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10/1/2012
1
EE415/515 Fundamentals of Semiconductor Devices
Fall 2012
Lecture 3: Density of States, Fermi Level
(Chapter 3.4-3.5/4.1)
Density of States• Need to know the density of electrons, n, and holes, p, per unit volume
• To do this, we need to find the density of permitted energy states and then the probability (given by the Fermi function) that these states are occupied by an electron
• See Neamen Sections 3.4 and 3.5 for derivations, and compare these with alternative approaches given below
ECE415/515 Fall 2012 2J.E. Morris
10/1/2012
2
Density of States
ECE415/515 Fall 2012 3
•See figure in “k-space.”
•Also: p=ħ k=(h/2π)k, i.e.•Direct linear relationship between (wave-vector) k-space and (momentum) p-space.
•The Neamen derivation (next slide) is based in k-space.•See also a slightly different approach based in p-space in following slide.
J.E. Morris
Density of States
ECE415/515 Fall 2012 4
Consider electron confined to crystal (infinite potential well) of dimensions a(volume V= a3)
It has been shown that k=nπ/a, so ∆k=kn+1-kn=π/a
Each quantum state occupies volume (π/a)3 in k-space.
Number of quantum states in range k to k+dk is 4πk2.dk
and the number of electrons in this range k to k+dk is
gT(k)dk = 2(1/8).4πk2dk/ (π/a)3 = (k2dk/ π2)a3
where “2” comes from spin degeneracy, and “1/8” because E=(ħk)2/2m,
so negative k values do not add more energy states,
i.e. use kx, ky, kz >0 quadrant only
Converting k-space to energy ………..
J.E. Morris
10/1/2012
3
Density of States
ECE415/515 Fall 2012 5
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J.E. Morris
Density of States (alternate)
ECE415/515 Fall 2012 6
Heisenberg says ∆p∆x=h,
(but see ħ in Neamen section 2.1.3 p30, and ħ/2 in Streetman)
and for crystal volume V=L3=LxLyLz
∆px∆x=h, ∆py∆y=h, ∆pz∆z=h,
and hence for a single electron
∆px ∆py ∆pz=h3/V for ∆x=Lx, etc
is the “volume” occupied in p-space by one electron.
Number of allowed conduction electron states g(p)dp from p to p+dp
[see Fig 3.26 (slide 3) in k-space with p = ħk]
is the volume of the p-space “shell” containing all states from p to p+dp
[4πp2dp]
divided by the p-space volume of each electron, i.e. ……
J.E. Morris
10/1/2012
4
Density of states
ECE415/515 Fall 2012 7
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J.E. Morris
(Compare 1D, 2D systems with 3D)
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10/1/2012
5
And similarly for holes:-
ECE415/515 Fall 2012 9
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Now that we have the densities of states, we need the probabilities that these are filled.
J.E. Morris
Ex 3.3 Calculate the density of quantum states (/cm3) for a free electron over the energy range of (a) 0 ≤ E ≤ 2eV & (b) 1 ≤ E ≤ 2eV
ECE415/515 Fall 2012 10J.E. Morris
10/1/2012
6
Ex 3.3 Calculate the density of quantum states (/cm3) in Si from (Ev-kT) to Ev at T=300 K
ECE415/515 Fall 2012 11J.E. Morris
Fermi-Dirac statistics (see also text “derivation”)
ECE415/515 Fall 2012 12
First, let’s find the Maxwell-Boltzmann distribution
Consider a system of identical particles with random collisions
Assume 2 particles of initial energies E1 & E2 collide with energy transfer δ to give final energies E3 & E4
Conservation of Energy requires that:
If p(E)=probability that particle has energy E, then:
Equilibrium: equal numbers of forward & reverse collisions for constant energy distribution, i.e.
)()( 214321 EEEEEE
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J.E. Morris
10/1/2012
7
MB continued & FD intro
Result suggests an exponential distribution, i.e.
Check: Aexp-α(Ei+δ)=λ. Aexp-αEi if λ=exp-αδ
For Fermi-Dirac statistics, repeat as for MB, but with the Pauli Exclusion Principle included, i.e.
Can only get transfer of energy E1, E2 → E3, E4
if E1 & E2 are occupied AND if E3 & E4 are unoccupied
ECE415/515 Fall 2012 13
)(.)( ii EpEp
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J.E. Morris
Fermi functionSo the collision probability is
and applying Equilibrium and Energy Conservation again:
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ECE415/515 Fall 2012 14J.E. Morris
10/1/2012
8
MB/FD/(BE) comparisons
Maxwell-Boltzmann Bose-Einstein Fermi-Dirac
Particles distinguishable
Particles indistinguishable (wave functions overlap – cannot find individual particle)
Any number of particles can be in the same energy state
Only one particle in any energy state
(Pauli) but note spin degeneracy
Particles are statistically independent
Atoms, molecules, etc BosonsPhotons, phonons,
4He nuclei (integral spins)
FermionsElectrons, protons,
neutrons, 3He nuclei (½-odd-integral spins)
ECE415/515 Fall 2012 15
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J.E. Morris
Fermi distributions
Electrons in solids obey Fermi-Dirac statistics due to the Pauli exclusion principle (each state can have only one electron – but remember spin!)
Probability density function gives the ratio of filled to total allowed states at a given energy.
Using statistical mechanics to count states we find the Fermi-Dirac distribution function:• f(E) = {1 + exp[(E-Ef)/kT]}-1
• k is Boltzmann’s constant = 8.62x10-5 eV/K
= 1.38x10-23J/K
• EF is the Fermi level, [where f(EF)=½]
ECE415/515 Fall 2012 16J.E. Morris
10/1/2012
9
The Fermi-Dirac distribution function.
ECE415/515 Fall 2012 17
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J.E. Morris
At T = 0 K
ECE415/515 Fall 2012 18J.E. Morris
10/1/2012
10
At T > 0 K
ECE415/515 Fall 2012 19J.E. Morris
Fermi levels
The Fermi distribution gives the probability that an available state will be occupied.
• Within the band gap there are no available states.
• Even with non-zero f(E) there is no occupancy in the gap. (An energy state at the Fermi level would have a probability of ½ of being occupied, if there was one.)
At 0K every available state up to the Fermi level is occupied and every state above the Fermi level is empty.
Fermi function is “symmetric” about the Fermi level
For holes:
ECE415/515 Fall 2012 20
kT
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J.E. Morris
10/1/2012
11
ECE415/515 Fall 2012 21
Fermi function symmetry:electrons and holes
J.E. Morris
ECE415/515 Fall 2012 22
Approximate FD → MB for (E-EF)>>kT
See Example 3.6:f(EF+3kT) ≈ 5% for 300K
So f(E) varies quickly
Example 3.8:(fMB-fFD)/FFD=5%
at E-EF=kT ln(1/0.05) ≈ 3kT
Note f(E) = {1 + exp[(E-Ef)/kT]}-1
≈ exp -[(E-Ef)/kT] for
“(E-EF)>>kT” condition where
exp[(E-Ef)/kT] >> 1
J.E. Morris
10/1/2012
12
Ex 3.6 Assume EF=EC-0.3eV and T=300 K. Determine probability of state occupied at (a) E=EC+kT/4 & (b) E=EC+kT
ECE415/515 Fall 2012 23J.E. Morris
ECE415/515 Fall 2012 24
Ex 3.7 Assume EF=EC-0.3eV. Determine T at which the probability of state at E=EC+0.025eV is occupied is 8x10-6.
J.E. Morris
10/1/2012
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Ex.3.8 Calculate the energy, in terms of kT and EF, at which the difference between the FD and MB distributions is 2% of the FD value.
ECE415/515 Fall 2012 25J.E. Morris
ECE415/515 Fall 2012 26
Electron & hole densities
n(E) = gC(E).f(E)
p(E) = gV(E).[1-f(E)]
J.E. Morris
10/1/2012
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Effective density of states & Fermi integral
ECE415/515 Fall 2012 27
band valencein the holesfor )(gfor result similar a giveseatment similar trA
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J.E. Morris
Effective density of states
ECE415/515 Fall 2012 28
kT
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NC is the effective density of states as if all electrons at conduction band edge EC. Only limiting assumption is that EC-EF>>kT; if so, result holds for intrinsic and extrinsic. (Note: EF→EC for N-type, but approx usually still OK.)
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J.E. Morris
10/1/2012
15
Electron Concentrations
Simplify: represent states throughout the conduction band by an effective density of states Nc located at Ec.
n0 = Ncf(Ec)
If Ec – EF > 3kT then
f(Ec) ~ e-(Ec-EF)/kT (Maxwell-Boltzmann)
kT = 0.026 eV at 300K, so usually OK
n0 = Nc e-(Ec-EF)/kT ; Nc = 2[2πmn*kT/h2]3/2
Termed non-degenerate.
If EF within 3kT of either band edge or inside band degenerate and we need to use the full Fermi function.
ECE415/515 Fall 2012 29J.E. Morris
Hole Concentrations
Concentration of holes in valence band
p0 = Nv [1 – f(Ev)]
If EF – Ev > 3kT then Fermi function reduces to Maxwell Boltzmann and
p0 = Nv exp[ -(EF – Ev)/kT ]
Nv = [2π mp*kT/h2]3/2
p0 = ∫gV(E)[1-f(E)]dE
1-f(E)=1-[1+exp((E-EF)/kT)]-1
=[exp (E-EF)/kT)]/[1+exp((E-EF)/kT)]
=1/[1+exp((EF-E)/kT)] ≈ exp-(EF-E)/kT
ECE415/515 Fall 2012 30J.E. Morris
10/1/2012
16
Ex 4.1 Determine the probability that a quantum state at E=EC+kT is occupied by an electron, and calculate the electron concentration in GaAs at T=300 K if EF=EC-0.25eV.
ECE415/515 Fall 2012 31J.E. Morris
(Nc from Table 4.1)
Ex 4.2 (a) Calculate the equilibrium hole concentration in Si at T=250K if EF=EV+0.27eV. [NV(300K)=1.04x1019/cm3] (b) What is p0(250K)/p0(400K)? (See Example 4.2)
ECE415/515 Fall 2012 32J.E. Morris
10/1/2012
17
Intrinsic Concentrations EF = EFi for intrinsic material
n0=ni = Ncexp[-(Ec-EFi)/kT] pi = Nvexp[-(EFi-Ev)/kT]
n0 p0 = constant.
Plug in above => n0 p0 = ni2 = [NcNv]exp[-(EC-EV)/kT]
ni = [NcNv]1/2 exp[-Eg/2kT]
= 2[2πkT/h2]3/2(mnmh)3/2 exp[-Eg/2kT]
Remember n0p0 = ni2 and ni = 1.5x1010 cm-3 for Si
You’ll use both often! Also for Si: 6 different conduction band valleys (2 each x,y,z)
Note: varies in different sources
See Appendix F for more detail (complications) on effective mass
ECE415/515 Fall 2012 33
2
12
3* 6)( tn mmm
*nm
J.E. Morris
Ex 4.3 (a) Calculate the intrinsic carrier concentration in GaAs at T=400 K and T=250 K, assuming Eg=1.42eV. (b) What is ni(400K)/ni(250K)?
ECE415/515 Fall 2012 34J.E. Morris
10/1/2012
18
Intrinsic concentrations
In general, if EF ≠ EFi, then re-write these as
n0 = ni exp[(EF – EFi)/kT]
p0 = pi exp[(EFi – EF)/kT]
Also used frequently and helpful for visualization
With each problem consider what information you have and what you are looking for.
Law of Mass Action:
• n0p0=ni2
• Independent of EF, (i.e. of doping)
• So applies to both intrinsic and extrinsic at equilibrium
ECE415/515 Fall 2012 35J.E. Morris
Temperature Dependence
ni T3/2 exp[-Eg/2kT]
ECE415/515 Fall 2012 36
J.E. Morris
10/1/2012
19
Physical constants (@300 K unless noted)
Boltzmann’s const (k)
1.38 * 10-23 J/K 8.62 * 10-5 ev/K
Planck’s constant (h) 6.63*10-34 J-s 4.14*10-15 eV-s
Room temp value of kT
0.0259 eV Electron rest mass (m0)
9.11 * 10-31 kg
Electronic charge (q)
1.60 * 10-19 C Permittivity of free space (ε0)
8.85*10-14 F/cm
Density of states Nc Nv
2.8*1019 cm-3
1.04*1019 cm-3
Relative permittivity of silicon (εr)
11.9
Speed of light (c) 3*1010 cm/s Electron volt (eV) 1.6*10-19 J Band Gap Si Eg 1.11 ev Band gap Ge Eg 0.67 eV
ECE415/515 Fall 2012 37J.E. Morris
Effective density of states & masses
ECE415/515 Fall 2012 38
See Appendix B for m* values
J.E. Morris
10/1/2012
20
Intrinsic EF position
ECE415/515 Fall 2012 39
.** .*
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.*
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J.E. Morris
Ex 4.4 Find the position of EFi at T=300 K wrt the band-gap center for (a) GaAs, and (b) Ge.
ECE415/515 Fall 2012 40J.E. Morris
10/1/2012
21
Summary: Fermi function
Fermi function describes probability of a given state being occupied at temperature T.
f(E) = {1 + exp[(E-EF)/kT]}-1
For intrinsic semiconductors EF is mid-gap.
If E – EF > 3kT (non-degenerate, with EF in the forbidden gap) then
f(Ec)~e-(E-EF)/kT
Use effective density of states Nc located at Ec or Nv located at Ev. (still non-degenerate)
n0 = Nc e-(Ec-EF)/kT ; Nc = 2[2πmn*kT/h2]3/2
p0 = Nv e-(Ev-EF)/kT; Nv = [2π mp*kT/h2]3/2
ECE415/515 Fall 2012 41J.E. Morris