PHYSICS OF SEMICONDUCTOR
DEVICES
TOPIC : SEMICONDUCTOR IN EQUILIBRIUM

Density of States function, g(E)

Fermi-Dirac Distribution function, f(E)

Distribution Function and Fermi Energy

Equilibrium Distribution of Electrons and Holes

n0 and p0 Equation

Intrinsic carrier concentration

Fermi level for Intrinsic Semiconductor

Extrinsic Semiconductor

Position of the Fermi Level of the Extrinsic Semiconductor

Non-Degenerated Semiconductor

Variation of EF with Doping Concentration and with Temperature

Compensated Semiconductor

Statistics of donors and acceptors

Formula to remember

Question Bank

Solved problems

Assignment
SEMICONDUCTOR IN EQUILIBRIUM
Density of States Function g (E)
The conduction in the crystal depends on the number of charge carriers. The number of
charge carrier that contributes to the conduction process is the function of the number of
available energy or quantum states.
The density of these allowed energy state is called the density of state function. This
helps to calculate the hole and electron concentration.
4π(2m)3/2
E
h3
g(E) = number of quantum states per unit volume per unit energy
It is given by g(E) =
Where,
Density of allowed electronic energy states in the conduction band is
g c (E) =
4π(2m*n )3/2
h3
E-E c
As the energy of the electron in the conduction band decreases, the number of available
quantum states also decreases.
Similarly, density of allowed quantum state in the valence band is
g v (E) =
4π(2m*p )3/2
h3
E v -E
It is important to note that quantum states do not exist
within the forbidden energy band.
So
g(E) = 0
for Ev < E < Ec
The density of energy states in the conduction band
and valence band as a function of energy is shown in
the figure.
Fermi – Dirac Probability Function :
In the crystal, the electrical behaviour will be determined by statistical behaviour of a
large number of electrons. Electrons in the crystal obey Fermi – Dirac probability
function where the particles are distinguishable but only one particle is permitted in
each quantum state. The particles are assumed to be non – interacting.
fF (E) = Fermi – Dirac distribution function =
Fermi – Dirac probability function
= Probability that a quantum state at the energy E will be occupied by an
electron.

Filled quantum state
Total quantum state
Fermi – Dirac distribution fF (E) is given by f F (E) =
1
 E-E F 
1+ exp 
 kT 
Where EF = Fermi energy
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SEMICONDUCTOR IN EQUILIBRIUM
Boltzmann approximation
If
(E - EF)
>> kT
E-E F
kT
Then

 E-E F 
exp 

 kT 

 - (E-E F ) 
f F (E) = exp 

 kT 
>> 1
Then
f F (E) =
>> 1
1
 (E-E F ) 
exp 

 kT 
This become Boltzmann approximation
Distribution function and Fermi Energy
We know that
1.
f F (E) =
Let T = 0 K and if
Then
1
 E-E F 
1+ exp 

 kT 
E < EF
 E-E F 
exp 
 = exp ( -  ) = 0
 kT 
So
fF (E) = 1
Thus the entire electron have energy below the Fermi energy at T=0 K in their
lowest possible energy states.
2.
Let T = 0K and if
Then
E > EF
 E-E F 
exp 
 = exp ( +  ) = 
 kT 
FF(E)
So
fF (E) = 0
Thus no electron have energy above the
Fermi energy at T=0 K
A plot for the Fermi probability function
versus energy at T = 0 K
3. Let T > 0K and
Then
EF
E
E = EF
 E-E F 
exp 
 = exp ( 0 ) = 1
 kT 
So f F (E) = 1
2
1
if
1.0
Thus the probability of a state being occupied at E = EF is
1
2
Fermi – Dirac probability function varies
with temperature and is shown in the figure
at different temperature.
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SEMICONDUCTOR IN EQUILIBRIUM
2
A plot for the Fermi – Dirac probability
function and the Maxwell – Boltzmann
approximation is shown in the figure.
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SEMICONDUCTOR IN EQUILIBRIUM
Equilibrium Distribution of Electrons and Holes
In an intrinsic semiconductor at T = 0K, all the energy state in the conduction band is
empty with electrons and all the energy state in the valence band is filled with electrons.
However, as the temperature starts to increase, the valence electron will gain thermal
energy. Some electrons in the valence band may gain sufficient energy and jump to the
conduction band. This creates an empty space (called the holes) in the valence band
corresponding to each electron jumping from valence band to the conduction band.
Thus in an intrinsic semiconductor the number of electrons in the conduction band is
equal to the number of holes in the valence band.
If n(E) is the distribution of electrons in the conduction band with respect to energy,
then it is given by
n(E) = g c (E) . f F (E)
Where g c (E)  density of allowed quantum states in the conduction band
and
f F (E)  Fermi - Dirac probability function i.e the probability
of a quantum state being occupied by an electron
Similarly, the distribution of hole p(E) in the valence band with
respect to energy is given by
p(E) = g v (E) . [ 1 - f F (E) ]
Where g v (E)  density of allowed quantum
states in the valence band
and 1 - f F (E)  Probability that a quantum state
is not occupied by an electron
Graphical representation of n(E) , p(E) , g c (E) , g v (E) , f F (E) are shown in the
adjoining graph.
Calculation of thermal equilibrium concentration of electrons (n0 )
n0 is the thermal equilibrium concentration of electron. We can find it out by integrating
the distribution of electron, n(E), in the conduction band
n(E)=g c (E) . f F (E)
We know that
Where g c (E)  density of allowed quantum states in the conduction band
and
 Fermi - Dirac probability function i.e the probability
f F (E)
of a quantum state being occupied by an electron

Now,
n0 
 n(E) dE
Ec

 n0 

g(E) f F (E) dE
Ec
Where E c = minimum amount of energy corresponding to conduction band
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SEMICONDUCTOR IN EQUILIBRIUM

=>
4π(2m*n )3 2
h3

n0 
Ec
E-E c
1
dE
 E-E F 
1+exp 

 kT 
-------
------ (1)
If
 Ec -EF 
>> kT
 E-EF 
>> kT
Then
( For conduction band electron, we have E > E c )
Then Fermi –Dirac distribution reduces to Boltzmann approximation, i.e.
1

 E-E F 
1 + exp 

 kT 
fF  E  =
1
 E-E F 
exp 

 kT 
 -  E-E F  
= exp 

 kT 
-------
------ (2)
Putting equation (2) in equation (1) we get

n0 

Ec
4π(2m*n )3 2
h3
 -(E-E F ) 
E-Ec exp 
 dE
 kT 
-------
------ (3)
Equation (3) can be solved by substituting the following
η=
Let
E - Ec
kT
-------
------ (4)
Then
dη =
dE
kT

dE = kTdη
-------
------ (5)
Again as E  E c , then η  0
------ (6)
And
-------
as E  , then η  
Again from eqn (4) we have,

E - Ec = η1 2
η=
E - Ec
kT
kT1 2
-------
------ (7)
Putting the above values in eqn (3) we get

n0 

0



n0 

0

n0 

0
4π(2m*n )3 2 1 2
 -( E- E F ) 
η (kT)1 2 exp 
 kT dη
h3
 kT 
4π(2m*n )3 2 1 2
 -(E- Ec ) - (Ec - E F ) 
η (kT)3 2 exp 
 dη
3
h
kT


* 32
4π(2mn )
 - (Ec - E F ) 
η1 2 (kT)3 2 exp (-η )exp 
 dη
h3
KT


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SEMICONDUCTOR IN EQUILIBRIUM

4π(2m*n kT )3 2
 - (Ec - E F ) 
1 2 -η
exp 
  η e dη
3
h
kT

0

n0 

n0 =

 2πm*n kT 
n0 = 2 

2
 h


 - (E c - E F ) 
n 0 = N c exp 

kT


4π(2m*n kT )3 2
 - (Ec - E F )  1
exp 
2 π
3
kT
h


32
( integral is a gama function )
 - (E c - E F ) 
exp 

kT


-------
------ (8)
32
 2π m*n kT 
Where, N c = 2 
 is called the effective density of states in the conduction
h2


band.
*
If m n  m 0 = 9.1  10 -31 kg and T = 300 K ( say )
Then Nc = 2.5 x 1019 cm-3
Similarly we can get thermal equilibrium concentration of holes in the valence band (p0)
 - (E F - E v ) 
p0 = N v exp 

kT


-------
------ (9)
32
 2π m*p kT 

h2


Where, Nv = 2 

is called the effective density of states in the Valence
band.
The order of Nv is also 1019 cm-3
The effective density of state functions Nc and Nv are constant for a given
semiconductor material at a fixed temperature.
From eqn (8) and (9) it is clear that n0 and p0 depends on:


Effective density of states function
Fermi energy level
Intrinsic carrier concentration
For intrinsic semiconductor, we have,
n 0 = p 0 = n i ( say )
and E F = E Fi = intrinsic fermi energy
We know that
and
 - (E c - E Fi ) 
n 0 = n i = N c exp 

kT


 - (E Fi - E v ) 
p 0 = n i = N v exp 

kT


So,
 - (E c - E Fi ) 
 - (E Fi - E v ) 
n 0 p0 = n 2i = N c N v exp 
 exp 

kT
kT





 - (E c - E v ) 
n 0 p 0 = n 2i = N c N v exp 

kT


----------------- (1)
----------------- (2)
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SEMICONDUCTOR IN EQUILIBRIUM

 - Eg 
n 0 p0 = n 2i = N c N vexp 

 kT 
----------------(3)
Where, (E c - E v ) = E g = Bandgap Energy
ni is strong function of temperature. At a constant temperature, the value of ni for a
semiconductor material is constant and is independent of the Fermi energy.
ni = 1.5 x 1010 cm-3
Silicon at T = 300 K
Eg = 1.12 eV
Fermi level for Intrinsic Semiconductor
We know that the thermal equilibrium concentration of electrons and holes are:
and
 - (E c - E Fi ) 
n 0 = N c exp 

kT


 - (E Fi - E v ) 
p0 = N v exp 

kT


------------------ (1)
------------------ (2)
Where E Fi = intrinsic fermi energy
But we know that at thermal equilibrium for intrinsic semiconductor, we have,
n 0 = p0

 - (E c - E Fi ) 
N c exp 

kT


= N v exp  - (E Fi - E v ) 

kT

Taking natural logarithm on both sides we get
 - (E c - E Fi ) 
lnN c + 
 =
kT


 - (E Fi - E v ) 
lnN v + 

kT



kT lnN c - E c + E Fi = kT lnN v - E Fi + E v

2E Fi = (E c +E v ) + kT ln

E Fi =


Nv
Nc
N
1
1
(E c +E v ) +
kT ln v
2
2
Nc
E Fi = E midband energy
 m* 
1
+ kT ln  *p 
m 
2
 n
E Fi = E midband energy +
3
 m*p 
3
kT ln  * 
m 
4
 n
2
----------------- (3)
From equation (3) following things are obtained
Then, E Fi = E midgap
1.
If
mp = mn
Thus Fermi Energy level is exactly at the center of the band gap.
2.
If
mp > mn
Then Fermi Energy level EFi is slightly above the center of the
band gap.
3.
If
mp < mn
Then Fermi Energy level EFi is slightly below the center of the
band gap.
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SEMICONDUCTOR IN EQUILIBRIUM
We know that Nc and Nv depends on mn and mp
We have seen from the above that the intrinsic Fermi energy level shifts
away from the band with larger density of states. This is due to maintain
equal number of electron and holes.
4.
Extrinsic Semiconductor
n- type semiconductor
1.
Intrinsic + pentavalent = n-type
2.
Electrons are the majority charge carrier
3.
Holes are the minority charge carrier
4.
n 0 > p0
5.
n 0 > n i and p 0 < n i
6.
E F > E Fi
p- type semiconductor
1
Intrinsic + trivalent = p-type
2
Holes are the majority charge carrier
3
Electrons are the minority charge carrier
4
p 0 > n0
5
p 0 > n i and n 0 < n i
6
E Fi > E F
Position of the Fermi level of Extrinsic Semiconductor
We know that
 - (E c - E F ) 
n 0 = N c exp 

kT


-------------- (1)
Taking logarithm on both the sides we get
 - (E c - E F ) 
lnn 0 = lnN c + 

kT


N 
E c - E F = kT ln  c 
 n0 
In n-type semiconductor we have,

Nd > ni
Thus
and

(E c - E F )
= lnNc - ln(n 0 )
kT
-------------- (2)
n 0  Nd
N 
E c - E F = kT ln  c 
 Nd 
-------------- (3)
Equation (3) represents the information regarding EF for n-type semiconductor with
respect to EC. It is clear that EF lies below EC.
Similarly, in p-type semiconductor we have,
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SEMICONDUCTOR IN EQUILIBRIUM
N 
E F - E v = kT ln  v 
 Na 
---------------(4)
Equation (4) represents the information regarding EF for p-type semiconductor with
respect to Ev. It is clear that EF lies above Ev.
Fermi Energy Level of Extrinsic Semiconductor w.r.t EFi :
We know that

 - (E c - E F ) 
n 0 = N c exp 

kT


(E
E

c
Fi ) + (E F -E Fi ) 
n 0 = N c exp 

kT



 - (E c - E Fi ) 
 (E F -E Fi ) 
n 0 = N c exp 

 .exp 
kT
 kT 



 (E - E ) 
n 0 = n i exp  F Fi 
 kT 
---------------------- (5)

n 
E F - E Fi = kT ln  o 
 ni 
---------------------- (6)
Equation (6) represents the information regarding EF for n-type semiconductor with
respect to EFi. It is clear that EF lies above EFi.
Similarly, for holes concentration we have

 - (E F -E Fi ) 
p 0 = n i exp 

kT


---------------------- (7)
p 
E Fi - E F = kT ln  o 
 ni 
---------------------- (8)
Equation (8) represents the information regarding EF for p-type semiconductor with
respect to EFi. It is clear that EF lies below EFi.
From equations (6) & (8) we have
If
E F > E Fi
If E F < E Fi
then n 0 > n i
from eqn. (6)
then n 0 < n i
from eqn. (6)
and
from eqn. (8)
and p0 > n i
from eqn. (8)
p0 < n i
thus n 0 > p 0
This is n-type semiconductor
thus p0 > n 0
This is p-type semiconductor
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SEMICONDUCTOR IN EQUILIBRIUM
Non- Degenerated Semiconductor

In extrinsic semiconductor number of impurity atom added is small compared to
the semiconductor atom.

Small number of impurity atoms are spread far enough apart so that there is no
interaction between donor electrons ( in case of n-type )

Due to the presence of impurity a discrete, non-interacting donor energy states in
n-type semiconductor is create. Similarly a discrete non-interacting accepter
energy state is created in p-type semiconductor.

These types of semiconductor are referred as non-degenerated semiconductor.
N-type Semiconductor
P-type Semiconductor
Degenerated Semiconductor
When concentration of impurity atom increases then

Distance between the impurity atom decreases

There will be an interaction between the charge carriers ( for example donor
electrons in n-type)

Single discrete donor energy level ( for n-type) and single discrete acceptor energy
level (for p-type) will split into a band of energies.

Donor states widen and may overlap the bottom of the conduction band. Acceptor
state may overlap the top of the valance band.

In n-type semiconductor when concentration of electrons in the conduction band
exceeds the density of states Nc then EF will lie within the conduction band.

In p-type semiconductor when concentration of holes in the valence band exceeds
the density of states Nv then EF will lie within the valence band.

Energy states below EF are mostly filled with electrons and energy state above E F
are mostly empty.

In degenerated n-type semiconductor, the states between EF and Ec are mostly
filled with electrons. So electron concentration is very large

In degenerated p-type semiconductor, the states between Ev and EF are mostly
empty. So hole concentration is very large
Variation of EF with Doping Concentration
We know that
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SEMICONDUCTOR IN EQUILIBRIUM
N 
E c - E F = kT ln  c 
 Nd 
As Nd increases, (E c - E F ) decreases. Thus EF shifts
towards the conduction band for n-type
semiconductor.
Similarly, EF shifts towards the valence band for ptype
semiconductor
as
acceptor
impurity
concentration increases.
In general as doping level increases the Fermi level
shifts towards that band.
The variation of Fermi energy with the variation of
doping concentration is shown in the figure.
Variation of EF with Temperature
We know that, for n-type semiconductor the position of the Fermi level is given by
N 
E c - E F = kT ln  c 
 Nd 
and
n 
E F - E Fi = kT ln  o 
 ni 
*
As the temperature increases, then Nc
increases and (E c - E F ) increases. Thus EF
moves away from the conduction band.
*
As the temperature increases, then ni
increases and EF moves closer to intrinsic
Fermi level.
*
At higher temperature, semiconductor
material loses its extrinsic characteristics
and begins to behave more like an intrinsic
semiconductor.
*
At low temperature Fermi level goes above
Ed for n-type material and below Ea for p-type material.
*
At T= 0K, All energy states below EF are full and all energy states above EF are
empty.
The variation of Fermi energy with the variation of temperature is shown in the figure.
Compensated Semiconductor
When a semiconductor contains both donor and acceptor impurities atom then it called
a compensated semiconductor.
If
Nd>Na
then it is called n-type compensated semiconductor we have
If
Na>Nd
then it is called p-type compensated semiconductor we have
If
Na = Nd,
then it is a completely compensated semiconductor. It will show
the characteristics of intrinsic semiconductor.
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SEMICONDUCTOR IN EQUILIBRIUM
Statistics of Donors and Acceptors
Due to the presence of donor electrons, a donor level is created just below the
conduction band. The energy corresponds to the donor level is Ed. When donor
electrons receive energy it jump to the conduction band and become ionized. These
electrons are treated as free electrons. The corresponding amount of energy is called the
ionization energy.
Thus,
ionization energy = Ed – Ec
Where,
Ec = Minimum amount of energy corresponds to conduction band
Ed = Energy corresponds to the donor level
The density of the electrons occupying the donor level ( nd ) is given by
nd = Nd - Nd+
Nd = Concentration of the donor atoms or donor electrons
Nd+ = Concentration of the ionized donors (i.e the number of electron jumped to the
conduction band.
Again, n d =
If
--------------(1)
 Ed - EF  >> kT , then eqn. (1) become
nd 
If
Nd
1
 E - EF 
1 + exp  d

2
 kT 
   Ed - E F  
Nd
= 2 N d exp 

kT
1
 E - EF 


exp  d

2
 kT 
 Ed - EF  >> kT ,
--------------(2)
then also Boltzmann approximation valid. So we can write
   Ec - E F  
--------------(3)
n 0  Nc exp 

kT


nd
Now
represents relative number of electrons in the donor state compared with
nd + n0
the total number of electrons
   Ed - EF  
2N exp

kT

   Ed - EF  
2N d exp 
 + Nc exp
kT


So,
nd
=
nd + n0

nd
=
nd + n0
d


   Ec - EF  


kT


1
   Ec - Ed  
Nc
1+
exp 

2N d
kT


--------------(4)
Where  Ec - Ed  = Ionization energy of the donor electrons
Case : I Freeze – out
At absolute zero ( T = 0K ), all the donor electrons are in their lowest possible energy
state i.e all the donor state must contain an electrons. No electron from the donor state is
thermally elevated into the conduction band. This effect is called freeze – out.
Thus at freeze – out we have nd = Nd ,
and
Nd+ = 0
Putting this condition in eqn. (1) we get
13 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
nd =
Nd
1
 E - EF 
1 + exp  d

2
 kT 

Nd =
Nd
1
 E - EF 
1 + exp  d

2
 kT 

1
 E - EF 
exp  d
= 0
2
 kT 

 E - EF 
exp  d
 = 0 = exp  -  
 kT 

 Ed

EF > Ed
- E F  =  ve
Thus the Fermi energy level must be above the donor energy level at absolute zero. In
case of p-type semiconductor at absolute zero temperature, the impurity atom will not
contain any electron, so Fermi energy level must be below the acceptor energy state, so
that the Fermi energy level must be below the acceptor energy state.
Case : II Complete ionization
At room temperature, all the electrons from the donor level
are thermally elevated to the conduction band for n-type
semiconductor and become ionized. This is called complete
ionization.
So for complete ionization we have
nd = Nd - Nd+ = 0


Nd = Nd+
Donor Impurity concentration = Ionized Donor Impurity
For p-type semiconductor acceptor atom has accepted an
electron from the valence band. So a hole is created at the
valence band for each acceptor impurity atom. So for
complete ionization we have
pa = Na - Na- = 0


Na = N -a
Acceptor Impurity concentration = Ionized Acceptor Impurity
Where pa is the density of the holes occupying the acceptor level and is given by
pd =
Now
Na
= N a - N a1
 EF - Ea 
1+
exp 

4
 kT 
pa
represents relative number of holes in the acceptor state compared with
pa + p0
the total number of holes
pa
=
pa + p0
1
   Ea - E v  
N
1 + v exp 

4N a
kT


14 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
Equilibrium Electron and Hole concentration using charge neutrality
Due to the presence of donor electrons, a donor level is created just below the
conduction band. The energy corresponds to the donor level is Ed. When donor
electrons receive energy it jump to the conduction band and become ionized. These
electrons are treated as free electrons. The corresponding amount of energy is called the
ionization energy.
Thus,
ionization energy = Ed – Ec
Where,
Ec = Minimum amount of energy corresponds to conduction band
Ed = Energy corresponds to the donor level
The density of the electrons occupying the donor level ( nd ) is given by
nd = Nd - Nd+
Nd = Concentration of the donor atoms or donor electrons
N = Concentration of the ionized donors (i.e the number of electron jumped to the
conduction band.
+
d
For p-type semiconductor the occupies the acceptor level which is just above the
valence band. The acceptor atom has accepted an electron from the valence band. So a
hole is created at the valence band for each acceptor impurity atom and we say that
holes are ionized ( N-a )
So,
pa = Na - Napa = density of the holes in the accepter level
Na = Concentration of the impurity acceptor atoms
N = Concentration of the ionized acceptors
From the charge neutrality condition we have
Density of negative charges = Density of positive charges
-
a

n0 + Na- = p0 + Nd+

n 0 + Na

= p0 + N d
n 0 + Na =

n 0 + (N a - pa ) = p0 + (Nd - n d )
(During complete ionization)
n i2
+ Nd
n0
( p0 =
n i2
)
n0
n02 - (Nd - Na ) n0 - ni2 = 0
This is a quadratic equation of n0. The solution is given by
n0 =
(N d - N a ) 
(N d - N a ) 2 4  n i2
2
Negative sign is not acceptable.
So,
n0 =
(Nd - Na )
2
2
 Nd - Na 
2

 + ni
2


-------------------- (1)
Similarly for p0, we have
p0 =
(Na - Nd )
2
2
 Na - Nd 
2

 + ni
2


Equation (1) and (2) represents the n0
compensated semiconductor.
-------------------- (2)
and p0 concentrations for and extrinsic
15 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
Formulae to Remember
1.
4π(2m )
E-E c
h3
g c (E) = Density of allowed electronic energy states in the conduction band
* 3/2
n
g c (E) =
4π(2m*p )3/2
2.
g v (E) =
3.
f F (E)
4.
fF (E) = Fermi – Dirac distribution function
= Fermi – Dirac probability
function
= Probability that a quantum state at the energy E will be occupied by an
electron.
Where,
EF
= Fermi energy
Probability that a state is being empty of electron is given by
E v -E
h3
g v (E) = Density of allowed quantum state in the valence band

1
 E-E F 
1 + exp 

 kT 
1  f F (E) =1 
5.
1
1

E-E

E - E
F 
1+exp 
1+exp  F

 kT 
 kT 
If
(E - EF) >> kT
1
Then f F (E) =
 (E-E F ) 
exp 
 kT 
=
 - (E-E F ) 
exp 
 kT 
This become Boltzmann
approximation
6.
If n(E) is the distribution of electrons with respect to energy in the conduction
band then
n(E) = g c (E) . f F (E)
Where g c (E)  density of allowed quantum states in the conduction band
and f F (E)  Fermi - Dirac probability function i.e the probability
7.
of a quantum state being occupied by an electron
If p(E) is the distribution of holes with respect to energy in the valance band
then
p(E) = g v (E) . [ 1 - f F (E) ]
Where g v (E) = density of allowed quantum states in the valence band
8.
and 1 - fF (E) = Probability that a quantum state is not occupied by an electron
Concentration of electrons in the conduction band by assuming Boltzmann
approximation is given by
 - (E c - E F ) 
n 0 = N c exp 

kT


*
Where, N c = 2  2π m2n kT 
h


32
= effective density of states in the conduction
band.
16 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
9.
Concentration of holes in the valence band by assuming Boltzmann
approximation is given by
 - (E F - E v ) 
p 0 = N v exp 

kT


32
 2 π m*p k T 

h2


Where, N v = 2 

10.
Valence band.
Intrinsic carrier concentration is given by
 - Eg 
Where, (E c - E v ) = E g = Bandgap Energy
n 0 p0 = n 2i = Nc N v exp 

kT


Silicon at T = 300 K
11.
is called the effective density of states in the
ni = 1.5 x 1010 cm-3
Eg = 1.12 eV
Position of the intrinsic fermi level with respect the center of the band gap is
given by
E Fi = E midband energy +
 m*p
3
kT ln  *
m
4
 n



mn = effective mass of the electron mp = effective mass of the holes
Where E Fi = intrinsic fermi energy
12.
Position of the fermi level in the n-type semiconductor in terms of N c and n 0 is
given by
N 
Ec - E F = kT ln  c 
 n0 
13.
Position of the fermi level in the n-type semiconductor in terms of N d and N c
N 
Ec - E F = kT ln  c 
 Nd 
is given by
14.
Since, in n-type semiconductor we have, 
N > n

and n 0  Nd
i
 d

Position of the fermi level in the p-type semiconductor in terms of N a and N v
15.
N 
is given by E F - E v = kT ln  v 
 Na 
Concentration of electrons in the conduction band in terms of n i is given by
 (E -E ) 
n 0 = n i exp  F Fi 
 kT 
16.
Concentration of holes in the valence band in terms of n i is given by
 - (E F -E Fi ) 
p 0 = n i exp 

kT


17.
Position of the fermi level in the n-type semiconductor in terms of n o and n i is
given by E F - E Fi = kT ln  n o 
 ni 
17 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
18.
Position of the fermi level in the p-type semiconductor in terms of p o and n i is
given by E Fi - E F = kT ln  po 
 ni 
18 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
Numerical
Density of state functions & Fermi-Dirac Distribution Function
1.
Determine the total number of energy states in the GaAs between Ec and Ec + kT
at T = 300 K. For GaAs m*n = 0.067 m0, m*p = 0.48 m0
Ans.:
We know that
4π(2m*n )3/2
g c (E) =
E-E c
h3
E + kT
4π(2m*n )3/2 c
 g T (E) =
 E-Ec dE
h3
Ec
E c + kT
3
4π(2m*n )3/2  2  
2
g T (E) =
   E-E c  
h3
Ec
3
* 3/2
3
4π(2m n )
2
. .  kT  2
 g T (E) =
3
3
h
g T (E) =
2.
4π  2 x 0.067 x 9.11 x 10-31 
 6.625 x 10 
-34 3
.
3
2
.  0.0259 x 1.6 x 10-19  2
3
3.28 x 1017 cm -3
g T (E) = 3.28 x 10 23 m -3
=
Find the ratio of the effective density of states in the conduction band at Ec + kT
to the effective density of states in the valence band at Ev - kT.
gc
gv
Ans.:
=
4π(2m*n )3/2
h3
4π(2m*p )3/2
E-E c
gc
gv
(a)
(b)
Ans. (a)

E v -E
h3
3.
3/2
 m*
=  *n
m
 p
gc
gv
 m*
=  *n
m
 p
3/2



Ec  kT - Ec

E v - E v + kT
3/2



If EF = Ec, find the probability of a state being occupied at E = Ec + kT
If EF = Ev, find the probability of a state being empty at E = Ev - kT
f F (E) =
1
1
1
=
=
 0.269
E-E
E
+
kT
-E
1+exp
1

 c

F 
c 
1+exp 
1+exp 


kT
 kT 


(b)
1 - f F (E) = 1 
1
1
1
 0.269
= 1
=1
1+exp  -1
 E-E F 
 E v -kT-E v 
1+exp 
1+exp 


kT
 kT 


4.
Determine the probability that an energy level is occupied by an electron if the
state is above the Fermi level by (a) kT
(b) 5 kT (c) 10 kT
1
1
Ans.: (a)
Here E – EF = kT
f (E) =
=
= 0.269
F
 E-E F 
1+ exp 

 kT 
1+ exp 1
19 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
(b)
Here E – EF = 5kT, So
f F (E) =
5.
1
1
=
= 6.69 x 10-3
E-E
1+exp
5

 
F 
1+exp 

 kT 
Determine the probability that an energy level is empty of an electron if the state
is below the Fermi level by (a) kT
(b) 5 kT
(c) 10 kT
Ans.: 1 - f (E) = 1 
F
(a)
1
=
 E-E F 
1+exp 

 kT 
E – EF = kT,
1
 E F -E 
1+exp 

 kT 
1 - f F (E) =
1
= 0.269
1+exp 1
(b) and (c)
Similarly
Show that the probability of an energy state being occupied ΔE above the Fermi
energy is the same as the probability of a state being empty ΔE below the Fermi
level.
Ans.: The probability of a state at E1 = EF + ΔE , being occupied is
6.
1
 E -E 
1+exp  1 F 
 kT 
f F (E1 ) =
1
 E 
1+exp 
 kT 

The probability of a state at E2 = EF - ΔE , being empty is
1 - f F (E 2 ) = 1 
7.
Hence
f F (E1 ) = 1 - f F (E 2 )
(a)
Determine for what energy above EF ( in terms of kT ) the Fermi – Dirac
probability function is within 1% of the Boltzmann approximation.
(b) Give the value of the probability function at this energy.
Ans.: (a)



 Boltzmann  -  Fermi - Dirac 
 Fermi - Dirac 
E - E 
1+exp  1 F 
 kT 
E - E 
exp  1 F 
 kT 
8.
= 0.01
- 1 = 0.01


1
 E1 -E F 
exp 

 kT 
1
 E1 - E F 
1+exp 

 kT 
1
E - E 
1+exp  1 F 
 kT 
-
1
 E1 - E F 
exp 

 kT 
= 0.01
= 0.01
E - E 
E - E 
 exp  1 F   100
1 = 0.01 x exp  1 F 
 kT 
 kT 
E1 - E F
 ln(100)
 E1 - E F  kT ln(100)
kT

(b)
1
1
1
=

 E 2 -E F 
 EF - E2 
 E 
1+exp 
1+exp 
1+exp 


 kT 
 kT 
 kT 
f F (E1 ) =
E1 - E F  4.6 kT
1
 E -E 
1+exp  1 F 
 kT 

 E1  E F  4.6 kT
1
1

 0.0099  0.01
4.6
kT
1+exp  4.6


1+exp 

 kT 
Calculate the temperature at which there is a 1% probability that a state 0.30 eV
below the Fermi level will be empty of an electron.
20 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
Ans.:



1
1

E-E

E - E
F 
1+exp 
1+exp  F


 kT 
 kT 
1
1
 0.99 
1 - 0.01 =
 - 0.30 
 - 0.30 
1+exp 
1+exp 
 kT 
 kT 
1
 - 0.30 

 1.0101
1+exp 
0.99
 kT 
1 - f F (E) = 1 
 - 0.30 
exp 
 0.0101
 kT 
0.30
= ln (99)

kT

kT
1
 0.30 
exp 

 99

0.0101
 kT 
0.30

=
= 0.06529
ln(99)
T = 756 K
9.
Assume the Fermi energy level is exactly in the center of the band gap energy of a
semiconductor at T = 300 K
(a) Calculate the probability that an electron occupies an energy state in the bottom
of the conduction band for Si, Ge and GaAs.
(b) Calculate the probability that an energy state in the top of the valence band is
empty for Si, Ge and GaAs.
Ans.:
At E = E midgap
f F (E) =
f F (E) =
1
 E-E F 
1+exp 

 kT 

1
 Eg

1+exp  2
kT





1
 Eg 
1+exp 

 2kT 
1
= 4.07 x 10-10
 1.12 
1+exp 

 2(0.0259) 
1
f F (E) =
= 2.93 x 10-12
 0.66 
1+exp 

 2(0.0259) 
For Si, E g = 1.12 eV
f F (E) =
For Ge, E g = 0.66 eV
For GaAs, E g = 1.42 eV
f F (E) =
1
 1.42 
1+exp 

 2(0.0259) 
= 1.42 x 10-12
(b)
Similarly
10. Calculate the temperature at which there is a 10-6 probability that an energy state
0.55 eV above the Fermi energy level is occupied by an electron.
Ans.:
1
6
f F (E) =


10

 0.55 
1+exp 
 kT 
1
 0.55 
 0.55 
1+ exp 
=
 106
 exp 
 106
6

10
 kT 
 kT 
0.55
0.55
= ln 106
 kT 
 T = 461 K
kT
ln 106
11. Calculate the energy range ( in eV) between f F (E) = 0.95 and f F (E) = 0.05 for
E F = 7.0eV at (a) T 300 K
(b)
500 K
21 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
Ans.: At
E = E2
So,
f F (E 2 ) = 0.05
1
 E2  EF 
1+exp 

 kT 
1
 E  EF 
1+exp  2
  0.05  20
kT


f F (E 2 ) = 0.05 

 E  EF 
exp  2

 kT 

By symmetry, at
 19
E = E2

E2  EF
 ln 19
kT
-------(i)
1 - f F (E1 ) = 0.05
E F - E1
 ln 19
kT
So,
--------(ii)
Now from eq (i) + (ii)
E 2 - E1
 2 ln 19
kT
(a)
At
E 2 - E1 
T = 300 K,
E 2 - E1 
(b)

At
kT = 0.0259 eV,
2 (0.0259) ln 19 = 0.1525 eV
T = 500 K,
E 2 - E1 
2 kT ln 19
kT = 0.04317 eV,
2 (0.04317) ln 19 = 0.254 eV
The Semiconductor in Equilibrium
For Silicon
N C = 2.8 × 1019 cm -3
19
-3
10
-3
, N V = 1.04 × 10 cm , n i = 1.5 × 10 cm
12. Calculate the probability that a state in conduction band is occupied by an electron
and calculate the thermal equilibrium electron concentration in Silicon at T =
300K. Given that Fermi energy is 0.25 eV below the conduction band.
Answer: The probability that an energy state E = Ec is occupied by an electron is
given by
1
 6.43 x 10-5
1

f F (E) =
 E-E F 
1+ exp 

 KT 
 0.25 
1  exp 
 0.0259 
Electron concentration is given by
 -  E C -E F  
=>
n 0 = N C exp 

KT


 - 0.25 
n 0 = 2.8 * 1019 exp 

 0.0259 
n 0 = 1.8 *1015 cm -3
13. Calculate the thermal equilibrium hole concentration in silicon at T = 400 K.
Given that Fermi energy is 0.27 eV above the valence band energy.
3
Answer: N v  T 2
 N v 400 K
 N v 300 K
=>
Again
=
 400 


 300 
3
3
 400  2
=>  N v 
=  N v 300 K 

400 K
2
400
 Nv 400 K = 1.04 x 1019  
300


kT
=
0.0259
eV
 300 K
 300 
3
2
=>
 Nv 400 K
= 1.6 x 1019 cm-3
22 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
 kT 400 K
Thus
 kT 300 K
=
 400 
 400 

 = 0.0259 x 
 =0.03453 eV
300


 300 
Now the hole concentration is
 -  E -E  
 - 0.27 
15
-3
p0 = N V exp  F V  = 1.6 *1019 exp 
 =6.43 x 10 cm
KT
0.03453




14. Calculate the intrinsic concentration in Gallium Arsenide ( GaAs) at 300 K and
450 K. Given that N C = 4.7 × 1017 cm -3 N V = 7.0 × 1018 cm-3
Answer:
 kT300 K
= 0.0259 eV
T = 300 K
At
 -E g 
n i2 = NC N V exp 

 kT 
=>
 -1.42 
n i2 = (4.7x1017 ) (7.0 x 1018 ) exp 
=5.09 x 1012 cm -6
 0.0259 
n i = 2.26 x 10 6 cm -3
At
T = 450 K
 450 

 = 0.0259 x
 300 
 kT 450 K
=
 kT 300 K
 NC 450 K
 450 
= (4.7 x 10 ) 

 300 
3
17
2
=>
 450 

 =0.03885 eV
 300 
 NV 450 K
3
 450 
= (7.0 x 10 ) 

 300 
18
3
2
3
 450  2
 450  2
 1.42 
18
n = (4.7 x 10 ) 
(7.0
x
10
)


 exp 
 300 
 300 
 0.0259 
=>
n i2 = 1.48 x 10 21 cm -6
n i = 3.85 x 1010 cm -3
2
i
17
15. Find the intrinsic carrier concentration in Silicon at T = 200 K
Answer: n 2 = N N exp  -E g 
i
n
2
i
C
=


 kT 
V
19
(2.8 x 10
 200 
) 

 300 
3
2
19
(1.04 x 10
 200 
) 

 300 
3
2




1.12


exp 

 200  
 0.0259 


 300  

n i = 7.68 x 10 4 cm -3
16. Calculate the position of the intrinsic Fermi Level with respect to the center of the
band gap in Silicon at T = 300 K. Given that mn* = 1.08 m0 and mp* = 0.56 m0
Answer: We know that
E Fi - E midband energy =
E Fi - E midband energy =
 m*p 
3
kT ln  * 
m 
4
 n 
 0.56m0 
3
 0.0259  ln 
 = - 0.0128 eV = - 12.8 meV. Thus
4
 1.08m0 
intrinsic Fermi level in Silicon is 12.8 meV below the mid energy.
23 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
17. The intrinsic carrier concentration in silicon is no greater than
n i = 1.0 x 1012 cm -3 . Assume Eg =1.12 eV. Determine the maximum temperature
allowed for the Silicon
Answer: n i2 = NC N V exp  -E g 
 kT 
1.0x10 
12
2
=
 T 
(2.8 x 10 ) 

 300 
19
3
2
 T 
(1.04 x 10 ) 

 300 
19
3
2




1.12

exp 
 T 

0.0259



 300  


T = 381 K
18. The magnitude of the product g C (E) f (E) in the conduction band is a function of
energy. Assume the Boltzmann approximation is valid. Determine the energy with
respect to E C at which the maximum occurs.
Answer: We know that
3
 -  E - EF  
4 (2m*n ) 2
g (E) f(E) =
E-E
exp
C



g C (E) f(E)
=
4 (2m*n )
h3
g C (E) f(E)

E - EC


C
h3
3
2
E - EC
 -  E - EC  
 -  EC - EF  
exp 
 exp 

kT
kT




 -  E - EC  
exp 

kT


- x 
x exp 
 kT 
g C (E) f(E) 


kT
E - EC  x
Let
For g C (E) f(E) to be maximum we must have


d 
 - x 
x exp 
 = 0
dx 
 kT  

 - x   -1 
 - x  1 
x exp 

 + exp 

 = 0
 kT   kT 
 kT   2 x 
- x
1
1

+
= 0
=
kT
2 x
2 x


x =
E
=

kT
2
EC +
x
kT
E - EC =
kT
2
kT
2
19. Assume the Boltzmann approximation in a semiconductor is valid. Determine the
ratio of n(E) = g C (E) f (E) at E = EC + 4 kT to that E = E C + kT
Answer:
g C (E) f(E)
Thus

n(E1 )
n(E 2 )

n(E1 )
n(E 2 )

E - EC
2
 -  E - EF  
exp 

kT


 -  E1 - E C  
exp 

kT
E1 - E C
n(E1 )


=
n(E 2 )
 -  E2 - EC  
E2 - EC
exp 

kT


4kT
 -E1  E 2 
=
exp 

kT
kT


2
= 2 2
1

exp   4 
2

= 2 2
exp
 - 3.5 
24 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM

n(E1 )
n(E 2 )
= 0.0854
20. Two semiconductor materials have exactly the same properties except that
material A has a band gap energy of 1.0 eV and material B has a band gap energy
of 1.2 eV. Determine the ratio of ni of material A to that of material B at T = 300
K.
Answer:
2
i
2
i
n (E1 )
n (E 2 )
=
 n(E1 )
n(E 2 )
 -E gA
exp 
 kT
 -E gB
exp 
 kT




 = exp  -  E gA  E gB  


kT





 -  E gA  E gB  
 = 47.5
= exp 


2kT


21. The carrier effective masses in a semiconductor are m*n = 0.62 m 0 and
m*p = 1.4 m0 Determine the position of the intrinsic Fermi level with respect to
the center of the band gap at T = 300 K.
Answer:
E Fi E Fi -
E midband energy =
E midband energy =
 m*p 
3
KT ln  * 
m 
4
 n 
 1.4m0 
3
 0.0259  ln 
 = + 0.0158 eV
4
 0.62m0 
22. Calculate EFi, with respect to the center of the band gap in Silicon for T = 200 K.
Answer:
E Fi - E midband energy =
N 
 1.04 x 1019 
1
1
kT ln  V  = kT ln 
19 
2
2
 2.8 x 10 
 NC 
EFi - Emidband energy = - 0.495  kT 
23. The electron concentration in Silicon at T = 300 K is n0 = 5 x 104. (a) Determine
p0. Is this n or p – type material? (b) Determine the position of the Fermi level
with respect to the intrinsic Fermi level.
Answer: n0 = 5 x 104
at T = 300 K
1.5 x 1010 
n2
p0 = i =
n0
5 x 104
2
 4.5 x 1015 cm 3
As p0 > n 0 the material is p-type semiconductor.
We know that
 -  E F -E Fi  
p0 = n i exp 
 =>
kT


p  
E Fi -E F = kT ln  0 
 ni 
 4.5 x 1015 
E Fi -E F = 0.0259 ln 
= 0.3266 eV
10 
 1.5 x 10 
24. Determine the values of n0 and p0 for Silicon at T = 300 K if the Fermi energy is
0.22 eV above the valence band energy.
Answer: E F -E V = 0.22 eV
at T = 300 K
 -  E F -E V  
 -  0.22  
19
15
-3
p0 = N V exp 

KT
 = 1.04 x 10

exp 
 = 2.13 x 10
 0.0259 
cm
E C - E F = 1.12 - 0.22 = 0.90 eV
25 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
 -  E C -E F  
 -  0.90  
18
4
-3
n 0 = N C exp 
 = 2.8 x 10 exp 
 = 2.27 x 10 cm
kT


 0.0259 
25. The value of p0 in Silicon at T = 300 K is 1015 cm-3. Determine EC – EF and n0.
Answer: p0 = 1015 cm-3
at T = 300 K
 -  E F -E V  
p0 = N V exp 

=>
kT


N 
1.04 x 1019 
E F - E V = kT ln  V  = 0.0259 ln 
 = 0.24 eV
1015


 p0 
=> EC  EF  1.12  0.24  0.88 eV
26. For a particular semiconductor, Eg = 1.50 eV. m*p = 10 m*n at T = 300 K, and ni =
1.0 x 105 cm-3.
(a)
Determine the position of the intrinsic Fermi energy level with respect to
the center of the band gap.
(b) Impurity atoms are added so that the Fermi energy level is 0.45 eV below the
center of the band gap.
(i) Are acceptor or donor atoms added?
(ii) What s the concentration of impurity atoms added?
Answer: (a)
E Fi - E midband energy =
(b)
 m*  3
3
 kT  ln  *p  =  0.0259  ln 10 = + 0.0447 eV
4
 mn  4
Impurity added so that E midband energy - E Fi = 0.45 eV
(i)
So it is p-type semiconductor. Thus acceptor impurity is added.
(ii)
E Fi - E F = 0.0447 + 0.45 = 0.4947 eV
 -  E F -E Fi  
 0.4947 
5
13
-3
p0 = n i exp 
 = 10 exp 
 = 1.97 x 10 cm
KT
0.0259



p 0 = N a = 1.97 x 1013 cm -3
ASSIGNMENT
1.
What is the quantum state within the forbidden energy band?
2.
When do the Fermi – Dirac distribution changes to Boltzmann approximation?
3.
What is the physical significance of Fermi – Dirac distribution function fF(E)?
4.
What is the value of fF(E) at 0K for E > EF and E < EF?
5.
At absolute zero a semiconductor acts as an insulator. Justify it by using Fermi –
Dirac distribution function.
6.
fF(E) represents the probability of a state being occupied by an electron. Then
what will be the probability of a state being empty?
7.
Define allowed energy band.
8.
Define density of state function.
9.
What is Fermi energy?
10. What is Boltzmann approximation?
11. Fermi – Dirac distribution function varies with temperature. Represent it
graphically.
26 | P a g e
SEMICONDUCTOR IN EQUILIBRIUM
12. What is direct and indirect band gap semiconductor?
13
Write the equation for n(E) as a function of the density of states function for
conduction band and the Fermi probability distribution function.
14
Write the equation for p(E) as a function of the density of states for valence band
and the Fermi probability distribution function.
15
Write the equations for n0 and p0 in terms of Fermi energy by assuming
Boltzmann approximation.
16
What is the value of intrinsic carrier concentration in silicon at T=300K?
17
Under what condition would the intrinsic Fermi level be at the mid-gap energy?
18
What is donor impurity and acceptor impurity?
19
Sketch a graph of n0 versus temperature for an n-type material.
20
Sketch graphs of the Fermi energy versus donor impurity concentration.
21
Sketch graphs of the Fermi energy versus temperature.
22
Write the condition of which Fermi-Dirac probability function changes to
Boltzmann approximation.
23
What is the order of Nc and Nv for most of the semiconductor material at
T=300K?
24
Discuss that the intrinsic Fermi energy level is a function of mn and mp
25
What are donor and acceptor level?
26
What is ionization energy?
27
Justify that in n-type semiconductor the position of the Fermi energy level is
above the Fermi energy level for intrinsic semiconductor.
28
Justify that in p-type semiconductor the position of the Fermi energy level is
below the Fermi energy level for intrinsic semiconductor.
29
What is a non-degenerate semiconductor?
30
What is a compensated semiconductor?
31
How is the Fermi energy level change with impurity concentration?
32
How is the Fermi energy level change with temperature?
27 | P a g e