eg because more energetic photons can excite Large hυ 3kT electrons from populated regions of the VB to numerous available states E c 2 deep in the CB. From Principles of Electronic Materials and Devices, Second Edition , S.O. Kasap (© McGraw-Hill, 2002) Eg http://Materials.Usask.Ca

hυ Eg E v

VB

Fig. 5.35: Optical absorption generates electron hole pairs. Energetic electrons must loose their excess energy to lattice vibrations until their average energy is (3/2)kT in the conduction band.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Luminescence Thermalization CB

Photoluminescence (PL) Trapping Excited State E c Et E g Luminescent Center hυ E g Recombination VB

Fig. 5.38: Optical absorption generates an EHP. The electron thermalizes and then becomes trapped at a local center and thereby removed from the CB. Later it becomes detrapped and wanders in the CB again. Eventually it is captured by a luminescent center where it recombines with a hole emitting a photon. Traps therefore delay recombination.

From Principles of Electronic Materials and Devices, Second Edition, S .O . K asap (© M cG raw -H ill, 2002) http://Materials.Usask.Ca direct Indirect UV 100-400 nm 12.4-3.10 eV Violet 400-425 nm 3.10-2.92 eV Blue 425-492 nm 2.92-2.52 eV Green 492-575 nm 2.52-2.15 eV Yellow 575-585 nm 2.15-2.12 eV Orange 585-647 nm 2.12-1.92 eV Red 647-700 nm 1.92-1.77 eV Near IR 10,000-700 nm 1.77-0.12 eV

Red Violet

Orange Blue

Yellow Green Photoconductivity Conductivity is dependent on the intensity of the incident Eg electromagnetic radiation h hων≥EEgg

E = hν = hc/λ, c = λ(m)ν(sec -1)

Band Gaps: Si - 1.11 eV (Infra red) Ge 0.66 eV (Infra red) GaAs 1.42 eV (Visible red) ZnSe 2.70 eV (Visible yellow) SiC 2.86 eV (Visible blue) GaN 3.40eV (Blue) AlN 6.20eV (Blue-UV) BN 7.50eV (UV)

Total conductivity σ = σe + σh = nqμe + pqμh For intrinsic semiconductors: n = p & σ = nq(μe + μh) Photoluminescence (PL)

Excitation and relaxation Semiconductor Devices: Light-related Three major methods for light to interact with a material: Absorption: incoming photon creates electron-hole pair (solar cell). Spontaneous Emission: electron-hole pair spontaneously decays to eject photon (LED). Stimulated Emission: incoming photon stimulates electron-hole pair to decay and eject another photon, i.e. one photon in → two photons out (LASER). hc Energy = EE21− λ E2 1 2 1 1 2 2 3 E1 Absorption Spontaneous Stimulated Emission Emission Light-emitting diode (LED) Converts electrical input to light output: electron in → photon out Device with spontaneous light emission as a result of injection of carriers across a p-n junction Light source with long life, low power, compact design. Applications: traffic and car lights, large displays. LEDs are p-n junction devices constructed of (GaAs), gallium arsenide phosphide (GaAsP), or gallium phosphide (GaP). Silicon and germanium are not suitable because those junctions produce heat and no appreciable IR or visible light. The junction in an LED is forward biased and when electrons cross the junction from the n- to the p-type material, the electron-hole recombination process produces some photons in the IR or visible in a process called electroluminescence. An exposed semiconductor surface can then emit light.

When the applied forward voltage on the diode of the LED drives the electrons and holes into the active region between the n-type and p-type material, the energy can be converted into infrared or visible photons. This implies that the electron-hole pair drops into a more stable , releasing energy on the order of electron volts by emission of a photon. The red extreme of the visible spectrum, 700 nm, requires an energy release of 1.77 eV to provide the quantum energy of the photon. At the other extreme, 400 nm in the violet, 3.1 eV is required. Light Emitting Diode (LED) Electron energy p n+ p n+ Ec eVo E Eg c E Eg F EF ≈ hυ E Ev g eVo Ev Distance into device V Electron in CB Hole in VB (a) (b) Fig. 6.43: (a) The energy band diagram of a p-n+ (heavily n-type doped) junction without any bias. Built-in potential Vo prevents electrons from diffusing from n+ to p side. (b) The applied bias reduces Vo and thereby allows electrons to diffuse, be injected, into the p-side. Recombination around the junction and within the diffusion length of the electrons in the p-side leads to photon emission.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Light output

p n+ Epitaxial layers n+ Substrate

Fig. 6.44: A schematic illustration of one possible LED device structure. First n+ is epitaxially grown on a substrate. A thin p layer is then epitaxially grown on the first layer. Ec EN From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Eg

Ev

(b) N doped GaP (a) GaAs1-yPy y < 0.45

Fig. 6.45: (a) Photon emission in a direct bandgap semiconductor. (b) GaP is an indirect bandgap semiconductor. When doped with nitrogen there is an electron recombination center at EN. Direct recombination between a captured electron at EN and a hole emits a photon.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca p p (a) AlGaAs GaAs AlGaAs

~ 0.2 μm Ec ΔEc Electrons in CB No bias eVo 2 eV EF 1.4 eV E E F c Ev (b) 2 eV Holes in VB Ev

With forward bias (c)

n+ p p (d)

AlGaAs GaAs AlGaAs

Fig. 6:46: (a) A double heterostructure diode has two junctions which are between two different bandgap semiconductors (GaAs and AlGaAs). (b) A simplified energy band diagram with exaggerated features. EF must be uniform. (c) Forward biased simplified energy band diagram. (d) Forward biased LED. Schematic illustration of photons escaping reabsorption in the AlGaAs layer and being emitted from the device. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Solar Cell Converts light input to electrical output: photon in → electron out (generated electrons are “swept away” by E field of p-n junction) Renewable energy source!

If light (with E>Eg) generates free electrons and holes in the depleted region, the electric field makes these carriers move. The electrons generated by light move from the p-side to the n-side. The holes generated by light move from the n-side to the p-side. Light is converted to electrical energy. contact metal contact n-side − − − − − − + + + + + + p-side E

I “load”, e.g., motor − V + The Solar Cell

front metal grid n-type region sun

pn junction

electron current

rear metal contact p-type region + external circuit - Solar cell under illumination

V=IR Conventional current _ + n-type p-type + _ + - - - _ + + _ + + - + _ + - - + - + + _ _ _ + + sun + _ + + + + - - _ - + + + _ _ + - + + _ - + + - - + Key point In an operating diode: – The voltage is in the forward direction – The current is in the forward direction In an operating solar cell: – The voltage is in the forward direction – The current is in the reverse direction

Consider a pn junction with a very narrow and more heavily doped n‐region. The illumination is through the thin n‐side. The depletion region W or the space charge layer (SCL) extends primarily into the p‐side. There is a built‐in field Eo in this depletion layer. The electrode attached to the n‐side must allow illumination to enter the device. As the n‐side is very narrow, most of the photons are absorbed within the depletion region (W) and within the neutral p‐side. Neutral Neutral n-region Eo p-region

Diffusion Drift Long λ

Medium λ L e Back electrode

Short λ Finger electrode Lh

Depletion region

ln W lp

Voc

Fig. 6.49: The principle of operation of the solar cell (exaggerated features to highlight principles)

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca LASER LASER = Light Amplification by Stimulated Emission of Radiation

• Laser creates inverted population of electrons in upper energy levels and then stimulates them to all coherently decay to lower energy levels. GaAs Laser • Applications: fiber optics, CD player, machining, medicine, etc. e.g. GaAs laser: 25% efficiency, 100 yr lifetime, mm size, IR to visible

Population Inversion and Optical Pumping The number of atoms in the valence band (lower energy) is more than that in the excited state (conduction band). According to Boltzmann, the ratio of atoms in the energy states j and i at a temperature T is given by E j − ()−EE ij N kT − j e kT E == e N − i i e kT

〈〉 NNEEAs ijij

Any method by which the atoms are directly and continuously excited from the ground state to the excited state (such as optical absorption) will eventually reach equilibrium with the de‐exciting processes of spontaneous and stimulated emission. At best, an equal population of the two states, N1 = N2 = N/2, can be achieved, resulting in optical transparency but no net optical gain. To emit photons which are coherent (in same phase), the number of atoms in the higher energy state must be greater than that in the ground state (lower energy). The process of making population of atoms in the higher energy state more than that in the lower energy state is known as ‘population inversion’. The method by which a population inversion is affected is called ‘optical pumping’. In this process atoms are raised to an excited state by injecting into system photon of frequency different from the stimulating frequency. Population inversion can be understood with the help of 3‐energy level atomic systems.

E2 Excited State

E1 Meta Stable State

E0 Ground Atoms State Atoms E2 Excited State

hν E Meta Stable State hν 1 hν E Ground State Pumping 0 E2 E2

E1 E1 Atoms hν’ Atoms hν’ E hν’ E0 0 Rapid fall after 10-8 s After Stimulated Emission

E0, E1 and E2 represent ground state, meta‐stable state (temporarily stable state) and excited state respectively.

The atoms by induced absorption reach excited state E2 from E0. They stay there only for 10‐8 seconds. After this time they fall to meta‐stable state where they stay for quite a longer time (10‐3 seconds). Within this longer time more number of atoms get collected in the meta‐stable state which is large than that at lower energy level. Thus population inversion is achieved. In atomic systems such as chromium, neon, etc, meta‐stable states exist. Diode Laser: is a variant of LED in which its special construction help to produce stimulated radiation as in laser. In conventional solid state or gas laser, discrete atomic energy levels are involved whereas in semiconductor lasers, the transitions are associated with the energy bands. In forward biased p‐n junction of LED, the higher energy level (conduction band) is more populated than the lower energy level (valence band), which is the primary requirement for the population inversion.

When a photon of energy hν = Eg impinges the device, while it is still in the excited state due to the applied bias, the system is immediately stimulated to make its transition to the valence band and gives an additional photon of energy hν which is in phase with the incident photon.

+ Roughened hν surface Ec P hν Optically N hν P flat side N Laser beam Ev - The perpendicular to the plane of the junction are polished. The remaining sides of the diode are roughened. When a forward bias is applied, a current flows. Initially at low current, there is spontaneous emission (as in LED) in all the directions. Further, as the bias is increased, a threshold current is reached at which the stimulated emission occurs. Due to the plane polished surfaces, the stimulated radiation in the plane perpendicular to the depletion layer builds up due to multiple reflections in the cavity formed by these surfaces and a highly directional coherent radiation is emitted. Diode lasers are low power lasers used as optical light source in optical communication. •A pn junction in a direct bandgap material will produce light when forward biased. However, re‐absorption (photon recycling) is likely and thus should be avoided. •Use of quantum wells in the “active region” (region where minority carriers are injected and recombine from the “majority carrier” (source of holes) and cathode (source of electrons) results in minimal re‐absorption since the emitted light is below the bandgap of the cladding layers (higher bandgap regions). •The also strongly confines the electrons and holes to the same region of the material enhancing the probability of recombination and thus enhancing the radiation efficiency (light power out/electrical power in).

The transistor The transistor was invented in 1947 by three American physicists at the Bell Telephone Laboratories, John Bardeen, Walter H. Brattain, and William B. Shockley. It proved to be a viable alternative to the vacuum tube, and by the late 1950s supplanted the latter in many applications. “Transistor” is short for “Transfer Resistor” or modulation of the resistance between two terminal by applying an electrical signal to a third terminal. Capable of two primary types of function: (1)It amplify an electrical signal. (2)Serve as switching devices (on/off) in computers for the processing and storage of information There are two major types: (A)Bipolar Junction Transistor (BJT) (B)Metal Oxide Semiconductor Field Effect Transistor (MOSFET) BJT It is composed of two p-n junctions arranged back-to-back in either the n-p-n or the p-n-p confguration. p-n-p: A very thin n-type base region is sandwiched in between the p-type emitter and collector regions. Region emitter-base junction is forward bias (junction 1), whereas a reverse bias voltage is applied across the base-collector junction (junction 2). Since the emitter is p-type and junction 1 is forward biased, large numbers of holes enter the base region. These injected holes are minority carriers in the n- type base and some will combine with the majority electrons. If the base is extremely narrow, most of the holes will be swept through the base without recombination, then across junction 2 and into the p-type collector. The holes become part of the emitter-collector circuit. A small increase in input voltage within the emitter-base circuit produces a large increase in current across junction 2. The large increase in collector current is also reflected by a large increase in voltage across the load resistor. n-p-n Depletion Mode p-type The MOSFET Metal-Oxide Semiconductor Field- Depletion Effect Transistor Mode n-type Depletion mode p-type It consists of two small islands of p-type semiconductor, that are created within a substrate of n-type silicon. The islands are joined by a narrow p-type channel. Appropriate metal connections (source and drain) are made to these islands; an insulting layer of silicon dioxide is formed by the surface oxidation of the silicon. A final connector (gate) is then fashioned onto the surface of this insulating layer. An electric field imposed on the gate varied the conductivity of the channel. If the electric field is positive, it will drive charge carriers (holes) out of the channel, reducing the electrical conductivity. Thus a small change in the gate field will produce a large variation in current between the source and the drain. Primary difference between BJT and MOSFET is that the gate current is small in comparison to the base current in the BJT. are used where the signal sources to be amplified can not sustain an appreciable current.