SEMICONDUCTORS
Applied Physics
UNIT IV
CARRIER CONCENTRATION
How to Find the Carrier Concentration?
Current is the rate at which charge flows. The charge carriers in a
semiconductor are electrons and holes. To determine the electrical
properties and analyze device behavior, we often need to know the carrier
concentration i.e., number of electrons and holes per cm3 of the material.
For heavily doped extrinsic semiconductor,
n0  N D  p0
(for n-type) and
p0  N a  n0
(for p-type)
However, we don’t know anything about the minority carrier concentration.
Nor do we know about the temperature dependence of the carriers.
In this section, we wish to determine the concentration of electrons and
holes in the conduction and valence bands.
Electrons and holes are distributed over the empty conduction band and
valence band states, respectively, as function of energy. The distribution
depends upon the temperature. So it is important to find the electron and
hole distribution over the required energy range, in order to determine the
carrier concentration.
2
THE CARRIER CONCENTRATION
Equilibrium Electron and Hole Concentrations
To calculate the number of electrons in the CB and holes in
the VB we must know
1) How many states are available
2) What is the probability of occupancy
We thus introduce g(E)dE, DENSITY OF STATES (cm-3),
available for occupancy in the energy range dE.
3
Carrier Concentration
Electron Distribution
The distribution (with respect to energy) of electrons in the conduction
band is given by the density of allowed quantum states, gc(E), times the
probability, f(E), that an electron occupies a state.
nE   g c E f E 
where f(E) is the Fermi-Dirac probability function and gc(E) is the density of
quantum states in the conduction band.
Hole Distribution
The distribution (with respect to energy) of holes in the band is given by
the density of allowed quantum states, gv(E), times the probability, 1-f(E),
that a state is Not occupied by an electron.
pE   g v E 1 f E 
 Both g(E) and f(E) are in terms of per unit volume per unit energy.
4
Density of States
Density of States
h is the Planck’s constant
h  6.625 1034 J sec
Exercise: Determine the total number of energy states in Si between Ec
and Ec + kT at T=300 K. [ m *  1.08m ].
n
e
5
The Carrier Density Distribution
The Carrier Density Integral
nE  g c E f E
The carrier density integral. Shown are the density of states, gc(E), the
density per unit energy, n(E), and the probability of occupancy, f(E). The
carrier density, no, equals the crosshatched area.
6
THE FERMI-DIRAC DISTRIBUTION
The Fermi-Dirac Distribution Function
The Fermi-Dirac distribution function, which provides the probability of
occupancy of energy levels by electrons, is based on following premises:
- Particles (electrons) are indistinguishable, and only one particle
is allowed in each quantum state.
- At absolute zero temperature (T = 0 K), the energy levels are all
filled up to a maximum level, called the Fermi level. Beyond the
Fermi level, all states are empty.
- At higher temperature, there is a gradual transition between
the completely filled states and the completely empty states.
 Mathematically, Fermi-Dirac distribution function is given by,
f (E) 
1
1  e ( E E F ) / kT
and gives the probability that an available state at energy E will be
occupied by an electron at absolute temperature T.
The quantity EF is called the Fermi level, and k is Boltzmann’s constant
(k = 1.38x10-23 J/K = 8.62x10-5 eV/K )
7
The Fermi-Dirac Distribution
The Fermi-Dirac Distribution Function
 Consider the situation where T = 0 K.
For E < E F,
For E > EF,
f (E) 
f (E) 
1
1 e( E E
F
) / kT
1
1 e( EE
F
) / kT


1
1 eE / kT 0
1
1 eE / kT 0


1
1

1 e
1
0

1 e
 Thus for T = 0 K,
For energies E < EF, all states are filled.
For energies E > EF, all states are empty.
Fermi Probability
function for T = 0 K.
8
The Fermi-Dirac Distribution
The Fermi-Dirac Distribution Function
 Consider the situation where T > 0 K.
For E = EF,
f (E) 
1
1 e( E E
F
) / kT
1
1


1 e0 / kT 2
thus an energy state at the Fermi level has a probability of ½ of
being occupied by an electron.
In general, for temperatures T > 0 K, there is a nonzero probability
that some energy states above EF will be occupied by electrons
and some energy states below EF will be empty.
9
The Fermi-Dirac Distribution
The Fermi-Dirac Distribution Function
As the temperature climbs, we
notice that more states below
EF are empty and more states
above EF are filled.
Fermi Probability function vs.
Energy for T > 0 K.
 For E - EF >> kT,
Fermi-Dirac distribution function can be approximated to,
f (E) 
1
1 e( EE
F
) / kT

1
e( E E
F
) / kT
 e( EE
F
) / kT
10
The Fermi Level
The Fermi Level – Intrinsic Semiconductor
For intrinsic semiconductor, since electrons and holes are
generated in pairs, n0 = p0 for all temperature. This leads to,
f EC   1 f (EV )
( E E ) / kT
e
1
1
1
1



1 e( E E ) / kT 1 e( E E ) / kT 1 e ( E E ) / kT
1 e( E E ) /kT
V
C
F
V
F
F
V
F
V
F
 e ( E E ) / kT  e ( E E ) / kT  EC  E F  (EV EF )
C
 EF 
F
V
F
1
EC  EV   E Fi
2
i stands for intrinsic, and EFi is the Fermi level for intrinsic
semiconductor.
This shows that for intrinsic semiconductor, Fermi level lies
exactly halfway between the CB and the VB.
11
The Fermi Level
The Fermi Level – Intrinsic Semiconductor
EC
EF = EFi
EV
Energy band diagram of an intrinsic semiconductor.
Fermi level lies exactly halfway between
the CB and the VB.
12
The Density of States and the Carrier
Concentration
Carrier Concentration – Intrinsic Semiconductor
The figure shows a possible
distribution of density of
states and electron and
hole concentrations for an
EF near the mid gap energy
(Fermi
probability
function superimposed):
nE
gc Ef E
Ec
pE
gv E1f E
Ev
The shaded areas under the curves in the figure is then
the total density of electrons in the conduction band (CB) and
the total density of holes in the valence band (VB).
13
The Carrier Density and the Effecttive
Density of States
Equilibrium Electron and Hole Concentrations
The electron concentration can be calculated by integrating the
carrier density distribution from EC


n0  g cE . f E .dE NC f EC
EC
 2m*ekT 

N C  2
2
 

32
Where,
is called the effective
density of states in the CB.
And f (EC) is the occupation propability of available states at the
conduction band edge EC by an electron.
Likewise, the equilibrium hole concentration in the VB is given by,

Ev
p0  gv 
E. 1 f E .dE N

where
*
 2mkT
p
NV  2
 2





32
V
1f E V
is called the effective
density of states in the VB.
14
The Carrier Concentration Revisited
Equilibrium Electron and Hole Concentrations
From slide 6, the equilibrium electron concentration in CB is
n0 N C  f EC  N C

1
1 eE
C
E F  kT
For EC – EF >> KT, the electron concentration n0 can be written as,
n0  N C e E  E  kT
C
F
Likewise, the equilibrium hole concentration in the VB is given by,
p0  NV e E
F
 EV
 kT
15
The Fermi Level: Extrinsic Semiconductor
Fermi Level for Extrinsic Semiconductor
It can be concluded that
for n-type material EF is closer to EC
and as
(EC-EF)
n
p
Similarly for p-type material, EF is closer to EV
and as
(EF-EV)
p
n
EC
EF moves toward EC
for n-type
EF moves toward EV
for p-type
EF = EFi,
for intrinsic
semiconductor
EV
16
The Effecttive Density of States
Effective Density of States
Note that the effective density of states is temperature dependent and can be obtain
from:
where Nc(300 K) is the effective density of states at 300 K.
17
THE PROBABILITY CALCULATION
Exercise: Assume the Fermi energy level is 0.30 eV below the conduction
band energy and T = 300 K.
Determine the probability of a state being occupied by an electron at Ec.
f (E) 
1
1 e( E E
F
) / kT

1
1 e( E
C
E F ) / kT

1
1 e0.03eV / 0.0259eV
 9.323106
Exercise: Assume the Fermi energy level is exactly in the center of the
bandgap energy of a semiconductor at T = 300K..
Calculate the probability that an energy state in the bottom of the conduction
band (at Ec) is occupied by an electron for Si..
For Si, Eg = 1.12 eV. Since, E - EF = EC - EF = 0.56 >> kT, we can use the
Boltzmann approximation:
f (E) 
Hence,
1
1 e( E E
f (E)  e
F
) / kT
( E E F ) / kT

e
1
e( E E
F
) / kT
( EC E F ) / kT
 e( EE
e
F
) / kT
0.56eV
0.02586eV
 3.9381010
18
The Fermi Level: Extrinsic Semiconductor
Exercise: Calculate the thermal equilibrium electron and hole concentration
in silicon at T = 300K, where the Fermi energy level is 0.22 eV below the
conduction band energy Ec. (EF = Ec – 0.22 eV). Assume Nc=2.8x1019 cm-3
and Nv=1.04x1019 cm-3
E C E F kT
n0  N C e
For p-type material,
E F E V kT
p0 N eV
(EC-EF)
E F  EV  EC  0.22  EV  EC  EV  0.22
 Eg  0.22  1.12  0.22  0.90
19
p0  1.0410 e
0.9 0.0259
 8,427.9cm-3
(EC-EV)
Equilibrium hole concentration
EC
EF
(EF-EV)
EV
19
The Fermi Level: Extrinsic Semiconductor
Fermi Level Position: Extrinsic Semiconductor
We can now determine the position of the Fermi energy level as a
function of the doping concentrations.
For n-type semiconductor
n0  N C e E
C
 E F  kT
n-type semiconductor
 ND
 NC 
 EC  E F  kTln 

 ND 
p-type semiconductor
For p-type semiconductor
 NV 
EF  EV  kTln 

N
 A
20
The Fermi Level: Extrinsic Semiconductor
Carrier Concentration and the Fermi Level Position
N 
Ec  E F  kT ln  c 
 Nd 
EC
For n-type
N 
EF  E v  kT ln  v 
 Na 
EV
For p-type
21
MASS ACTION LAW
Mass Action Law
The product of equilibrium electron and hole concentrations can be written as:
n0 .p0  N C e E
C
 EF  kT
.NV e E
F
 EV
 kT
 N C N V e E  E  kT  N CN eV
C
For Intrinsic semiconductor:
n0  ni  N C e
ni . pi  N C e E
C
p 0  pi  N V e
EC  E Fi  kT
 E Fi  kT
.NV e E  E kT
Fi
V
 N CN eV
V
E Fi  E V
EC E V kT
E g kT
 kT
 N CN eV
E g kT
Which is same for the extrinsic semiconductor. So for any semiconductor,
at thermal equilibrium,
n p  n p  n2  p  n
0
0
i
i
i
i
i
This is known as the mass action law.
22
Temperature Dependence of Carrier
Concentration
Complete Ionization
At room temperature, the donor states are essentially completely ionized which
says that almost all donor impurity atoms have donated an electron to the
conduction band. n ≈ Nd
For the valence band, at room temperature, acceptors atoms are also
essentially completely ionized. This means that each acceptor atom has
accepted an electron from the valence band.
Energy band diagrams at room temperature for (a) n-type and
(b) p-type semiconductors.
23
Temperature Dependence of Carrier
Concentration
Freeze-Out
The opposite of complete ionization occurs at T = 0 K. At absolute zero
degrees, all electrons are in their lowest possible energy state. For n-type, each
donor state must contain an electron. nd = Nd
No electrons from the donor state are thermally elevated to the conduction
band; this effect is called freeze-out.
For p-type, each acceptor site will be empty. No electrons from the valence
band are thermally elevated to the acceptor states; this effect is also called
freeze-out.
Energy band diagrams at T = 0 K for (a) n-type and (b) p-type semiconductors.
24
Temperature Dependence of Carrier
Concentration
n0  ni
n0  Nd
At high enough
temperatures, the
material returns to its
intrinsic properties.
n i2  BT 3e E g
Freeze
out
kT
B=5.4x1031 forSi
Low Temp.
High Temp.
Electron concentration vs. Temperature for an n-type material
25