ELECTRONIC

CIRCUITS
-CHAPTER -1
SEMICONDUCTORS
 Semiconductors
A semiconductor is a material that is
between conductors and insulators in its
ability to conduct electrical current. A
semiconductor in its pure (intrinsic) state is
neither a good conductor nor a good
insulator. The most common single-element
semiconductors are silicon, germanium, and
carbon. Compound semiconductors such as
gallium arsenide are also commonly used.
The single-element semiconductors are
characterized by atoms with four valence
electrons.
 Covalent bond
Valence electrons

+4 +4 +4

Silicon ion

+4 +4 +4

+4 +4 +4

Intrinsic Semiconductor
At a very low temperature (0 K) the
crystal of semiconductor behaves as an
insulator.
An intrinsic (pure) silicon crystal at
room temperature derives heat (thermal)
energy from the surrounding air,
This causes some valence electrons to
gain sufficient energy to jump the gap
from the valence band into the
conduction band,
Electrons in the conduction band are
free electrons (conduction electrons)
The absence of the electron in the
covalent bond is represented by a small
circle and is called a hole.
For intrinsic semiconductor:

n= p = ni
Where:
P= hole concentration
n= electron concentration
ni = intrinsic concentration
electric field ε is obtained by:
The current density Ј that results from an

Ј = q (nµn + pµp ) ε =δ ε A\
m2
The conductivity is
δ = q (nµn + pµp ) (Ω-m)-1
For intrinsic semiconductors, p = n = ni
δi = q ni(µn + µp ) (Ω-m)-1
n-type Semiconductor
Covalent bond
Valence electrons

+4 +4 +4

Free electron

Silicon ion

+4 +5 +4

Pentavalent
impurity ion
+4 +4 +4
 n-type
Intrinsic semiconductor

Donor impurity

electron
hole

electrons majority
holes minority
Semiconductor n-type
p-type Semiconductor
Covalent bond
Valence electrons

+4 +4 +4

Free hole

Silicon ion

+4 +3 +4

trivalent
impurity ion
+4 +4 +4
 p-type
Intrinsic semiconductor

Acceptor impurity

electron
hole

electrons minority
holes majority
Semiconductor p-type
 The Mass-Action Law
Under thermal equilibrium, the
product of the free negative and
positive concentrations is a
constant independent of the
amount of donor and acceptor
impurity doping.

n p = ni2
 Charge Densities in a Semiconductor
Since the semiconductor is electrically
neural, the magnitude of the total
positive charge density (ND +p) must
equal that of the negative concentration
(NA +n), Hence;
(ND +p)= (NA +n)
• For n-type
NA= 0, n>>p n≈ND, p=n 2/n ≈n 2/N
i i D
• For p-type
ND= 0, p>>n p≈NA, n=ni2/p ≈ni2/NA
 Example
(a) Determine the concentration of free electrons and
holes in a sample of germanium at 300oK which has a
concentration of donor atoms equal to 2x10 14
atoms/cm3 and a concentration of acceptor atoms
equal to 3x1014 atoms/cm3. Is this p- or n-type
germanium? In other words, is the conductivity due
primarily to holes or to electrons? (n i for Ge at
300oK=2.5x1013 atoms/cm3)
(b) Repeat part a for equal donor and acceptor
concentration of 1015 atoms/cm3. Is this p- or n-type
germanium?
(c) Repeat part a for donor concentration of 10 16
atoms/cm3 an acceptor concentration of 10 14
atoms/cm3
 Solution
(a)
ND +p = NA +n

P + (ND - NA ) – n = 0

n p = ni2 n = ni2 /p
P + (ND - NA ) – (n
( i2 /p) = 0
Multiplying both sides by p
P2 + (ND - NA )p – ni2 = 0
Choose the “+” sign since p > 0 and substituting
ni =2.5x1013 atoms/cm3 one obtains

n = P + (ND - NA ) =1.06 x 1014 – 1 x 1014

=0.06 x 1014 electrons/cm3
Therefore, the sample is p-type
 ∵ ND = NA
(b)
∴
P=n
n2
= p 2
= n i
2

or
n = p = ni = 2.5 x 1013 /cm3
The sample is an intrinsic semiconductor
(c)
Since ND >> NA
n ≈ ND =1016 electrons/cm3
p = ni2 / n = 6.25x1026/1016
= 6.25x1010 holes/cm3
The sample is n-type
Example
(a) A sample of germanium is doped to the
extent of 3x1014 donor atoms/cm3 and 2x1014
acceptor atoms/cm3. At the temperature of
the sample the resistivity of pure (intrinsic)
germanium is 60 Ω-cm. Assume that the
value of the mobility of holes and electrons
is approximately the same as at 300oK
(µp=1800cm2/V.s and µn=1800cm2/V.s). If
the applied electric field intensity is 2V/cm,
find the total conduction current density.
 Solution

electric field ε is obtained by:

The current density Ј that results from an

Ј = q (nµn + pµp ) ε =δ ε A\
cm 2
To find n and p, we first find ni

1 1
 
 q ( n n  p p )
For intrinsic germanium, p = n = ni
 1
i 
qn i (  n   p )
1
ni 
q i (  n   p )
1

1.6 10 60 3800  1800
 19

13 3
1.86 10 cm
From mass action law and neutral
equation
ND +p = NA +n

n+ (NA - ND ) –p = 0

n p = ni2 p= ni2 /n
n+ (NA - ND ) – (n
( i2 /n) = 0
Multiplying both sides by n
n2 + (NA - ND )n – ni2 = 0
 2 2
 ( N A  N D )  ( N A  N D )  4n i
n
2
14 28 13 2
110  10  4 (1.86 10 )

2
14 3
1.0335 10 electrons / cm
p = n + (NA - ND )
=1.0335 x1014 – 1 x 1014
= 0.0335 x 1014 holes/cm3
J q(n n  p p )
 19 14 14
1.6 10 ( 1.0335 10  3800  0.0335 10 1800 ) 2
2
0.1276 A / cm
 -CHAPTER -2
THE p-n JUNCTION DIODE
The p-n Junction Diode
When donor impurities are introduced
into one side and acceptors into the other
side of a single crystal of a semiconductor, a
p-n junction is formed.
junction

)A( )K( )A(

p n )K(
Anode Cathode

symbol structure
The p-n Junction Diode at
Equilibrium junction
P-type n-type

hole electron

Hole concentration Electron concentration

Electron concentration
Hole concentration

Charge concentration across the junction

 junction
p-type n-type
+ +
+ +
+ +
+ +
+ +
+ +

hole electron
Depletion
Acceptor ion + Donor ion
region
 Depletion region
When the pn junction is formed, the n region
loses free electrons as they diffuse across the
junction. This creates a layer of positive ions
near the junction. As the electrons move
across the junction, the p region loses holes as
the electrons and holes combine. This creates
a layer of negative ions near the junction.
These two layers of positive and negative
charges form the depletion region, space-
charge, or transition region.
 contact (Barrier) Potential
The potential difference that exists across
the depletion region between the positive
and the negative ions. The forces between
the opposite charges form an electric field.
This electric field is a barrier to the free
electrons and holes, and energy must be
expended to move the free charge through
the electric field. That is, external energy
must be applied to get the free charge to
move across the barrier of the electric field
in the depletion region.
The p-n Junction as a rectifier

Forward
Bias
Junction
P-type n-type
+
A K +
+
+
If
Holes current
Depletion Electrons current
region
- +

- +
The width of the transition region will decrease.
The holes cross the junction from the p-type
into the n-type and the electrons cross the
junction from the n-type into the p-type.
The resultant current crossing the junction is
the sum of the holes and electrons currents.
The junction potential will decrease (VJ=Vo–Vd)
The forward-bias voltage Vd is always less than
the built-in potential barrier Vo. If Vo= Vd , the
barrier disappear and the current could be
arbitrarily large.
Reverse Bias

junction
-type p n-type
+
A K +
+
+
Ir
Depletion
region
+ -

+ -
 The width of the transition region will increase.
Both the holes in the p-type and the electrons in
the n-type into the n-type will move away from
the junction.
The junction potential will increase (VJ=Vo +Vd)
Where:- VJ = the junction potential
Vo = the contact potential
Vd = the applied potential
A small current called reverse saturation
current does flow because a small number of
hole-electron pairs are generated within the
crystal as a result of thermal energy.
The reverse saturation current (Io) will increase with
increasing temperature.
- The volt-ampere characteristics

- I is the current flow in the junction

- V is the voltage across the junction
- Io is the diode reverse saturation current
- η is a constant =1 for Ge , 2 for Si
-VT is the volt equivalent of temp.
VT =T / 11600
= 0.026V= 26mV at room temp. (300oK)
Characteristics of the Semiconductor
Diode Forward current
IF, mA

Ge Si

breakdown voltage
VBR(Si) VBR(Ge)
Reverse voltage Forward voltage

VR, V VF, V
V 0.2 0.6V
IO

Reverse current
IR, µA
The Cutin voltage V𝜸
The cutin, offset, break-point, or threshold
voltage is defined as the voltage below
which the current is very small (less than 1
percent of maximum rated value).

V𝜸⋍ 0.2 for Ge

V𝜸⋍ 0.6 for Si
•The piecewise linear diode characteristics

ID, mA
The Diode Equivalent Circuit
A A

V𝜸
Rr
Rf

K K 0 V𝜸 VD, V
Reverse bias Forward bias
The piecewise linear diode
characteristics
The Diode as a Circuit
Element

- VD +

A K

ID
Vi RL Vo
The Load Line and Operating Point
Applying Kirchhoff's voltage law to the circuit, we get

V i = V D + I D RL
At VD = 0 ID=Vi / RL

At ID = 0 V D = Vi

The point of intersection A of the load line and the

characteristic curve gives the current IA that will flow through
the diode and the voltage across the diode
 ID, mA

Vi /RL Characteristic curve

IDA A (operating point)

load line

0 VDA Vi VD,V

VL = ID RL = Vi - VD
Example:
the following figure represents the diode characteristics curve of
the previous circuit, if RL = 50 Ω and Vi = 1.5 V, find:
(a) the current passing through the diode
(b) the voltage drop across the diode
(c) the voltage drop across the load resistance

ID, mA

30
25
20
15
10
5
0 1.5 1.0 0.5 VD, V
Solution:

Vi = VD + ID RL
1.5V = VD + 50 ID
At VD= 0
ID=Vi / RL= 1.5V / 50 Ω = 0.03A = 30 mA

At ID = 0

VD = Vi = 1.5V
 ID, mA

30 Characteristic curve
25
20
15=
IDA A (operating point)
10
load line
5

0 1.5 1.0 0.5 VD, V

VDA VL

Vi
Example:

V𝜸=0.6V find:
For the following circuit, Rf = 20 Ω, Rr= 1M Ω and

(a) the current passing through the diode.

(b) the voltage drop across the diode

VD
- +

ID
Vi = 6V RL= 100 Ω
:Solution
:For forward bias )1
VD
A- V𝜸 Rf +K

+
If
Vi = 6V RL= 100 Ω

-
If=(Vi –V𝜸)/(Rf +RL)
)= 6V-0.6 V))/ 20 Ω + 100
Ω
V )=45mA
=(V +I R )=(0.6V +0.045×20)=1.5
D 𝜸 f f
:For reverse bias )2
VD
+ -
A Rr K
-
Ir
Vi = 6V RL= 100 Ω

+
Ir=(Vi)/(Rr +RL)
)= 6V))/ 1MΩ + 100 Ω )=6µA
VD= Ir Rr= 6µA x 1MΩ=6V
•The Ideal Diode

Ideal Diode Equivalent Circuit ID, mA

A A V𝜸 =0
Rf =0
Rr =∞

K K 0 VD, V
Reverse bias Forward bias
piecewise linear characteristics of ideal diode
.
Varactor
diodes
The transition region is not a constant but varies
with applied voltage
The larger the reverse voltage, the larger in the
transition region width W, and hence the smaller the
capacitance CT.
Diodes made especially for the applications which
are based on the voltage-variable capacitance are
called varactors. varicaps, or voltacaps.
Breakdown (Zener) diode
a semiconductor diode usually used as a
voltage regulator because its resistance breaks
down at a predetermined voltage level (VZ), at
which time it maintains a constant voltage
while producing a sudden large increase in
current.
ZENER DIODE CHARACTERISTIC CURVE

IF
A K
Breakdown
voltage
Vz
VR VF
IZK
Reverse
Breakdown
region

IZM
IR
Zener diode equivalent circuit
∆VZ
K K 0
IZ IZ
VR IF
IZK
VZ
rZ 
I Z
+

VZ ∆IZ

A A IZM

equivalent circuit IR
The ideal Zener diode equivalent circuit

K K ∆V
VZ = 0 0
IZ
VR IF
IZK
+

VZ
∆IZ
VZ
rZ  =0
I Z
A A IZM
equivalent circuit
IR
Example:
Find the current through the zener diode in the For the
following circuit. Use the ideal-zener- diode approximation

R=2KΩ

I
+
Vin=50V VZ = 30V
-
Solution:

R=2KΩ

I
+ +
Vin=50V VZ = 30V
- -

I =(Vi –VZ)/(R)

=(50V –30V)/(2000Ω) = 10mA

Example:

R=20Ω, rZ=0, and RL=200Ω. Voltage Vi varies

For the voltage regulator shown, assume that V Z=20V,

between 24 and 30V.

(a)Specify the maximum and minimum current for the
zener diode.
(b)Determine the maximum power dissipated in resistance
R and in the zener diode (PR(max) and PZ(max))
R

+ It IZ IL
Vin DZ RL VL = VZ
-
Solution:
(a)
It = IZ + IL
IZ(max) = It(max) - IL

IZ(min) = It(min) - IL
V in(min)  V Z 24V  20 V
I t (min)    0 .2 A
R 20
V in(max)  V Z 30V  20 V
I t (max)    0 .5 A
R 20
 VZ 20 V
IL    0 .1 A
R L 200

 I Z (max)  0 .5 A  0 .1 A  0 .4 A
 I Z (min)  0 .2 A  0 .1 A  0 .1 A
(b)
PR(max) = It(max)2 R = (0.5A)2 (20Ω) = 5W

PZ(max) = VZ IZ(max) = (20V)(0.4A) = 8W

 Example:

Vin=30V, R= 50Ω, rZ=0, and IZK = 5 mA, IZM = 200 mA.

For the voltage regulator shown, assume that V Z=20V,

Determine the variation in RL over which the load voltage

is still regulated at the zener voltage.

+ It IZ IL
Vin DZ RL VL = VZ
-
Solution:
(a)

Vin  V Z 30V  20 V
It   200 mA
R 50 
I L (max)  I t  I Z (min)
 200mA  5mA 195mA
VZ 20 V
 R L (min)   103
I L (max) 195m A
 I L (min)  I t  I Z (max)
 200mA  200mA  0mA
VZ 20 V
 R L (max)   
I L (min) 0mA

 R L  103   
2.2 Diode Circuits
2.2.1 Rectifier Circuits
Power
transformer
+
AC Voltage
Diode
voltage Filter regulat Load
source
rectifie
r or

Block diagram of an electronic power supply

The Half-Wave
Rectifier
Vi --+ Vo
+
I = I0 A
+
0 RL 0
t0 t1 t2 t0 t1 t2
+

Vo

Vpo

Slop =1

0 0 v𝜸 Vi
Rectified output voltage Voltage transfer characteristics
 PEAK OUTPUT VOLTAGE

VPo = Vpi - V𝜸
where:
VPo = Peak output voltage = = Vo(max)
VPi =Peak output voltage =Vi(max)
V𝜸=cutin –voltage = 0.6 V for si
= 0.3V for Ge
= 0.0V for ideal diode
 Average Value of the Half-Wave Output
Voltage

Vpo
Vave
Area
0
2π

VP
Vave  o

 Peak Inverse Voltage
(PIV)

PIV at tp
Vi +

tp I=0A
0 RL
+

VPi-

PIV = VPi
 Half-Wave Rectifier with
Transformer-Coupled Input
Voltage
N1 N2

Vi VI VS RL Vo

 N2 
VS   VI
 N1 
VPo = VS – V𝜸 PIV = VR(max) =VS
 Full-Wave
Rectifiers

‫موحد موجة‬
0 Vin ‫كاملة‬
Vout 0

VP Vi

Vave  o
 VH.W
2V P
Vave  o VF.W
 2π
 The Full-Wave Center-Tapped
Rectifier

N1 N2 D1

VS/2
Vin V1

VS/2 RL

D2
 D1
N1 N2 +- - +

Vi +
Vo
VS/2 I
+ 0
0
t0 t1 t2 +
VS/2 I RL
+
- + D +-
2
 Vo

Slop = -1 Slop =1

-v𝜸 0 v𝜸 Vi

Voltage transfer characteristics

PEAK OUTPUT VOLTAGE

Vpo

0
V
S
VPo  - V
2
 Peak Inverse Voltage
(PIV)
N1 N2 D- 1 +
+
Vi
VS/2
Vo
0 t t 0
0 1 t2 +
V /2S

RL

+
D2 -
-(VS/2) VB-)VS/2(+

 PIV  VS  V  PIV  2V po  V
 The Full-Wave Bridge
Rectifier
N1 N2
Vo
Vi +
I D1
D3 0
t0 t1 t2
0 VS +
t0 t1 t2
D2 RL Vo
D4
+

Vo Vs - 2V
Peak Inverse Voltage
(PIV)
N1 N2
+ +
D3
+ V𝜸
+
PIV D1
VI VS +
D2 D4
+ + RL Vo
V𝜸 PIV
VPo
-V𝜸

PIV  VPo  V
Capacitor
Filter
+ D+
Vi
+ I I
+
0 C VC RL
t0 t1 t2
+ 0
t0 t1 t2

0
 •ripple factor

Vr(PP)

VP(rect) VDC

0V

Vr PP 
r
VDC
  1 
Vr ( PP )   VP ( rect )
f R C
  L 
1
V DC V P ( rect )  Vr ( pp )
2
OR
 1 
VDC   1  VP ( rect )
 2 f  RL C 
 fo =

fo = fH.W= fi fo = fF.W= 2fi

VP(rect) =

VP 2
VPo = VP2 – Vꭚ V Po  - VB V Po V P 2 - 2V B
2
Example:
.For the circuit shown , find the ripple factor

1 : 10

D3 D1
V 115
VP1 VP2
Hz 60
D2 D4 RL
C 2.2K Ω
50µF
Solution:
V P 1  2V rms  (1 .414 )(115 ) V 163 V

 N2   1
VP 2  V P1  163 V 16 .3V
 N1   10 
V P ( rect ) V P 2  1 .4 V 16.3V - 1.4V 14.9V
 1 
Vr ( PP )  V P ( rect )
 f  RLC 
 1 
 14.9V 1.13V
 (120Hz)(2.2K  )(50 F) 
  1 
VDC  1  VP ( rect )
 2 f  RL C 
 1 
 1  14.9V
 2(120Hz)(2.2K  )(50 F) 
14.3V
Vr ( PP ) 1.13V
r  0.079
V DC 14.3V
7.9%=
Example
A half-wave rectifier produces an 50 V average
output voltage from a 60 Hz ac source. If a 100
µF filter capacitor is used, determine the ripple
factor for a load resistance of 1KΩ. What
minimum PIV rating must the diode has?
Solution:
VP rec 
 Vave 50V 

VP ( rec ) 50 3.14 157V
Vr PP 
r
VDC
  1   1 
Vr ( PP )  VP ( rect )  6 
94.2V 18.84V
 f  RLC   50 1000 100 10 

VDC VP ( rect )  0.5Vr ( PP ) 94.2V  0.5 18.84V 84.78V

Vr PP  18.84
r  0.222 22.2%
VDC 84.78

PIV = VPi =VP(rec) + Vɤ= 94.2V +0.6V=94.8V

2.2.2 Clipper Circuits
Clipper circuits, also called limiter circuits, are used
to eliminate portions of a signal that are above or
below a specified level
For a single-diode clipper circuit

R
+ +

Vi Vo
+
VB
-
- -
When Vi <VB + V𝜸 , the diode D is off and Vo = Vi

When Vi >VB + V𝜸 , the diode D is on and Vo = VB + V𝜸

Vi

VB+V𝜸
VO

t
 R
+ + VO
VB-V𝜸

Vi Vo t
+
VB
- Vi
- -

When Vi >VB - V𝜸 , the diode D is off and Vo = Vi

When Vi <VB -V𝜸 , the diode D is on and Vo = VB -V𝜸

 R
+ + Vi

Vi Vo t
-
VB -VB+V𝜸
+ - VO
-

When Vi <-VB +V𝜸, the diode D is off and Vo = Vi

When Vi >-VB+V𝜸, the diode D is on and Vo=-VB +V𝜸

 R
+ +

VO
Vi Vo
- t

-(VB+V𝜸)
VB
+ -
- Vi

When Vi<-(VB+V𝜸), the diode D is on and Vo=-(VB +V𝜸)

When Vi >-(VB +V𝜸), the diode D is off and Vo = Vi

For a double-diode clipper circuit

R
+ +

D1 D2
Vi Vo
+ -
VB1 VB2
- +
-
-

When -(VB2 +V𝜸) < Vi <(VB1 +V𝜸), D1 &D2 are off

and Vo = Vi
When Vi < -(VB2+V𝜸), D1 is off & D2 is on and
Vo=-(VB2 +V𝜸)
When Vi > (VB1+V𝜸), D1 is on & D2 is off and
Vo= (VB1 +V𝜸)

Vi

VB1+V𝜸
VO

-(VB2+V𝜸)
 2.2.3 Clamping Circuits
Generally, whenever a node becomes connected through a
conducting diode to some reference voltage V R, we say that
the node has been clamped to VR.
D1
VR1
D2
VR2

+
Device VO
-

D1 clamps the output VO and does not allow it to rise above VR2.
D2 clamps the output VO and does not allow it to fall below VR2.
2.2.4 Diode Logic Circuits
Two-input diode OR logic circuit

D1
V1
ID1
D2 VO
V2
ID2 I R
V1 V2 D1 D2 ID1 ID2 I VO

0 0 off off 0 0 0 0

5V 0 on off VO/R 0 VO/R 4.3V

0 5V off on 0 VO/R VO/R 4.3V

5V 5V on on I/2 I/2 VO/R 4.3V

Two-input diode AND logic circuit

+5V

D1 I R
V1
ID1 VO
D2
V2
ID2

Assume a diode cut-in voltage of V𝜸= 0.7V

V1 V2 D1 D2 ID1 ID2 I VO

0 0 on on I/2 I/2 5-0.7/R 0.7V

5V 0 off on 0 I 5-0.7/R 0.7V

0 5V on off I 0 5-0.7/R 0.7V

5V 5V off off 0 0 0 5V
 2.2.5 Light-Emitting Diodes
The light-emitting diode (LED) is a good example of an
optoelectronic device. When forward-biased, this semiconductor diode
gives of light.
The advantages of LED over an incandescent lamp are.
1. Long life (more than 20 years)
2. Low voltage (1 to 2 volts typical)
3. Fast on-off switching (nanoseconds)
A free electron is at a higher energy level than a hole. Because of
this, after free electrons cross the junction, they radiate energy when
they fall into the holes. This energy comes off as heat and light.
A silicon diode is opaque, so non of the light escapes. But an LED
uses a semitransparent material that lets some of the light out.
Different materials radiate as follows
1. GaAs infrared radiation (invisible)
2 GaAsP red or yellow light
3. GaP red or green light

p n

structure symbol
LED arrays
Seven-segment LED array is a shape formed by LEDs, often
to display numbers, letters, or other symbols

Seven-segment LED array

 2.2.6 photodiodes
A reverse-biased diode has a small current because of its minority
carriers
When the diode is in a glass package, strong light does change the
amount of reverse current
The incoming photons create free electrons and holes and hence
increase the number of minority carriers

I
A