The document discusses regulated DC power supplies and their components. It explains that a regulated DC power supply consists of a step-down transformer, rectifier, filter, and voltage regulator. The transformer steps down AC voltage, the rectifier converts it to DC, the filter smooths the output, and the regulator sets the output to a fixed voltage. It then discusses half-wave and full-wave rectifiers in detail, deriving their key parameters such as DC output voltage and current, ripple factor, and efficiency.
Three Phase Controlled Rectifier Study in Terms of firing angle variationsIDES Editor
This paper introduce topology of three phase
controlled rectifiers and proposed an accurate Statistical
method to calculate their input current harmonic components,
and calculate THD and harmonic currents with accurate
simulation in various firing angles, then investigate influence
of load variations in terms of firing angle variations on
harmonic currents. Finally a harmonic current database of
rectiûers is obtained in terms of firing angle and load
variations.
Three Phase Controlled Rectifier Study in Terms of firing angle variationsIDES Editor
This paper introduce topology of three phase
controlled rectifiers and proposed an accurate Statistical
method to calculate their input current harmonic components,
and calculate THD and harmonic currents with accurate
simulation in various firing angles, then investigate influence
of load variations in terms of firing angle variations on
harmonic currents. Finally a harmonic current database of
rectiûers is obtained in terms of firing angle and load
variations.
Introduction
Working Principle
Step Down and Step Up Cycloconverter
Single phase to single phase cycloconverter
Mid-Point and Bridge type cycloconverter
Advantages and disadvantages
Applications
Design, Modeling and control of modular multilevel converters (MMC) based hvd...Ghazal Falahi
Modular multilevel converter (MMC) is a relatively new and promising topology, which has gained a lot of interest in industry in the recent years due to its modular design and easy adaption for applications that require different power and voltage level, such as power transmission through HVDC. This presentation investigates the operation of MMC based HVDC systems and proposes new solutions to improve the performance of the system by using new devices and improving the control strategies.
The manual is useful for PG students belongs to ME power Electronics and Drives
By
M.MURUGANANDAM. M.E.,(Ph.D).,MIEEE.,MISTE,
Assistant Professor & Head / EIE,
Muthayammal Engineering College,
Rasipuram,
Namakkal-637 408.
Cell No: 9965768327
The power electronics device which converts DC power to AC power at required output voltage and frequency level is known as inverter. Multilevel inverter is to synthesize a near sinusoidal voltage from several levels of dc voltages. In order to maintain the different voltage levels at appropriate intervals, the conduction time intervals of MOSFETS have been maintained by controlling the pulse width of gating pulses. In this paper single phase to three phase power conversion using PWM technique. The simulation is carried out in MATLAB/Simulink environment which demonstrate the feasibility of proposed scheme.
Soft Switched Multi-Output Flyback Converter with Voltage DoublerIJPEDS-IAES
A novel multi-output voltage doubler circuit with resonant switching
technique is proposed in this paper. The resonant topology in the primary
side of the flyback transformer switches the device either at zero voltage or
current thus optimizing the switching devices by mitigating the losses. The
voltage doubler circuit introduced in the load side increases the voltage by
twice the value thereby increasing the load power and density. The proposed
Multi-output Isolated Converter removes the need for mutiple SMPS units
for a particular application. This reduces the size and weight of the
converters considerably leading to a greater payload. This paper aims at
optimizing the proposed converter with some design changes. The results
obtained from the hardware prototype are given in a comprehensive manner
for a 3.5W converter operating at output voltages of 5V and 3.3V at 50 kHz
switching frequency. The converter output is regulated with the PI controller
designed with SG3523 IC. The effects of load and line regulation for ±20%
variations are analyzed in detail.
Power Amplifiers: Classification of amplifiers, Class A power Amplifiers and their analysis, Harmonic Distortions,
Class B Push-pull amplifiers and their analysis, Complementary symmetry push pull amplifier, Class AB power
amplifier, Class-C power amplifier, Thermal stability and Heat sinks, Distortion in amplifiers.
Tuned Amplifiers : Introduction, Q-Factor, small signal tuned amplifier, capacitance single tuned amplifier, double
tuned amplifiers, effect of cascading single tuned amplifiers on band width, effect of cascading double tuned
amplifiers on band width, staggered tuned amplifiers, stability of tuned amplifiers, wideband amplifiers
Introduction
Working Principle
Step Down and Step Up Cycloconverter
Single phase to single phase cycloconverter
Mid-Point and Bridge type cycloconverter
Advantages and disadvantages
Applications
Design, Modeling and control of modular multilevel converters (MMC) based hvd...Ghazal Falahi
Modular multilevel converter (MMC) is a relatively new and promising topology, which has gained a lot of interest in industry in the recent years due to its modular design and easy adaption for applications that require different power and voltage level, such as power transmission through HVDC. This presentation investigates the operation of MMC based HVDC systems and proposes new solutions to improve the performance of the system by using new devices and improving the control strategies.
The manual is useful for PG students belongs to ME power Electronics and Drives
By
M.MURUGANANDAM. M.E.,(Ph.D).,MIEEE.,MISTE,
Assistant Professor & Head / EIE,
Muthayammal Engineering College,
Rasipuram,
Namakkal-637 408.
Cell No: 9965768327
The power electronics device which converts DC power to AC power at required output voltage and frequency level is known as inverter. Multilevel inverter is to synthesize a near sinusoidal voltage from several levels of dc voltages. In order to maintain the different voltage levels at appropriate intervals, the conduction time intervals of MOSFETS have been maintained by controlling the pulse width of gating pulses. In this paper single phase to three phase power conversion using PWM technique. The simulation is carried out in MATLAB/Simulink environment which demonstrate the feasibility of proposed scheme.
Soft Switched Multi-Output Flyback Converter with Voltage DoublerIJPEDS-IAES
A novel multi-output voltage doubler circuit with resonant switching
technique is proposed in this paper. The resonant topology in the primary
side of the flyback transformer switches the device either at zero voltage or
current thus optimizing the switching devices by mitigating the losses. The
voltage doubler circuit introduced in the load side increases the voltage by
twice the value thereby increasing the load power and density. The proposed
Multi-output Isolated Converter removes the need for mutiple SMPS units
for a particular application. This reduces the size and weight of the
converters considerably leading to a greater payload. This paper aims at
optimizing the proposed converter with some design changes. The results
obtained from the hardware prototype are given in a comprehensive manner
for a 3.5W converter operating at output voltages of 5V and 3.3V at 50 kHz
switching frequency. The converter output is regulated with the PI controller
designed with SG3523 IC. The effects of load and line regulation for ±20%
variations are analyzed in detail.
Power Amplifiers: Classification of amplifiers, Class A power Amplifiers and their analysis, Harmonic Distortions,
Class B Push-pull amplifiers and their analysis, Complementary symmetry push pull amplifier, Class AB power
amplifier, Class-C power amplifier, Thermal stability and Heat sinks, Distortion in amplifiers.
Tuned Amplifiers : Introduction, Q-Factor, small signal tuned amplifier, capacitance single tuned amplifier, double
tuned amplifiers, effect of cascading single tuned amplifiers on band width, effect of cascading double tuned
amplifiers on band width, staggered tuned amplifiers, stability of tuned amplifiers, wideband amplifiers
“Microcontroller Based Substation Monitoring system with gsm modem”.Priya Rachakonda
• The system is used for transmitting the message to predefined number about the
status of electrical parameters such as voltage, current, temperature etc., to improve
the quality of power.
• Studied about the protection, monitoring and control of a power system.
Cosmetic shop management system project report.pdfKamal Acharya
Buying new cosmetic products is difficult. It can even be scary for those who have sensitive skin and are prone to skin trouble. The information needed to alleviate this problem is on the back of each product, but it's thought to interpret those ingredient lists unless you have a background in chemistry.
Instead of buying and hoping for the best, we can use data science to help us predict which products may be good fits for us. It includes various function programs to do the above mentioned tasks.
Data file handling has been effectively used in the program.
The automated cosmetic shop management system should deal with the automation of general workflow and administration process of the shop. The main processes of the system focus on customer's request where the system is able to search the most appropriate products and deliver it to the customers. It should help the employees to quickly identify the list of cosmetic product that have reached the minimum quantity and also keep a track of expired date for each cosmetic product. It should help the employees to find the rack number in which the product is placed.It is also Faster and more efficient way.
Final project report on grocery store management system..pdfKamal Acharya
In today’s fast-changing business environment, it’s extremely important to be able to respond to client needs in the most effective and timely manner. If your customers wish to see your business online and have instant access to your products or services.
Online Grocery Store is an e-commerce website, which retails various grocery products. This project allows viewing various products available enables registered users to purchase desired products instantly using Paytm, UPI payment processor (Instant Pay) and also can place order by using Cash on Delivery (Pay Later) option. This project provides an easy access to Administrators and Managers to view orders placed using Pay Later and Instant Pay options.
In order to develop an e-commerce website, a number of Technologies must be studied and understood. These include multi-tiered architecture, server and client-side scripting techniques, implementation technologies, programming language (such as PHP, HTML, CSS, JavaScript) and MySQL relational databases. This is a project with the objective to develop a basic website where a consumer is provided with a shopping cart website and also to know about the technologies used to develop such a website.
This document will discuss each of the underlying technologies to create and implement an e- commerce website.
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
Hierarchical Digital Twin of a Naval Power SystemKerry Sado
A hierarchical digital twin of a Naval DC power system has been developed and experimentally verified. Similar to other state-of-the-art digital twins, this technology creates a digital replica of the physical system executed in real-time or faster, which can modify hardware controls. However, its advantage stems from distributing computational efforts by utilizing a hierarchical structure composed of lower-level digital twin blocks and a higher-level system digital twin. Each digital twin block is associated with a physical subsystem of the hardware and communicates with a singular system digital twin, which creates a system-level response. By extracting information from each level of the hierarchy, power system controls of the hardware were reconfigured autonomously. This hierarchical digital twin development offers several advantages over other digital twins, particularly in the field of naval power systems. The hierarchical structure allows for greater computational efficiency and scalability while the ability to autonomously reconfigure hardware controls offers increased flexibility and responsiveness. The hierarchical decomposition and models utilized were well aligned with the physical twin, as indicated by the maximum deviations between the developed digital twin hierarchy and the hardware.
Welcome to WIPAC Monthly the magazine brought to you by the LinkedIn Group Water Industry Process Automation & Control.
In this month's edition, along with this month's industry news to celebrate the 13 years since the group was created we have articles including
A case study of the used of Advanced Process Control at the Wastewater Treatment works at Lleida in Spain
A look back on an article on smart wastewater networks in order to see how the industry has measured up in the interim around the adoption of Digital Transformation in the Water Industry.
Industrial Training at Shahjalal Fertilizer Company Limited (SFCL)MdTanvirMahtab2
This presentation is about the working procedure of Shahjalal Fertilizer Company Limited (SFCL). A Govt. owned Company of Bangladesh Chemical Industries Corporation under Ministry of Industries.
Runway Orientation Based on the Wind Rose Diagram.pptx
Edc unit 2
1. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 1
UNIT – II RECTIFIERS & FILTERS
1. Explain about Regulated DC power supply.
For the operation of most of the electronics devices and circuits, a d.c. source is required.
So it is advantageous to convert domestic a.c. supply into d.c.voltages. The process of
converting a.c. voltage into d.c. voltage is called as rectification.
This is achieved with i) Step-down Transformer, ii) Rectifier, iii) Filter and iv) Voltage
regulator circuits.
These elements constitute d.c. regulated power supply shown in the figure below.
Fig. Block diagram of Regulated D.C. Power Supply
Transformer – steps down 230V AC mains to low voltage AC.
Rectifier – converts AC to DC, but the DC output is varying.
Smoothing – smooth the DC from varying greatly to a small ripple.
Regulator – eliminates ripple by setting DC output to a fixed voltage.
The block diagram of a regulated D.C. power supply consists of step-down transformer,
rectifier, filter, voltage regulator and load.An ideal regulated power supply is an electronics
circuit designed to provide a predetermined d.c. voltage Vo which is independent of the load
current and variations in the input voltage ad temperature. If the output of a regulator circuit is
a AC voltage then it is termed as voltage stabilizer, whereas if the output is a DC voltage then it
is termed as voltage regulator.The elements of the regulated DC power supply are discussed as
follows:
TRANSFORMER:
A transformer is a static device which transfers the energy from primary winding to
secondary winding through the mutual induction principle, without changing the frequency.
The transformer winding to which the supply source is connected is called the primary,
while the winding connected to the load is called secondary. If N1,N2 are the number of turns of
the primary and secondary of the transformer then 2
1
N
N
is called the turns ratio of the
transformer.
The different types of the transformers are
1) Step-Up Transformer
2) Step-Down Transformer
3) Centre-tapped Transformer
The voltage, current and impedance transformation rations are related to the turns ratio of
the transformer by the following expressions.
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2. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 2
Voltage transformation ratio : 2 2
1 1
V N
V N
Current transformation ratio : 2 2
1 1
I N
I N
Impedance transformation ratio :
2
2
1
L
in
Z N
Z N
2. Explain Rectifier Characteristics.
RECTIFIER:
Any electrical device which offers a low resistance to the current in one direction but a
high resistance to the current in the opposite direction is called rectifier. Such a device is capable
of converting a sinusoidal input waveform, whose average value is zero, into a unidirectional
waveform, with a non-zero average component.
A rectifier is a device which converts a.c. voltage (bi-directional) to pulsating d.c.
voltage (Uni-directional).
Characteristics of a Rectifier Circuit:
1. Load currents: They are two types of output current. They are average or d.c. current
and RMS currents.
i) Average or DC current: The average current of a periodic function is
defined as the area of one cycle of the curve divided by the base.
It is expressed mathematically as
2
0
1
( )
2dcI id t ; where mI sini t
ii) Effective (or) R.M.S. current: The effective (or) R.M.S. current squared of
a periodic function of time is given by the area of one cycle of the curve
which represents the square of the function divided by the base.
It is expressed mathematically as
1
2
2
2
0
1 ( )
2rmsI i d t
2. Load Voltages: There are two types of output voltages. They are average or D.C.
voltage and R.M.S. voltage.
i) Average or DC Voltage: The average voltage of a periodic function is
defined as the areas of one cycle of the curve divided by the base.
It is expressed mathematically as
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3. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 3
2
0
1
( )
2dcV Vd t ; where m sinV V t
(or)
Ldc dc
V I R
ii) Effective (or) R.M.S Voltage: The effective (or) R.M.S voltage squared of
a periodic function of time is given by the area of one cycle of the curve
which represents the square of the function divided by the base.
1
2 2
2
0
1 ( )
2rmsV V d t
(or) rms rms LV I R
3. Ripple Factor ( ) : It is defined as ration of R.M.S. value of a.c. component to the d.c.
component in the output is known as “Ripple Factor”.
'
rms
dc
V
V
Where 2 2'
rms dcrmsV V V
2
1rms
dc
V
V
4. Efficiency ( ): It is the ratio of d.c output power to the a.c. input power. It signifies,
how efficiently the rectifier circuit converts a.c. power into d.c. power.
It is given by
dc
ac
P
P
5. Peak Inverse Voltage (PIV):It is defined as the maximum reverse voltage that a diode
can withstand without destroying the junction.
6. Regulation: The variation of the d.c. output voltage as a function of d.c. load current is
called regulation. The percentage regulation is defined as
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4. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 4
% Regulation = 100%no load full load
full load
V V
V
For an ideal power supply, % Regulation is zero.
Using one or more diodes in the circuit, following rectifier circuits can be designed.
1. Half - Wave Rectifier
2. Full – Wave Rectifier
3. Bridge Rectifier
3. Explain Half-wave Rectifier and derive all parameters.
Half-Wave Rectifier:
A Half – wave rectifier is one which converts a.c. voltage into a pulsating voltage using
only one half cycle of the applied a.c. voltage.
Fig. a Basic structure of Half-Wave Rectifier
The half-wave rectifier circuit shown in above figure consists of a resistive load, a
rectifying element i.e., p-n junction diode and the source of a.c. voltage, all connected is series.
The a.c. voltage is applied to the rectifier circuit using step-down transformer.
Fig.b Input and output waveforms of a Half wave rectifier
The input to the rectifier circuit,
m sinV V t
Where Vm is the peak value of secondary a.c. voltage.
Operation:
For the positive half-cycle of input a.c. voltage, the diode D is forward biased and hence
it conducts. Now a current flows in the circuit and there is a voltage drop across RL. The
waveform of the diode current (or) load current is shown in figure.
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5. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 5
For the negative half-cycle of input, the diode D is reverse biased and hence it does not
conduct. Now no current flows in the circuit i.e., i=0 and Vo=0. Thus for the negative half-
cycle no power is delivered to the load.
Analysis:
In the analysis of a HWR, the following parameters are to be analyzed.
i) DC output current ii) DC Output voltage
iii) R.M.S. Current iv) R.M.S. voltage
v) Rectifier Efficiency ( ) vi) Ripple factor ( )
vii) Regulation viii) Transformer Utilization Factor (TUF)
ix) Peak Factor (P)
Let a sinusoidal voltage Vi be applied to the input of the rectifier.
Then m sinV V t
Where Vm is the maximum value of the secondary voltage.
Let the diode be idealized to piece-wise linear approximation with resistance Rf in the
forward direction i.e., in the ON state and Rr(=∞) in the reverse direction i.e., in the OFF state.
Now the current „i‟ in the diode (or) in the load resistance RL is given by
mI sini t for 0 t
i=0 for 2t
where mI m
Lf
V
R R
i) Average (or) DC Output Current (Iav or Idc):
The average dc current Idc is given by
dc
I
2
0
1 ( )
2
id t
m
2
0
sin ( )
1 0 ( )
2
I td t d t
m 0
1 I ( cos )
2
t
m
1 I ( 1 ( 1)
2
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6. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 6
mI (or) 0.318 mI
Substituting the value of mI , we get I
Lf
m
dc
R R
V
If RL>>Rf then I m
dc
L
V
R
= 0.318 m
L
V
R
ii) Average (or) DC Output Voltage (Vav or Vdc):
The average dc voltage is given by
Ldc dcV I R = mI
LR =
Lf
m L
R R
V R
Lf
m L
dc R R
V R
V
If RL>>Rf then m
dc
V
V = 0.318 mI m
dc
V
V
iii) R.M.S. Output Current (Irms):
The value of the R.M.S. current is given by
rmsI
2
1
2
2
0
1 ( )
2
i d t
1
2
22
2
0
sin ( )
1 0 ( )
2
1I
2
.m
t d t d t
1
22
0
1 cos ( )
2
I
2
m t d t
1
22
0
I 1( ) sin
4 2
m t t
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GRIET-ECE G.Surekha Page 7
1
2 2
sin0
I sin20
4 2
m
1
2 2I
4
m
mI
2
mI
2rmsI (or)
2 Lf
m
rms
R R
VI
iv) R.M.S. Output Voltage (Vrms):
R.M.S. voltage across the load is given by
rms rms LV I R =
2 Lf
m L
R R
V R
=
2 1 f
L
m
R
R
V
If RL >> Rf then
2
m
rms
V
V
v) Rectifier efficiency ( ) :
The rectifier efficiency is defined as the ration of d.c. output power to the a.c.
input power i.e.,
dc
ac
P
P
2
2
2
m L
Ldcdc
I R
I RP
2
2
4rms
m
L Lf fac
I
I R R R RP
2
2 22
4 4m
m
L L
L fL f
dc
ac
R R
R RR R
P I
P I
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GRIET-ECE G.Surekha Page 8
2
1 0.406
11
4
ff
LL
RR
RR
%
40.6
1 f
L
R
R
Theoretically the maximum value of rectifier efficiency of a half-wave rectifier is 40.6%
when
f
L
R
R
= 0.
vi) Ripple Factor ( ) :
The ripple factor is given by
2
1rms
dc
I
I
(or)
2
1rms
dc
V
V
2
2
/
/ 1m
m
I
I
=
2
1
2
= 1.21
1.21
vii) Regulation:
The variation of d.c. output voltage as a function of d.c. load current is called regulation.
The variation of Vdc with Idc for a half-wave rectifier is obtained as follows:
mI /I m
dc
Lf
V
R R
But Ldc dcV I R
dc
V m L
Lf
RV
R R
1 fm
Lf
RV
R R
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GRIET-ECE G.Surekha Page 9
dc f
m RV I
dc dc f
mV RV I
This result shows that Vdc equals mV
at no load and that the dc voltage decreases linearly
with an increase in dc output current. The larger the magnitude of the diode forward resistance,
the greater is this decrease for a given current change.
viii) Transformer Utilization Factor (UTF):
The d.c. power to be delivered to the load in a rectifier circuit decides the rating of the
transformer used in the circuit. So, transformer utilization factor is defined as
( )
dc
ac rated
P
TUF
P
The factor which indicates how much is the utilization of the transformer in the circuit is
called Transformer Utilization Factor (TUF).
The a.c. power rating of transformer
= Vrms Irms
The secondary voltage is purely sinusoidal hence its rms value is
1
2
times maximum
while the current is half sinusoidal hence its rms value is
1
2
of the maximum.
( )ac rated
P mI
22
mV mI
2 2
mV
The d.c. power delivered to the load
2
dc LI R
m
2
I
LR
( )
dc
ac rated
P
TUF
P
m
2
I
LR
m
2 2
ImV
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GRIET-ECE G.Surekha Page 10
2
2
2
2 2I
I
m
m
L
L
R
R mIm LV R
= 0.287
TUF 0.287
The value of TUF is low which shows that in half-wave circuit, the transformer is not
fully utilized.
If the transformer rating is 1 KVA (1000VA) then the half-wave rectifier can deliver
1000 X 0.287 = 287 watts to resistance load.
ix) Peak Inverse Voltage (PIV):
It is defined as the maximum reverse voltage that a diode can withstand without
destroying the junction. The peak inverse voltage across a diode is the peak of the negative half-
cycle. For half-wave rectifier, PIV is Vm.
x) Form factor (F):
The Form Factor F is defined as
F = rms value / average value
Im/ 2
Im/
F
0.5Im
0.318Im
1.57F
F=1.57
xi) Peak Factor (P):
The peak factor P is defined as
P= Peak Value / rms value
/ 2
m
m
V
V
= 2
P = 2
Disadvantages of Half-Wave Rectifier:
1. The ripple factor is high.
2. The efficiency is low.
3. The Transformer Utilization factor is low.
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11. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 11
Because of all these disadvantages, the half-wave rectifier circuit is normally not used as
a power rectifier circuit.
4.Explain Full Wave Rectifier and derive all the parameters.
Full – Wave Rectifier:
A full-wave rectifier converts an ac voltage into a pulsating dc voltage using both half
cycles of the applied ac voltage. In order to rectify both the half cycles of ac input, two diodes
are used in this circuit. The diodes feed a common load RL with the help of a center-tap
transformer.
A center-tap transformer is the one which produces two sinusoidal waveforms of same
magnitude and frequency but out of phase with respect to the ground in the secondary winding
of the transformer.
The full wave rectifier is shown in the figure below.
Fig. Full-Wave Rectifier.
Fig. wave forms of the input voltage, the o/p voltage across the load Resistor
Operation:
During positive half of the input signal, anode of diode D1 becomes positive and at the
same time the anode of diode D2 becomes negative. Hence D1 conducts and D2 does not
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12. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 12
conduct. The load current flows through D1 and the voltage drop across RL will be equal to the
input voltage.
During the negative half cycle of the input, the anode of D1 becomes negative and the
anode of D2 becomes positive. Hence, D1 does not conduct and D2 conducts. The load current
flows through D2 and the voltage drop across RL will be equal to the input voltage.
It is noted that the load current flows in the both the half cycles of ac voltage and in the
same direction through the load resistance.
Analysis:
Let a sinusoidal voltage Vi be applied to the input of a rectifier. It is given by Vi=Vm
sinωt
The current i1 though D1 and load resistor RL is given by
I sinm1
i t for 0 t
0
1
i for 2t
Where Im
Vm
R R
Lf
Similarly, the current i2 through diode D2 and load resistor RL is given by
2
0i for 0 t
I sinm2
i t for 2t
Therefore, the total current flowing through RL is the sum of the two currents i1 and i2.
i.e., iL = i1 + i2.
1. Average Value:
2. R.M.S load Current Ir.m.s
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GRIET-ECE G.Surekha Page 13
3. DC output voltage Vdc
4.Ripple Factor
5. Regulation:
The variation of d.c. output voltage as a function of d.c. load current is called regulation.
%Regulation= 100
V V
no load full load
V
full load
6.Rectification Efficiency:
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14. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 14
7. TRANSFORMER UTILIZATION FACTOR (TUF)
8. PIV = 2 Vm
9. Advantages
1) Ripple factor = 0.482 (against 1.21 for HWR)
2) Rectification efficiency is 0.812 (against 0.405 for HWR)
3) Better TUF (secondary) is 0.574 (0.287 for HWR)
4) No core saturation problem
Disadvantages:
1) Requires center tapped transformer.
3. Explain about Bridge rectifier using neat diagram.
A bridge rectifier makes use of four diodes in a bridge arrangement to achieve full-wave
rectification. This is a widely used configuration, both with individual diodes wired as shown
and with single component bridges where the diode bridge is wired internally.
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15. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 15
Current flow in the bridge rectifier
For both positive and negative swings of
the transformer, there is a forward path
through the diode bridge. Both conduction
paths cause current to flow in the same
direction through the load resistor,
accomplishing full-wave rectification.
While one set of diodes is forward biased,
the other set is reverse biased and effectively
eliminated from the circuit.
4. Compare HWR,FWR,Bridge rectifier.
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16. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 16
5.Explain about all Filters.
CAPACITOR FILTER WITH HWR
CAPACITOR FILTER WITH FWR.
Inductor Filter with Half Wave Rectifier
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17. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 17
Inductor Filter with Full Wave Rectifier
COMPARISON OF FILTERS: -
1) A capacitor filter provides Vm volts at less load current. But regulation is poor.
2) An Inductor filter gives high ripple voltage for low load currents. It is used for
high load currents
3) L – Section filter gives a ripple factor independent of load current. Voltage
regulation can be improved by use of bleeder resistance
4) Multiple L – Section filter or filters give much less ripple than the single L –
Section Filter.
Problems from previous external question papers:
1. A diode whose internal resistance is 20Ω is to supply power to a 100Ω load from
110V(rms) source pf supply. Calculate (a) peak load current (b) the dc load current (c)
the ac load current (d) the percentage regulation from no load to full load.
Solution:
Given a half-wave rectifier circuit
Rf=20Ω, RL=100Ω
Given an ac source with rms voltage of 110V, therefore the maximum amplitude of
sinusoidal input is given by
Vm = 2 Vrms = 2 x 110 = 155.56V.
(a) Peak load current : Im
Vm
R R
Lf
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18. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 18
155.56
120
Im = 1.29A
(b) The dc load current : I
Im
dc
= 0.41A
(c) The ac load current : I
2
Im
rms = 0.645A
(d) Vno-load =
Vm =
155.56
= 49.51 V
Vfull-load =
Vm I R
dc f
= 41.26 V
% Regulation= 100
V V
no load full load
V
full load
= 19.97%
2. A diode has an internal resistance of 20Ω and 1000Ω load from 110V(rms) source pf
supply. Calculate (a) the efficiency of rectification (b) the percentage regulation from
no load to full load.
Solution:
Given a half-wave rectifier circuit
Rf=20Ω, RL=1000Ω
Given an ac source with rms voltage of 110V, therefore the maximum amplitude
of sinusoidal input is given by
Vm = 2 Vrms = 2 x 110 = 155.56V.
(a) % Efficiency ( ) =
40.6
20
1
100
=
1.02
40.6
= 39.8%.
(b) Peak load current : Im
Vm
R R
Lf
=
155.56
1020
= 0.1525 A
= 152.5 mA
The dc load current : I
Im
dc
= 48.54 mA
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19. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 19
Vno-load =
Vm =
155.56
= 49.51 V
Vfull-load =
Vm I R
dc f
= 49.51 – (48.54 x10-3
x 20)
= 49.51 – 0.97
= 48.54 V
% Regulation= 100
V V
no load full load
V
full load
=
49.51 48.54
100
48.54
= 1.94 %
3. An a.c. supply of 230V is applied to a half-wave rectifier circuit through transformer of
turns ration 5:1. Assume the diode is an ideal one. The load resistance is 300Ω.
Find (a) dc output voltage (b) PIV (c) maximum, and (d) average values of power
delivered to the load.
Solution:
(a) The transformer secondary voltage = 230/5 = 46V.
Maximum value of secondary voltage,
Vm = 2 x 46 = 65V.
Therefore, dc output voltage,
65VmV
dc
= 20.7 V
(b) PIV of a diode : Vm = 65V
(c) Maximum value of load current,
Im
Vm
R
L
=
65
300
= 0.217 A
Therefore, maximum value of power delivered to the load,
Pm = Im
2
x RL = (0.217)2
x 300 = 14.1W
(d) The average value of load current,
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20. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 20
20.7
I =
300
V
dc
dc R
L
= 0.069A
Therefore, average value of power delivered to the load,
Pdc = Idc
2
x RL = (0.069)2
x 300 = 1.43W
4 . A 230V, 60 Hz voltage is applied to the primary of 5:1 step down center tapped
transformer used in a FWR having a load of 900 . If the diode resistance and
secondary coil resistance together has a resistance of 100 , determine:
a) DC voltage across the load
b) DC current flowing through the load
c) DC power delivered to the load
d) PIV across each diode
e) Ripple voltage and frequency.
Given:
AC input – 230V, 60 Hz RL = 900 RS + Rf = 100
(a)
DC voltage across load = ?
Voltage secondary of transformer = 230/5 = 46 V.
Each of half = 23 volts,(rms); Vrms=23V; Vm = ? Vrms = Vmax (0.707)
(b) DC current IDC = Imax (0.636)
= 32.53 (0.636) = 20.69 mA.
VRL = IDC.RL = 20.69 x 10-3
x 900 = 18.62 volts
(c) DC power Pdc = Vdc. Idc = 18.62 x 20.69 x 10-3
= 3.85 m.w.
(d) PIV across each diode = Vmax x 2 = 32.53 x 2 = 65.06 volts
(e) Ripple voltage = ?
Ripple factor =Ripple Voltage/Load Voltage
V = .VRL = 0.483 x 18.62 = 8.99 volts
Ripple frequency = 2 x Input source frequency
=2 x 60=120 Hz
5. Draw the circuit diagram of a FWR using center tapped transformer to obtain an
output DC voltage of 18V at 200 mA and VDC no load equals 20V. Find the transformer
ratings.
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21. EDC-UNIT2 Question&answer
GRIET-ECE G.Surekha Page 21
Solution:
VDC = 18V VDCNL = 20V. Idc = 200 mA VDCNL – VDCFL = IDC (RS + Rf)
20 – 18 = 200 x 10-3
(RS+Rf) or RS + Rf = 2 / 200 x 10-3 = 10
Transformer rating is Input 220 V Ac.
Output 22 – 0 – 22V (RMS)
DC current 200 mA.
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