ECE 3080 - Dr. Alan DoolittleGeorgia Tech
Lecture 4
Density of States and Fermi Energy Concepts
Reading:
(Cont’d) Notes and Anderson2 sections 2.8-2.13
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Density of States Concept
In lower level courses, we state that “Quantum Mechanics” tells us that the number of available states in a cubic cm per unit of energy, the density of states, is given by:
eVcm
StatesofNumber
unit
EEEEmm
Eg
EEEEmm
Eg
vvpp
v
ccnn
c
3
32
**
32
**
,)(2
)(
,)(2
)(
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Density of States Concept
Thus, the number of states per cubic centimeter between energy E’ and E’+dE is
otherwiseandand
0,EE’ if )dE(E’g,EE’ if )dE(E’g
vv
cc
But where does it come from?
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Where Does the Density of States Concept come from?
Approach:
1. Find the smallest volume of k-space that can hold an electron. This will turn out to be related to the largest volume of real space that can confine the electron.
2. Next assume that the average energy of the free electrons (free to move), the fermienergy Ef, corresponds to a wave number kf. This value of kf, defines a volume in k-space for which all the electrons must be within.
3. We then simply take the ratio of the total volume needed to account for the average energy of the system to the smallest volume able to hold an electron ant that tells us the number of electrons.
4. From this we can differentiate to get the density of states distribution.
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Derivation of Density of States Concept
First a needed tool: Consider an electron trapped in an energy well with infinite potential barriers. Recall that the reflection coefficient for infinite potential was 1 so the electron can not penetrate the barrier.
After Neudeck and Pierret Figure 2.4a
2
222
n
22
2
22
2
2
22
22
2 Eand sin)(
3...,2 1,nfor a
nk 0kaAsin 0(a)
0 B 0)0( :ConditionsBoundary
2or E 22k where
cossin)( :Solution General
0
02
2
man
axnAx
mkmE
kxBkxAx
kx
Exm
EVm
nn
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Derivation of Density of States Concept
What does it mean?
After Neudeck and Pierret Figure 2.4c,d,e
2
222
n 2 Eand sin)(
man
axnAx nn
A standing wave results from the requirement that there be a node at the barrier edges (i.e. BC’s: (0)=(a)=0 ) . The wavelength determines the
energy. Many different possible “states” can be occupied by the
electron, each with different energies and wavelengths.
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Derivation of Density of States Concept
What does it mean?
After Neudeck and Pierret Figure 2.5
2
222
n 2 Eand sin)(
man
axnAx nn
Recall, a free particle has E ~k2. Instead of being continuous in k2, E is discrete in n2! I.e. the energy values
(and thus, wavelengths/k) of a confined electron are quantized (take on only certain values). Note that as the dimension of the “energy well”
increases, the spacing between discrete energy levels (and discrete k
values) reduces. In the infinite crystal, a continuum same as our free
particle solution is obtained.
Solution for much larger “a”. Note: offset vertically for clarity.
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Derivation of Density of States Concept
We can use this idea of a set of states in a confined space ( 1D well region) to derive the number of states in a given volume (volume of our crystal).
Consider the surfaces of a volume of semiconductor to be infinite potential barriers (i.e. the electron can not leave the crystal). Thus, the electron is contained in a 3D box.
After Neudeck and Pierret Figure 4.1 and 4.2
mkmE
kzyx
Exm
EVm
2or E 22k where
cz0 b,y0 a,x0...for
0
02
2
22
2
22
2
2
2
2
2
2
22
22
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Derivation of Density of States Concept
Using separation of variables...
2222
22
22
2
22
2
2
22
2
2
2
2
2
22
2
2
2
2
2
0)()(
1 ,0)(
)(1 ,0)(
)(1
0)()(
1)()(
1)()(
1
get... we(2)by dividing and (1) into (2) Inserting)()()(),,( (2)
0 (1)
zyx
zz
zy
y
yx
x
x
z
z
y
y
x
x
zyx
kkkkwhere
kz
zz
ky
yy
kx
xx
kz
zzy
yyx
xx
zyxzyx
kzyx
Since k is a constant for a given energy, each of the three terms on the left side must individually be equal to a constant.
So this is just 3 equivalent 1D solutions which we have already done...
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Derivation of Density of States Concept
Cont’d...
2
2
2
2
2
222
nz ny, nx,
zz
yy
xx
2 Eand
3...,2 1,nfor ,a
nk ,a
nk ,
ank where
sinsinsin),,()()()(),,(
cn
bn
an
m
zkykxkAzyxzyxzyx
zyx
zyx
zyx
0
a
b
c
kx
ky
kzEach solution (i.e. each combination of nx, ny, nz) results in a volume of “k-space”. If we add up all possible combinations, we would have an infinite solution. Thus, we will only consider states contained in a “fermi-sphere” (see next page).
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Derivation of Density of States Concept
Cont’d...f
22
f Eenergy electron average for the valuemomentum a defines 2
E mk f
Volume of a single state “cube”: V3
state single
Vcba
Volume of a “fermi-sphere”: 34V 3
sphere-fermi
fk
A “Fermi-Sphere” is defined by the number of states in k-space necessary to hold all the electrons needed to add up to the average energy of the crystal (known as the fermi energy).
“V” is the physical volume of the crystal where as all other volumes used here refer to volume in k-space. Note that: Vsingle-state is the smallest unit in k-space. Vsingle-state is required to “hold” a single electron.
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Derivation of Density of States Concept
Cont’d...k-space volume of a single state “cube”: V
3
state single
Vcba
k-space volume of a “fermi-sphere”: 34V 3
sphere-fermi
fk
Number of filled states in a fermi-sphere: 2
3
3
3
sin 3
4
34
21
21
212N
f
f
stategle
spherefermi kV
V
kxxx
VV
Correction for allowing 2 electrons per state (+/- spin)
Correction for redundancy in counting identical states resulting
from +/- nx, +/- ny, +/- nz. Specifically, sin(-)=sin(+) so the state would be the same. Same as
counting only the positive octant in fermi-sphere.
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Derivation of Density of States Concept
Cont’d...
Number of filled states in a fermi-sphere:
31
2
f2
3 3k 3
N
VNkV f
Ef varies in Si from 0 to ~1.1 eV as n varies from 0 to ~5e21cm-3
density electron theis n where2
3 E 2
3
2
E3
222
f
32
22
22
f mn
mV
N
mk f
23
222
3 233
N
mEVkV f
Thus,
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Derivation of Density of States Concept
Cont’d...
E
V
mmEV
32
2
21
22
2mm G(E)
223
23
G(E)
VdEdN
per volumeenergy per states of # G(E)
23
222
3 233
N
mEVkV f
Applying to the semiconductor we must recognize m m* and since we have only considered kinetic energy (not the potential energy) we have E E-Ec
cEE 32
** 2mm G(E)
Finally, we can define the density of states function:
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Probability of Occupation (Fermi Function) Concept
Now that we know the number of available states at each energy, how do the electrons occupy these states?
We need to know how the electrons are “distributed in energy”.
Again, Quantum Mechanics tells us that the electrons follow the “Fermi-distribution function”.
)(~
,tan1
1)( )(
crystaltheinenergyaverageenergyFermiEand
KelvinineTemperaturTtconsBoltzmankwheree
Ef
F
kTEE F
f(E) is the probability that a state at energy E is occupied
1-f(E) is the probability that a state at energy E is unoccupied
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Probability of Occupation (Fermi Function) Concept
At T=0K, occupancy is “digital”: No occupation of states above EF and complete occupation of states below EF
At T>0K, occupation probability is reduced with increasing energy.
f(E=EF) = 1/2 regardless of temperature.
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Probability of Occupation (Fermi Function) Concept
At T=0K, occupancy is “digital”: No occupation of states above EF and complete occupation of states below EF
At T>0K, occupation probability is reduced with increasing energy.
f(E=EF) = 1/2 regardless of temperature.
At higher temperatures, higher energy states can be occupied, leaving more lower energy states unoccupied (1-f(E)).
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
0.00
0.20
0.40
0.60
0.80
1.00
1.20
0 0.2 0.4 0.6 0.8 1 1.2
f(E)
E [eV]
3 kT3 kT3 kT3 kT
+/-3 kT
Ef=0.55 eV
How do electrons and holes populate the bands?Probability of Occupation (Fermi Function) Concept
For E > (Ef+3kT):
f(E) ~ e-(E-Ef)/kT~0
For E < (Ef-3kT):
f(E) ~ 1-e-(E-Ef)/kT~1
T=10 K, kT=0.00086 eV
T=300K, kT=0.0259
T=450K, kT=0.039
kTEE F
eEf )(
1
1)(
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
But where did we get the fermi distribution function?Probability of Occupation (Fermi Function) Concept
After Neudeck and Pierret Figure 4.5
Consider a system of N total electrons spread between S allowable states. At each energy, Ei, we have Si available states with Ni of these Si states filled.
We assume the electrons are indistinguishable1.
1A simple test as to whether a particle is indistinguishable (statistically invariant) is when two particles are interchanged, did the electronic configuration change?
Constraints for electrons:(1) Each allowed state can accommodate at most, only one electron (neglecting
spin for the moment)(2) N=Ni=constant; the total number of electrons is fixed(3) Etotal=EiNi; the total system energy is fixed
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
But where did we get the fermi distribution function?Probability of Occupation (Fermi Function) Concept
!N ! iii
ii NS
SW
How many ways, Wi, can we arrange at each energy, Ei, Ni indistinguishable electrons into the Si available states.
i ii
i
ii NS
SWW!N ! i
When we consider more than one level (i.e. all i’s) the number of ways we can arrange the electrons increases as the product of the Wi’s .
If all possible distributions are equally likely, then the probability of obtaining a specific distribution is proportional to the number of ways that distribution can be constructed (in statistics, this is the distribution with the most (“complexions”). For example, interchanging the blue and red electron would result in two different ways (complexions) of obtaining the same distribution. The most probable distribution is the one that has the most variations that repeat that distribution. To find that maximum, we want to maximize W with respect to Ni’s . Thus, we will find dW/dNi = 0. However, to eliminate the factorials, we will first take the natural log of the above...
... then we will take d(ln[W])/dNi=0. Before we do that, we can use Stirling’s Approximation to eliminate the factorials.
i
i!Nln! ln!lnln iii NSSW
or
or
Ei
Ei
Ei
Ei+2
Ei+1
Ei
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
But where did we get the fermi distribution function?Probability of Occupation (Fermi Function) Concept
Using Stirling’s Approximation, ln(x!) ~ (xln(x) – x) so that the above becomes,
iii
iiii
ii
NlnNlnlnln terms,like Collecting
NNlnNlnlnln
!Nln! ln!lnln
iiiiii
iiiiiiiii
iii
NSNSSSW
NSNSNSSSSW
NSSW
Now we can maximize W with respect to Ni’s . Note that since d(lnW)=dW/W when dW=0, d(ln[W])=0.
i
ii
ii
i
1ln0ln
1Nln1lnln
1Nln1lnlnln
ii
i
iii
iiii
dNNSWd
dNNSWd
NSNW
dNWd
To find the maximum, we set the derivative equal to 0...
(4)
1lnln used have weWhere xdx
xxd
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
But where did we get the fermi distribution function?Probability of Occupation (Fermi Function) Concept
From our original constraints, (2) and (3), we get...
ii
ii
0
0
iitotalii
ii
NdEENE
NdNN
Using the method of undetermined multipliers (Lagrange multiplier method) we multiply the above constraints by constants – and – and add to equation 4 to get ...
i
i
0
0
ii
i
NdE
Nd
i allfor 01ln that requires which
1ln0lni
ii
i
iii
i
ENS
dNENSWd
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
But where did we get the fermi distribution function?Probability of Occupation (Fermi Function) Concept
This final relationship can be solved for the ratio of filled states, Ni per states available Si ...
kT
EEi
ii
Ei
ii
ii
i
Fi
i
eNSEf
eNSEf
ENS
1
1)(
kT1 and
kTE-
have wetors,semiconduc of of case thein1
1)(
01ln
F
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Probability of Occupation
We now have the density of states describing the density of available states versus energy and the probability of a state being occupied or empty. Thus, the density of electrons (or holes) occupying the states in energy between E and E+dE is:
otherwise
and
and
0
,EE if dEf(E)]-(E)[1g
,EE if dEf(E)(E)g
vv
cc
Electrons/cm3 in the conduction band between Energy E and E+dE
Holes/cm3 in the valence band between Energy E and E+dE
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Band Occupation
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Intrinsic Energy (or Intrinsic Level)
…Equal numbers of electrons and holes
Ef is said to equal Ei (intrinsic energy) when…
ECE 3080 - Dr. Alan DoolittleGeorgia Tech
How do electrons and holes populate the bands?Additional Dopant States
Intrinsic: Equal number of electrons and holes
n-type: more electrons than holes
p-type: more holes than electrons