Lecture 4
Density of States and Fermi Energy Concepts
Reading:
(Cont’d) Notes and Anderson2 sections 2.8-2.13
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ECE 3080 - Dr. Alan Doolittle
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:
mn* 2mn* ( E  Ec )
gc (E) 
, E  Ec
2 3
 
gv (E) 
m*p 2m*p ( Ev  E )
 
2
3
, E  Ev
 Number of States 


3
cm

unit  
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eV
ECE 3080 - Dr. Alan Doolittle
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
g c (E’ )dE if E’  E c
and ,
g v (E’ )dE if E’  E v
and ,
0 otherwise
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But where does it come from?
ECE 3080 - Dr. Alan Doolittle
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 fermi
energy 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.
Georgia Tech
ECE 3080 - Dr. Alan Doolittle
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.
2
 

 
 2  V    E
 2m

2 2

 E  0
2
2m x
2
 k 2  0
2
x
General Solution :  ( x )  A sinkx   B coskx 
2
2k 2
2mE

where k 
or E 
2

2m

Boundary Conditions :
 ( 0)  0  B  0
n
 (a)  0  Asin ka   0  k 
for n  1,  2,3...
a
n 2 2  2
 nx 
n ( x )  An sin
 and E n 
2ma 2
 a 
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After Neudeck and Pierret Figure 2.4a
ECE 3080 - Dr. Alan Doolittle
How do electrons and holes populate the bands?
Derivation of Density of States Concept
What does it mean?
n 2 2  2
 nx 
n ( x )  An sin
 and E n 
2ma 2
 a 
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.
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After Neudeck and Pierret Figure 2.4c,d,e
ECE 3080 - Dr. Alan Doolittle
How do electrons and holes populate the bands?
Derivation of Density of States Concept
What does it mean?
n 2 2  2
 nx 
n ( x )  An sin
 and E n 
2ma 2
 a 
Solution for much larger “a”. Note:
offset vertically for clarity.
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.
After Neudeck and Pierret Figure 2.5
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ECE 3080 - Dr. Alan Doolittle
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.
 2 2

 
  V    E
m
2


2  2
 E  0

2
2m x
2 2 2
2
k



0
z 2
x 2
y 2
...for 0  x  a, 0  y  b, 0  z  c
where k 
2


2k 2
2mE
or E 
2

2m
After Neudeck and Pierret Figure 4.1 and 4.2
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ECE 3080 - Dr. Alan Doolittle
How do electrons and holes populate the bands?
Derivation of Density of States Concept
Using separation of variables...
(1)
(2)
2 2 2
2
k



0
2
2
2
x
y
z
 ( x, y , z )  x ( x )y ( y ) z ( z )
Inserting (2) into (1) and dividing by (2) we get...
2
1  2 z ( z )
1  2 x ( x )
1  y ( y )
2
k



0
2
2
2
x ( x ) x
y ( y ) y
z ( z ) z
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.
2

y ( y )
1  2 x ( x )
1
1  2 z ( z )
2
2
 k x  0,
 k y  0,
 k z2  0
2
2
2
x ( x ) x
y ( y ) y
z ( z ) z
where k 2  k x2  k y2  k z2
So this is just 3 equivalent 1D solutions which we have already done...
Georgia Tech
ECE 3080 - Dr. Alan Doolittle
How do electrons and holes populate the bands?
Derivation of Density of States Concept
Cont’d...
 ( x, y , z )  x ( x ) y ( y ) z ( z )
 ( x, y , z )  A sink x x  sin k y y sin k z z 
n y
n
n x
, ky 
, k z  z , for n  1,  2,3...
where k x 
a
a
a
2
 2  2  n x2 n y n z2 


and E nx, ny, nz 
2m  a 2 b 2 c 2 
Each solution (i.e. each
 kz
combination of nx, ny, nz)
c
results in a volume of “kspace”. If we add up all
possible combinations, we
would have an infinite

solution. Thus, we will only
0
b
consider states contained in a
“fermi-sphere” (see next
ky
page).
kx 
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a
ECE 3080 - Dr. Alan Doolittle
How do electrons and holes populate the bands?
Derivation of Density of States Concept
Cont’d...
Ef 
 2 k 2f
2m
Volume of a
Volume of a
 defines a momentum value for the average electron energy E f
3
       
single state “cube”: Vsingle state          
 a  b  c   V 
4
“fermi-sphere”: Vfermi-sphere    k 3f 
3

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).
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“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 kspace. Vsingle-state is
required to “hold” a
single electron.
ECE 3080 - Dr. Alan Doolittle
How do electrons and holes populate the bands?
Derivation of Density of States Concept
Cont’d...
k-space volume of a single state “cube”:
k-space volume of a “fermi-sphere”:
Number of filled states
V fermi  sphere
N
in a fermi-sphere:
Vsin gle state
4

Vfermi-sphere    k 3f 
3

4
3 
  k f  V k3
3
1 1 1
f
2 x  x  x    3  
3 2
 
2 2 2
 4
V 
Correction for
allowing 2 electrons
per state (+/- spin)
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Vsingle state
3
       
         
 a  b  c   V 
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 Doolittle
How do electrons and holes populate the bands?
Derivation of Density of States Concept
Cont’d...
Number of filled states
in a fermi-sphere:
 3 N 

 
V



2m
2
2
Ef 
 2 k 2f
2m
2
N
Vk
3
3
f
2
 3 N 

 k f  
 V 
2
1
3
3
 3 n 
 Ef 
2m
2
2
2
3
where n is the electron density
Ef varies in Si from 0 to ~1.1 eV as n varies from 0 to ~5e21cm-3
Thus,
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N
V k 3f
3 2
 V  2mE 
  2  2 
 3   
3
2
ECE 3080 - Dr. Alan Doolittle
How do electrons and holes populate the bands?
Derivation of Density of States Concept
Cont’d...
N
V k 3f
3 2
 V  2mE 
  2  2 
 3   
3
2
Finally, we can define the density of states function:
 dN 


dE


G(E)  # of states per energy per volume 
 3  V  2mE  12  2m 
  2  2   2  
 2  3      
G(E) 
G(E) 
m 2m
 23
V
V
E
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
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m * 2m *
G(E) 
 23
E  Ec
ECE 3080 - Dr. Alan Doolittle
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”.
f (E) 
1
( E  EF )
where k  Boltzman cons tan t , T  Temperature in Kelvin
kT
1 e
and E F  Fermi energy (~ average energy in the crystal )
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
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ECE 3080 - Dr. Alan Doolittle
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.
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ECE 3080 - Dr. Alan Doolittle
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)).
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ECE 3080 - Dr. Alan Doolittle
How do electrons and holes populate the bands?
Probability of Occupation (Fermi Function) Concept
1
f (E) 
1 e
( E  EF )
kT
1.20
+/-3 kT
T=10 K,
kT=0.00086 eV
3 kT
3 kT
3 kT 3 kT
T=300K,
kT=0.0259
1.00
Ef=0.55 eV
f(E)
0.80
T=450K,
kT=0.039
For E < (Ef-3kT):
0.60
f(E) ~ 1-e-(E-Ef)/kT~1
For E > (Ef+3kT):
0.40
f(E) ~ e-(E-Ef)/kT~0
0.20
0.00
0
0.2
0.4
0.6
0.8
1
1.2
E [eV]
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ECE 3080 - Dr. Alan Doolittle
But where did we get the fermi distribution function?
Probability of Occupation (Fermi Function) Concept
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.
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
1A simple
test as to whether a particle
is indistinguishable (statistically
invariant) is when two particles are
interchanged, did the electronic
configuration change?
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After Neudeck and Pierret Figure 4.5
ECE 3080 - Dr. Alan Doolittle
But where did we get the fermi distribution function?
Probability of Occupation (Fermi Function) Concept
How many ways, Wi, can we arrange at each energy, Ei, Ni indistinguishable electrons into
the Si available states.
E
Si
Wi 
Si  N i ! N i !
or
or
i
Ei
Ei
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 .
Ei+2
Ei+1
Si
W   Wi  
i
i S i  N i  ! N i !
Ei
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...
lnW    ln Si !  ln Si  N i !  lnN i !
i
... then we will take d(ln[W])/dNi=0. Before we do that, we can use Stirling’s Approximation to
eliminate the factorials.
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ECE 3080 - Dr. Alan Doolittle
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,
ln W    ln S i !  ln S i  N i !  ln N i !
i

ln W    S i ln S i   S i  S i  N i  ln S i  N i   S i  N i   N i ln N i   N i
i
Collecting like terms,
ln W    S i ln S i   S i  N i  ln S i  N i   N i ln N i 
i
Now we can maximize W with respect to Ni’s . Note that since d(lnW)=dW/W when dW=0,
d(ln[W])=0.
 ln W 
d ln W 
dN i

i
N i
  ln Si  N i   1  ln N i   1
i
d ln W    ln S i  N i   1  ln N i   1dN i
i
To find the maximum, we set the derivative equal to 0...
(4)
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  Si

d ln W   0   ln  1 dN i
i   Ni

Where we have used
d x ln  x 
 ln x   1
dx
ECE 3080 - Dr. Alan Doolittle
But where did we get the fermi distribution function?
Probability of Occupation (Fermi Function) Concept
From our original constraints, (2) and (3), we get...
N
i
 d N   0
N 
i
i
E N
i
i
i
 E d N   0
 E total 
i
i
i
i
Using the method of undetermined multipliers (Lagrange multiplier method) we multiply the
above constraints by constants – and – and add to equation 4 to get ...
   d N   0
i
i
   E d N   0
i
i
i

  Si


 1    Ei  dN i
d ln W   0   ln
i   Ni


 Si

 1    Ei  0 for all i
which requires that ln
 Ni

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ECE 3080 - Dr. Alan Doolittle
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 ...
S

ln i  1    Ei  0
 Ni


S
1
f ( Ei )  i 
N i 1  e  Ei
in the case of of semiconductors, we have
EF
1
and  
kT
kT
S
1
f ( Ei )  i 
N i 1  e  Ei  EF  kT
 -
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ECE 3080 - Dr. Alan Doolittle
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:
Electrons/cm3 in the conduction
band between Energy E and E+dE
Holes/cm3 in the valence band
between Energy E and E+dE
g c (E) f(E) dE
if E  E c
and ,
g v (E)[1 - f(E)] dE if E  E v
and ,
0 otherwise
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ECE 3080 - Dr. Alan Doolittle
How do electrons and holes populate the bands?
Band Occupation
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ECE 3080 - Dr. Alan Doolittle
How do electrons and holes populate the bands?
Intrinsic Energy (or Intrinsic Level)
Ef is said to equal
Ei (intrinsic
energy) when…
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…Equal
numbers of
electrons and
holes
ECE 3080 - Dr. Alan Doolittle
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
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ECE 3080 - Dr. Alan Doolittle