1. EE 5340 Semiconductor Device Theory Lecture 4 - Fall 2010
Professor Ronald L. Carter

2. First Assignment Send e-mail to ronc@uta.edu
On the subject line, put “5340 ”
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Your Name*: _______________________
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* Your name as it appears in the UTA Record - no more, no less
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3. Second Assignment
Please print and bring to class a signed copy of the document appearing at
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4. QM density of states (cont.)
So density of states, gc(E) is (Vol in p-sp)/(Vol per state*V) = 4pp2dp/[(h3/LxLyLz)*V]
Noting that p2 = 2mE, this becomes gc(E) = {4p(2mn*)3/2/h3}(E-Ec)1/2 and E - Ec = h2k2/2mn*
Similar for the hole states where Ev - E = h2k2/2mp*
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5. Fermi-Dirac distribution fctn
The probability of an electron having an energy, E, is given by the F-D distr fF(E) = {1+exp[(E-EF)/kT]}-1
Note: fF (EF) = 1/2
EF is the equilibrium energy of the system
The sum of the hole probability and the electron probability is 1
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6. Fermi-Dirac DF (continued)
So the probability of a hole having energy E is 1 - fF(E)
At T = 0 K, fF (E) becomes a step function and 0 probability of E > EF
At T >> 0 K, there is a finite probability of E >> EF
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7. Maxwell-Boltzman Approximation
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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8. Electron Conc. in the MB approx.
Assuming the MB approx., the equilibrium electron concentration is
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9. Electron and Hole Conc in MB approx
Similarly, the equilibrium hole concentration is po = Nv exp[-(EF-Ev)/kT]
So that nopo = NcNv exp[-Eg/kT]
ni2 = nopo, Nc,v = 2{2pm*n,pkT/h2}3/2
Nc = 2.8E19/cm3, Nv = 1.04E19/cm3 and ni = 1.45E10/cm3
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10. Calculating the equilibrium 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) = [4p(2mn*)3/2(E-Ec)1/2]/h3
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11. Equilibrium con- centration for no
Earlier quoted the MB approximation no = Nc exp[-(Ec - EF)/kT],(=Nc exp hF)
The exact solution is no = 2NcF1/2(hF)/p1/2
Where F1/2(hF) is the Fermi integral of order 1/2, and hF = (EF - Ec)/kT
Error in no, e, is smaller than for the DF: e = 31%, 12%, 5% for -hF = 0, 1, 2
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12. Equilibrium con- centration for po
Earlier quoted the MB approximation po = Nv exp[-(EF - Ev)/kT],(=Nv exp h’F)
The exact solution is po = 2NvF1/2(h’F)/p1/2
Note: F1/2(0) = 0.678, (p1/2/2) = 0.886
Where F1/2(h’F) is the Fermi integral of order 1/2, and h’F = (Ev - EF)/kT
Errors are the same as for po
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13. Figure (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 M&K)
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14. Degenerate and nondegenerate 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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15. Figure 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 M&K)
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16. 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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17. Donor ionization (continued)
Further, if Ed - EF > 2kT, then nd ~ 2Nd exp[-(Ed-EF)/kT], e < 5%
If the above is true, Ec - EF > 4kT, so no ~ Nc exp[-(Ec-EF)/kT], e < 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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18. Figure 1. 9 Electron concentration vs
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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19. Classes of semiconductors
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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20. Equilibrium concentrations
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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21. Equilibrium conc (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/Nd2 + … ]
So no = Nd, and po = ni2/Nd in the limit of Nd >> Na and Nd >> ni
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22. Example calculations For Nd = 3.2E16/cm3, ni = 1.4E10/cm3
no = Nd = 3.2E16/cm3
po = ni2/Nd , (po is always ni2/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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23. Position of the Fermi 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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24. 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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25. 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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26. Locating Efi in the bandgap
Since Ec - Efi = kT ln(Nc/ni), and Efi - Ev = kT ln(Nv/ni)
The 1st equation minus the 2nd gives Efi = (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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27. Sample calculations
Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv), so at 300K, kT = 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 = meV ln(280/3), Ec - EF = eV =117meV ~3x(Ec - ED) what Nd gives Ec-EF =Ec/3
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28. Equilibrium electron conc. and energies
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29. Equilibrium hole conc. and energies
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30. vx = axt = (qEx/m*)t, and the displ
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 tcoll, a collision occurs which “resets” the velocity to <vx(tcoll)> = 0, then <vx> = qExtcoll/m* = mEx
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31. Carrier mobility (cont.)
The response function m is the mobility.
The mean time between collisions, tcoll, may has several important causal events: Thermal vibrations, donor- or acceptor-like traps and lattice imperfections to name a few.
Hence mthermal = qtthermal/m*, etc.
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32. Carrier mobility (cont.)
If the rate of a single contribution to the scattering is 1/ti, then the total scattering rate, 1/tcoll is
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33. Figure (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 with the following values of the parameters [3] (see table on next slide).
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34. Parameter Arsenic Phosphorus Boron
μmin
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
Figure (cont. M&K)
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35. Drift Current
The drift current density (amp/cm2) is given by the point form of Ohm Law
J = (nqmn+pqmp)(Exi+ Eyj+ Ezk), so
J = (sn + sp)E = sE, where
s = nqmn+pqmp defines the conductivity
The net current is
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36. Net silicon extr resistivity (cont.)
Since
r = (nqmn + pqmp)-1, and
mn > mp, (m = qt/m*) we have
rp > rn
Note that since
1.6(high conc.) < rp/rn < 3(low conc.), so
1.6(high conc.) < mn/mp < 3(low conc.)
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37. Figure (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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38. 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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