Unit-I: RECTIFIERS AND FILTERSLectures required
VM Umale
Topic
Books
L3
Rectifiers and Filters
Rectifiers-
T1,T2
L2
Filters-
T2, T1
Dept. of Electronics and Telecommunication Engineering
1
Subject: Electronic Devices and Circuits
(a) Course Objectives:
 To provide an overview of the principles and operation of electronic
devices.
 To explore use of electronic devices for various applications in
electronic circuits.
 To analyze various electronic circuits.
(b) Course Outcomes(Proposed):
After successfully completing the course, the students will be able to
• CO1: Comprehend the knowledge of semiconductor devices(Diode,
BJT, JFET, MOSFET, UJT), rectifiers, filters, amplifiers &
oscillator circuits
• CO2: Understand basics of Diode, BJT, JFET, MOSFET, UJT,
Rectifier, filters, Amplifiers, Oscillators with analysis of their
characteristics and operational parameters.
• CO3: To understand Biasing, feedback concept, topologies and
their applications.
• CO4: Implement and analyze various electronic circuits.
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Dept. of Electronics and Telecommunication Engineering
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Subject: Electronic Devices and Circuits
Unit-1
PN junction diode: Formation of p-n junction, biasing the diode, current
equation and V-I characteristics of diode, static and dynamic resistance,
Analysis of Half Wave Rectifier (HWR), Full Wave Rectifier (FWR),
introduction to filters C, L,LC and CLC filters, working of diode as a
Switch, Zener diode and its application as voltage regulator.
Unit-2
Waveshaping: Analysis of RC low pass, and high pass filters for
Sinusoidal, Step, Pulse, Square signal, analysis of clipping and clamping
circuits using diodes.
Unit-3
Bipolar Junction Transistors: Operation of PNP and NPN transistor, CB,
CE and CC configurations with characteristics and parameters, transistor as
a switch, Transistor switching times, dc load line, transistor biasing
methods, bias stability, Introduction to voltage divider biased CE amplifiers
using h-parameter model.
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Dept. of Electronics and Telecommunication Engineering
3
Subject: Electronic Devices and Circuits
Unit-4
Feedback amplifiers: Feedback concept, effects of negative feedback,
basic feedback topologies Sinusoidal oscillators: Barkhausen’s criteria,
Hartley, Colpitts, RC Phase shift, Wein bridge and crystal oscillators.
Unit-5
Multistage Amplifiers: Need of multistage, direct coupled amplifier, RC
coupled amplifier, transformer coupled amplifier, emitter follower,
Darlington emitter follower, bootstrapping principle (analysis not
expected).
Unit-6
JFET: Theory, construction and characteristics: parameters (μ, gm & rd)
MOSFET: Theory, construction and characteristics of enhancement &
depletion type MOSFET. UJT: Theory, construction and characteristics;
UJT as relaxation oscillator.
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Dept. of Electronics and Telecommunication Engineering
4
Subject: Electronic Devices and Circuits
Text Books:
 David Bell: Electronic Devices and Circuits, Oxford University Press,
2010.
 Milliman and Halkias: Integrated Electronics, TMH, New Delhi.
References:





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Robert L.Boylestad, “Electronic Devices and Circuit theory”, Pearson Edu.
Floyd, “Electron Devices” Pearson Asia 5th Edition, 2001.
Donald A Neamen, “Electronic Circuit Analysis and Design” TMH, 3rd Ed.
R1: Allen Mottershead “Electronic Devices and Circuits, An Int”., PHI
R2: Dr. R. S. Sedha , “A textbook of- Applied Electronics”, S. Chand
Dept. of Electronics and Telecommunication Engineering
5
Subject: Electronic Devices and Circuits
Unit-1
PN junction diode: Formation of p-n junction, biasing the diode,
current equation and V-I characteristics of diode, static and dynamic
resistance, Analysis of Half Wave Rectifier (HWR), Full Wave Rectifier
(FWR), introduction to filters C, L,LC and CLC filters, working of diode
as a Switch, Zener diode and its application as voltage regulator.
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Dept. of Electronics and Telecommunication Engineering
6
Subject: Electronic Devices and Circuits
Unit1: PN diode-
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Part-I
 Formation of p-n junction
 Biasing the diode
 Junction Voltages and current equation
 V-I characteristics of diode
 Static and dynamic resistance
Part-II
 Rectifiers:
 HWR,
 FWR,
 Theory of filters
 C, L,
 LC and CLC
 Analysis of C-input filter,
 Zener diodes & its application as voltage regulator,
Dept. of Electronics and Telecommunication Engineering
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Rectifiers
Prerequisite
 Fundamentals of PN Junction Diode
 Biasing and Operation of the Diode
 VI Characteristics of Diode

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Ideal and Practical Parameters
Dept. of Electronics and Telecommunication Engineering
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Subject: Electronic Devices and Circuits
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Dept. of Electronics and Telecommunication Engineering
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Rectifiers
 Importance of rectifiers
 Characteristics of rectifiers
 Classification(Types) of rectifiers
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
Operation of the circuits

Analysis of rectifier circuits(various terms,
factors or parameters (with
mathematical expression)etc

Merits/Demerits/Applications

Comparison among rectifiers

Numerical
Dept. of Electronics and Telecommunication Engineering
10
Rectifiers:
 Analysis of rectifier circuits(various terms, factors
or parameters(with mathematical expression) etc
 DC(Av.) Value of Current (Idc) & Voltage(Vdc)
 Ac(rms) value of current(Irms)& Voltage(Vrms)
 Ripple factor, Form factor
 DC /AC power output/Input(Pdc/Pac)
 Ratio of rectification(Efficiency)
 PIV and TUF
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Rectifiers
 Importance of rectifiers
DC power supply

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Rectifiers
 Importance of rectifiers
 Characteristics of rectifiers
 Classification(Types) of rectifiers
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
Operation of the circuits

Analysis of rectifier circuits(various terms,
factors or parameters (with
mathematical expression)etc

Merits/Demerits/Applications

Comparison among rectifiers

Numerical
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Rectifiers
Characteristics of rectifiers:
 Output waveforms
 Efficiency
 Ripple factor
 Regulation
 Peak voltage/Current
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Dept. of Electronics and Telecommunication Engineering
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Rectifiers
Characteristics of rectifiers
 Waveforms of the Load current/Load Voltage
 Rectifier converts ac to pulsating dc. Important to analyze
the nature of the current through load which ultimately
determines the waveforms of the load voltage
 Regulation of the output Voltage
 As load current changes, load voltage changes, so study the
effect of changes in input voltage and or load current on the
load voltage
 Rectifier efficiency
 How efficiently the rectifier circuit converts ac power into
dc power
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Dept. of Electronics and Telecommunication Engineering
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Rectifiers
Characteristics of rectifiers…..
 Peak value of current in the rectifier circuit
 Maximum value of an ac current in the rectifier circuit, this
decides the rating of the rectifier circuit element which is
diode.
 Peak value of voltage across the rectifier
element in Reverse biased(PIV)
 When diode is not conducting, the reverse voltage gets
applied across the diode. The peak value of such voltage
decides the peak inverse voltage i.e. PIV rating of a diode or
rectifying element
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Types of Rectifiers
Half wave Rectifier
Full wave Rectifier
2 Diodes (using CT Transformer)
4 Diodes ( Bridge)
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Rectifiers
 Half wave rectifier:

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Half wave Rectifiers
 Nature of waveforms:

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Rectifiers
 Full wave rectifier:
 2 Diodes (With Centre Tap Transformer)
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Full wave Rectifiers2 Diodes (With Centre Tap Transformer)
Current flow during
positive half cycle:
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Current flow during
Negative half cycle:
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Full wave Rectifiers
 Nature of waveforms:

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Rectifiers
 Full wave rectifier: 4 Diodes ( Bridge type)
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Bridge Rectifier: Operation of the Circuit
Current flowing
during Positive
Half Cycle
Current flowing
during Negative
Half Cycle
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Dept. of Electronics and Telecommunication Engineering
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Full wave Rectifier
 Nature of waveforms:

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Rectifiers
 Importance of rectifiers
 Characteristics of rectifiers
 Classification(Types) of rectifiers
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
Operation of the circuits

Analysis of rectifier circuits(various terms,
factors or parameters (with
mathematical expression)etc

Merits/Demerits/Applications

Comparison among rectifiers

Numerical
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Rectifiers : Analysis of rectifier circuits
 Half wave Rectifier
 Full wave Rectifier(with CTT, 2 diodes)
 Full wave Rectifier(Bridge type, 4 diodes)
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Analysis of rectifier circuits :HWR
 Idc , Vdc
 Irms, Vrms
 Im, Vm
 r = Ripple factor, f=Form factor
 Pdc , Pac
 η = Efficiency (Ratio of rectification)
 Regulation
 PIV
TUF
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Dept. of Electronics and Telecommunication Engineering
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Analysis of rectifier circuits :HWR
 Vs = Vmsinωt,
N2/N1 = Vs/ Vp
 Is = IL = Imsinωt,
 ω = 2πf
 Idc – DC or Average value of load current:
It is defined as the area under the
curve over one complete cycle i.e.
from 0 to 2π
1 2
Idc 
 ILd (t )
2
0
Idc = Im/ π
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Vdc = Vm/ π
Dept. of Electronics and Telecommunication Engineering
29
Analysis of rectifier circuits :HWR

Average or DC value of Current(Idc) or Voltage(Vdc)
1 2
Idc 
 ILd (t )
2 0
Where IL = Imsinωt
 sin = -cos
1 2
Idc 
 Im sin (t )d (t )
2 0
signal is available for half cycle
1 
Idc 
 Im sin (t )d (t )
2 0
Idc = Im/π
Vdc = Idc*RL
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Vdc = Vm/π
Im = Vm/Rs+Rf+RL
Rs+Rf << RL
Dept. of Electronics and Telecommunication Engineering
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Analysis of rectifier circuits :HWR

AC or rms value of Current(Irms) or Voltage(Vrms):
(rms means squaring, finding mean and then getting sq. root)
2
Irms 
1 / 2  (Im sin t ) 2 d (t )
0
2
Irms 
1 / 2  (Im 2 sin 2 td (t ))
0
Where IL = Imsinωt
sin2 = (1-cos2ωt)/2
signal is available for half cycle

Irms 
1 / 2  (Im 2 sin 2 td (t ))
0
Irms  Im
1 / 2 {t / 2  sin(2t ) / 4}0
Irms = Im/2
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Vrms = Vm/2
Im = Vm/Rs+Rf+RL
Dept. of Electronics and Telecommunication Engineering
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Analysis of rectifier circuits :FWR
Ripple factor- The pulsating components present in the output
of rectifier unit is called ripples and the measure of such
ripples present in the output is referred as Ripple factor(γ). It
tells how smooth is the output, smaller the γ, closer is the
output to a pure DC
rms value of the ac component(Iac = I’rms)
γ=
Average or dc component(Idc)
=
( Irms 2  Idc 2 )
Idc 2
=
Irms2
1
2
Idc
γ=
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2
4
1
Irms = Im/ 2
Idc = Im/π
γ = 1.21 or 121%
Dept. of Electronics and Telecommunication Engineering
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Analysis of rectifier circuits :HWR
DC Power output – Pdc
AC Power input – Pac
Pdc = Vdc * Idc
Pac = Vrms * Irms
= Idc2 RL
= Irms2 ( Rs+Rf + RL)
= (Im/π)2 RL
= (Im/2)2 ( Rs+Rf + RL)
where Im = Vm/ ( Rs+Rf + RL)
where Im = Vm/ ( Rs+Rf + RL)
Vm2
Vm2 RL
Pac =
Pdc =
2
4
π ( Rs+Rf + RL)
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Ratio of rectification or Efficiency (η )
Pdc = (Im/π)2 RL
2 ( R +R + R )
(I
m
/2)
4
s
f
L
Pac
η=
π2 (( Rs+Rf )/ RL)+1)
4 RL
=
π2 ( Rs+Rf + RL)
( Rs+Rf ) << RL)
4
2 = 0.4050r 40.5%
=
η
=
4/
π
π2 ( (Rs+Rf )/ RL)+1)
33
Dept. of Electronics and Telecommunication Engineering
Analysis of rectifier circuits :HWR
Transformer Utilization Factor(TUF)- It indicates how much
is the utilization of the transformer in the circuit, it is defined
as the ratio of dc power delivered to the load to the rated ac
power from secondary ( ac power rating of the transformer)
dc power delivered to the load(Idc2 RL)
TUF =
rated ac power(Vrms * Irms )
=
(Im/π)2 RL
=
(Vm / 2 )* Im /2
(Im/π)2 RL
=
(Vm / 2 )* Im /2
TUF =
2 2
(π)2
Irms = Im/2
Idc = Im/π
Vm = Im(Rs+Rf + RL)
But Rs+Rf << RL
Vm = ImRL
TUF = 0.287
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Dept. of Electronics and Telecommunication Engineering
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Disadvantages of HWR
1. Ripples in HWR is 121%, which is quite high
2. The maximum efficiency(η) is 40.8% is even
less than 50%, HWR is quite inefficient
3. TUF = 0.287, the transformer is not fully
utilized
4. The dc current flowing through secondary
winding of the transformer which may cause
dc saturation of the core of the transformer
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Dept. of Electronics and Telecommunication Engineering
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Analysis of rectifier circuits :HWR
Parameter
HWR
Idc
Im/π
Vdc
Vm/π
Irms
Im/2
Vrms
Vm/2
f=Form factor
1.57
Irms /Idc = Vrms/Vdc
r = Ripple factor
1.21
( Irms / Idc) 2  1
Pdc = Idc2 RL
( Im / π)2 RL
Pac = Irms2(Rs+Rf+RL)
Im2(Rs+Rf+RL)/4
Efficiency- η= Pdc/Pac
4 /π2 = 40.6%
PIV
2Vm
0.287
TUF
 Im = Vm/Rs+Rf+RL
Regulation(VNL- VFL)/VFL (RS+Rf)/RL ,~Rf/RL
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2
Analysis of rectifier circuits :FWR
iL = Imsinωt ….
0  ωt  π
iL = -Imsinωt ….
1 2
π  ωt  2π
Idc 
2
 ILd (t )
0
signal is available for both half cycle
1 
Idc 
[  Im sin (t )d (t )    Im sin(t )d (t )]
2 0

2
Idc = 2Im/π
Where IL = Imsinωt
 sin = -cos
Vdc = 2Vm/π
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Dept. of Electronics and Telecommunication Engineering
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Analysis of rectifier circuits :FWR
Vdc = 2Vm/π------ No load voltage
Average DC load voltage
Vdc = Idc RL
= 2ImRL / π
= 2VmRL / π(Rs+Rf+ RL)
= 2Vm / π [(Rs+Rf /RL) +1]…..full load voltage
Rs+Rf << RL (Rs+Rf /RL) << 1
Vdc = 2Vm/π------ No load voltage
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Dept. of Electronics and Telecommunication Engineering
38
Analysis of rectifier circuits :FWR

AC or rms value of Current(Irms) or Voltage(Vrms)
2
1 / 2  (Im sin t ) d (t )
Irms 
2
signal is available for both half cycle
0


2 / 2  (Im2 sin 2 td (t ))
0


1 /   [(1  cos 2t ) / 2 ) d (t )]
Where IL = Imsinωt
sin2 ωt = (1-cos2ωt)/2
0
 Im
Irms  Im
1 / 2 [t ]0  [sin(2t ) / 2]0
(1 / 2 ) *[(  0)  (0  0)]
Irms  Im / 2
Irms = Im/ 2
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Vrms = Vm/ 2
Im = Vm/Rs+Rf+RL
Dept. of Electronics and Telecommunication Engineering
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Analysis of rectifier circuits :FWR
Ripple factor- The pulsating components present in the output
of rectifier unit is called ripples and the measure of such
ripples present in the output is referred as Ripple factor(γ). It
tells how smooth is the output, smaller the γ, closer is the
output to a pure DC
rms value of the ac component(Iac = I’rms)
γ=
Average or dc component(Idc)
=
( Irms 2  Idc 2 )
Idc 2
=
Irms2
1
2
Idc
γ=
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2
8
1
Irms = Im/ 2
Idc = 2Im/π
γ = 0.48 or 48%
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Analysis of rectifier circuits :FWR
DC Power output – Pdc
AC Power input – Pac
Pdc = Vdc * Idc
Pac = Vrms * Irms
= Idc2 RL
= Irms2 ( Rs+Rf + RL)
= (2Im/π)2 RL
= (Im/ 2 )2 ( Rs+Rf + RL)
where Im = Vm/ ( Rs+Rf + RL)
where Im = Vm/ ( Rs+Rf + RL)
Vm2
4Vm2 RL
Pac =
Pdc =
2
2 ( Rs+Rf + RL)
π ( Rs+Rf + RL)
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Ratio of rectification or Efficiency (η )
Pdc = (2Im/π)2 RL
2 ( R +R + R )
(I
m
/
)
2
8
s
f
L
Pac
η=
π2 (( Rs+Rf )/ RL)+1)
8 RL
=
π2 ( Rs+Rf + RL)
( Rs+Rf ) << RL)
8
2 = 0.812 0r 81.2%
η
=
8/
π
=
π2 ( (Rs+Rf )/ RL)+1)
41
Dept. of Electronics and Telecommunication Engineering
Analysis of rectifier circuits :FWR
Voltage regulation = %R
%R = [(Vdc)NL-(Vdc)FL*]/ (Vdc)FL
(Vdc)NL = (2Vm/π), (Vdc)FL = Idc RL
%R = [(2Vm/π) - Idc RL /Idc RL]*100
where Vm = Im*( Rs+Rf + RL)
and Idc = 2Im / π
%R =
%R =
%R =
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2Im/π(Rs+Rf + RL) - 2Im/π* RL]*100
2Im/π* RL
(Rs+Rf + RL) – RL]
*100
RL
(Rs+Rf )
*100
RL
( Rs << RL)
Rf
%R = R *100
L
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2
Analysis of rectifier circuits :FWR
Transformer Utilization Factor(TUF)- It indicates how much
is the utilization of the transformer in the circuit, it is defined
as the ratio of dc power delivered to the load to the rated ac
power from secondary ( ac power rating of the transformer)
dc power delivered to the load(Idc2 RL)
TUF =
rated ac power(Vrms * Irms )
Irms = Im/
=
=
=
TUF =
(2Im/π)2 RL
(Vm / 2 )* (Im / 2 )
(2Im/π)2 RL
(Im.RL / 2 )* Im / 2
4*2
(π)2
TUF (sec) = 0.812
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2
Idc =2Im/π
Vm = Im(Rs+Rf + RL)
But Rs+Rf << RL
Vm = ImRL
TUF(pri) = 2(TUF of HWR) = 2*0.287 = 0.574
Average TUF for FWR = (0.574+0.812) /2
= 0.693
Dept. of Electronics and Telecommunication Engineering
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Analysis of rectifier circuits :FWR
Parameter
FWR
Idc
2Im/π
Vdc
2Vm/π
Irms
Im/ 2
Vrms
Vm/ 2
f=Form factor
1.11
Irms /Idc = Vrms/Vdc
r = Ripple factor
0.48
( Irms / Idc) 2  1
Pdc = Idc2 RL
(2Im / π)2 RL
Pac = Irms2(Rs+Rf+RL)
Im2(Rs+Rf+RL)/2
Efficiency- η= Pdc/Pac
8 /π2 = 81.2%
PIV
Vm
(0.574+0.812)/2
TUF
 Im = Vm/R
+RL (R’S+Rf)/RL ,~Rf/RL
s+R
Regulation(V
NL- VFL)/
VfFL
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Dept. of Electronics and Telecommunication Engineering
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2
Comparison of HWR, FWR(CT), FWR(Bridge) Ckts
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Comparison of HWR, FWR(CT), FWR(Bridge)
Ckts
Parameter
HWR
FWR(CT)
FWR(Bridge)
Idc
Im/π
2Im/π
2Im/π
Vdc
Vm/π
2Vm/π
2Vm/π
Irms
Im/2
Im/ 2
Vrms
Vm/2
Vm / 2
Im/ 2
Vm / 2
f=Form factor
1.57
1.11
1.11
( Irms / Idc) 2  1
1.21
0.48
0.48
Pdc = Idc2 RL
( Im / π)2 RL
(2Im / π)2 RL
(2Im / π)2 RL
Pac = Irms2(Rs+Rf+RL)
Im2(Rs+Rf+RL)/4 Im2(Rs+Rf+RL)/2
Irms /Idc = Vrms/Vdc
r = Ripple factor
Efficiency- η= Pdc/Pac 4 /π2 = 40.6%
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8 /π2 = 81.2%
Im2*(Rs+Rf+RL)/2
8 /π2 = 81.2%
PIV
2Vm
Vm
Vm
TUF
 Im = Vm/Rs+Rf+RL
%Reg
0.287
(0.574+0.812)/2
0.812
(Rs +Rf)/RL
(Rs/2 +Rf)/RL
(Rs +2Rf)/RL 46
Dept. of Electronics and Telecommunication Engineering
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Comparison of HWR, FWR(CT), FWR(Bridge)
Ckts
Parameterio
HWR
FWR(CT)
FWR(Bridge)
1.57
f=Form factor
1.11
1.11
Im/ 2
Im/ 0.48
2
Vm
2 / 2
Irms /Idc = Vrms/Vdc
r = Ripple factor
1.21
Efficiency- η= Pdc/Pac 4 /π2 = 40.6%
0.48
Vm / 2
8 /π2 = 81.2%
8 /π = 81.2%
PIV
2Vm
Vm
Vm
TUF
0.287
(0.574+0.812)/2
0.812
1
F
2
2f
4
2f
(Rs +Rf)/RL
(Rs/2 +Rf)/RL
(Rs +2Rf)/RL
( Irms
)2 1
No.
of/ Idc
Diodes,
frequency
%Reg
 Im = Vm/Rs+Rf+RL
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2
Comparison between HWR and FWR(CT)
 DC load current or Average value of load current in FWR
is double to that in HWR, similarly for VDC
 Ripple frequency is double in FWR(2f) as compare to
HWR(f), smaller value of filter is suitable in FWR as
compared to HWR and hence cost of filter circuit is less ,
also ripples are less in the output of FWR(48%) than that of
HWR(121%)
 No net DC current flowing through secondary of
transformer in FWR so losses are less in FWR as compared
to HWR, so danger of DC saturation of secondary winding
is removed in FWR
 DC power output is four times larger in FWR than HWR,
so ratio of rectification or efficiency of FWR is double as
compared to HWR
48
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Dept. of Electronics and Telecommunication Engineering
2
Advantages of Bridge rectifier Circuit
 The current in both pri. And sec. of the power transformer flows
for the entire cycle and hence for a given power output, power
transformer of a small size may be used so circuit cost reduced
 No centre tap transformer is required in the transformer sec.,
hence ac voltage can directly be applied to the bridge
 The current in the sec of the transformer is in opposite direction
in two half cycles. Hence net dc current flowing through sec.
winding is zero which reduces the losses and danger of saturation
 Due to pure ac current in the sec. of the transformer , the
transformer gets utilized effectively and hence the circuit is
suitable for application where large power is required
 As two diodes conduct in series in each half cycle, inverse voltage
appearing across diodes get shared .So PIV rating of diodes is
reduced or circuit can be used for high voltage application.
VM Umale
Dept. of Electronics and Telecommunication Engineering
49
2
Assignment on topic- rectifier circuits
2U1
Q1- a diode is used as rectifier and supplied from a 230V, 50Hz ac mains with a
step down transformer ratio of 3:1 to a restive load of 5KΩ. The diode forward
resistance is of 50Ω while transformer secondary resistance is 5 Ω.
a) What is the no-load dc output voltage of the rectifier?
b) What is the dc output voltage when full load draws a dc current of 100mA?
c) What is the % Regulation of this power supply.
d) Also Determine maximum, Average and rms values of current and voltage,
ripple factor, dc output power, ac input power, ratio of rectification in % ,
Transformer utilization factor(TUF) and PIV.
VM Umale
Q2- A full wave rectifier circuit is fed from step down transformer having centre
tapped secondary winding. The rms voltage from either end of secondary to
centre tap is 18V. The diode forward resistance is 20Ω and that of the half
secondary winding resistance is 2Ω , for a restive load of 1KΩ.
a) What is the internal resistance this power supply?
b) What is the % regulation of this power supply?
c) Also Calculate maximum, Average and rms values of current and voltage,
ripple factor, dc output power, ac input power, % Efficiency (η) ,
Transformer utilization factor(TUF of Primary, Secondary and Average) and
50
PIV.
Dept. of Electronics and Telecommunication Engineering
Assignment on topic- rectifier circuits
2U2
Q1- A half wave rectifier consist of a diode having a dynamic resistance of 1Ω
as its operating current and a transformer whose open circuit secondary voltage
is 12.6V, 50Hz. It has a secondary resistance of 3Ω
a) What is the no-load dc output voltage of the rectifier?
b) What is the dc output voltage when full load draws a dc current of 100mA?
c) What is the % Regulation of this power supply.
d) Also Determine maximum, Average and rms values of current and voltage,
ripple factor, dc output power, ac input power, % Efficiency (η), Transformer
utilization factor(TUF) and PIV.
Q2- A full wave rectifier circuit is fed from step down transformer having centre
tapped secondary winding. The rms voltage from either end of secondary to
centre tap is 22V. The diode forward resistance is 20Ω and that of the half
secondary winding resistance is 2Ω , for a restive load of 2KΩ.
a) What is the internal resistance this power supply?
b) What is the % regulation of this power supply?
c) Also Calculate maximum, Average and rms values of current and voltage,
ripple factor, dc output power, ac input power, ratio of rectification in % ,
TUF of Primary, Secondary and Average) and PIV.
VM Umale
Dept. of Electronics and Telecommunication Engineering
51
γαβδηπλΩθω
THANKS A LOT
VM Umale
Dept. of Electronics and Telecommunication Engineering
52
Filter circuits:
 Capacitor Filter circuit(C )
 Inductor Filter circuit (L)
 Choke Input Filter(LC or L )
 Capacitor Input filter( CLC or π)
VM Umale
Dept. of Electronics and Telecommunication Engineering
53
Analysis of rectifier circuits :HWR
VM Umale
Dept. of Electronics and Telecommunication Engineering
54
Thanks a lot
vm umale
Dept. of Electronics and Telecommunication Engineering
55