1. Homogeneous Semiconductors
Dopants
Use Density of States and Distribution Function to:
Find the Number of Holes and Electrons.

2. Energy Levels in Hydrogen Atom

3. Energy Levels for Electrons in a Doped Semiconductor

4. Assumptions for Calculation

8. Density of States (Appendix D) Energy Distribution Functions (Section 2.9) Carrier Concentrations (Sections )

9. GOAL: The density of electrons (no) can be found precisely if we know
1. the number of allowed energy states in a small energy range, dE: S(E)dE
“the density of states”
2. the probability that a given energy state will be occupied by an electron: f(E)
“the distribution function”
no = bandS(E)f(E)dE

10. S(E) = (1/2p2) (2mdse*/2)3/2(E - EC)1/2
For quasi-free electrons in the conduction band: 1. We must use the effective mass (averaged over all directions) 2. the potential energy Ep is the edge of the conduction band (EC)
S(E) = (1/2p2) (2mdse*/2)3/2(E - EC)1/2
For holes in the valence band: 1. We still use the effective mass (averaged over all directions) 2. the potential energy Ep is the edge of the valence band (EV)
S(E) = (1/2p2) (2mdsh*/2)3/2(EV - E)1/2

11. Energy Band Diagram E(x) Eelectron S(E) Ehole x
conduction band
valence band EC EV x
E(x)
Eelectron
S(E)
Ehole
note: increasing electron energy is ‘up’, but increasing hole energy is ‘down’.

12. Reminder of our GOAL:
The density of electrons (no) can be found precisely if we know
1. the number of allowed energy states in a small energy range, dE: S(E)dE
“the density of states”
2. the probability that a given energy state will be occupied by an electron: f(E)
“the distribution function”
no = bandS(E)f(E)dE

13. Fermi-Dirac Distribution
The probability that an electron occupies an energy level, E, is
f(E) = 1/{1+exp[(E-EF)/kT]}
where T is the temperature (Kelvin)
k is the Boltzmann constant (k=8.62x10-5 eV/K)
EF is the Fermi Energy (in eV)

14. f(E) = 1/{1+exp[(E-EF)/kT]}EF E
T=0 oK
T1>0
T2>T1
0.5
All energy levels are filled with e-’s below the Fermi Energy at 0 oK

15. Fermi-Dirac Distribution for holes
Remember, a hole is an energy state that is NOT occupied by an electron.
Therefore, the probability that a state is occupied by a hole
is the probability that a state is NOT occupied by an electron:
fp(E) = 1 – f(E) = 1 - 1/{1+exp[(E-EF)/kT]}
={1+exp[(E-EF)/kT]}/{1+exp[(E-EF)/kT]} /{1+exp[(E-EF)/kT]}
= {exp[(E-EF)/kT]}/{1+exp[(E-EF)/kT]}
=1/{exp[(EF - E)/kT] + 1}
for the last line, multiply by 1 = exp(-(E-Ef)/kT)/exp(-(E-Ef)/kT)

16. The Boltzmann Approximation If (E-EF)>kT such that exp[(E-EF)/kT] >> 1 then,
f(E) = {1+exp[(E-EF)/kT]}-1  {exp[(E-EF)/kT]}-1
 exp[-(E-EF)/kT] …the Boltzmann approx.
similarly, fp(E) is small when exp[(EF - E)/kT]>>1:
fp(E) = {1+exp[(EF - E)/kT]}-1  {exp[(EF - E)/kT]}-1
 exp[-(EF - E)/kT]
If the Boltz. approx. is valid, we say the semiconductor is non-degenerate.

17. Putting the pieces together: for electrons, n(E)
f(E) 1 EF E
T=0 oK
T1>0
T2>T1
0.5 EV EC
S(E) E
n(E)=S(E)f(E)

18. Putting the pieces together:
for holes, p(E)
fp(E)
T=0 oK 1
T1>0
T2>T1
0.5
S(E) E EV EF EC
p(E)=S(E)f(E)
hole energy

19. Finding no and po the effective density of states
in the conduction band

20. Energy Band Diagram intrinisic semiconductor: no=po=ni
E(x)
conduction band EC
n(E)
EF=Ei
p(E) EV
valence band x
where Ei is the intrinsic Fermi level

21. Energy Band Diagram n-type semiconductor: no>po
E(x)
conduction band EC EF
n(E)
p(E) EV
valence band x

22. Energy Band Diagram p-type semiconductor: po>no
E(x)
conduction band EC
n(E)
p(E) EF EV
valence band x

23. A very useful relationship
…which is independent of the Fermi Energy
Recall that ni = no= po for an intrinsic semiconductor, so
nopo = ni2
for all non-degenerate semiconductors.
(that is as long as EF is not within a few kT of the band edge)

24. The intrinsic carrier density
is sensitive to the energy bandgap, temperature, and m*

25. The intrinsic Fermi Energy (Ei)
For an intrinsic semiconductor, no=po and EF=Ei
which gives
Ei = (EC + EV)/2 + (kT/2)ln(NV/NC)
so the intrinsic Fermi level is approximately
in the middle of the bandgap.

26. no = 0.5(ND-NA)  0.5[(ND-NA)2 + 4ni2]1/2
Space-charge Neutrality Consider a semiconductor doped with NA ionized acceptors (-q) and ND ionized donors (+q).
positive charges = negative charges
po + ND = no + NA
using ni2 = nopo
ni2/no + ND = no+ NA
ni2 + no(ND-NA) - no2 = 0
no = 0.5(ND-NA)  0.5[(ND-NA)2 + 4ni2]1/2
we use the ‘+’ solution since no should be increased by ni
no = ND - NA in the limit that ni<<ND-NA

27. Similarly, po= NA - ND if NA-ND>>ni
T (K)
no/(ND-NA)
po/(NA-ND) 1
po= no= ni
intrinsic material
freeze-out
insufficient energy to ionize the dopant atoms
carrier conc. vs. temperature
300
100
600

28. Degenerate Semiconductors …the doping concentration is so high that EF moves within a few kT of the band edge (EC or EV) EC EV +
ED1
Eg0
Eg(ND)
for ND > 1018 cm-3 in Si
impurity band
High donor concentrations cause the allowed donor wavefunctions to overlap, creating a band at EDn
First only the high states overlap, but eventually even the lowest state overlaps.
This effectively decreases the bandgap by
DEg = Eg0 – Eg(ND).

29. Degenerate Semiconductors
As the doping conc. increases more, EF rises above EC EV
EC (intrinsic)
available impurity band states EF
DEg
EC (degenerate) ~ ED
filled impurity band states
apparent band gap narrowing:
DEg* (is optically measured) -
Eg* is the apparent band gap:
an electron must gain energy Eg* = EF-EV

30. Electron Concentration in degenerately doped n-type semiconductors
The donors are fully ionized: no = ND
The holes still follow the Boltz. approx. since EF-EV>>>kT
po = NV exp[-(EF-EV)/kT] = NV exp[-(Eg*)/kT]
= NV exp[-(Ego- DEg*)/kT]
= NV exp[-Ego/kT]exp[DEg*)/kT]
nopo = NDNVexp[-Ego/kT] exp[DEg*)/kT]
= (ND/NC) NCNVexp[-Ego/kT] exp[DEg*)/kT]
= (ND/NC)ni2 exp[DEg*)/kT]

31. Summary non-degenerate: nopo= ni2 degenerate n-type:
nopo= ni2 (ND/NC) exp[DEg*)/kT]
degenerate p-type:
nopo= ni2 (NA/NV) exp[DEg*)/kT]