AP handout by Dr.T.Rajani, Asst.
Prof(Physics)
Unit-III 1
Semiconductors
• Contents
Intrinsic semiconductors-Carrier concentration, dependence of Fermi level on Carrier -
concentration and temperature, Extrinsic Semiconductors (Qualitative), Continuity equation -
Carrier generation and recombination, Carrier transport: diffusion and drift currents, Hall Effect,
Hall Experiment, Measurement of Hall mobility, Resistivity, carrier density using Hall effect.
• WIT-WIL
“Semiconductor” is a common word in technological world. Most of electronic equipment are
made with semiconductors. To manufacture transistors, amplifiers LEDs, Laser diodes we use
semiconductors. Conductivity will be changed when concentration changed in a semiconductor.
Energy gap of semiconductor diode will decide the colour of output LED or laser diode. Here we
study concentration of carriers in semiconductors and about Fermi level.
• Introduction
Certain substances like Germanium, Silicon, Carbon are neither good conductors like
copper nor insulators like glass. In other words the resistivity of these materials lies in between
conductors and Insulators. They are extensively used in electronic circuits. “A semi-conductor is
a substance which has resistivity (10-4 to 0.5 ῼ m) in between Conductors and Insulators.
• Characteristics of a Semi-Conductor
The main features of pure semi-conductor are
1. The resistivity of a semiconductor is more than metal conductor but less than insulators.
2. Semiconductors have negative temperature coefficient of resistance i.e the resistance of a
semiconductor decreases with increase in temperature and vice-versa.
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
2
3. One of the most important characteristic is semi-conductor doping.It is the process
of adding very small well controlled amounts of impurities into a semi-conductor. It
enables the control of resistivity and other properties over a wide range of values.
• Bonds In Semi-Conductors
The best examples of semi-conductors are Ge, Si, Carbon, Selenium etc. There are four valence
electrons in both Ge, Si. Such atoms do not usually gain (or)lose four valence electrons but share
them with neighboring atoms. The union of atoms sharing the four valence electrons is called a
covalent bond. These filled covalent bonds does not have the individual free electrons present in
metal conductors. Thus semiconductors have much more high resistance than the conductors.
The crystal structure forming covalent bonds make it possible to add impurities The main idea
of adding impurities is to alter Electrical Characteristics
• Energy Band description of Semiconductor
In Semi-Conductors,the energy band is very narrow.The energy provided at room temperature is
sufficient to lift the electrons from valence band to conduction band.Some electrons do jump the
gap and go into the conduction band.Thus at room temperature semi-conductors are capable of
conducting partial amount of current.Thus a semi-conductor has
1) A partially filled Conduction band
2) A partially filled Valence band
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
3) A very narrow energy gap (̴ 1eV) between them. 3
• Types of Semi-Conductors
There are two types of Semi-Conductors
1) Intrinsic Semi-conductor 2)Extrinsic Semi-conductor
• Intrinsic Semi-Conductor
A semi-conductor in its natural pure form is called Intrinsic Semi-Conductor. The oxides of both
Ge and Si are reduced chemically to produce the elements with 100% purity.
• Mechanism of Current Flow in Intrinsic Semi-Conductor
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
Consider energy band diagram of semi-conductor. At 0K, there are no electrons in conduction
band and the valence band is completely filled, because all the electrons in valence band are tightly
bound due to covalent bonding. At this temperature, the covalent bonds are very strong and there
are no free electrons .Due to non-availability of electrons, the semi-conductor behaves as an
Insulator.
When the temperature is raised, some of the covalent bonds in the semi conductor breaks due to
the thermal energy supplied. The breaking of these bonds sets those electrons free which are
engaged in the formulation of those bonds. These free electrons constitute tiny current when a
potential difference is applied across the semi conductor crystal. As the temperature is raised ,some
of the valence electrons acquire sufficient energy to enter into the conduction band and thus
becomes free electrons. Under the influence of electric field, these free electrons constitute electric
current. Also when a valence electron jumps from valence band to conduction band, a hole is
created in valence band. These holes also constitute current. The direction of hole current is same
as conventional current but opposite to electron flow.
✓ NOTE: A hole has the same amount of positive charge equal to an electron but with opposite
polarity. This will be present only in semi-conductors because of un filled covalent bonds.
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
• Carrier Concentration In Intrinsic Semi-Conductor In Thermal 5
Equilibrium
A semi-conductor in an extremely pure form is known as intrinsic semiconductor. In an intrinsic
semiconductor at absolute zero, valence band is completely filled and conduction band is empty.
Because of non-availability of electrons the semi conducting material behaves as an insulator.
When temperature is raised above absolute zero, hole-electron pairs are created. The free electrons
are produced due to the breaking up of some covalent bonds due to thermal energy. At the same
time holes are created in covalent bonds. Under the influence of electric field, conduction through
the semiconductor is by both free electrons and holes. Therefore the total current inside the
semiconductor is the sum of currents due to free electrons and holes.
Let 1) Electrons in conduction band behave as the particles with an effective mass me
2) The number of conduction electrons per cubic meter whose energy lie between E and E+dE
is given by
𝑑𝑛𝑐 = 𝑍(𝐸)𝑓(𝐸)𝑑𝐸------------------(1)
where Z(E) dE is the density of states in the energy interval E and E+dE and f(E) is the electron
occupancy probability i.e the probability that a state of energy E is occupied by an electron.
Here
4𝜋 ∗ 2
3 1
𝑍(𝐸) = (2𝑚 𝑒 ) (𝐸 − 𝐸𝑐 2 − − − − − − − −(2)
)
ℎ3
Where me* is the effective mass of electron.
1
𝑓(𝐸) = (𝐸−𝐸𝑓 ) − − − − − − − − − − − (3)
1+𝑒 𝑘𝑇
where K=Boltzmann Constant, T=Temperature in Kelvin, Ef= Fermi level.
For possible temperatures ; E-Ef >>KT hence above equation can be written as
1 −(𝐸−𝐸𝑓 ) (𝐸𝑓 −𝐸)
𝑓(𝐸) = (𝐸−𝐸𝑓 )
=𝑒 𝑘𝑇 =𝑒 𝑘𝑇 − − − − − − − − − − − (4)
𝑒 𝑘𝑇
The electrons in conduction band are having energies lying from Ec to ∞ while the electrons in
valence band have energies lying from -∞ to Ev.
The concentration of electrons in the conduction band is given by
∞
𝑛𝑐 = ∫Ec 𝑍(𝐸)𝑓(𝐸)𝑑𝐸 − − − − − − − −(5)
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
where Ec is the energy at the bottom of conduction band. 6
Combining (2),(4),(5) then the density of electrons in conduction band is
∞ 𝐸𝑓 − 𝐸
4𝜋 ∗ 2
3 1
𝑛𝐶 = ∫ (2𝑚 𝑒 ) (𝐸 − 𝐸𝑐 ) 2 exp( ) 𝑑𝐸
Ec ℎ3 𝑘𝑇
4𝜋 3 ∞ 1 𝐸𝑓 − 𝐸
∗ 2
𝑛𝐶 = (2𝑚 𝑒 ) ∫ (𝐸 − 𝐸𝑐 ) 2 exp( ) 𝑑𝐸
ℎ3 Ec 𝑘𝑇
𝐸𝑓 ∞
4𝜋 ∗ 2
3 1 −𝐸
𝑛𝐶 = 3
(2𝑚 𝑒 ) exp ( ) ∫ (𝐸 − 𝐸 )
𝑐 2 exp 𝑑𝐸 − − − − − − − −(6)
ℎ 𝑘𝑇 Ec 𝑘𝑇
To solve this integral let us put
𝐸 − 𝐸𝑐 = 𝑥
𝐸 = 𝐸𝑐 + 𝑥
𝑑𝐸 = 𝑑𝑥
𝐸𝑓 ∞
4𝜋 ∗ 2
3 1 −(𝐸𝑐 + 𝑥)
𝑛𝐶 = 3 (2𝑚𝑒 ) exp ( ) ∫ (𝑥)2 exp 𝑑𝑥
ℎ 𝑘𝑇 0 𝑘𝑇
𝐸𝑓 − 𝐸𝑐 ∞
4𝜋 ∗ 2
3 1 −𝑥
𝑛𝐶 = 3
(2𝑚 𝑒 ) exp ( ) ∫ (𝑥)2 exp 𝑑𝑥 − − − − − − − −(7)
ℎ 𝑘𝑇 0 𝑘𝑇
Using gamma function it can be shown that
1
∞ 1 −𝑥 3 𝜋2
∫ (𝑥)2 exp 𝑑𝑥 = (𝑘𝑇)2
0 𝑘𝑇 2
Hence, the equation (7) can be written as
1
4𝜋 3 𝐸𝑓 − 𝐸𝑐 3 𝜋2
𝑛𝐶 = 3 (2𝑚𝑒∗ )2 exp ( ) (𝑘𝑇)2
ℎ 𝑘𝑇 2
i.e the number of electrons per unit volume of the material is given by
3⁄
2𝜋𝑚𝑒∗ 𝑘𝑇 2 𝐸𝑓 − 𝐸𝑐
𝑛𝐶 = 2 ( ) exp ( )
ℎ2 𝑘𝑇
Calculation of density of holes:
Let dp be the number of holes in energy interval E and E+dE in the valance band
𝑑𝑝 = 𝑍(𝐸) (1 − 𝑓(𝐸))𝑑𝐸------------------(1)
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
where N(E) dE is the density of states in the energy interval E and E+dE and (1-f(E)) is the hole 7
(absence of electron) occupancy probability i.e the probability that a state of energy E is occupied
by a hole.
Here
4𝜋 ∗ 2
3 1
𝑁(𝐸) = (2𝑚 ℎ ) (𝐸𝑣 − 𝐸) 2 − − − − − − − (2)
ℎ3
Where mh* is the effective mass of hole.
1
1 − 𝑓(𝐸) = (1 − (𝐸−𝐸𝑓 )
− − − − − −(3)
1+𝑒 𝑘𝑇
where K=Boltzmann Constant, T=Temperature in Kelvin, Ef= Fermi level.
For possible temperatures ; E-Ef >>KT hence above equation can be written as
(𝐸−𝐸𝑓 )
1 − 𝑓(𝐸) = 𝑒 𝑘𝑇 − − − − − − − −(4)
The holes in valance band are having energies lying from -∞ to Ev
The concentration of holes in the valance band is given by
Ev
𝑝𝑣 = ∫−∞ 𝑁(𝐸)(1 − 𝑓(𝐸))𝑑𝐸 − − − − − − − −(5)
where Ev is the energy at the highest level of valance band.
Combining (2),(4),(5) then the density of holes in valance band is
𝐸𝑣 𝐸 − 𝐸𝑓
4𝜋 ∗ 2
3 1
𝑝𝑣 = ∫ 3
(2𝑚 ℎ ) (𝐸𝑣 − 𝐸) 2 exp( ) 𝑑𝐸
−∞ ℎ 𝑘𝑇
4𝜋 3 𝐸𝑣 1 𝐸 − 𝐸𝑓
∗ 2
𝑝𝑣 = (2𝑚 ℎ ) ∫ (𝐸𝑣 − 𝐸) 2 exp( ) 𝑑𝐸
ℎ3 −∞ 𝑘𝑇
−𝐸𝑓 𝐸𝑣
4𝜋 ∗ 2
3 1 𝐸
𝑝𝑣 = 3
(2𝑚 ℎ ) exp ( ) ∫ (𝐸𝑣 − 𝐸)2 exp ( ) 𝑑𝐸 − − − − − −(6)
ℎ 𝑘𝑇 −∞ 𝑘𝑇
To solve this integral let us put
𝐸𝑣 − 𝐸 = 𝑥
𝐸 = 𝐸𝑣 − 𝑥
𝑑𝐸 = −𝑑𝑥
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
𝐸𝑣 1 𝐸 0 1
𝐸𝑣 − 𝑥 8
∫ (𝐸𝑣 − 𝐸)2 exp( ) 𝑑𝐸 = ∫ 𝑥 2 exp ( ) (−𝑑𝑥)
−∞ 𝑘𝑇 ∞ 𝑘𝑇
3 −𝐸 1
4𝜋 0 𝐸𝑣 −𝑥
𝑝𝑣 = (2𝑚ℎ∗ )2 exp ( 𝑘𝑇𝑓 ) ∫∞ 𝑥 2 exp ( ) (−𝑑𝑥)
ℎ3 𝑘𝑇
𝐸𝑣 − 𝐸𝑓 ∞ 1
4𝜋 ∗ 2
3 −𝑥
𝑝𝑣 = 3 (2𝑚ℎ ) exp ( ) ∫ 𝑥 2 exp ( ) (𝑑𝑥) − − − − − −(7)
ℎ 𝑘𝑇 0 𝑘𝑇
Using gamma function it can be shown that
1
∞ 1 −𝑥 3 𝜋2
∫ (𝑥)2 exp 𝑑𝑥 = (𝑘𝑇)2
0 𝑘𝑇 2
Hence, the equation (7) can be written as
1
4𝜋 3 𝐸𝑣 − 𝐸𝑓 3 𝜋2
𝑝𝑣 = 3 (2𝑚ℎ∗ )2 exp ( )(𝑘𝑇)2
ℎ 𝑘𝑇 2
i.e the number of holes per unit volume of the material is given by
3⁄
2𝜋𝑚ℎ∗ 𝑘𝑇 2 𝐸𝑣 − 𝐸𝑓
𝑝𝑣 = 2 ( ) exp ( )
ℎ2 𝑘𝑇
Intrinsic carrier concentration:
In intrinsic semiconductors concentration of electrons and holes are equal
Hence, nc=pv=ni is called intrinsic carrier concentration.
Therefore
2𝜋𝑘𝑇 3 3 𝐸𝑣 − 𝐸𝑐
𝑛𝑖2 = 𝑛𝑐 𝑝𝑣 = 4 ( 2 ) (𝑚ℎ∗ 𝑚𝑒∗ )2 exp ( )
ℎ 𝑘𝑇
2𝜋𝑘𝑇 3 3 −𝐸𝑔
= 4 ( 2 ) (𝑚ℎ∗ 𝑚𝑒∗ )2 exp ( )
ℎ 𝑘𝑇
Where Ec-Ev=Eg is the forbidden energy gap.
Hence
3⁄
2𝜋𝑘𝑇 2 3 −𝐸𝑔
𝑛𝑖 = 2 ( 2 ) (𝑚ℎ∗ 𝑚𝑒∗ )4 exp ( )
ℎ 2𝑘𝑇
• Fermi level in intrinsic semiconductor:
In intrinsic semiconductors concentration of electrons and holes are equal
Hence, nc=pv, using the above equations,
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
2𝜋𝑚𝑒∗ 𝑘𝑇
3⁄
2 𝐸𝑓 − 𝐸𝑐 2𝜋𝑚ℎ∗ 𝑘𝑇
3⁄
2 𝐸𝑣 − 𝐸𝑓 9
2( ) exp ( ) = 2( ) exp ( )
ℎ2 𝑘𝑇 ℎ2 𝑘𝑇
3⁄ 𝐸𝑓 − 𝐸𝑐 3 𝐸𝑣 − 𝐸𝑓
(𝑚𝑒∗ ) 2 exp ( ) = (𝑚ℎ∗ ) ⁄2 exp ( )
𝑘𝑇 𝑘𝑇
3⁄
2𝐸𝑓 𝑚ℎ∗ 2 𝐸𝑣 + 𝐸𝑐
exp ( ) = ( ∗) exp ( )
𝑘𝑇 𝑚𝑒 𝑘𝑇
Taking logarithms on both sides
2𝐸𝑓 3 𝑚ℎ∗ 𝐸𝑣 + 𝐸𝑐
( ) = 𝑙𝑛 ∗ + ( )
𝑘𝑇 2 𝑚𝑒 𝑘𝑇
i.e.,
3𝑘𝑇 𝑚ℎ∗ 𝐸𝑣 + 𝐸𝑐
𝐸𝑓 = 𝑙𝑛 ∗ + ( ) − − − −(1)
4 𝑚𝑒 𝑘𝑇
if we assume that me*= mh*
𝐸𝑣 + 𝐸𝑐
𝐸𝑓 = − − − − − −(2)
2
Figure: Temperature dependence of
Fermi level in intrinsic semiconductor
* *
(a):at T=0K (b): T>0K and mh >me
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
Thus Fermi level is located half way between the valance and conduction bands and its position is 10
independent of temperature. Since mh* is greater than me*, Ef is just above the middle, and rises
slightly with increasing temperature as shown in above figure.
• Carrier concentration in extrinsic semiconductor:
As already mentioned the carrier concentration and hence the conductivity can be enhanced many
orders in semiconductor by doping technology. Depending on the nature of the impurities majority
charge carriers may be free electrons as in the n-type (or) holes as in the p-type semiconductor.
Suppose the doped atoms are donors (n-type) so that free electron concentration nc is increased to
n=k nc, k>1. The hole concentration pv (=nc before doping) decreases in the same proportion to p=
pv/k as we have seen that in any type of semiconductor.
n.p = constant, at a given temperature
total concentration is,
nc+ pv = 2ni =2pi before doing
then after doping
n + p = k nc + pv/k = (k+1/k) ni
in intrinsic nc= pv = ni
(k+1/k) ni > 2 ni since k>1
Here the decreased number of holes is less than added number of electrons.
From the above equations. By proper doping number of charge carriers increased and then
conductivity also increases.
Temperature dependence of Fermi level in N-type semiconductor:
In energy level diagram of n-type semiconductor, Ed is the donor energy level. At low temperatures
number of electrons in conduction band must be equal to number of donors. By taking this into
consideration and solve the equation. Then,
𝐸𝑐 + 𝐸𝑑 𝑘𝑇 𝑁𝑑
𝐸𝑓 = ( )+ 𝑙𝑜𝑔 3⁄
2 2 2𝜋𝑚𝑒∗ 𝑘𝑇 2
2( )
( ℎ2 )
At T=0K
𝐸𝑐 + 𝐸𝑑
𝐸𝑓 = ( )
2
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
i.e., at 0K Fermi level lies exactly lies middle of the donor level and bottom of conduction band 11
Ec. As ‘T’ increases Fermi level moves downward and behaves as intrinsic semiconductor at high
temperatures as shown in below figure.
Temperature dependence of Fermi level in P-type semiconductor:
In energy level diagram of P-type, Ea is the acceptor energy level. At low temperature density of
holes in valance band is equal to density of ionized acceptors. When the calculation done based on
this the final equation will be
𝐸𝑐 + 𝐸𝑎 𝑘𝑇 𝑁𝑎
𝐸𝑓 = ( )− 𝑙𝑜𝑔 3⁄
2 2 2𝜋𝑚ℎ∗ 𝑘𝑇 2
2( )
( ℎ2 )
At 0K
𝐸𝑐 + 𝐸𝑎
𝐸𝑓 = ( )
2
i.e at 0K Fermi level lies exactly middle of Ev and Ea as temperature increases, it moves upward
and behaves like intrinsic at high temperature as shown here.
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
12
Acceptor
level
Figure: Temperature
dependence of Fermi level in P-
Type Semiconductor
• Hall Effect:
“When a current-carrying semiconductor is kept in a magnetic field, the charge carriers of the
semiconductor experience a force in a direction perpendicular to both the magnetic field and the
current. At equilibrium, a voltage appears at the semiconductor edges.”
In semiconductor it is not possible to assert the sign of charges from the direction of the external
current because it is the same whether positive charge flow in that direction (or) negative charge
flow in opposite direction.
Hall effect, discovered by E.H Hall in 1879, helps to resolve the dilemma on sign of the charges.
It is the following experiment result.
When slab carrying current ‘I’ is kept in transverse magnetic field (B), is a potential difference
will develop along a direction perpendicular to both I and B.
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
13
Consider a rectangular slab carrying current ‘I’ in X-direction. if place it in a magnetic field B
which is in the Y-direction, a potential difference Vpq will develop between the faces P and Q
which are perpendicular to Z-direction.
This effect on charge carriers due to magnetic field force. This can be written as, q(V X B), where
‘q’ is charge.
The charges initially drifting such that the current ‘I’ is along the X-direction, will be deflected
along Z-direction under the action of force due to magnetic field.
If the charges responsible for I are positive, they will deflect towards P surface as shown in figure
(a) and potential Vp at P will be greater than potential Vq at Q.
If they are negative, then they will drift in opposite direction of I, they will also deflect towards
‘P’ as shown in figure (b) then Vp will be lower than Vq.
Thus the sign of Vpq=Vp-Vq = VH will determine the sign of the current carrying charges ‘Vpq’ is
called “Hall voltage”.
The corresponding electric field is called “Hall electric field (EH)”. which is perpendicular to ‘V’
and B plane.
𝑉𝑝𝑞
𝐸𝐻 = − − − − − (1)
𝑑
Where ‘d’ is the spacing between surfaces P and Q.
EH will oppose the sideway deflection of the charges. Eventually, at equilibrium is reached
further deflection of charges prevented.
This condition occurs when,
̅̅̅̅
𝑞𝐸 ̅ ̅
𝐻 + 𝑞 (𝑉 × 𝐵 ) = 0 − − − −(2)
̅̅̅̅
𝐸𝐻 = −𝑉̅ . 𝐵̅ sin 𝜃
where ‘θ’ is angle between V and B i.e 90o.
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
̅̅̅̅
𝐸𝐻 14
𝑉̅ = − − − − − − − − (3)
𝐵̅
Further current is due to electrons, we have.
𝑗 = 𝑛𝑒𝑉̅ − − − − − −(4)
Substitute (3) in (4), then
̅̅̅̅
𝐸𝐻
𝑗 = −𝑛𝑒
𝐵̅
̅̅̅̅
𝐸𝐻 1
= − = 𝑅𝐻
𝑗 𝐵̅ 𝑛𝑒
Where ‘RH’ is called Hall coefficient.
1 1
We can write , 𝑅𝐻 = − 𝑛𝑒 for electrons, 𝑅𝐻 = 𝑛𝑒 for holes.
• Applications of Hall Effect:
(a) To measure type of semiconductor:
If Hall coefficient is Positive then it is P- type semiconductor, if it is negative then it is N-
type semiconductor.
(b) To measure concentration of charge carriers:
We know that Hall coefficient can be written as
1
𝑅𝐻 =
𝑛𝑒
1
𝑛=
𝑅𝐻 𝑒
Concentration of charge carriers can be measured by Hall coefficient.
(c) To measure mobility of charge carriers:
We know that Conductivity can be written as
𝜎 =𝑛𝑒𝜇
𝜎
𝜇= = 𝜎𝑅𝐻
𝑛𝑒
Using conductivity and Hall coefficient mobility can be measured.
(d) To measure magnetic flux density:
Unit-V: Semiconductors
AP handout by Dr.T.Rajani, Asst.Prof(Physics)
Using a semiconductor sample of known RH the magnetic flux can be measured by 15
following formula.
𝑉𝐻 . 𝑡
𝐵=
𝑅𝐻 𝐼
Questions:
1. Obtain expression for concentration of charge carriers in intrinsic semiconductor.
2. Explain the variation of Fermi level with temperature in intrinsic semiconductor?
3. Discuss the carrier concentration in extrinsic semiconductors?
4. What is Hall effect? Obtain expression for Hall coefficient?
5. Mention the applications of Hall effect?
*****
Unit-V: Semiconductors
