Introduction to Electronics

Mrinal K Mandal
mkmandal@ece.iitkgp.ernet.in
Department of E & ECE
I.I.T. Kharagpur. 721302.
www.ecdept.iitkgp.ernet.in
Introduction to Electronics
1

Books
Electric Circuit Theory - Van Valkenburg, Prentice Hall.
Electronic Circuits Analysis and Design – Donald A. Neamean.
Digital Logic and Computer Design – M. Morris Mano, Prentice
Hall.
Reference books:
1. Microelectronic Circuits, A.S. Sedra and K. C. Smith, Oxford
University Press.
2. Introduction to Microelectronics, B. Razavi.
3. Electronic Devices and Circuit Theory – Robert L. Boylestad,
Prentice Hall.
4. The Art of Electronics, Paul Horowitz and Winfield Hill.
5. Electronic Devices and Circuits, David A. Bell.
Department of Electronics & Electrical Communication Engineering, I.I.T. Kharagpur mkmandal@ece.iitkgp.ernet.in
2

Some Important Points
•Attendance.
•Risk of deregistration: if attendance is lower than a certain
percentage.
•Total marks distribution = surprise class test + attendance + mid-
sem + end-sem evaluation results.
•n+1 surprise class tests: best n results will be considered.
•Register in Intinno paathsaala.
3

Signal
Generation of discrete time signal
by sampling a continuous signal.
4
Signal is a function that conveys information about the behavior or attributes of
some phenomenon.
• In the physical world, any quantity exhibiting variation in time or variation in space
(such as an image) is potentially a signal that might provide information on the
status of a physical system, or convey a message.
• Popular forms: audio, video, speech, image, communication, geophysical, sonar,
radar, medical and musical signals.
• In electronics engineering: time varying voltage/ current/ electromagnetic waves.
• Analog and digital signals.
Digital approximation of the
analog signal.

Signal
RMS value of a function:
5
( )
2
0
1
lim
T
rms
T
f f t dt
T→∞
= ∫ ( )
2
1
2
2 1
1
lim .
T
rms
T
T
f f t dt
T T→∞
=
− ∫
signal
transducer
Electronic
signal
Signal
processing
Transmitter
Electromagnetic
wave
Electromagnetic
wave
Receiver
Signal
processing
Electronic
signal
transducer
signal
Transmitter.
Receiver.

Function Generator
Signal generator
Different periodic waveforms
rms values:
6
Effect of thermal noise
V(t)
t

7
Cathode Ray Oscilloscope

8
Electronic Circuits
Electronics in daily life.
Nokia phone circuit board.

9
Electronic Circuits
Electronic warfare and communication systems.

10
Passive Components
1. Resistor
Thin film carbon resistor Adjustable
wire wound
Rheostat
Resistors of different power
dissipation factors
Circuit symbol

11
Resistor Colour Coding
Thin film carbon resistor
•Band A - first significant figure (left side),
•band B - second significant figure,
•band C - the decimal multiplier,
•band D (if present) - tolerance of value in percent
(gold - ±5%, silver - ±10%, no band -20%).
• Inside material - A mixture of finely
powdered carbon and an insulating
material, usually ceramic. A resin holds the
mixture together. The resistance is
determined by the ratio of the powdered
ceramic to the carbon.
Black Brown Red Orange Yellow Green Blue Violet Gray White
0 1 2 3 4 5 6 7 8 9
Colour codes:
•Resistance – AB10^C ±D%
Example: yellow-violet-red-gold  4.7k Ω ±5%, between 4,465 Ω and 4,935 Ω.

12
2. Inductor
Passive Components
Different form of inductors
Magnetic field lines
• Inductor stores energy in magnetic field.
• Voltage leads the current.
• Unit is Henry (H).

13
Circuit Components
3. Capacitor
Electrolytic capacitor Ceramic capacitor Polyester capacitor
Polarized capacitor
symbol
Non-polarized
capacitor symbol
• Capacitor stores energy in electric field.
• Current leads the voltage.
• Unit is Farad (F).
•Ceramic capacitor marking – AB10^C pF ±10%.
•Example – 154 means 15×10000 pF±10%.
+

Active Components
Diode Bipolar transistor
Field effect transistor Operational amplifier
14

Concept of Ground
• Ground: a common reference point in any electrical circuit that may or may not
physically connected to the Earth.
• High power circuits: exposed metal parts are connected to ground to prevent
user contact with dangerous voltage if electrical insulation fails.
Connections to ground limit the build-up of static electricity when handling
flammable products or electrostatic-sensitive devices.
• In some power transmission circuits, the earth itself can be used as one
conductor of the circuit, saving the cost of installing a separate return conductor.
Signal ground Earth groundChassis ground
• In portable electronic devices, a large conductor attached to one side of the
power supply acts as a "ground”.
15

Concept of Ground
• Planet earth is not a good conductor of dc voltage.
A B
1kΩ 1kΩ
RAB = ?
C D
1kΩ 1kΩ
Planet
earth
RCD = ?
A typical earthing electrode. Printed circuit board (PCB)
PCB
ground
16

Thevenin’s Theorem
Any two-terminal linear, bilateral network containing impedances and energy
sources can be represented by an independent voltage source VTh and a single
impedance ZTh.
VTh is the open circuit output voltage, ZTh is the impedance viewed at the terminals
when all independent energy sources are replaced by their internal impedances.
 VTh calculation:
Calculate the no load output voltage. It is equal to VTh.
 RTh calculation:
Remove if any load.
Replace all sources by their input impedances.
Compute the total resistance between the load terminals.
+
-
Vth
Rth
A
B
A
B
Black
box
17

Example
Obtain the Thevenin’s equivalence of the following circuit.
•VTh calculation:
•RTh calculation:
1 kΩ
+
-
7.5 V
A
B
2 kΩ
18

Norton’s Theorem
Any two-terminal linear, bilateral network containing impedances and energy
sources can be represented by an independent current source IN in parallel with a
single impedance ZN (admittance YN).
IN is the short-circuit current between the terminals, ZN is the impedance viewed at
the terminals when all independent energy sources are replaced by their internal
impedances.
 IN calculation:
Short the output terminals and calculate current through it.
 RN calculation:
The same as RTh.
A
B
Black
box IN
RN
A
B
19

Example
Obtain the Norton’s equivalence of the previous circuit.
•IN calculation:
•RN calculation:
20

Source Transformation
+
-
Vth
Rth
A
B
A
B
Black
box
IN
RN
A
B
Voltage source Current source
Th N
Th N N
Th
N
Th
R R
V I R
V
I
R
=
=
=
Calculations:
21
Voltage source Current source

Example
Obtain the Thevenin’s equivalent of the following circuit.
•RTh calculation:
RTh = {(2||2) + 1} || 2 Ω
= 1Ω
•VTh calculation:
 
 
22

Capacitor Circuit
Current leads the phase of input voltage by 900.
Current and voltage wave forms.
( )
( )
( )
( )0 0
0
0
cos [ sin ]
sin
2
sin
2
dv t
i t
dt
dv t
C
dt
C V t for v t V t
C V t
I t
ω ω ω
π
ω ω
π
ω
∞
=
= =
 
= + 
 
 
= + 
 
Instantaneous current:
( ) ( ) ( ) 2
0
1
sin 2
2
p t v t i t CV tω ω= =
Instantaneous power expended in charging:
Energy delivered in time interval t1: ( ) ( ) ( )
1
2
1 0 1
0
1
1 cos2
4
t
tW t p t dt CV tω= = −∫
Energy delivered in n half-cycles: ( )2
0
1
1 cos2 0
4
nW CV nπ= − =
23

Capacitor Circuit
Capacitive reactance:
1 1
[where ].Cx s j
j C sC
σ ω
ω
= = = +
Joule loss due to an ideal capacitor is zero.
Representation of non-ideal capacitor.
Series and parallel connections:
1 2 ...eq nC C C C= + + + 1 2
1 1 1 1
...
eq nC C C C
= + + +
24

Inductor Circuit
( )
( )
( )
( )0 0
0
cos [ sin ]
cos .
di t
v t
dt
di t
L
dt
L I t for i t I t
V t
ω ω ω
ω
∞
=
= =
=
Current lags the phase of input voltage by 900.
[where ].Lx j L sL s jω σ ω= = = +Inductive reactance:
Instantaneous voltage:
( ) ( ) ( ) 2
0
1
sin 2
2
p t v t i t LI tω ω= =
Instantaneous power expended in charging:
Energy delivered in time interval t1: ( ) ( ) ( )
1
2
1 0 1
0
1
1 cos2
4
t
tW t p t dt LI tω= = −∫
Energy delivered in n half-cycles: ( )2
0
1
1 cos2 0
4
nW LI nπ= − =
25

Inductor Circuit
Joule loss due to an ideal inductor is zero.
Representation of non-ideal inductor.
Series and parallel connections:
1 2 ...eq nL L L L= + + +
1 2
1 1 1 1
...
eq nL L L L
= + + +
26

RC Circuit
Frequency domain analysis:
0, 1
, 0
f gain
f gain
∴ → =
→ ∞ =
RC circuit
(Lowpass filter)
Half-power points:
1 1
.
2 2
out out
in in
P v
P v
= ⇒ =
( )
( )
2
2
3
3 3
1 1
21
1 2
1 1
,
2 2
d
d dB
CR
CR
f
RC RC
ω
ω
ω
ω ω
π π
=
+
⇒ + =
⇒ = = ∴ = =B
B
•f3dB: cut-off frequency (fc).
( )
2
1
1
1 1
1 1
1
| | (transfer function)
1
c in
c
in
c
in
j C
v v
R j C
v
v j CR sCR
v
v CR
ω
ω
ω
ω
=
+
∴= =
+ +
∴ =
+
27
LPF
Log100.5 = -0.301

CR Circuit
( )
2
1
1 1
1
R in
R
in
R
in
R
v v
R j C
v j CR sCR
v j CR sCR
v CR
v CR
ω
ω
ω
ω
ω
=
+
∴= =
+ +
∴ =
+
0, 0
, 1
f gain
f gain
∴ → =
→ ∞ =
CR circuit
Half-power points:
( )
2
3
3
3
1
21
1
,
1
.
2 2
d
d
dB
CR
CR
RC
f
RC
ω
ω
ω ω
ω
π π
=
+
⇒ = =
⇒ = =
B
B
The same as RC circuit.
CR circuit
(High pass filter)
28
HPF

Comparison of CR & RC Frequency Responses
RC circuit
CR circuit
Low pass filter High pass filter
29
θ
VR = RI
V = ZI VC = jxCI
Phasor diagram
I

LR & RL Circuits
0, 1
, 0
f gain
f gain
∴ → =
→ ∞ =
RL circuit
( )
22
L in
L
in
j L
v v
R j L
v L
v R L
ω
ω
ω
ω
=
+
∴ =
+
0, 0
, 1
f gain
f gain
∴ → =
→ ∞ =
( )
22
R in
R
in
R
v v
R j L
v R
v R L
ω
ω
=
+
∴ =
+
Output voltage across the inductor: Output voltage across the resistor:
HPF LPF
3 3/ , .
2
d dB
R R
rad S f Hz
L L
ω
π
= =B
Frequency domain analysis:
•Half-power points for the both cases:
30
Phasor diagram
θ
VR = RI
V = ZI
VL = jxLI
I

Time Domain Analysis
31
Steady-state response: ( ) ( )Lts
t
f t f t
→∞
=
Transient response: response before the steady is achieved.
Laplace transform:
0
[ ( )] ( ) ( ) , ( ) , :realand positive, :constant
Re( )>
st at
f t F s f t e dt f t K e a K
s a
−
∞
−
= = <∫L
Unit step function: ( ) 1, 0
0, 0
U t t
t
= ≥
= ≤
0
1
[ ( )] st
U t e dt
s−
∞
−
= =∫L
Inverse Laplace transform:
-1 1
( ) [ ( )] ( )
2
iT
st
iT
f t F s e F s ds
i
γ
γπ
+
−
= = ∫L
U(t)
t
1

Time Domain Analysis
32
Differentiation: [ ] ( ) (0 ) , (0 ) is the initial value of ( )
dy
sY s y y y t
dt
− −= −L
Integration:
0
1
[ ] ( )
t
y dt Y s
s−
=∫L
1
[ ]at
e
s a
=
−
LExponential function:
For a sinusoidal wave in steady state:
1
sC
sL
Capacitive reactance Inductive reactance

Time Domain Analysis of RC/CR Circuits
RC circuit
t = 0
V
0
Applying Kirchhoff’s voltage law
...(1)
Taking Laplace transform,
...(2)
Considering vc(0-) = 0,
...(3)
33
0
1
(0 ) ( )
t
cv i dt Ri V U t
C
− + + =∫
(0 ) ( )
( )cv I s V
R I s
s Cs s
−
+ + =
1
( )
1 1
V C V
I s
sCR R s CR
 
∴ = =  
+ + 
Taking inverse Laplace transform,
( ) ( )t CRV
i t e U t
R
−
∴ =
...(4)
Voltage across the resistor:
t RC
Rv i R V e−
= = ...(5)

Time Domain Analysis of RC/CR Circuits
...(6)
• Time constant of the circuit:
Time taken to drop the voltage across the resistor to V/e.
...(7)
[Euler’s number e = 2.71828…]
34
Voltage across the capacitor: ( )1 t RC
C Rv V v V e−
= − = −
1 2 c
RC
f
τ
π
=
=
• Rise time:
Time taken to reach the capacitor voltage from 10% to 90% of the final value.
Τr = 2.3 RC – 0.1RC
= 2.2 RC
~ 0.35 × 2πRC
= 0.35/ fc. ...(8)
1
put t RC
e e− −
 = 

Step response of a capacitor. Step response of the resistor.
Time Domain Response of RC/CR Circuits
35
vR / vin = 0.905 at t = 0.1τ
= 0.990 at t = 0.01τ
= 0.05 at t = 3τ
= 0.007 at t = 5τ
vC / vin = 0.95 at t = 3τ
= 0.993 at t = 5τ
( )1 t RC
Cv V e−
= −
t RC
Rv V e−
=

Example
Draw the output voltage waveform for the following circuit and calculate the rise
time.
Solutions:
At t = 0, the capacitor is shorted, so V0 = 0 and I0 = 10 mA.
Now,
Time constant = 1k × 1n = 1 μS.
•Rise time = 2.2 RC
= 2.2 1k × 1n
= 2.2 μS.
Output voltage waveform.
36
• Calculate the time when the output voltage is half
of that of the input. (Ans – 0.693 μS)
( )1 t RC
Cv V e−
= −

Time Domain Response of RC/CR Circuits
RC circuit
Low PRF (RC<<Ton) High PRF (RC>>Ton)
Charging phase:
37
( )1 t RC
Cv V e−
= −
t RC
Rv V e−
=
Rectangular pulse:
( ) ( )onV U t U t T− −  
Ton
Discharging phase:
t RC
Rv V e−
= −
t RC
Cv V e−
=

RC Integrator
Consider the output across the capacitor at high frequency i.e. f >>1/Ton.
Integrator circuit
Loop current is
The frequency condition, gives
...(1)
...(2)
Now, voltage across the capacitor is
...(3)
...(4)
At high frequency, the voltage across the capacitor is proportional to the time
integration of the input voltage.
Low pass filter at high frequency
38
1
inv
i
R j Cω
=
+
1 C Rω <<
inv
i
R
≈
0
1 t
Cv i dt
C
= ∫
0
1 t
C inv v dt
RC
∴ ≈ ∫

RC Integrator Waveforms
Integrator circuit
vc
vc
at very high
frequency
at medium
frequency
Input:
Output:
39
1:5 approximation:
τ ≥ 5/PRF (=5T)

RC Differentiator
Consider the output across the resistor at low frequency i.e. f <<1/ Ton.
Differentiator circuit
Loop current is
The frequency condition, gives
...(1)
...(2)
•The capacitor has enough time to charge up until vc
is nearly equal to the source voltage.
Now, voltage across the resistor is ...(3)
...(4)
At low frequency, the voltage across the resistor is proportional to the time
differentiation of the input voltage.
High pass filter at low frequency.
40
1
inv
i
R j Cω
=
+
1R Cω<<
1
in
in c
v i
i v v
j C j Cω ω
≈ ⇒ ≈ =
c
R
dv
v iR C R
dt
= =
in
R
dv
v RC
dt
∴ ≈

RC Differentiator Waveforms
Some other waveforms.
41
[Differentiation]
1:5 approximation:
1/PRF (=T) ≥5τ

RL/LR Circuits
42
RL circuit
t = 0
V
0
Applying Kirchhoff’s voltage law
...(1)
...(2)
Considering iL(0-) = 0,
...(3)
( )
di
L Ri V U t
dt
+ =
( ) ( ) ( )0
V
L s I s i R I s
s
−− + =  
( )
1 1
( )
V V
I s
s R sL R s s R L
 
∴ = = − 
+ + 
Taking inverse Laplace transform,
( )( ) 1 , 0Rt LV
i t e t
R
−
∴ = − ≥
...(4)
( )1 , .Rt L Rt L
R L Rv i R V e v v v Ve− −
∴ = = − = − = ...(5)
/
1 2 c
L R
f
τ
π
=
=
Time constant:

Problem
A positive square wave of amplitude 10 V and PRP of 1 kHz is applied to the
following circuit. Draw the vR and vC waveforms for R = 1 kΩ, C = 10 nF.
Solution:
Time period of the input wave: 1 mS.
Ton = 0.5 mS
Check for 1:5 approximation:
Time constant = 1 k ×10 n Sec
= 10 μS.
43
( )max 1
10V
ONT RC
Cv V e−
= −
=
0.5 10 50t mτ µ∴= =

Semiconductor
A semiconductor is a material which has electrical conductivity between that of
a conductor such as (copper, 5.96x107 S/m) and an insulator (glass, ~10-13 S/m).
In semiconductors
•The conductivity increases with increasing temperature.
•Current conduction occurs via free “electrons” and "holes", charge carriers.
•Doping greatly increases the number of charge carriers.
•Contains excess holes p-type, and excess free electrons n-type.
Simplified band-diagrams (a graphic representation of the energy levels)
44

Intrinsic and Extrinsic Semiconductors
Intrinsic semiconductor:
When the number of impurities is extremely low so that it can be considered as a
pure semiconductor material.
Example:
Semiconductor carriers/cc at room temp
Ga As 1.7× 106
Si 1.5× 1010
Ge 2.5× 1013
Ge atomSi atom
Electron configuration: 2, 8, 18, 4.Electron configuration: 2, 8, 4.
45

Intrinsic Semiconductor
Covalent bond of the Si atoms at t = 0 K
Creation of electron-hole pair because of thermal agitation at t >0K.
electron
hole
•For intrinsic semiconductor, electron density ni = hole density pi.
•Si about 1 free
electron for every 1012
atoms at 300K)
46

Extrinsic Semiconductors
A semiconductor with doping material is called extrinsic semiconductor.
1. n-type material:
Created by introducing impurity elements that have five valance electrons ( eg.
antimony, arsenic, phosphorus etc).
•Impurities with five valance electrons are called donor atoms.
2. p-type material:
Created by introducing impurity elements that have three valance electrons ( eg.
boron, gallium, indium etc).
•Impurities with three valance electrons are called acceptor atoms.
47

Extrinsic Semiconductors
Majority carriers and minority carriers:
Excess carriers are majority carriers.
•In n-type: electrons are majority carriers and holes are minority carriers.
•In p-type: holes are majority carriers and electrons are minority carriers.
• For extrinsic semiconductors, np = ni
2, n: electron concentration, p: hole
concentration in the extrinsic material, ni: intrinsic free electron density.
48
•The resulting material is electrically neutral.

Drift and Diffusion Currents
49
•Charge carriers moves under the influence of electric field.
•They cannot travel in a straight line.
•Drift current depends on carrier concentration, carrier mobility and the applied
electric field.
E
Drift current
atom
electron
hole
Electrical conductivity ( )n pe n pσ µ µ= + n: electron concentration/ cc
p: hole concentration/ cc
μn: electron mobility cm2/V-S
μp: hole mobility cm2/V-S
(Ω-cm)-1

Drift and Diffusion Currents
50
•Charge carriers (one type) when injected from an external source produces
concentration gradient.
•Current density is proportional to the concentration gradient.
Diffusion current
Diffusion
where

p-n Junction: Semiconductor Diode
51
p-n diode
Formation of depletion region
• Depletion region contains an equal number of immobile ionized atoms on both
sides of the pn junction.
• The barrier voltage opposes the flow of majority carriers across the junction but
assists the flow of minority carriers across the junction.

p-n Junction: Semiconductor Diode
52

Reverse-Bias Condition
pn junction under reverse-bias condition
•The holes in the p-side and the electrons in the n-side are attracted by the
electrodes  the depletion region widens  barrier voltage increases.
•Minority carriers are generated on each side causes a reverse saturation current.
•Reverse saturation current is almost independent of applied voltage but increases
with increasing temperature.
53

Forward-Bias Condition
pn junction under forward-bias condition
•The holes in the p-side region and the electrons in the n-side region are driven
toward the junction  width of the depletion region is reduced  barrier voltage
decreases.
•With increasing forward-bias voltage, the barrier voltage finally disappears 
current increases suddenly.
54
Diode current can be given by Shockley equation: ( )0 1D TV nV
DI I e= −
n = 1 for Ge and 2 for Si, VT = kT/e = 26 mV at room temp, T – temperature (K),
k = Boltzman’s const (1.38x10-23 J/K), e = electronic charge (1.6x10-9 C)

I-V Characteristics
I-V characteristics of a resistor. I-V characteristics of a diode.
55
Temperature coefficient for a pn junction: -1.8 mV/0C for Si and -2.02 mV/0C for
Ge.

Diode Breakdown Under Reverse Bias
56
1. Avalanche breakdown:
• Thermo-generated electron-hole pair accelerates under the applied electric field.
•Under the right circumstances, they gain sufficient energy to knock other
electrons free (by collision), creating more electron-hole pairs.
•The procedures is regenerative avalanche of carriers  conductivity increases
suddenly.
Ein + Eext
Avalanche breakdown Avalanche of carriers
Etotal

Diode Breakdown Under Reverse Bias
57
2. Zener breakdown:
• High electric field (~107 V/m) enables tunnelling of electrons from the valence to
the conduction band of a semiconductor (tear out a covalent bond)  a large
number of free minority carriers suddenly increases the reverse current.
• High doping concentration is required.
• Occurs at lower reverse bias voltage (~ 6 V).
Diode characteristicsZener breakdown
•Usually, Rbreakdown <RFB
•Applications: voltage regulator, clipper circuit, voltage shifter etc
ID
VD
Zener
breakdown
Avalanche
breakdown
p n
Symbol
Ein + Eext
Iz

Diode Modelling
Ideal diode response. Ideal diode as a switch. More accurate: use to solve problems.
58
ID
VD
ID
VD
Si-diode Ge-diode Zener-diode Tunnel-diode Power-diode
dc and ac resistances:
VD
Diode characteristics
ID
ΔID
ΔVD
dc resistance, , ac resitance, .D D
D d
D D
V V
R r
I I
∆
=
∆
A good approximation for Vi >12 V (error <5%)

Diode Load Line
59
Vi
R
VDID
Load line
( ) − 0 exp 1 ...(1)D D TI I V nV
Diode with a resistance
Diode current:
VD
ID
Vi /R
Vi
IDQ
VDQ
Q-point
Load lineShockley diode model:
• Difficult to solve for VD from (1) and (2).
• Plot them in a graph and obtain the solution.
Applying KVL,
[ ]
= +
=− =+ form ...(2)
i D D
i D
D
V I R V
V V
I y mx C
R R
So, from (1) and (2),
( ) − = − 0 exp 1 . ...(3)i D
D T
V V
I V nV
R R

Diode Analysis
60
Diode characteristics
vD
iD
IDQ
VDQ
t
t
Diode circuit
ac comp.,
dc comp.,
total instantaneous comp.
d
D
D
i
I
i
→
→
→
( )
( )( )
( )( ) ( ) ( )
( ) ( )
( )
0
0
0 0
2
0
0
exp 1
exp 1
exp exp exp
exp
linear approximati
1 [assume ], 1 ...
2!
[ exp , ]
on
D D T
DQ d T
DQ d T DQ T d T
x
DQ T d T d T
DQ d DQ DQ T d DQ d T
i I v nV
I V v nV
I V v nV I V nV v nV
x
I V nV v nV v V e x
I i I I V nV i I v nV
−  
 = + −
 
≈ + =
= × + << = + + +
=+ ==

vD
iD
IDQ
iD
[ ]d T DQ d d dr V I v i r= ∴=
Diffusion resistance
Vi
R
VDID
vi

Diode Modelling: Small Signal
61
Small signal approximation:
( )
( )
exp
1 [for 1 ].
D DQ d T
x
DQ d T d T
i I v nV
I v nV v V e x
=
≈ + << ⇒ ≈ +
[ ]
( ) ( )
2 2 3 3
2 3
sin .
sin sin
1 sin ... where
2! 3!
1 1
Now, sin 1 cos2 , sin 3sin sin3 ...
2 4
d
D DQ T
v a t
b t b t
i I b t b a nV
t t t t t
ω
ω ω
ω
ω ω ω ω ω
=
 
∴= + + + + = 
 
=− =−
• There are many frequency components those cannot be neglected.
• The output signal waveform is not an exact replica of that of the input
 non-linear device.
Problem with large signal:
Mathematical models used to approximate the actual behaviour of real diodes to
enable calculations and circuit analysis.

Diode Modelling: Large Signal
62
Vi
R
VDID
Iterative Solution:
( )
( )
 − 
⇒ + =
 
⇒ = + 
 
 −  
⇒ = + =−   
  
 −
⇒ + 
 
0
0
0
0
0
exp 1
1 exp
1
1 put
1 ...(3)
D D T
D
D T
D
D T
i D i D
D T D
i D
D T
I I V nV
I
V nV
I
I
V nV ln
I
V V V V
V nV ln I
RI R R
V V
V nV ln
RI
Guess a starting value of VDQ on R.H.S. and compute
the function. Guess a second value and continue
unless it converges.
Load line
VD
ID
Vi /R
Vi
IDQ
VDQ
Q-point
Load line
Graphical solution:
Plot the two equations on a graph paper (computer) and note down the value.

Diode Modelling: Large Signal
63
Vi
R
VDID
Piecewise linear model:
Piecewise linear approximation.
VD
ID
Vi /R
Vi
IDQ
VDQ
Load line
A function is broken down into several linear segments.
Equivalent model.
ID
VD
Diode characteristics.
(Forward bias) (Reverse bias)
Vγ rD
• rD represents the 1/slope at Q-point.
26 mV( at room temp)
.T
D
DQ DQ
V
r
I I
≈ ≈

Diode Circuit
64
vi RL
vD
id
+ -
+
-
vL
Half-wave rectifier:
Half-wave rectifier circuit.
vi RLid
+
-
vL
Vγ
vi RLid
+
-
vL
Equivalent circuits.
(in forward bias) (in reverse bias)
A rectifier converts alternating current to direct current.
Response to solve problems.
(Consider the ideal diode model if nothing is specified)
ID
VD
Ideal
diode
ID
VD

Half-Wave Rectifier
65
Peak Inverse Voltage (PIV): peak voltage in reverse bias condition that a rectifier
can block. It is limited by diode break down voltage.
vi RLid
+
-
vL
vi RLid
+
-
vL
(in forward bias)
(in reverse bias)
Voltage waveforms considering ideal diode.
vi (V) t (mS)
T/2 T 3T/2
vL (V) t (mS)
T/2 T 3T/2
vD (V) t (mS)
T/2 T 3T/2
Vm
Vm
-Vm

Half-Wave Rectifier
66
sini mv V tω=
vi (V) t (mS)
T/2 T 3T/2
Vγ
vL (V) t (mS)
T/2 T 3T/2
Vγ
vD (V) t (mS)
T/2 T 3T/2
Vγ
Vm
mV Vγ−
Voltage waveforms.
vi RLid
+ -
+
-
vL
dv
vi RLid
+
-
vL
(in forward bias)
(in reverse bias)
( ) ( )
( )
| |d peak i peak L D
m L
i v V R R
V V R
γ
γ
= − +
≈ −
-Vm

67
( ) ( )0
1
cos sinn n
n
v t V a n t b n tω ω
∞
=
=+ +∑
Fourier series:
Any periodic function can be represented as a summation of sinusoidal functions.
Representation of a square wave
according to Fourier series.
Half-Wave Rectifier: Analysis
• Rectified output voltage can be
represented by a Fourier series.
• It contains a dc term + many
sinusoidal (time varying)
components.
• Sinusoidal signal is represented by
its rms value.
t (mS)f (t)
1
2
6

68
[ ]
2
0
0
1
sin where
2
1
sin 0
2
dc av m
m
m
V V V d t
V d
V
π
π
θ θ θ ω
π
θ θ
π
π
= = =
+
=
∫
∫
dc part:
ac part:
2
2 2
0
2
0
1
sin
2
1 cos2
2 2
2
rms m
m
m
v V d
V
d
V
π
π
θ θ
π
θ
θ
π
 
=  
 
 −
=  
 
=
∫
∫
vL (V) t (mS)
T/2 T 3T/2
Vm
2T 5T/2
Output voltage waveforms considering an
ideal diode.
• rms value of a 12 V DC voltage is 12 V.
• DC contribution could not be avoided.

69
Ripple voltage and ripple factor:
( ) 2 2
2 2
2
4
0.1487
0.385
r rms dc
m m
m
m
v rms v V
V V
V
V
π
 = − 
 
= − 
 
=
=
Vdc
t (mS)
(V)
vr|p-p
vt = Vdc + vr (ripple voltage) (V)
Ripple voltage
ripple voltage(rms)
Ripple factor = 100%
dc voltage
100%
0.385
100% 121%
r
dc
m
m
v
V
V
V π
×
= ×
= × =
2
2
2
2
0.406
dc L
dc ac
rms L
m
m
V R
P P
v R
V
V
η
π
= =
 
= 
 
=
Efficiency:

Half-Wave Rectifier
70
• In the circuit, both of the diodes have a
cut in voltage of 0.7 V. Calculate the
current components.
10 0.7
330
28.18
2 14.09
R R
D R
I V R
mA
I I mA
−
= =
=
= =
Solution:
vi RL
vD
id
+ -
+
-
vL
• A sinusoidal source vi = 12sin100πt V is used
in the half-wave rectifier circuit with RL = 1
kΩ. The diode has a cut-in voltage of 0.7 V.
Calculate the PRV and the power rating of
the diode.
Answer: PRV = 12 V.
Power rating = Imax x Vdmax = 7.91 mW.
10 V ID
IR
330 Ω
D1 ID D2
N.B.: If the diodes have different cut-in
voltages, the diode with lower cut-in
voltage will be switched on.

Half-Wave Rectifier : Experiment
71
Transformer:
• Controllable voltage ratio.
• Impedance matching.
2
1
s
i
v N
v N
=
Half-wave rectifier.
RL
vD+ -
+
-
vLLine voltage
(220 V, 50 Hz)
Center tapped
transformer
(12-0-12 V)
Primary
Secondary
Center tapped transformer
Voltage regulation:
Voltage regulation 100%NL FL
FL
V V
V
−
= ×
VNL: open circuited output voltage
VFL: output voltage with minimum
load resistance.

Full-Wave Rectifier
72
Full-wave rectifier.
Voltage waveforms considering ideal diodes.
vi (V)
(Diode 1)
t (mS)T/2 T 3T/2
iD2 (mA) t (mS)T/2 T 3T/2
vL (V) t (mS)T/2 T 3T/2
vi (V)
(Diode D2)
t (mS)T/2 T 3T/2
1 1 2m sV v=
2 2sv
1 2s Lv R
2 2sv
2 2s Lv R
1 2sv
D1 D1
D2
RL
vD1+ -
+
-
vLvs1
vs2
+
+
-
-
vD2+ -
D1
D2
1 12sv Vγ−
For non-ideal diode
Consecutive peaks have different
values if
• cut-in voltages are different.
• vs1 ≠ vs2

73
Full-Wave Rectifier: Analysis
2
0
0
1
sin
2
2
sin 0
2
2 2
dc av m
m
m dc half wave
V V V d
V d
V V
π
π
θ θ
π
θ θ
π
π −
= =
+
= =
∫
∫
dc part:
ac part:
2
2 2
0
2
0
1
sin
2
1 cos2
2
2 2
2 2
rms m
m
m rms
half wave
v V d
V
d
V v
π
π
θ θ
π
θ
θ
π
−
 
=  
 
 −
= × 
 
=
∫
∫
vL (V) t (mS)
T/2 T 3T/2
Vm
2T 5T/2
2
2
2
8
0.812 2
dc L
full wave dc ac
rms L
half wave
V R
P P
v R
η
π
η
−
−
= =
=
= =
Efficiency:
Output of a full-wave rectifier.

74
Full-Wave Rectifier: Analysis
vL (V) t (mS)
T/2 T 3T/2
Vm
2T 5T/2
Ripple voltage and ripple factor:
( ) 2 2
2 2
2
4
2
0.0947
0.308
r rms dc
m m
m
m
v rms v V
V V
V
V
π
 = − 
 
= − 
 
=
=
Ripple voltage
Ripple factor 100%
0.308
100% 48%
2
r
dc
m
m
v
V
V
V π
= ×
= × =
Output of a full-wave rectifier.
C
vD1+ -
+
-
v= ?vs1
vs2
+
+
-
-
vD2+ -
D1
D2
• Vm should not exceed PIV.

Full-Wave Rectifier
75
RL
vD1+ -
+
-
vLvs1
vs2
+
+
-
-
vD2+ -
D1
D2
vi (V) t (mS)T/2 T 3T/2
vL (V) t (mS)T/2 T 3T/2
vi (V) t (mS)T/2 T 3T/2
1 1 2m sV v=
2 2sv
1 2s Lv R
2 2sv
2 2s Lv R
1 2sv
D1
D2

Bridge Rectifier
76
Equivalent circuit in negative cycle.
vL
D1
D2
D3
D4
RLLine voltage
(220 V, 50 Hz)
Bridge rectifier circuit.
Equivalent circuit in positive cycle.
vL
D1
D2
D3
D4
RL
+
-
vL
D1
D2
D3
D4
RL
+
-
vi (V) t (mS)
T/2 T 3T/2
iD2 (mA)
vL (V)
iD1 (mA)
mV
D1, D2
D3, D4
m LV R
m LV R
m LV R−
mV
D1, D2
1 12 2m sV v Vγ= −For non-ideal diode:

Rectifier With Filter
77
• Rectified output contains a dc + many sinusoidal terms (amplitude of the
sinusoidal components decreases as the frequency increases).
• A series inductor attenuates and a shunt capacitor bypasses high frequency
signals.
Shunt capacitor
as a filter.
C L
R
C C
L
C C
Rectifier.
RLvs1
vs2
+
+
-
-
D1
D2
Low
Pass
Filter
ac
source
Series inductor
as a filter.
LC π filter.RC π filter.
Different types of filters.Rectifier with a filter.

Rectifier With Filter: Half-Wave
78
Half-wave rectifier.
RL
vD+ -
+
-
vLLine voltage
(220 V, 50 Hz)
C
Voltage waveforms considering ideal diode.
vi (V) t (mS)
2T
vL (V) t (mS)
T 2T
Vm
Vm
T
Vmexp(-t/RLC)
T1 T2
Analysis:
• Charging (T1) time constant
≈ (RD + Rind ) x C
≈ 0
• Discharging (T2) time constant
= RL xC
( )
( )
( )( )
( ) ( )
( )
min
When the capacitor discharges:
exp /
exp /
1 1 /
/ /
/
L c m L
L m L
r m Lp p
m L L
m L
v v V t R C
v V T R C
v V T R C
V T R C I T C
V f R C
−
= = −
≈ −
≈ − −
= =
=
• Time period T = T1 + T2
Output voltage:

Rectifier With Filter: Half-Wave
79
Half-wave rectifier with filter.
RL
vD+ -
+
-
vLLine voltage
(220 V, 50 Hz)
C
• A repetitive surge current IS
(during T1) flows to recharge the
capacitor.
• IS is spike shaped. For simplified
analysis, it is modeled by a
rectangular pulse.
• Average input current to rectifier
is equal to the average load
current:
Surge current:
1
2
1
1
L Sav
S L av
I T I T
T
I I
T
× = ×
 
⇒= + 
 
Surge limiting resistor.
Rs
Surge current through the diode.
vL (V)
t (mS)
T
Vm
T1 T2
t (mS)
Id (mA)
Surge current
Triangular wave approximation: ( ) | 2 3r r p pv rms v −=

Problem
80
Half-wave rectifier with filter.
RL
vD+ -
+
-
vLLine
voltage
C
As shown in the figure, a half-wave rectifier
dc power supply is to provide 10 V (dc) to a
1 kΩ load. Calculate the capacitance
required so that the peak-to-peak ripple
voltage does not exceed ±10% of the
average output voltage. Source frequency is
50 Hz. What should be the amplitude of the
input voltage?
Solutions:
( )
2
1
20
20
10% of 10V 9V - 11V
11 V
= / 110 .
r p p
m
m L r p p
T mS
f
T T mS
v
V
C V f R v Fµ
−
−
= =
≈ =
=± ≈
∴ =
=
vL (V)
t (mS)
T
Vm
T1 T2
Triangular wave approximation: ( ) | 2 3r r p pv rms v −=
10 V

Rectifier With Filter: Full-Wave
81
Analysis:
• Charging (T1) time constant
= RD xC
= 0
• Discharging (T2) time constant
= RL xC
( )
( )( )
( )
( ) ( )
min
exp / 2
1 1 / 2
/ 2
/ 2 2
L m L
r m Lp p
m L
m L L
v V T R C
v V T R C
V T R C
V f R C I f C
−
≈ −
≈ − −
=
= =
• Time period T/2 = T1 + T2
Output voltage:
vi (V) t (mS)
2T
vL (V) t (mS)
T 2T
Vm
Vm
T
Vmexp(-t/RLC)
T1 T2
T/2 3T/2 5T/2
RL
vD1+ -
+
-
vLvs1
vs2
+
+
-
-
vD2+ -
D1
D2
C

Problem
82
Calculate the peak-to-peak ripple voltages for a half-wave and a full-wave rectifiers
with a capacitor filter. (RL = 1 kΩ, C = 470 μF, vi = 12 V, 50 Hz)
Solutions:
T = 1/50 = 20 mS
Discharging time constant = RLC = 470 mS.
T1 = 0 approximation is valid.
• Full-wave rectifier:
( )
( )
/
12 2 50 1 470
0.722 ( 0.707 using exp. function)
r m Lp p
v V f R C
k
V V
µ
−
=
= × ×
= =
( )
( )
/ 2
12 2 50 1 470
0.361 ( 0.357 using exp. function)
r m Lp p
v V f R C
k
V V
µ
−
=
= × ×
= =
• Half-wave rectifier:
Becomes a complex problem
for RL = 1 kΩ, C = 47 μF.
( )
( )
2
2
3 4
4
Half-wave: sin
Full-wave: sin
L
L
T T R C
m
T T R C
m
t V e
t V e
ω
ω
− +
− +
=
=
Solve for T2 first:
Triangular wave
approximation:
2 3r rp p rms
v v−
=

Voltage Regulator
83
• Automatically maintains a constant voltage level.
• Output voltage is independent of load or source voltage variation.
A dc voltage regulator using Zener diode.
1. The diode must be fired (in break down mode).
2. Diode current must be limited to avoid burn out.
Zener breakdown
ID (mA)
-VD (V)
VZ
RZ = 0
Iz|allowed
Iz|allowed /10
Operating
region
Vi
Rs
Vz
I
RL
+
-
VLIZ
IL
Conditions for proper operations:
( )
(1)
(2)
z L
i s z
i s L
I I I
V I R V
V I R R
= +
= +
= +
When the diode operates in breakdown region.
When the diode is not in breakdown region.×

Voltage Regulator
84
A dc voltage regulator using Zener diode.
( )
min
min
min
min
min
without the diode
Calculate for the voltage divider formed
by and
1
L Z
L
s L
L
i Z
L s
i
s L
z
V V
V
R R
R
V V
R R
V
R R
V
>
⇒ × >
+
 
⇒ < −  
 
max
max
max
wherez z z z zallowed allowed
i z
z allowed
s
i z
s
z z
I I I P V
V V
I
R
V V
R
P V
 ≤ = 
−
⇒ ≤
−
⇒ ≥
Condition 1: diode in breakdown Condition 2: avoid diode burn out
Vi
Rs
Vz
I
RL
+
-
VLIZ
IL
Both Vi and RL vary:
min max
min max
max min
to
to
to
i i i
L L L
L L
V V V
R R R
I I
⇒
⇒
Take RLmax
as infinite if
unspecified.

85
Voltage Regulator
• In the following circuit, Vi can vary between 9
and 12 V, and RL between 1 kΩ and infinity. A
Zener diode with Vz = 6 V and PD = 400 mW
is used to design the voltage regulator.
Choose a suitable value of Rs to avoid diode
burn out.
Vi
Rs
Vz
I
RL
+
-
VLIZ
IL
Solutions:
The diode must be fired:
min
min
min max
max
max
|
|
| |
1
9 6
1 |
| 0.5
L
i z
L s
s
s
R
V V
R R
k
k R
R k
× ≥
+
⇒ × ≥
+
⇒ ≤ Ω
Maximum allowed Zener current:
/ 66.67z allowed D zI P V mA= =
max
min
min min
|
|
12 6
| | 90 .
66.67
i z
s
z allowed
s s
V V
R
I
R R
m
−
≥
−
⇒ ≥ ⇒ ≥ Ω
The range of Rs is 90 Ω <Rs <500 Ω.

86
Voltage Regulator
100%, Noloadoutput voltage
Fullloadoutput voltage.
NL FL
NL
FL
FL
V V
Voltage regulation V
V
V
−
= × →
→
100%.L
i
V
Source regulation
V
∆
= ×
∆
ΔVL change in output voltage for a
change of ΔVi input voltage.
Ripple rejection ratio = output ripple voltage/ input ripple voltage.
Vi
Rs
VzIZ
I
Vi
Rs
VzIZ
I
Vi
Rs
Vz
IZ
I Rz
Voltage regulator
without the load.
Equivalent circuit
(ideal Zener).
Equivalent circuit

87
Clipping Circuits
Series clipper:
• A clipper (limiter) clips off an unwanted portion of a waveform.
• Often used to protect a circuit from a large amplitude signal.
• Example: half-wave rectifier.
RL
v0
+
-
vin
+
-
RL
v0
+
-
vin
+
-
RL
v0
+
-
vin
+
-
0 V
5.3 V
0 V
6 V
Negative series clipper. Output voltage.Input voltage.
Equivalent circuit when the
diode is in forward bias.
Equivalent circuit when the
diode is in reverse bias.
ID
VD
Diode characteristics.
OFF ON

88
Series Clippers
Positive clipper:
RL
v0
+
-
vin
+
-
0 V
-5.3 V
0 V
6 V
Positive series clipper. Output voltage.Input voltage.
Series noise clipper:
Series noise clipper. Output voltage.Input voltage.
RL
v0
+
-
vin
+
-
0 V
6 V
0.7 V
-0.7 V
Dead zone
0 V
5.3 V
• Unwanted lower level noise can be eliminated by a series clipper circuit.

89
Shunt Clippers
0 V
V1 V
Input voltage.
Positive shunt clipper:
0 V
V1 V
Input voltage.
Negative shunt clipper:
Positive shunt clipper.
RL
v0
+
-
vin
+
-
R1
IL
Vγ
Negative shunt clipper.
RL
v0
+
-
vin
+
-
R1
IL
Vγ
Output voltage.
(V1 – ILR1) V
0 V
Vγ−
Output voltage.
-(V1 – ILR1) V
0 V
Vγ
• Diode is in reverse bias condition: shunt branch has no effect on the output
voltage.

90
Biased Clipping Circuits
Series clipper:
Input voltage.
RL
v0
+
-
vin
+
-
Series clipper with
negative shift.
VγBV
0 V
V1 V
Output voltage.
0 V
V1 V
( )1 BV V Vγ− +
( )in BV V Vγ= +
• Check the diode biasing condition for Vin = 0 V.
• Replace the diode by its equivalent model.
• Determine the value of Vin required to change the above biasing condition.
• Calculate the output voltages for max./min. values of Vin.
• Check for zero crossing (calculate the value of Vin for which the output is zero).
Solution steps:
No voltage values are specified: consider 1 .BV V Vγ> >

91
Input voltage.
RL
v0
+
-
vin
+
-
Series clipper with
positive shift.
VγBV
0 V
V1 V
Input voltage. Positive shunt clipper.
0 V
V1 V
• Consider ideal diode if the type of the diode (Si, Ge etc) or the cut-in voltage is
unspecified.
• If unspecified consider R1 = 0.
R1
vin RL
v0
+
-
+
-
IL
vB
Output voltage.
0 V
V1 V
( )in BV V Vγ= +
F.B.
R.B.
1 1LV I R−
Output voltage.
0 V
V1 V
( )1 BV V Vγ+ −
( )in BV V Vγ= −

92
RL
v0
+
-
+
-
R1
IL
vB
vin
Input voltage.
0 V
V1 V
Output voltage.
0 V
V1 V
( )in BV V Vγ= −
R.B.
F.B.
Negative shunt clipper
with positive shift.
RL
v0
+
-
+
-
R1
IL
vB
vin
Input voltage.
0 V
V1 V
Output voltage.
0 V
V1 V
in BV V Vγ=− +
R.B.
F.B.
Positive shunt clipper
with negative shift.

93
RL
v0
+
-
+
-
R1
IL
vB
vin
Input voltage.
0 V
V1 V
Output voltage.Negative shunt clipper.
Input voltage.
0 V
V1 V
Output voltage.Negative shunt clipper.
RL
v0
+
-
+
-
R1
vB1
vin
vB2
D1 D2
0 V
V1 V
( )2 2in BV V Vγ=− +
D1 in F.B.
D2 in F.B.
( )1 1γ= +in BV V V
0 V
V1 V
( )in BV V Vγ=− +
R.B.
F.B.

94
Clampers
• A clamping circuit (dc restorer) shifts the entire signal by a dc value.
• Change the dc voltage level but does not affect its shape.
• Always use a capacitor.
Negative voltage clamping circuit.
0 V
V1 V
Input voltage.
v0
0 V
V1 V
-2V1 V
vin
v0
Output voltage
(ideal diode).
Some important points:
• Identify the charging and discharging path of the capacitor.
• Diode in forward bias: input voltage appears across the capacitor (charging).
• Diode in reverse bias: capacitor holds the voltage (discharging).
• Charging time constant >>discharging time constant.
• Total input swing = total output swing.
RL
+
-
vin
+
-
C
IL
Vγ
vC+ -

95
Positive Voltage Clamper
Positive voltage clamping circuit.Input voltage. Output voltage.
v0RL
+
-
vin
+
-
C
IL
Vγ
vC+ -
0 V
12 V
-12 V
-0.7 VRL
+
-
11.3 V +-
-12 V 0.7 V
23.3 V
-0.7 V
During the negative cycle. During the positive cycle.
23.3 VRL
+
-
11.3 V +-
12 V
5Tτ ≥
• Consider capacitor charging step first.

96
Biased Clamping Circuits
v0 = 4.3 V
+
-
5 V
0.7 V
-12 V
16.3 V- +
v0 = 28.3 V
+
-
5 V
0.7 V
12 V
16.3 V- +
Positive shunt clamper.Input voltage. Output voltage.
v0RL = 10 kΩ
+
-
vin
+
-
C = 4.7 μF IL
Vγ
5 V
0 V
12 V
2 mS
-12 V
28.3 V
2 mS
4.3 V
Equivalent circuit in negative
half-cycle.
Equivalent circuit in positive
half-cycle.
• Charging time const. = 0, discharging time const. = 47 mS<< time period = 2 mS.

97
Biased Clamping Circuits
v0 = -4.3 V
+
-
-5 V
0.7 V12 V
16.3 V+ -
v0 = -28.3 V
+
-
-5 V
-12 V
16.3 V+ -
Negative shunt clamper.Input voltage. Output voltage.
v0RL = 10 kΩ
+
-
vin
+
-
C = 4.7 μF IL
Vγ
-5 V
0 V
12 V
2 mS
-12 V
Equivalent circuit in positive
half-cycle.
Equivalent circuit in negative
half-cycle.
-28.3 V
2 mS
-4.3 V

Problems
98
Q1. (i) What type of filter is it? Calculate the
cutoff frequency of the filter.
(ii) The switch is closed at t = 0, calculate the
output voltage at t = 0 and at t = 10 mS.
10 kΩ
+
-
vo
1 μF
12 V 1 kΩ
(i) Lowpass filter.
Cutoff frequency = 175.1 Hz.
(ii) At t = 0, the capacitor is shorted.
Vo = 0 V.
Now, time constant = Req×C = 0.909 mS.
t ≥ 5τ
Vo = 1.09 V.
909 Ω
+
-
vo1 μF
10 kΩ
+
-
vo1 μF12 V
10 kΩ
• Calculate the time constant of the following
circuit.
The output is open-circuited.
time constant = R×C = 10 mS.

Problems
99
R
+
-
vcC Cvi
+
-
Q2. In the circuit, the capacitors are fully charged at t =
0- so that vi = 12 V (C = 47 μF, and R = 1 kΩ).
(i) The switch is closed at t = 0, calculate the time
when vc = 6 V.
(ii) Calculate the minimum power rating of the resistor.
(i) Ceq = 94 μF.
ln
65.2 .
eqt RC
C o
C
eq
o
v V e
v
t RC
V
t mS
−
=
⇒ =−
⇒ =
(ii). ( )
2
2
Power rating
12
1
1
144 .
peaki R
k
k
mW
=
 
= 
 
=

Problems
100
Q3. In the following circuit, a current source ii =
sin(2πft) mA with internal resistance Ri = 10 kΩ is
connected to a RC circuit. Calculate the output
voltages (magnitudes) at f = 10 kHz and 100 kHz.
Given that C = 2.2 nF, and RL = 10 kΩ.
Transform the current source into a voltage source vi.
( )
( )
( )
( )
( )
22 2
22 2
3
10
4
100
10sin 2 V.
Now,
1
10 , 1.382, 7.643 0.47
4.7sin 20 10 V.
13.82
100 , 0.4996 5sin 20 10 V.
1 764.3
L
L
L
L
L
i i i
R L
i
i L
R
L i L
i
R kHz
R
R kHz
i
v i R ft
v CR
v C R R
v
at kHz CR C R R
v
v t
v
at kHz v t
v
π
ω
ω
ω ω
π
π
∴ = =
=
+ +
∴ = + = ⇒ =
∴ = ×
∴ = = ⇒ = ×
+
Ri
+
-
v0
C
RL
ii

Problems
101
Q4. In the following circuit, a pulse of height V and
width a is applied at t = 0. Find an expression for the
current.
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( )
( ) ( )
0
.
1
Now, applying KVL, 0
0
1
1 1
Assuming 0 0,
1 1 1
Taking inverse Laplace transfo
in
t
c
c as
as as
c
v t V U t U t a
v i dt Ri V U t U t a
C
v I s V
R I s e
s Cs s
e V e
v I s VC
CRs R s CR s CR
−
−
− −
− −
−
∴ = − −  
+ += − −  
∴
 + + = − 
 −
= = = − 
+ + + 
∴
∫
( ) ( ) ( )
( )
rm,
.
t a CRt CRV
i t e U t e U t a
R
− −−
 = − − 
R
+
-
vcCV

Problems
102
Q5. The circuit is in steady state. The switch is
closed at t = 0. Find an expression for the vc.
( )
( )
( )
( ) ( )
( )
( )
( )
( )
0
2
0 .
3
But, for >0, looking from the capacitor terminal,
the Thevenin's voltage = .
2
1
Now, applying KVL, 0
2 2
0
,
2 2
2
,
2 3 2
3
2
c
t
c
c
v V
t
V
R V
i v i dt
C
v I sR V
I s
s Cs s
I sR V V
I s
s Cs s
V R
I s
s C
−
−
−
−
∴ =
+ + =
+ + =
⇒ + + =
⇒ =−
+
∫
( )
.
R
R
+
-
vc
C
V
R
R
R/2
+
-
vc
C
V/
2
Equivalent circuit for t >0.

Problems
103
( )
( ) ( ) ( )
( )
( ) ( )
( ) ( )
( ) ( )
( )
( ) ( )
( )
( )
0
2
Capacitor voltage for >0,
0 32
,
3 2
2 3
.
22
Now, expanding into partial fractions,
2 3
22
2 3
.
6
.
2 6 2
Taking in
c
c
s
s CR
c
t
v I s V RCV
V s
s Cs s s s CR
Vs V RC A B
s s CRs s CR
Vs V RC V
A
s CR
Vs V RC V
B
s
V V
V s
s s CR
−
=
= −
∴
= + = −
+  
+
= = +
++  
+
=
+  
+
=
∴ = +
+  
∴
( ) 2
verse Laplace transform,
, (for 0).
2 6
t CR
c
V V
v t e t−
=+ >
R
+
-
vc
C
V
R
R

Problems
104
( )
( ) ( ) ( )
( )
( ) ( )
( ) ( )
( ) ( )
( )
( ) ( )
( )
( )
0
2
Capacitor voltage for >0,
0 32
,
3 2
2 3
.
22
Now, expanding into partial fractions,
2 3
22
2 3
.
6
.
2 6 2
Taking in
c
c
s
s CR
c
t
v I s V RCV
V s
s Cs s s s CR
Vs V RC A B
s s CRs s CR
Vs V RC V
A
s CR
Vs V RC V
B
s
V V
V s
s s CR
−
=
= −
∴
= + = −
+  
+
= = +
++  
+
=
+  
+
=
∴ = +
+  
∴
( ) 2
verse Laplace transform,
, (for 0).
2 6
t CR
c
V V
v t e t−
=+ >
R
+
-
vc
C
V
R
R

Additional Questions
105
Q1. In the following circuit, charge on the capacitor is zero for t <0. R1 = 10 kΩ, R2 =
10 kΩ R3 = 1 kΩ C = 10 μF.
(i) At t = 0 Sec, the switch is closed. Find I1 and I2 at t = 0 and at t = 1 Sec.
(ii) The switch is reopened at t = 2 Sec. Find I1 and I2 at t = 2 Sec.
Answer:
( ) ( )
( ) ( )
1 1 2 3
2 3 2 3
(i) At 0, the capacitor isshorted.
0 9 || 0.825 mA. and
0 0.825 0.075 mA.
t
I R R R
I R R R
=
∴ = + =
= × + =
R1
+
-
vcC
9 V R2
R3I2
I1
( ) ( ) ( )1 2 1 2
= 60 1 Sec the capacitor isfully charged.
1Sec 1Sec 9 0.45 mA.
eqR C mS
I I R R
τ = << ⇒
∴ = = + =
( ) ( ) ( )1 2 2 3
(ii) At 2 Sec, left-hand part is open.
2 Sec 0 and 2 Sec 4.5 0.409 mA.
t
I I R R
=
∴ = = + =

106
Q2. A Series RC circuit is excited by a voltage source of variable frequency. The
output is taken across the R. Sketch the variation of the steady state transfer
function with angular frequency ω.
Hints: Obtain H(jω) and represent in magnitude and phase form.
ω
0.707
1.0
M(ω)
1/RC
Magnitude response
ω
900
θ(ω)
1/RC
450
Phase response
Q3. Initially the switch is connected to a and the circuit is
in steady state. The switch is moved from a to b at t = 0.
Find the current in the inductor. What is the power
dissipated in R at t = 0 and t = ∞?
R
iL
L6 V
a
b
Answer: 6 A, 0
6 A, 0
Rt L
i Re t
R t
−
=− ≥
<
36 W,
0 W.
RP R=
=

107
Q4. In the following circuit, the switch is closed at t = CR. Assuming that all currents
and voltages are zero at t = 0-, determine the output voltage for 0≤ t ≤∞.
Answer:
Q5. In the following circuit, the switch is connected to a. At t = 0, it is moved to
position b. Find an expression for the voltage v0 for t >0. Given that R1 = 2 kΩ, R2 = 1
kΩ, L1 = 2 mH, L2 = 1 mH.
Hints: ( )0 3 mA,i − =
( ) 6 3 2
0 3 10 , 0.t
v t e t− −
=− × >
v(t)
R
+
-
C
V0 U(t)
C
( )
( ) ( )0
0
0 , 0 ,
1 exp 2 V,
2
v t t CR
V
V e t CR CR CR t
e
= ≤ <
= − + − − ≤ < ∞  
R1
6 V
a b
R2L1
L2
vo
+
-( ) 0
0 0
1 2 20 0
Applying KCL at ,
1 1
0 0 0.
t t
b
v
i v dt v dt
L R L− −
−
   
+ + + + =   
   
   
∫ ∫
Answer:

108
Q6. In the following circuit, obtain the complex
transfer function and express them in magnitude
and phase forms. Obtain the 3 dB cutoff frequency
and the frequency at which the phase difference
between the output and input voltage is 450. (5+5)
R
+
-
vo
C
vin R
Q7. In the following circuit, the switch is
connected to a and the circuit is in steady state.
At t = 0, it is moved to position b. Find an
expression for the voltage v0 for t >0. Given that
R1 = 2 kΩ, R2 = 1 kΩ, L1 = 2 mH, L2 = 1 mH. (10)
R1
6 V
a b
R2L1
L2
vo
+
-
0
+1
-1
t
T/2 T 3T/2
f(t)
Q8. Express the following periodic function in
terms of step functions and determine the
corresponding Laplace transform. (5)

mkmandal@ece.iitkgp.ernet.in
Ph. – +91-3222-283550 (o)
Department of E. & E.C.E.
I.I.T. Kharagpur, 721302.
Thank you
?
109

Introduction to Electronics

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