SEMICONDUCTORS

Material which allows partial flow of electricity through it is called semiconductor

.Conductivity of semiconductor lies between conductors and insulators.
Silicon and Germanium are the examples for semiconductors. The energy gap for Si is
1.1eV and for Ge is 0.7eV.

INTRINSIC or PURE SEMICONDUCTOR:

Let us consider ‘Si’ with atomic no. 14 and valence is

4. All the silicon atoms form covalent bonds with the
neighboring Si atom and no electron is free for conduction
at temperature 0 k. Hence pure silicon acts as insulator at
absolute o k, as the temperature increases above 0 k, these
covalent bonds break and some electrons are released.
These electrons move in the crystal freely and responsible
for conductivity. So they are called free electrons.
Each electron leaves behind an empty space called
a hole which also acts as current carrier. These electrons and
holes move in opposite directions under the effect of external
field and constitute current.

ELECTRON CONCENTRATION IN THE CONDUCTION BAND OF INTRINSIC

SEMICONDUCTOR:

The no. of electrons per unit volume having energy in a range E and E+dE in the conduction
band of an intrinsic semiconductor is,

dn = Z(E)dE F(E) ----------(1)

where F(E) represents the Fermi distribution function gives the probability of occupation of
electron with energy E.
1
E−E F

F(E) =
1+exp
(K BT )
Z(E) is the density of states i.e. no. of available states per
unit volume of semiconductor.

4Π
3
Z(E) = h (2m)3/2 E1/2
1
E−E F
4Π
3
dn = h (2m)3/2 E1/2 .
1+exp
( K BT ) dE
For conduction band,

1
 SEMICONDUCTORS

1
E−E F
4Π
3
dn = h (2m*e)3/2 (E-Ec)1/2 .
1+exp
K BT ( ) dE -------------- (2)
¿
Where me is effective mass of electron in the conduction band.

E−E F

in the above equation, for conduction band,

exp
( KB T ) >> 1 so 1 can be neglected in the
denominator of the equ.(2).

−(E−E F )
4Π
3
dn = h (2m*e)3/2 (E-Ec)1/2
exp
K BT
-------------(3)
( )
To get the total no. of electrons per unit vol. in the conduction band is we have to integrate the
above equ. Between the bottom of the conduction to top of the conduction band.
∞
−(E−E F )
4Π
3
n = h (2m*e)3/2
∫
Ec
(E-Ec)1/2
exp
( K BT ) dE
∞
−(E−E F + E c−E c )
4Π
3
dn = h (2m*e)3/2
∫
Ec
(E-Ec)1/2
exp
( KB T ) dE
∞
( E F −Ec ) −(E−E c )
4Π
3
n = h (2m*e)3/2
exp
( KB T ) ∫
Ec
(E-Ec)1/2
exp
( KB T ) dE

( E−Ec )

put x =
( K BT ) , so that dE = KBTdx

Lower Limit: when E = Ec, x = 0 and

Upper Limit: when E = ∞, x = ∞
∞
( E F −Ec )
4Π
3
 n = h (2m*e)3/2
exp
( KB T ) ∫
0 e-x (xKBT)1/2 KBT dx
∞
( E F −Ec )
4Π
3
n = h (2m*ekBT)3/2
exp
( KB T ) ∫
0 e-x (x)1/2 dx
3
2m ¿e k B T ( E F −Ec )

n = 4Π
[ h2 ] 2
exp
( KB T ) √π
2

2
 SEMICONDUCTORS

¿ 3
2 Πme k B T
n= 2
[ h2 ] 2
(
exp −
(E c−E F )
KB T )
(E c−E F )

n = Nc
exp −
( KB T )
¿ 3
2 Πme k B T
Where Nc = 2
[ h 2 ] 2

HOLE CONCENTRATION IN THE VALENCE BAND OF INTRINSIC SEMI

CONDUCTOR:

The no. of holes per unit volume having energy in a range E and E+dE in the valence band of an
intrinsic semiconductor is,

dp = Z(E)dE [1-F(E)] ----------(1)

where [1-F(E) ]represents the probability of absence of electron in the particular energy level
with energy E.

E−E F
1
exp
( KB T )
E− E F E−E F

[1-F(E)] = 1-
1+exp
K BT ( ) =
1+exp
( K BT )
E−E F

For the valance band 1>> .

exp
( KB T )
So exponential term can be neglected in the denominator of the above equation.

E−E F

 [1-F(E)] =
exp
( KB T )
E−E F
4Π
3
dP = h (2m)3/2 E1/2
exp
( KB T ) dE
For valance band,

E−E F
4Π
3
dp = h (2m*h)3/2 (Ev-E)1/2
exp
KB T ( ) dE

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 SEMICONDUCTORS

To get the total no. of holes in the V.B. we have to integrate the above equation between the
limits bottom of the V.B. to top of the V.B.

Ev E−E F
4Π
3
p = h (2m*h)3/2
∫
−∞ (Ev-E)1/2
exp
( KB T ) dE

Ev E−E F + EV −E V
4Π
3
p = h (2m*h)3/2
∫
−∞ (Ev-E)1/2
exp
( K BT ) dE

E −E F Ev EV −E
4Π
3
p = h (2m*h)3/2
exp V
KB T ( ) ∫
−∞ (Ev-E)1/2
exp−
( KB T ) dE
EV −E

put
(K BT )
= x ; dE = -dx KBT

Lower Limit: when E = -∞, x = ∞.

Upper Limit: when E = Ev , x = 0.

E −E F 0
4Π
3
 p = h (2m*h)3/2
exp V
KB T ( ) ∫∞ e-x (xKBT)1/2 (-KBT dx)
∞
E −E F
4Π
3
¿
p = h ( mh KBT)3/2
exp V
KB T( ) ∫
0 e-x x1/2 dx

EV −E F
4Π
¿
p = h (2 mh KBT)3/2
3
exp
( KB T ) √Π
2

¿ 3
2mh πk B T

p=2
( h 2 ) 2
exp
( EV −E F
KB T ) or

E F −E V

p = Nv
exp−
( K BT ) ----------- (2)
¿ 3
2mh πk B T

Where Nv = 2
( h2 ) 2

Equ. (2) gives the no. of holes in the V.B of the intrinsic semi
conductor.

LOCATION OF FERMI LEVEL IN INTRINSIC SEMICONDUCTOR:

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 SEMICONDUCTORS

In intrinsic semiconductor no. of electrons in the C.B and no. of holes in the V.B are equal.

 n=p

(E c−E F ) E F −E V

Nc
exp −
( KB T ) = Nv
exp−
( K BT )
−E c + E F + E F −E v Nv
exp
( K BT ) = NC

2 EF ( EC +EV ) Nv
K BT - K BT = ln N C

(Ec+Ev ) K BT Nv
EF = 2 + 2 ln N C ----------- (1)
At T = 0 k,

(Ec+Ev )
EF = 2 -------- (2)
Fermi energy level lies exactly in the middle of the forbidden gap at absolute zero K.

INTRINSIC CARRIER CONCENTRATION (ni) [law of mass action]:

In the intrinsic semiconductor, n = p = ni .Where ni is known as intrinsic carrier concentration.

2
 np = ni
3
2 m ¿e πk B T ¿
2mh πk B T E F −E V
n2i =2
[ h2 ] 2
exp −
(
(E c−E F )
KB T ) ( 2 h2 ) (
exp−
K BT )
3
2 Πk B T −E c + E F −E F + E V )
n2i =4
[ h2 ] ( me m h )
¿ ¿
3
2
exp
( KB T )
3
2 Πk B T −(E c−E V )
n2i = 4 h2
3/2
[ ] ( m¿e m¿h )
3
2
exp
( KB T )
2 Πk B T −E g )

ni = 2
[ h2 ] ( m¿e m ¿h )
3
4
exp
( )
2 KB T
(since Ec-Ev = Eg)

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 SEMICONDUCTORS

This equ. Shows that for a given semiconductor the product of holes and electron concentration
at a given temp. is equal to square of the intrinsic semiconductor carrier concentration. This is
called law of mass action and holds both for intrinsic and extrinsic semiconductors.

CONDUCTIVITY OF INTRINSIC SEMICONDUCTORS:When the electric field is applied

to the semiconductor, charge carriers acquire velocity.

vd α E

vd = µE --------- (1)
where µ is called mobility of charge carriers.

Current density J = ne vd

J = neµE ---------- (2)

This is in the form of J = E
Where  = neµ --------- (3) is conductivity
For electrons n = neµe
For holes p = peµh

Where µe ,µh are mobilities of electrons and

holes respectively.

  = neµe + peµh

= (nµe + pµh )e

= ni(µe + µh)e --------- (4) where ni is called intrinsic carrier concentration.

3/2
2 Πk B T −E g )

= 2
[ h2 ] ( m¿e m ¿h )
3
4
exp
( )
2 KB T
(µe + µh)e
3/2
−E g ) 2 Πk B T
 = o
exp
( )
2 KB T
where o =2
[ h2 ] ( m¿e m ¿h )
3
4
(µe + µh)e
Eg
ln = ln o - 2 K B T -----------(4)
The above equ. gives the expression for conductivity of intrinsic semiconductor.
EXTRINSIC SEMICONDUCTORS:

To increase the conductivity of pure semiconductors

some impurities are added. This process is called
doping. When impurities are added to semiconductor
the available energy levels are altered. One or more

6
 SEMICONDUCTORS

energy levels are appeared in the band structure. Doping may create energy levels with in the
forbidden band.

N-type semiconductor:-

When pentavalent impurities such as phosphorous, Arsenic or Antimony is introduced into Si, or
Ge, four of its valence electrons form 4 covalent bonds with other 4 neighboring Si or Ge atoms
while the fifth valence electron loosely bound to its nucleus. A small amount of energy is
required to detach fifth electron from its nucleus and make it free
to conduct.
So pentavalent impurities are known as donor impurities.
The energy level corresponding to the fifth valence
electron lies in the band gap just below the C.B. edge as shown in
figure.
This level is called donor level.

ELECTRON CONCENTRATION IN N-TYPE

SEMICONDUCTOR:

The energy level diagram for n-type semiconductor is shown in fig. At 0k all donor levels are
unionized state that is all donor levels are occupied with electrons.
As temperature increases slightly some of the donors ionized and
contribute electrons to the conduction band. Also some of the
valence electrons may jump to the conduction band leaving hole in
valence band. The no. of holes produced quite small in this process.
Therefore Fermi level must lie near the middle of the donor level
and bottom of the conduction band.
Let there be Nd donors per unit volume occupying donor
levels with energy Ed. The electron concentration in the conduction
band is given by

(E c−E F )

n = Nc
exp −
( KB T )
---------- (1)
The electron concentration must be equal to the sum of concentration of ionized donors in donor
levels and concentration of thermally generated holes in valence band. i.e.
+
n=N d + p ---------------- (2)

If donors concentration is high, the holes generated can be

neglected.
+
 n  N d --------------- (3)
The concentration of ionized donors can be written as

7
 SEMICONDUCTORS
+
N d = Nd[1-F(Ed)]

= Nd
[ 1−
E −E
1+exp d F
K BT ( ) ]
Ed −E F

= Nd
[ exp

1+exp
( K BT

(
)
E d −E F
KB T ) ]
E F −E d

= Nd exp
[( −
K BT )] --------- (4)
E d −E F

In n-type semiconductor EF lies above the Ed, 1>>

exp
( KB T ) . So exponential term can be
neglected in the denominator of the above equation.

From equations (3) and (4), we get

Ec −E F E F −E d

Nc exp
[( −
K BT )] = Nd exp
[(
−
K BT )]
−Ec + E F + E F −Ed Nd
exp
( K BT ) = Nc

2 EF ( E c + Ed ) Nd
( KB T
−
K BT ) = ln N c

E c + Ed K BT Nd
EF = 2 + 2 ln N c --------- (5)

Substitute the value of EF in equ.(1)

8
 SEMICONDUCTORS

Nd

n = Nc exp
( −Ec E c +Ed
+
KB T 2 K BT
+
Nc
2
ln
)
Nd

n = Nc exp
( −2 E c +Ec + Ed
2 K BT
+
ln

1
2
Nc
)
Ed −E c Nd

n = Nc exp
( 2 KB T
+ ln
( ))
Nc
2

1
Nd Ed −E c

n = Nc
( )
Nc
2

exp
( 2 KB T )
Ed −E c

n = (Nc Nd)
1
2
exp
( )
2 KB T
or ----------- (6)
− ΔE
n = (Nc Nd)
1
2
exp
( )
2K B T
--------------- (7)

Where −ΔE =
Ed −Ec represents the ionization energy of donors.

P-type semiconductor:

When trivalent impurity such as aluminum, boron, gallium

or indium is added to pure silicon, it forms 3 covalent bonds with
the neighboring 3 silicon atoms while the fourth bond is not
completed due to the deficiency of one electron. Thus the trivalent
impurity atom has a tendency to accept one electron from
neighboring silicon atom to complete the fourth covalent bond. The
energy level corresponding to the electron deficiency that is ‘hole’
is located above the valence bond and is called acceptor level.
In this type of semiconductor majority charge carriers are
holes and minority charge carriers are electrons, called p-type
semiconductor.
CONDUCTIVITY OF EXTRINSIC SEMICONDUCTORS:

The expression for conductivity for n-type semiconductors is

9
 SEMICONDUCTORS

e = nee ---------- (1) and

For p-type material is p = neh --------- (2)

Where e and h are mobilities of electrons and holes.

Under the condition of thermal equilibrium electron and holes are uniformly distributed
in semiconductor and the average velocity of charge carriers is zero, no current
flows.
Conductivity is temperature dependent as shown in figure.
At low temp the conductivity increases with increase
of temperature.
This is due to increase in the no. of conduction
electrons due to ionization of donor impurities.
Conductivity reaches maximum value B in the graph
all donors is ionized.
Conductivity decreases further increase with
temperature. This is due to decrease of mobility
because of scattering of electrons from the periodic
potential field. A sharp rise in conductivity from C to D is due to large increase in intrinsic
conductivity.

DRIFT & DIFFUSION:

The net current that flows across semi conducting crystal has two components.
(i) Drift current
(ii) Diffusion current

Drift Current: When voltage is applied electrons attracted towards the positive
potentials and holes attracted towards the negative potential. This net movement of
charge carriers is called drift.
Due to the application of voltage charge carriers attain drift velocity V d , which is proportional to
the electric field E.

Vd  E

Vd = µE -------- (1)

Where µ is mobility of charge carriers.

The drift current density Je due to electrons is defined as the charge flowing across unit area per
unit time due to their drift under the influence of field is given by
Je(drift) = ne Vd or

Je(drift) = neµeE ------- (2)

10
 SEMICONDUCTORS

Where µe is mobility of electrons.

The drift current density due to holes in the valence band is
Jh(druft) = peµhE -------- (3)

So the total drift current is

J(drift) = Je(drift) + Jh(drift)

= e (nµe + pµh) E ----------- (4)

The above equation is applicable to intrinsic as well as extrinsic semiconductors. Drift current
depends upon two variables
(i) carrier concentration
(ii) electric field

Diffusion Current:

In addition to the drift motion, the chare carriers in semiconductor move by diffusion of charge
carriers from high concentration to low concentration region. Current produced by the diffusion
of the charge carriers is called diffusion current.
Suppose when light or temperature is incident on the semiconductor, additional electron
and hole pairs generated and they diffuse through out the semiconductor to restore the
equilibrium condition.

Let n, p be the excess charge of electron and holes respectively. According to Fick’s law,
diffusion current is proportional to rate of flow of excess charge.
∂
 rate of flow of excess charge ¿ ∂x (n) or
∂
Rate of flow of excess charge = -De ∂x (n)

 Je(diff.) = (-e) rate of flow of excess electrons

∂
= eDe ∂x
(n) ---------- (1)
Similarly diffusion current density due to holes is

Jh(diff.) = (e) rate of change of excess holes

∂
= -Dh ∂x (p) ------------ (2)

Total current density in semiconductor due to electrons is Je = Je(drift) + Je(diff.)

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 SEMICONDUCTORS

∂
= neµeE + eDe ∂x (n)
∂
= (nµeE + De ∂x
(n)) e ---------- (3)
Current density due to holes is

Jh = Jh(drift) + Jh(diff.)

∂
= peµhE + (-Dh ∂x (p))

∂
Jh = (pµhE - Dh ∂x
(p)) e ---------- (4)
EINSTEIN’S RELATION:

Einstein’s relation gives the direct relation between diffusion coefficient and mobility of charge
carriers.
At equilibrium condition drift current balances and opposite to the diffusion current .
∂n
 neµeE = - eDe ∂ x ---------- (1)
∂n
neµeE = -(1/µe) eDe ∂ x ---------- (2)

Einstein compared the movement of charge carriers with the gas molecules in a container.
According to Boltzmann’s statistics the concentrations of gas molecules can be written as

−Fx
n = C.exp
( ) KB T
where x is distance and F = eE is force acting on the charge carriers

−eEx −eE
∂n
∂ x = C.exp K B T ( ) .
( )
KB T
−eE
∂n
∂x = n.
KB T ( ) ----------- (3)
∂n
F=neE= K B T ∂x ----------- (4)

eE
 neµeE = neDe
( )
KB T

12
 SEMICONDUCTORS

De K BT Dh K BT
μe = e ---------- (2) for electrons μh = e --------- (3) for holes

HALL EFFECT:

Def:- When a semiconductor carrying current ‘i’ is

placed in a magnetic field which is perpendicular to the
direction of current, an electric field is developed across the
material in a direction perpendicular to both the current
direction and magnetic field direction. This phenomenon is
known as Hall Effect.
Explanation:
Consider a piece of semiconductor in which current passing along x-axis. When a magnetic field
B is applied along z-direction. An electric field is appeared along y-direction.
If the sample is p-type semiconductor holes move with velocity v in x-direction. As they move
across the semiconductor these holes experience a transverse force due to magnetic field. This
force drives the holes on the lower surface as shown in figure. As a result the lower surface
becomes positively charged and upper surface becomes negatively charged and creating Hall
field along y-direction.

If the sample is an n-type semiconductor majority

charge carriers are electrons, these electrons
experience a force ‘Bev’ in downward direction and
lower face gets negatively charged and upper face gets
positively charged which is shown in fig.b

Consider a rectangular slab of n-type semiconductor carrying current in positive x-

direction under the magnetic field electrons are deflected to the lower surface because of force
‘Bev’ due to magnetic field and upper surface gets positively charged because of this electric
field a force ‘eEH’ acts on electrons in upward direction. The two opposing forces ‘Bev’ and
‘eEH’ establish equilibrium. So

Bev = eEH

Bv = EH ---------- (1)

Let ‘J’ be the current density then

J
J = nev or v = ne --------- (2)
BJ
From (1) and (2), ne = EH -------- (3)

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 SEMICONDUCTORS

Hall Effect depends on the current density J and magnetic field B.

EH  JB
EH = RHJB ----------- (4) Where RH is Hall coefficient.
1
From (3) and (4), RH = - ne ----------- (5)
-ve sign is used because the electric field developed in –ve y-direction.

For p-type semiconductors,

1
RH = pe ---------- (6) where p is hole density.
Determination of Hall coefficient (RH):

If VH be the Hall voltage across the sample of thickness‘t’,

VH
EH = t --------- (7)
From (4) and (7),

VH
RHJB = t or VH = RHJBt ------------ (8)
I
If ‘b’ be the width of the sample then current density J = A

R H IxBxt
VH = bxt or

V H bxt
RH = IxB ---------- (9)

Significance of Hall Effect:

1. By means of Hall Effect we can assess the type of semiconductor whether it is n-type or p-
type. Hall coefficient is negative for n-type material.

2. Charge carrier concentration can be evaluated by means of Hall Effect.

1 1
RH = ne or n = eR H

3. Mobility of charge carriers can be calculated by means of Hall Effect.

1
 = ne and RH = ne

14
 SEMICONDUCTORS

 = RH

4. Hall Effect can be used to determine the power flow in electromagnetic wave

DIRECT AND INDIRECT BANDGAP SEMICONDUCTORS: According to the band

theory of solids, the energy spectrum of electrons consists of large number of allowed energy
bands and separated by forbidden regions. The lowest point of the C.B is called conduction band
edge and the highest point in V.B is called valence band valence band edge. The gap between
them is called band bap or forbidden gap. Based on the band gap semiconductors are classified
into two types.

(i) Direct band gap semiconductors and

(ii) Indirect band gap semiconductors

Direct band gap semiconductors:

Fig.a shows E-K curve for direct band gap semiconductor. In this case the maximum of the
valence band and the minimum of the conduction band occurs at the same value of the ‘K’.

In direct band gap semiconductors electrons in the C.B directly recombine with the holes in the
V.B.

Energy is released in the form of photons. So LED’s and Lasers diodes are prepared with them.

In direct band gap semiconductors life time of charge carries is very less. (i.e excited electrons
cannot stay long time in the higher energy states)

Direct band gap semiconductors are formed by compound semiconductors. Ex. InP, GaAs etc.

Indirect band gap semiconductors:

Fig.b shows E-K curve for direct band gap semiconductor. In this case the maximum of the
valence band and the minimum of the conduction band cannot occur at the same value of the ‘K’.

15
 SEMICONDUCTORS

In indirect band gap semiconductors electrons in the C.B do not directly recombine with the
holes in the V.B. Electrons are trapped in the energy gap called trapping centers.

Energy is released in the form of heat.

In indirect bandgap semiconductors life time of charge carries is longer. So they are used to
amplify the signals in diodes and transistors.

Inirect band gap semiconductors are formed by elemental semiconductors. Ex.Si, Ge.

16