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Home , Donor (semiconductors), Extrinsic semiconductor

Extrinsic Extrinsic semiconductors : impurity atoms dictate the properties Almost all commercial semiconductors are extrinsic Impurity concentrations of 1 atom in 1012 is enough to make extrinsic at room T! Impurity atoms can create states that are in the band gap. In most cases, the of a leads either to the creation of 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 in Semiconductor

For Si, add a group III element to “accept” an 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, , 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 !). 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 , , 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

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 no ≈ ND where no 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 po = ni / ND where po 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), po, will be approximately equal to the acceptor concentration, NA.

po = NA and the number of negative carriers in the p-type material, no, is given by 2 no = 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 a n-type silicon which contains 1016 electrons per cubic centimeter with an electron mobility of 1400 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. Example A Si sample is doped with 10-4 atomic% of P donors. Assuming complete ionisation of donors at room temperature, calculate the concentration and conductivity at room temperature. -3 2 -1 -1 2 -1 -1 [For Si: ρ = 2330 kg m , atomic weight = 28, μe = 0.15 m V s , μh = 0.05 m V s , 10 3 ni= 1.5x10 carriers per cm ] Solution: 1) Calculate the fraction of donor atoms (phosphorus atoms per silicon atom) where N – number of Si atoms per unit volume N Si D =10−6 N 2) Calculate the number of silicon atoms per unit volume Si −3 ρ 2330 .mKg 23 −1 28 3 NSi NAvogadro =×= −3 −1 ×× 106 .molatoms 105 −×= .matomsSi ASi ×1028 .molKg 22 −3 3) Calculate the number of donors atoms (phosphorus) ND 105 −×= .matomsP 4) As NA=0 and ND>>ni , then we can safely assume that no=ND and po is very small ~ zero

σ o qn ××= μe σ ()105 22 −×= matomsP 3 ().. 19C ××× ()1501061 sVm. -1-12 σ 1200Ω= .m −− 11 Consider the following equations: Mass Action: Neutrality: 2 =× npn ioo + = + NpNn DoAo

Solving the equations simultaneously Electron concentration (n-type semiconductor) Hole concentration (p-type semiconductor)

2 2 1 ⎡ 2 2 ⎤ 1 ⎡ ⎤ ()()+−+−= 4nNNNNn o 2 ()()DA DA +−+−= 4nNNNNp i o 2 ⎣⎢ AD AD i ⎦⎥ ⎣⎢ ⎦⎥ 2 2 ni ni no = po = po no Solving the problem again. NA=0

1 ⎡()()2 +−+−= 4nNNNNn 2 ⎤ o 2 ⎣⎢ AD AD i ⎦⎥

1 2 2 1 ⎡ 2 ++= 4nNNn 2 ⎤ ⎡ 22 ( 22 ) ( . ×+×+×= 10514105105 16 ) ⎤ o 2 ⎣⎢ D iD ⎦⎥ 2 ⎣⎢ ⎦⎥ 22 no ×= 105

2 16 2 ni (). ×1051 −310 po == 22 . ×= 10450 m no ×105

σ ()μ pnq ×+××= μhoeo σ ()(. 19C []1051061 22 −×××= matomsP 3 × 150 -1-12 .sVm.. )()()10450 10 ××+ () 0.05 σ 1200Ω= .m −− 11 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 electron density, n ?

2 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, no po = ni

And no + NA = po + ND 2 2 → no –ND no –ni = 0 11 -3 Solving this quadratic equations results in n0 = 1.02x10 [cm ] 2 20 11 and thus, p0 = ni / n0 = 2.25x10 / 1.02x10 9 -3 p0 = 2.2x10 [cm ]

Notice that, since ND>ni, the results would be very similar if we assumed 11 -3 no=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 (, 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 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 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

Electron and hole mobility in silicon The mobility of electrons and holes in silicon at room temperature is shown in the figure below.

Fig. Electron and hole mobility versus doping density for silicon The electron and hole mobilities have a similar doping dependence: For low doping concentrations the mobility is almost constant and is primarily limited by phonon scattering. At higher doping concentrations the mobility decreases due to ionized impurity scattering with the ionized doping atoms. The actual mobility also depends on the type of dopant. These are empirical relations obtained by fitting experimental values. The above figure is for phosphorous and boron doped silicon and is calculated using:

And Example A silicon wafer contains 1018 cm-3 phosphor atoms. Using the data in the table; calculate the resistivity and conductivity of the material. Repeat for arsenic and boron atoms. Solution Plugging the values from table

into the following equation

one obtains a mobility of 277 cm2/V-sec for phosphorus- doped material, 284 cm2/V-sec for arsenic-doped material and 153 cm2/V-sec for boron-doped material, corresponding to a resistivity of 22.6, 22.0 and 40.9 mΩcm and a conductivity of 44.3, 45.4 and 24.5 1/Ωcm. 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). G G G 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 (B). 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 Semiconductors (thermistors) p-n junctions () Transistors bipolar junction transistors (BJT’s) field effect transistors (FET) Optoelectronic and photonic devices: light emitting diodes (LED’s) semiconductor solar cells

Pentium 4 has 42 MILLION transistors in one processor! 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

current equal to ~Io. • Ability of p-n junction to pass Reverse Bias current in only one direction is known as “rectifying” behavior.