1. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/
Semiconductor Device Modeling and Characterization EE5342, Lecture 3-Spring 2005
Professor Ronald L. Carter
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4. Silicon Band Structure**
Indirect Bandgap
Curvature (hence m*) is function of direction and band. [100] is x-dir, [111] is cube diagonal
Eg = 1.17-aT2/(T+b) a = 4.73E-4 eV/K b = 636K
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5. Analogy: a nearly -free electr. model
Solutions can be displaced by ka = 2np
Allowed and forbidden energies
Infinite well approximation by replacing the free electron mass with an “effective” mass (noting E = p2/2m = h2k2/2m) of
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6. Si Energy Band Structure at 0 K
Every valence site is occupied by an electron
No electrons allowed in band gap
No electrons with enough energy to populate the conduction band
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7. Silicon Covalent Bond (2D Repr)
Each Si atom has 4 nearest neighbors
Si atom: 4 valence elec and 4+ ion core
8 bond sites / atom
All bond sites filled
Bonding electrons shared 50/50
_ = Bonding electron
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8. Si Bond Model Above Zero Kelvin
Enough therm energy ~kT(k=8.62E-5eV/K) to break some bonds
Free electron and broken bond separate
One electron for every “hole” (absent electron of broken bond)
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9. Band Model for thermal carriers
Thermal energy ~kT generates electron-hole pairs
At 300K
Eg(Si) = eV
>> kT = meV,
Nc = 2.8E19/cm3
> Nv = 1.04E19/cm3
>> ni = 1.45E10/cm3
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10. Donor: cond. electr. due to phosphorous
P atom: 5 valence elec and 5+ ion core
5th valence electr has no avail bond
Each extra free el, -q, has one +q ion
# P atoms = # free elect, so neutral
H atom-like orbits
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11. Bohr model H atom- like orbits at donor
Electron (-q) rev. around proton (+q)
Coulomb force, F=q2/4peSieo,q=1.6E-19 Coul, eSi=11.7, eo=8.854E-14 Fd/cm
Quantization L = mvr = nh/2p
En= -(Z2m*q4)/[8(eoeSi)2h2n2] ~-40meV
rn= [n2(eoeSi)h2]/[Zpm*q2] ~ 2 nm
for Z=1, m*~mo/2, n=1, ground state
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12. Band Model for donor electrons
Ionization energy of donor Ei = Ec-Ed ~ 40 meV
Since Ec-Ed ~ kT, all donors are ionized, so ND ~ n
Electron “freeze-out” when kT is too small
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13. Acceptor: Hole due to boron
B atom: 3 valence elec and 3+ ion core
4th bond site has no avail el (=> hole)
Each hole, adds --q, has one -q ion
#B atoms = #holes, so neutral
H atom-like orbits
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14. Hole orbits and acceptor states
Similar to free electrons and donor sites, there are hole orbits at acceptor sites
The ionization energy of these states is EA - EV ~ 40 meV, so NA ~ p and there is a hole “freeze-out” at low temperatures
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15. Impurity Levels in Si: EG = 1,124 meV
Phosphorous, P: EC - ED = 44 meV
Arsenic, As: EC - ED = 49 meV
Boron, B: EA - EV = 45 meV
Aluminum, Al: EA - EV = 57 meV
Gallium, Ga: EA - EV = 65meV
Gold, Au: EA - EV = 584 meV EC - ED = 774 meV
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16. Semiconductor Electronics - concepts thus far
Conduction and Valence states due to symmetry of lattice
“Free-elec.” dynamics near band edge
Band Gap
direct or indirect
effective mass in curvature
Thermal carrier generation
Chemical carrier gen (donors/accept)
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17. Counting carriers - quantum density of states function
1 dim electron wave #s range for n+1 “atoms” is 2p/L < k < 2p/a where a is “interatomic” distance and L = na is the length of the assembly (k = 2p/l)
Shorter ls, would “oversample”
if n increases by 1, dp is h/L
Extn 3D: E = p2/2m = h2k2/2m so a vol of p-space of 4pp2dp has h3/LxLyLz
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18. 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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19. 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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20. 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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21. 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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22. Electron Conc. in the MB approx.
Assuming the MB approx., the equilibrium electron concentration is
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23. 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 = 1E10/cm3
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24. Calculating the equilibrium no
The ideal 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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25. 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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26. 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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27. 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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28. 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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29. 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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30. 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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31. 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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32. Equilibrium conc n-type
For Nd > Na
Let N = Nd-Na, and (taking the + root) no = (N)/2 + {[N/2]2+ni2}1/2
For Nd+= Nd >> ni >> Na we have
no = Nd, and
po = ni2/Nd
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33. Equilibrium conc p-type
For Na > Nd
Let N = Nd-Na, and (taking the + root) po = (-N)/2 + {[-N/2]2+ni2}1/2
For Na-= Na >> ni >> Nd we have
po = Na, and
no = ni2/Na
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34. 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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35. 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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36. 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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37. Locating Efi in the bandgap
Since Ec - Efi = kT ln(Nc/ni), and Efi - Ev = kT ln(Nv/ni)
The sum of the two equations 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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38. 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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39. Equilibrium electron conc. and energies
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40. Equilibrium hole conc. and energies
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41. 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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42. References
1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986.
2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.
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