VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-2)
1. What is Hall Effect? Derive an expression for Hall coefficient and
discuss the applications of Hall Effect.
If a specimen (Metal or semi conductor)
carrying a current is placed in a
transverse magnetic field, an electric
field  is induced in the direction
perpendicular to both I & B. This
phenomenon is called ‘Hall Effect’.
At equilibrium the force due to magnetic
field is equal to force due to electric field
intensity (due to Hall
effect)
BeV
Or
Or
Or
Or

=
=
e
BV
-----------
(1)
(2)
But Electric field intensity  = V/d= VH / d
----------(4)
VH = d. 
Current density J = ne v =  v -----(5)
Where  is charge density
But current density J = I/A
-----(6)
J = I/w.d -----(7)
VH
= d Equation No. 4
Putting the value of  = B v from equation No. 2
= Bvd
-----(8)
VH
But J =  v equation No. 5
-----(9)
v = J/
Putting the value of v in equation No. 8
VH
= B.( J/).d
Putting the value of J from equation No. 7
I d
BI
VH
. 
= B.
d .w  W
B.I
VH
=
-----(10)
w
Hall coefficient RH is defined as RH = 1/ -----(11)
B. I . R H
V w
RH
Then VH
=
Or
= H
----W
B. I
Conductivity () = Charge density x mobility

=
.
1
(3)
(12)
VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-2)
or
=/
=
 RH ------
(13)
1/ - equation No. 11.
V w
But equation No. 12, RH  H is found valid for
B.I
RH
RH 
3
8
=
------
(14)
 8
And equation No. 13,  =  RH is modified as  = 
 3

 RH

2. What is a tunnel diode? With the help of energy band diagrams explain
the V-I characteristics of Tunnel Diode.
Impurity concentration in normal diode is 1 part in 108 . Impurity concentration
in Tunnel diode is1 part in 103 . The width of the depletion layer for normal
diode 5 microns 5 106 m . The Width of junction in Tunnel Diode is < 100 A0 
10-8m.
Normally an electron or hole must have energy greater than or equal to
potential energy barrier, to move to other side of the barrier. As the barrier is
very thin, instead of crossing the barrier carriers will tunnel through the
barrier which is called as “Tunneling”.
 Fermi level lies in conduction band in N type material as shown in Fig.523
(a).
 Fermi level lies in valence band in P type material
 Fermi level is at same energy level on both sides.
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VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-2)
Reverse Bias:
By reverse baising the tunnel diode, barrier height increases as shown in
fig.5.23(b). Fermi level on N side is lowered. Tunneling of electron from P to N
side is the result . (From filled sates to empty states). If we increase the reverse
bias, reverse current increases as shown in section 1 of (Fig.5.25).
Forward Bias:
Similarly for forward bias tunneling occurs from N to P type material as shown
in Fig.5.24.
Further increase in forward, the condition shown in fig.5.24(b)reached and
maximum current follows (section 2 of Fig.5.25)
 Further increase will reduce the current as shown in fig.5.24(1) till a
minimum current flows due to the condition shown in fig.5.24(d) section 3 of
5.25(a)
 Besides the above current due to tunneling normal diode current flows as
shown in dotted lines in fig.5.25(a). Resultant is the graph shown in 5.25(b).
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VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-2)
3)
Derive the expression for Transition capacitance ?
A)
Under reverse bias condition the majority carriers move away from the
junction, thereby uncovering more immobile charges. Hence the width of space
charged at junction increases by reverse voltage. When the diode is reverse
biased we observe the transition capacitance.
dQ
It is give by CT 
dV
Where dQ is B the increase in charge
caused by change in voltage dV.
Consider a step- graded junction i.e., one
region is heavily doped and another region is
lightly doped.
(1)
Suppose that p-region is heavily doped and
n- region is lightly doped.( NA>> ND)
(2)
The barrier width towards n-side is more
when compared to the barrier width towards pside.
(3)
A diffused junction is graded in which case the
donor and acceptor concentrations are functions of
distance across the junction.
Since the barrier is always neutral then
Wp NA = Wn ND
W = WP+Wn
W = Wn
-----
1
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VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-2)
The relationship between potential and charge density is given by the
poisons equation
d 2V  q N D

E
dx 2
Where V= reverse voltage , E is permittivity of semiconductor , ND = Donor
ion concentration , q = charge, x = width. By integrating we get,
x
q ND
dV

dx
dx w E

dV q N D
W  x

dx
E
2 

Wx  x 

2 

Suppose that at a reverse voltage V the barrier width is doubled then
x=W
q N D  2 W 2 
q N D  W 2 
V=
 V=
----- 2
W 
E 
2 
E  2 
q ND
V= 
W  x  dx 
E
q ND
V=
E
Let Q be the total charge with in the barrier ; A is the area of cross –
section of the barrier.
Total charge
Q = q ND AWn
Q = q ND AW
----- 3 (  W  Wn)
Transition capacitance
dQ
dW
 qN D A
CT =
----- 4
dV
dV
We know that from equation - 2
q N D  W 2 
dV qN DW
V=


E  2 
dW
E
dW
From equation 4
CT = qN D A .
dV

From equation 5
CT = qN D A .
qN DW
dW


dV
qN DW

CT =
-5
A
W
 The transition capacitance is given by CT =
A
.
W
4) Explain in detail about Breakdown mechanisms
A) When a diode is reverse biased the depletion layer increases to set up a
large potential barrier which prevents diffusion of majority carrier from one
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VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-2)
side to other. Thus there is no current due to the majority carriers. The
breakdown mechanisms are of two types.
1) Zener breakdown
2) Avalanche breakdown
Zener breakdown :
1.
2.
3.
4.
5.
6.
7.
This occurs when the P and N regions are heavily doped diodes. In this
diode depletion region is very small.
When reverse biasing the diode a very strong electric field exists across the
depletion region at near breakdown voltage levels.
As a result of heavy doping of P&N regions, the depletion region width
becomes very small.
For an applied voltage of 6 volts or less, the field across the depletion
region is very high is in the order of 2x 107 v/m.
The very high electric field breaks covalent bonds and create new electron
hole pairs which increases the reverse current.
For lightly doped diodes the zener breakdown voltage is quite high.
In the zener breakdown temperature coefficient is –ve. The breakdown
voltage decreases if the junction temperature increases.
Avalanche Breakdown :1. This occurs in lightly doped diodes where the depletion region is very wide
and electric field is very low .
2. As the applied reverse bias increases, the field across the junction
increases correspondingly.
3. Here the reverse voltage applied give high energy to the minority carriers.
4. The minority carriers with sufficient kinetic energy break up covalent
bonds in the crystal thus releasing the valence electron. This process is
called “Impact Ionization”.
5. The newly released valence electrons gain enough energy to breakup other
covalent bonds. This process is know as Avalanche multiplication.
6. In the avalanche breakdown temperature coefficient is +ve.
7. The breakdown voltage increases if the junction temperature increases.
5)
Explain the term diffusion capacitance and derive the expression.
A)
Diffusion Capacitance ( CD) ;-
The capacitance that exists in a forward biased junction is called a
diffusion or storage capacitance CD, whose value is usually much large then CT,
which exists in a reverse biased junction. This is also defined as the rate of
change of injected charge with applied voltage i.e.,
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VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-2)
dQ
where,
dQ represents the change in the number of minority
dV
carriers stored outside the depletion region when a change in voltage across
the diode dV, is applied.
CD 
Calculation of CD :Let us assume that P- material in one side of the diode is heavily doped in
comparison with the N- side. Since the holes move from the P to the N side, the
hole current I  IPn (0).
We know that I 
dI 
dQ
Q

 = life time of carriers i.e., holes & electrons
CD 
Q = Excess minority charge; We know that

The PN diode current equation is
vT
I = IO =  e v /   1


I ev / 
dI

 O v
dV
T
vT
=
 IO . e v / 
VT
dQ
dI
 .
dV
dV
= I+ IO.
I  IO
 vT
Here I indicates the diode current and IO Reverse saturation current.
As I >> IO
dI
I
 v 
dV  T
dI
I
 v
dV  T
Where  = a constant , 1 for Ge
2 for Si
VT –
VT
CD =
.
volt equivalent temperature i.e., thermal voltage
KT
T
=

q
11600
 Diffusion capacitance CD increases exponentially with forward bias or
proportional to diode forwards current. The values of CD range from 10 to 1000
PF, the large values being associated with the diode carrying a larger anode
current, I.
6.
Derive the Expression for diode current.
A. Let us derive the expression for the total
current as a function of applied voltage assuming
that the width of the depletion region is zero.
When the recombination at the junction are zero
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VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-2)
then the hole concentration at the junction boundary of P – region is equal to
the hole concentration at the junction boundary of n – region. When the
recombination starts the hole concentration in n-region decreases and becomes
zero. Therefore Drift current is also zero. Drift current due to hole is
dqP
dqP
JP = PqpE – q D P

0 = Pq pE – q D P
dx
dx
DP 1 dP
dqP
Pq pE = q D P
 E=
dx
 p P dx
According to Einstein’s equation
DP
p
E=
VT dP
P dx
E

Dn
n
VT ;
 dV 

E 

dx 

 dV VT dP

dx
P dx
dV  dP

VT
P
Let Pno be the hole concentration near the junction in n- region for an unbiased
diode and Pn(o) be the hole concentration near the junction in n- region when
the diode is biased.
Pn(o) is higher than Pno
v
1
 VT dV 
o
p n 0 

p no
1
dP
P

V
 ln pn o   ln  pno  
VT
V
p
 ln  p  p n o 
no
VT
 Pno  
V ln Pn o

 ln 

VT lnPno 
 Pno 
pno   pno eV / VT
This is termed as the law of junction
JP Diffusion =  q Dp
dP
dx
IP = JP Diffusion. A = -q Dp A
dP
dx
Let p n o  is the excess hole concentration when it is forward bias.
p n o  = pno   pno
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VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-2)
Let Lp be the diffusion current and at Lp the effective hole concentration is zero
(i.e., x2 = 0)
  pno   qDpA
dp P2  P1 O  pno   pn o 
 
 I P  qDpA 
( pno   pn )



dx x2  x1
Lp  O
Lp
Lp
 Lp 


pn o   pno   pno  pno eV / VT  pno  Pn o   Pno eV / VT  1
IP 


AqDp pno V / VT
e
1
Lp
Let npo be the electron concentration near the junction in p- region for an
unbiased diode.
In 
AqDn n po
Ln
e
V / VT

1
Total current  I = IP + In
 Aq Dp pno AqDn n po  V / VT
I 

1
e
Lp
Ln 





I  I o eV / VT  1
Where Io is the reverse saturation current Io 

Aq Dp pno Aq Dn n po

Lp
Ln

I  I o eV / VT  1
Where
I
saturation current
V
=

=
=
diode current;
Io
=
diode reverse
external voltage applied to the diode
a constant ; for Germanium
=1
for Silicon
=2
kT
T
( Thermal Voltage); k=Boltzman’s constant ( 1.38033 x 10-23 J/k)

q 11,600
q=Charge of the election ( 1.60219 x 10-19C); T = Temperature of the diode
junction (0k)
VT=
7) Explain the formation of depletion region in an open-circuited pnjunction with neat sketches?
9
VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-2)
10
VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-2)
Fig: Band diagram for a p-n junction under open-circuit conditions.
When P & N type material are put together, Fermi level readjusts due to
movements of charge carriers initially. At equilibrium Fermi level is same
in both P & N side.
However EF closer to EC in N type material and closer to Ev in P type
material. Conduction band in P type is at a higher level compared to that
in N type.
This creates a Energy Hill or Energy Barrier, for the electrons on N side,
denoted by Eo and given by:
EO  ECP  Ecn  EVP  Evn  E1  E2
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