EE 5340Semiconductor Device TheoryLecture 4 - Fall 2009

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc

L 04 Sept 04

First Assignment

• Send e-mail to ronc@uta.edu– On the subject line, put “5340 e-mail”– In the body of message include

• email address: ______________________• Your Name*: _______________________• Last four digits of your Student ID: _____

* Your name as it appears in the UTA Record - no more, no less

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L 04 Sept 04

Second Assignment

• Please print and bring to class a signed copy of the document appearing at

http://www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf

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Maxwell-BoltzmanApproximation• fF(E) = {1+exp[(E-EF)/kT]}-1

• For E - EF > 3 kT, the exp > 20, so within a 5% error, fF(E) ~ exp[-(E-EF)/kT]

• This is the MB distribution function

• MB used when E-EF>75 meV (T=300K)

• For electrons when Ec - EF > 75 meV and for holes when EF - Ev > 75 meV

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Electron Conc. inthe MB approx.• Assuming the MB approx., the

equilibrium electron concentration is

kTEE

expNn

dEEfEgn

Fcco

E

Eco F

max

c

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Electron and HoleConc in MB approx• Similarly, the equilibrium hole

concentration ispo = Nv exp[-(EF-Ev)/kT]

• So that nopo = NcNv exp[-Eg/kT]

• ni2 = nopo, Nc,v = 2{2m*n,pkT/h2}3/2

• Nc = 2.8E19/cm3, Nv = 1.04E19/cm3 and ni = 1.45E10/cm3

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Calculating theequilibrium no

• The idea is to calculate the equilibrium electron concentration no for the FD distribution, where

fF(E) = {1+exp[(E-EF)/kT]}-1

gc(E) = [42mn*)3/2(E-Ec)1/2]/h3

dEEfEgnF

max

c

E

Eco

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Equilibrium con-centration for no

• Earlier quoted the MB approximation no = Nc exp[-(Ec - EF)/kT],(=Nc exp F)

• The exact solution is no = 2NcF1/2(F)/1/2

• Where F1/2(F) is the Fermi integral of order 1/2, and F = (EF - Ec)/kT

• Error in no, is smaller than for the DF: = 31%, 12%, 5% for -F = 0, 1, 2

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Equilibrium con-centration for po

• Earlier quoted the MB approximation po = Nv exp[-(EF - Ev)/kT],(=Nv exp ’F)

• The exact solution is po = 2NvF1/2(’F)/1/2

• Note: F1/2() = 0.678, (1/2/2) = 0.886

• Where F1/2(’F) is the Fermi integral of order 1/2, and ’F = (Ev - EF)/kT

• Errors are the same as for po

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Figure 1.10 (a) Fermi-Dirac distribution function describing the probability that an allowed state at energy E is occupied by an electron. (b) The density of allowed states for a semiconductor as a function of energy; note that g(E) is zero in the forbidden gap between Ev and Ec.(c) The product of the distribution function and the density-of-states function. (p. 17 - M&K)

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Degenerate andnondegenerate cases• Bohr-like doping model assumes no

interaction between dopant sites• If adjacent dopant atoms are within 2

Bohr radii, then orbits overlap

• This happens when Nd ~ Nc (EF ~ Ec), or when Na ~ Nv (EF ~ Ev)

• The degenerate semiconductor is defined by EF ~/> Ec or EF ~/< Ev

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Figure 1.13 Energy-gap narrowing Eg as a function of electron concentration. [A. Neugroschel, S. C. Pao, and F. A. Lindhold, IEEE Trans. Electr. Devices, ED-29, 894 (May 1982).] taken from p. 25 - M&K)

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Donor ionization

• The density of elec trapped at donors is nd = Nd/{1+[exp((Ed-EF)/kT)/2]}

• Similar to FD DF except for factor of 2 due to degeneracy (4 for holes)

• Furthermore nd = Nd - Nd+, also

• For a shallow donor, can have Ed-EF >> kT AND Ec-EF >> kT: Typically EF-Ed ~ 2kT

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Donor ionization(continued)• Further, if Ed - EF > 2kT, then

nd ~ 2Nd exp[-(Ed-EF)/kT], < 5%

• If the above is true, Ec - EF > 4kT, sono ~ Nc exp[-(Ec-EF)/kT], < 2%

• Consequently the fraction of un-ionized donors is nd/no = 2Nd exp[(Ec-Ed)/kT]/Nc = 0.4% for Nd(P) = 1e16/cm3

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Figure 1.9 Electron concentration vs. temperature for two n-type doped semiconductors:(a) Silicon doped with 1.15 X 1016 arsenic atoms cm-3[1], (b) Germanium doped with 7.5 X 1015 arsenic atoms cm-3[2]. (p.12 in M&K)

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Classes ofsemiconductors• Intrinsic: no = po = ni, since Na&Nd << ni

=[NcNvexp(-Eg/kT)]1/2, (not easy to get)

• n-type: no > po, since Nd > Na

• p-type: no < po, since Nd < Na

• Compensated: no=po=ni, w/ Na- = Nd

+ > 0

• Note: n-type and p-type are usually partially compensated since there are usually some opposite- type dopants

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Equilibriumconcentrations• Charge neutrality requires

q(po + Nd+) + (-q)(no + Na

-) = 0

• Assuming complete ionization, so Nd

+ = Nd and Na- = Na

• Gives two equations to be solved simultaneously

1. Mass action, no po = ni2, and

2. Neutrality po + Nd = no + Na

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Equilibriumconc (cont.)• For Nd > Na (taking the + root) no

= (Nd-Na)/2 + {[(Nd-Na)/2]2+ni2}1/2

• For Nd >> Na and Nd >> ni, can use the binomial expansion, giving

no = Nd/2 + Nd/2[1 + 2ni2/Nd

2 + … ]

• So no = Nd, and po = ni2/Nd in the limit

of Nd >> Na and Nd >> ni

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Examplecalculations• For Nd = 3.2E16/cm3, ni = 1.4E10/cm3

no = Nd = 3.2E16/cm3

po = ni2/Nd , (po is always ni

2/no)

= (1.4E10/cm3)2/3.2E16/cm3

= 6.125E3/cm3 (comp to ~1E23 Si)

• For po = Na = 4E17/cm3,

no = ni2/Na = (1.4E10/cm3)2/4E17/cm3

= 490/cm3

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Position of theFermi Level• Efi is the Fermi level

when no = po

• Ef shown is a Fermi level for no > po

• Ef < Efi when no < po

• Efi < (Ec + Ev)/2, which is the mid-band

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EF relative to Ec and Ev• Inverting no = Nc exp[-(Ec-EF)/kT] gives

Ec - EF = kT ln(Nc/no) For n-type material: Ec - EF

=kTln(Nc/Nd)=kTln[(Ncpo)/ni2]

• Inverting po = Nv exp[-(EF-Ev)/kT] gives EF - Ev = kT ln(Nv/po) For p-type material: EF - Ev = kT ln(Nv/Na)

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EF relative to Efi

• Letting ni = no gives Ef = Efi ni = Nc exp[-(Ec-Efi)/kT], so Ec - Efi = kT ln(Nc/ni). Thus EF - Efi = kT ln(no/ni) and for n-type EF - Efi = kT ln(Nd/ni)

• Likewise Efi - EF = kT ln(po/ni) and for p-type Efi - EF = kT ln(Na/ni)

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Locating Efi in the bandgap • Since

Ec - Efi = kT ln(Nc/ni), andEfi - Ev = kT ln(Nv/ni)

• The 1st equation minus the 2nd givesEfi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv)

• Since Nc = 2.8E19cm-3 > 1.04E19cm-3 = Nv, the intrinsic Fermi level lies below the middle of the band gap

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Samplecalculations• Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv), so at

300K, kT = 25.86 meV and Nc/Nv = 2.8/1.04, Efi is 12.8 meV or 1.1% below mid-band

• For Nd = 3E17cm-3, given that Ec - EF = kT ln(Nc/Nd), we have Ec - EF

= 25.86 meV ln(280/3), Ec - EF = 0.117 eV =117meV ~3x(Ec - ED) what Nd

gives Ec-EF =Ec/3

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Equilibrium electronconc. and energies

o

v2i

vof

i

ofif

fif

i

o

c

ocf

cf

c

o

pN

lnkTn

NnlnkTEvE and

;nn

lnkTEE or ,kT

EEexp

nn

;Nn

lnkTEE or ,kT

EEexp

Nn

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Equilibrium hole conc. and energies

o

c2i

cofc

i

offi

ffi

i

o

v

ofv

fv

v

o

nN

lnkTn

NplnkTEE and

;np

lnkTEE or ,kT

EEexp

np

;Np

lnkTEE or ,kT

EEexp

Np

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Carrier Mobility

• In an electric field, Ex, the velocity (since ax = Fx/m* = qEx/m*) is

vx = axt = (qEx/m*)t, and the displ

x = (qEx/m*)t2/2

• If every coll, a collision occurs which “resets” the velocity to <vx(coll)> = 0, then <vx> = qExcoll/m* = Ex

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Carrier mobility (cont.)• The response function is the

mobility.• The mean time between collisions,

coll, may has several important causal events: Thermal vibrations, donor- or acceptor-like traps and lattice imperfections to name a few.

• Hence thermal = qthermal/m*, etc.

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Carrier mobility (cont.)• If the rate of a single contribution

to the scattering is 1/i, then the total scattering rate, 1/coll is

all

collisions itotal

all

collisions icoll

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by given is mobility total

the and , 11

Figure 1.16 (p. 31 M&K) Electron and hole mobilities in silicon at 300 K as functions of the total dopant concentration. The values plotted are the results of curve fitting measurements from several sources. The mobility curves can be generated using Equation 1.2.10 with the following values of the parameters [3] (see table on next slide).

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Figure 1.16 (cont. M&K)

Parameter Arsenic Phosphorus Boronμmin 52.2 68.5 44.9

μmax 1417 1414 470.5

Nref 9.68 X 1016 9.20 X 1016 2.23 X 1017

α 0.680 0.711 0.719

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αμμ

μμref

minmaxmin NN1

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Drift Current

• The drift current density (amp/cm2) is given by the point form of Ohm LawJ = (nqn+pqp)(Exi+ Eyj+ Ezk), so

J = (n + p)E = E, where

= nqn+pqp defines the conductivity

• The net current is SdJI

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Net silicon extrresistivity (cont.)• Since = (nqn + pqp)-1, and

n > p, ( = q/m*) we have

p > n

• Note that since1.6(high conc.) < p/n < 3(low conc.), so

1.6(high conc.) < n/p < 3(low conc.)

Figure 1.15 (p. 29) M&K Dopant density versus resistivity at 23°C (296 K) for silicon doped with phosphorus and with boron. The curves can be used with little error to represent conditions at 300 K. [W. R. Thurber, R. L. Mattis, and Y. M. Liu, National Bureau of Standards Special Publication 400–64, 42 (May 1981).]

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References

*Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989.

**Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago.

M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003.

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