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Power Electronics
Dr.
Ali Mohamed Eltamaly
Mansoura University
Faculty of Engineering

Chapter Four 113
Contents
Chapter 1
Introduction
1
1.1. Definition Of Power Electronics 1
1.2 Main Task Of Power Electronics
1
1.3 Rectification 2
1.4 DC-To-AC Conversion 3
1.5 DC-to-DC Conversion 4
1.6 AC-TO-AC Conversion 4
1.7 Additional Insights Into Power Electronics 5
1.8 Harmonics 7
1.9 Semiconductors Switch types 12
Chapter 2
Diode Circuits or Uncontrolled
Rectifier
17
2.1 Half Wave Diode Rectifier
17
2.2 Center-Tap Diode Rectifier
29
2.3 Full Bridge Single-Phase Diode Rectifier
35
2.4 Three-Phase Half Wave Rectifier
40
2.5 Three-Phase Full Wave Rectifier
49
2.6 Multi-pulse Diode Rectifier
56

Fourier Series 114
Chapter 3
Scr Rectifier or Controlled Rectifier
59
3.1 Introduction
59
3.2 Half Wave Single Phase Controlled Rectifier
60
3.3 Single-Phase Full Wave Controlled Rectifier
73
3.4 Three Phase Half Wave Controlled Rectifier
91
3.5 Three Phase Half Wave Controlled Rectifier With DC Load Current
95
3.6 Three Phase Half Wave Controlled Rectifier With Free Wheeling
Diode
98
3.7 Three Phase Full Wave Fully Controlled Rectifier
100
Chapter 4
Fourier Series
112
4-1 Introduction
112
4-2 Determination Of Fourier Coefficients
113
4-3 Determination Of Fourier Coefficients Without Integration
119

Chapter 1
Introduction
1.1. Definition Of Power Electronics
Power electronics refers to control and conversion of electrical power by
power semiconductor devices wherein these devices operate as switches.
Advent of silicon-controlled rectifiers, abbreviated as SCRs, led to the
development of a new area of application called the power electronics.
Once the SCRs were available, the application area spread to many fields
such as drives, power supplies, aviation electronics, high frequency
inverters and power electronics originated.
Power electronics has applications that span the whole field of
electrical power systems, with the power range of these applications
extending from a few VA/Watts to several MVA / MW.
"Electronic power converter" is the term that is used to refer to a power
electronic circuit that converts voltage and current from one form to
another. These converters can be classified as:
• Rectifier converting an AC voltage to a DC voltage,
• Inverter converting a DC voltage to an AC voltage,
• Chopper or a switch-mode power supply that converts a DC
voltage to another DC voltage, and
• Cycloconverter and cycloinverter converting an AC voltage
to another AC voltage.
In addition, SCRs and other power semiconductor devices are used as
static switches.
1.2 Rectification
Rectifiers can be classified as uncontrolled and controlled rectifiers, and
the controlled rectifiers can be further divided into semi-controlled and
fully controlled rectifiers. Uncontrolled rectifier circuits are built with
diodes, and fully controlled rectifier circuits are built with SCRs. Both
diodes and SCRs are used in semi-controlled rectifier circuits.
There are several rectifier configurations. The most famous rectifier
configurations are listed below.
• Single-phase semi-controlled bridge rectifier,
• Single-phase fully-controlled bridge rectifier,
• Three-phase three-pulse, star-connected rectifier,

2 Chapter One
• Double three-phase, three-pulse star-connected rectifiers with
inter-phase transformer (IPT),
• Three-phase semi-controlled bridge rectifier,
• Three-phase fully-controlled bridge rectifier, and ,
• Double three-phase fully controlled bridge rectifiers with IPT.
Apart from the configurations listed above, there are series-connected
and 12-pulse rectifiers for delivering high quality high power output.
Power rating of a single-phase rectifier tends to be lower than 10 kW.
Three-phase bridge rectifiers are used for delivering higher power output,
up to 500 kW at 500 V DC or even more. For low voltage, high current
applications, a pair of three-phase, three-pulse rectifiers interconnected by
an inter-phase transformer (IPT) is used. For a high current output,
rectifiers with IPT are preferred to connecting devices directly in parallel.
There are many applications for rectifiers. Some of them are:
• Variable speed DC drives,
• Battery chargers,
• DC power supplies and Power supply for a specific
application like electroplating
1.3 DC-To-AC Conversion
The converter that changes a DC voltage to an alternating voltage, AC is
called an inverter. Earlier inverters were built with SCRs. Since the
circuitry required turning the SCR off tends to be complex, other power
semiconductor devices such as bipolar junction transistors, power
MOSFETs, insulated gate bipolar transistors (IGBT) and MOS-controlled
thyristors (MCTs) are used nowadays. Currently only the inverters with a
high power rating, such as 500 kW or higher, are likely to be built with
either SCRs or gate turn-off thyristors (GTOs). There are many inverter
circuits and the techniques for controlling an inverter vary in complexity.
Some of the applications of an inverter are listed below:
• Emergency lighting systems,
• AC variable speed drives,
• Uninterrupted power supplies, and,
• Frequency converters.
1.4 DC-to-DC Conversion
When the SCR came into use, a DC-to-DC converter circuit was called a
chopper. Nowadays, an SCR is rarely used in a DC-to-DC converter.

Introduction 3
Either a power BJT or a power MOSFET is normally used in such a
converter and this converter is called a switch-mode power supply. A
switch-mode power supply can be one of the types listed below:
• Step-down switch-mode power supply,
• Step-up chopper,
• Fly-back converter, and ,
• Resonant converter.
The typical applications for a switch-mode power supply or a chopper
are:
• DC drive,
• Battery charger, and,
• DC power supply.
1.5 AC-TO-AC Conversion
A cycloconverter or a Matrix converter converts an AC voltage, such as
the mains supply, to another AC voltage. The amplitude and the
frequency of input voltage to a cycloconverter tend to be fixed values,
whereas both the amplitude and the frequency of output voltage of a
cycloconverter tend to be variable specially in Adjustable Speed Drives
(ASD). A typical application of a cycloconverter is to use it for
controlling the speed of an AC traction motor and most of these
cycloconverters have a high power output, of the order a few megawatts
and SCRs are used in these circuits. In contrast, low cost, low power
cycloconverters for low power AC motors are also in use and many of
these circuit tend to use triacs in place of SCRs. Unlike an SCR which
conducts in only one direction, a triac is capable of conducting in either
direction and like an SCR, it is also a three terminal device. It may be
noted that the use of a cycloconverter is not as common as that of an
inverter and a cycloinverter is rarely used because of its complexity and
its high cost.
1.6 Additional Insights Into Power Electronics
There are several striking features of power electronics, the foremost
among them being the extensive use of inductors and capacitors. In many
applications of power electronics, an inductor may carry a high current at
a high frequency. The implications of operating an inductor in this
manner are quite a few, such as necessitating the use of litz wire in place
of single-stranded or multi-stranded copper wire at frequencies above 50

4 Chapter One
kHz, using a proper core to limit the losses in the core, and shielding the
inductor properly so that the fringing that occurs at the air-gaps in the
magnetic path does not lead to electromagnetic interference. Usually the
capacitors used in a power electronic application are also stressed. It is
typical for a capacitor to be operated at a high frequency with current
surges passing through it periodically. This means that the current rating
of the capacitor at the operating frequency should be checked before its
use. In addition, it may be preferable if the capacitor has self-healing
property. Hence an inductor or a capacitor has to be selected or designed
with care, taking into account the operating conditions, before its use in a
power electronic circuit.
In many power electronic circuits, diodes play a crucial role. A normal
power diode is usually designed to be operated at 400 Hz or less. Many of
the inverter and switch-mode power supply circuits operate at a much
higher frequency and these circuits need diodes that turn ON and OFF
fast. In addition, it is also desired that the turning-off process of a diode
should not create undesirable electrical transients in the circuit. Since
there are several types of diodes available, selection of a proper diode is
very important for reliable operation of a circuit.
Analysis of power electronic circuits tends to be quite complicated,
because these circuits rarely operate in steady state. Traditionally steady-
state response refers to the state of a circuit characterized by either a DC
response or a sinusoidal response. Most of the power electronic circuits
have a periodic response, but this response is not usually sinusoidal.
Typically, the repetitive or the periodic response contains both a steady-
state part due to the forcing function and a transient part due to the poles
of the network. Since the responses are non-sinusoidal, harmonic analysis
is often necessary. In order to obtain the time response, it may be
necessary to resort to the use of a computer program.
Power electronics is a subject of interdisciplinary nature. To design and
build control circuitry of a power electronic application, one needs
knowledge of several areas, which are listed below.
• Design of analogue and digital electronic circuits, to build the
control circuitry.
• Microcontrollers and digital signal processors for use in
sophisticated applications.
• Many power electronic circuits have an electrical machine as
their load. In AC variable speed drive, it may be a reluctance

Introduction 5
motor, an induction motor or a synchronous motor. In a DC
variable speed drive, it is usually a DC shunt motor.
• In a circuit such as an inverter, a transformer may be connected
at its output and the transformer may have to operate with a
nonsinusoidal waveform at its input.
• A pulse transformer with a ferrite core is used commonly to
transfer the gate signal to the power semiconductor device. A
ferrite-cored transformer with a relatively higher power output
is also used in an application such as a high frequency inverter.
• Many power electronic systems are operated with negative
feedback. A linear controller such as a PI controller is used in
relatively simple applications, whereas a controller based on
digital or state-variable feedback techniques is used in more
sophisticated applications.
• Computer simulation is often necessary to optimize the design
of a power electronic system. In order to simulate, knowledge
of software package such as MATLAB, Pspice, Orcad,…..etc.
and the know-how to model nonlinear systems may be
necessary.
The study of power electronics is an exciting and a challenging
experience. The scope for applying power electronics is growing at a fast
pace. New devices keep coming into the market, sustaining development
work in power electronics.
1.7 Harmonics
The invention of the semiconductor controlled rectifier (SCR or
thyristor) in the 1950s led to increase of development new type
converters, all of which are nonlinear. The major part of power system
loads is in the form of nonlinear loads too much harmonics are injected to
the power system. It is caused by the interaction of distorting customer
loads with the impedance of supply network. Also, the increase of
connecting renewable energy systems with electric utilities injects too
much harmonics to the power system.
There are a number of electric devices that have nonlinear operating
characteristics, and when it used in power distribution circuits it will
create and generate nonlinear currents and voltages. Because of periodic
non-linearity can best be analyzed using the Fourier transform, these
nonlinear currents and voltages have been generally referred to as

6 Chapter One
“Harmonics”. Also, the harmonics can be defined as a sinusoidal
component of a periodic waves or quality having frequencies that are an
integral multiple of the fundamental frequency.
Among the devices that can generate nonlinear currents transformers
and induction machines (Because of magnetic core saturation) and power
electronics assemblies.
The electric utilities recognized the importance of harmonics as early
as the 1930’s such behavior is viewed as a potentially growing concern in
modern power distribution network.
1.7.1 Harmonics Effects on Power System Components
There are many bad effects of harmonics on the power system
components. These bad effects can derated the power system component
or it may destroy some devices in sever cases [Lee]. The following is the
harmonic effects on power system components.
In Transformers and Reactors
• The eddy current losses increase in proportion to the square of the
load current and square harmonics frequency,
• The hysterics losses will increase,
• The loading capability is derated by harmonic currents , and,
• Possible resonance may occur between transformer inductance and
line capacitor.
In Capacitors
• The life expectancy decreases due to increased dielectric losses
that cause additional heating, reactive power increases due to
harmonic voltages, and,
• Over voltage can occur and resonance may occur resulting in
harmonic magnification.
In Cables
• Additional heating occurs in cables due to harmonic currents
because of skin and proximity effects which are function of
frequency, and,
• The I2
R losses increase.
In Switchgear
• Changing the rate of rise of transient recovery voltage, and,
• Affects the operation of the blowout.
In Relays
• Affects the time delay characteristics, and,

Introduction 7
• False tripping may occurs.
In Motors
• Stator and rotor I2
R losses increase due to the flow of harmonic
currents,
• In the case of induction motors with skewed rotors the flux
changes in both the stator and rotor and high frequency can
produce substantial iron losses, and,
• Positive sequence harmonics develop shaft torque that aid shaft
rotation; negative sequence harmonics have opposite effect.
In Generators
• Rotor and stator heating ,
• Production of pulsating or oscillating torques, and,
• Acoustic noise.
In Electronic Equipment
• Unstable operation of firing circuits based on zero voltage
crossing,
• Erroneous operation in measuring equipment, and,
• Malfunction of computers allied equipment due to the presence of
ac supply harmonics.
1.7.2 Harmonic Standards
It should be clear from the above that there are serious effects on the
power system components. Harmonics standards and limits evolved to
give a standard level of harmonics can be injected to the power system
from any power system component. The first standard (EN50006) by
European Committee for Electro-technical Standardization (CENELEE)
that was developed by 14th European committee. Many other
standardizations were done and are listed in IEC61000-3-4, 1998 [1].
The IEEE standard 519-1992 [2] is a recommended practice for power
factor correction and harmonic impact limitation for static power
converters. It is convenient to employ a set of analysis tools known as
Fourier transform in the analysis of the distorted waveforms. In general, a
non-sinusoidal waveform f(t) repeating with an angular frequency ω can
be expressed as in the following equation.
( )∑
∞
=
++=
1
0
)sin()cos(
2
)(
n
nn tnbtna
a
tf ωω (1.1)

8 Chapter One
where ∫=
π
ωω
π
2
0
)(cos)(
1
tdtntfan (1.2)
and ∫=
π
ωω
π
2
0
)(sin)(
1
tdtntfbn (1.3)
Each frequency component n has the following value
)(sin)(cos)( tnbtnatf nnn ωω += (1.4)
fn(t) can be represented as a phasor in terms of its rms value as shown in
the following equation
njnn
n e
ba
F ϕ
2
22
+
= (1.5)
Where
n
n
n
a
b−
= −1
tanϕ (1.6)
The amount of distortion in the voltage or current waveform is
qualified by means of an Total Harmonic Distortion (THD). The THD in
current and voltage are given as shown in (1.7) and (1.8) respectively.
1
2
1
2
1
2
*100*100
s
nn
sn
s
ss
i
I
I
I
II
THD
∑
≠
=
−
= (1.7)
1
2
1
2
1
2
*100*100
s
nn
sn
s
ss
v
V
V
V
VV
THD
∑
≠
=
−
= (1.8)
Where THDi & THDv The Total Harmonic Distortion in the current
and voltage waveforms
Current and voltage limitations included in the update IEE 519 1992
are shown in Table(1.1) and Table(1.2) respectively [2].
Table (1.1) IEEE 519-1992 current distortion limits for general distribution
systems (120 to 69kV) the maximum harmonic current distortion in percent of LI
Individual Harmonic order (Odd Harmonics)
LSC II / n<11 11≤ n<17 17≤ n<23 23≤ n<35 35≤ n< TDD
<20 4.0 2.0 1.5 0.6 0.3 5.0
20<50 7.0 3.5 2.5 1.0 0.5 8.0
50<100 10.0 4.5 4.0 1.5 0.7 12.0
100<1000 12.0 5.5 5.0 2.0 1.0 15.0
>1000 15.0 7.0 6.0 2.5 1.4 20.0

Introduction 9
Where; TDD (Total Demand Distortion) = ∑
∞
=2
2100
n
n
ML
I
I
,
Where MLI is the maximum fundamental demand load current (15 or
30min demand).
SCI is the maximum short-circuit current at the point of common
coupling (PCC).
LI is the maximum demand load current at the point of common
coupling (PCC).
Table (1.2) Voltage distortion limits
Bus voltage at PCC Individual voltage distortion (%) vTHD (%)
69 kV and blow 3.0 5.0
69.001 kV through 161kV 1.5 2.5
161.001kV and above 1 1.5
1.8 Semiconductors Switch types
At this point it is beneficial to review the current state of semiconductor
devices used for high power applications. This is required because the
operation of many power electronic circuits is intimately tied to the
behavior of various devices.
1.8.1 Diodes
A sketch of a PN junction diode characteristic is drawn in Fig.1.1. The
icon used to represent the diode is drawn in the upper left corner of the
figure, together with the polarity markings used in describing the
characteristics. The icon 'arrow' itself suggests an intrinsic polarity
reflecting the inherent nonlinearity of the diode characteristic.
Fig.1.1 shows the i-v characteristics of the silicon diode and
germanium diode. As shown in the figure the diode characteristics have
been divided into three ranges of operation for purposes of description.
Diodes operate in the forward- and reverse-bias ranges. Forward bias is a
range of 'easy' conduction, i.e., after a small threshold voltage level ( »
0.7 volts for silicon) is reached a small voltage change produces a large
current change. In this case the diode is forward bias or in "ON" state.
The 'breakdown' range on the left side of the figure happened when the
reverse applied voltage exceeds the maximum limit that the diode can
withstand. At this range the diode destroyed.

10 Chapter One
Fig.1.1 The diode iv characteristics
On the other hand if the polarity of the voltage is reversed the current
flows in the reverse direction and the diode operates in 'reverse' bias or in
"OFF" state. The theoretical reverse bias current is very small.
In practice, while the diode conducts, a small voltage drop appears
across its terminals. However, the voltage drop is about 0.7 V for silicon
diodes and 0.3 V for germanium diodes, so it can be neglected in most
electronic circuits because this voltage drop is small with respect to other
circuit voltages. So, a perfect diode behaves like normally closed switch
when it is forward bias (as soon as its anode voltage is slightly positive
than cathode voltage) and open switch when it is in reverse biased (as
soon as its cathode voltage is slightly positive than anode voltage). There
are two important characteristics have to be taken into account in
choosing diode. These two characteristics are:
• Peak Inverse voltage (PIV): Is the maximum voltage that a diode
can withstand only so much voltage before it breaks down. So if
the PIV is exceeded than the PIV rated for the diode, then the
diode will conduct in both forward and reverse bias and the diode
will be immediately destroyed.
• Maximum Average Current: Is the average current that the diode
can carry.
It is convenient for simplicity in discussion and quite useful in making
estimates of circuit behavior ( rather good estimates if done with care and
understanding) to linearize the diode characteristics as indicated in
Fig.1.2. Instead of a very small reverse-bias current the idealized model
approximates this current as zero. ( The practical measure of the
appropriateness of this approximation is whether the small reverse bias
current causes negligible voltage drops in the circuit in which the diode is
embedded. If so the value of the reverse-bias current really does not enter
into calculations significantly and can be ignored.) Furthermore the zero

Introduction 11
current approximation is extended into forward-bias right up to the knee
of the curve. Exactly what voltage to cite as the knee voltage is somewhat
arguable, although usually the particular value used is not very important.
1.8.2 Thyristor
The thyristor is the most important type of the power semiconductor
devices. They are used in very large scale in power electronic circuits.
The thyristor are known also as Silicon Controlled Rectifier (SCR). The
thyristor has been invented in 1957 by general electric company in USA.
The thyristor consists of four layers of semiconductor materials (p-n-p-
n) all brought together to form only one unit. Fig.1.2 shows the schematic
diagram of this device and its symbolic representation. The thyristor has
three terminals, anode A, cathode K and gate G as shown in Fig.1.2.The
anode and cathode are connected to main power circuit. The gate terminal
is connected to control circuit to carry low current in the direction from
gate to cathode.
Fig.1.2 The schematic diagram of SCR and its circuit symbol.
The operational characteristics of a thyristor are shown in Fig.1.3. In
case of zero gate current and forward voltage is applied across the device
i.e. anode is positive with respect to cathode, junction J1 and J3 are
forward bias while J2 remains reverse biased, and therefore the anode
current is so small leakage current. If the forward voltage reaches a
critical limit, called forward break over voltage, the thyristor switches
into high conduction, thus forward biasing junction J2 to turn thyristor
ON in this case the thyristor will break down. The forward voltage drop
then falls to very low value (1 to 2 Volts). The thyristor can be switched
to on state by injecting a current into the central p type layer via the gate
terminal. The injection of the gate current provides additional holes in the

12 Chapter One
central p layer, reducing the forward breakover voltage. If the anode
current falls below a critical limit, called the holding current IH the
thyristor turns to its forward state.
If the reverse voltage is applied across the thyristor i.e. the anode is
negative with respect to cathode, the outer junction J1 and J3 are reverse
biased and the central junction J2 is forward biased. Therefore only a
small leakage current flows. If the reverse voltage is increased, then at the
critical breakdown level known as reverse breakdown voltage, an
avalanche will occur at J1 and J3 and the current will increase sharply. If
this current is not limited to safe value, it will destroy the thyristor.
The gate current is applied at the instant turn on is desired. The
thyristor turn on provided at higher anode voltage than cathode. After
turn on with IA reaches a value known as latching current, the thyristor
continuous to conduct even after gate signal has been removed. Hence
only pulse of gate current is required to turn the Thyrstor ON.
Fig.1.3 Thyristor v-i characteristics
1.8.3 Thyristor types:
There is many types of thyristors all of them has three terminals but
differs only in how they can turn ON and OFF. The most famous types of
thyristors are:
1. Phase controlled thyristor(SCR)
2. Fast switching thyristor (SCR)
3. Gate-turn-off thyristor (GTO)
4. Bidirectional triode thyristor (TRIAC)
5. Light activated silicon-controlled rectifier (LASCR)
The electric circuit symbols of each type of thyristors are shown in
Fig.1.4.

Introduction 13
In the next items we will talk only about the most famous two types :-
Fig.1.4 The electric circuit symbols of each type of thyristors.
Gate Turn Off thyristor (GTO).
A GTO thyristor can be turned on by a single pulse of positive gate
current like conventional thyristor, but in addition it can be turned off by
a pulse of negative gate current. The gate current therefore controls both
ON state and OFF state operation of the device. GTO v-i characteristics is
shown in Fig.1.5. The GTO has many advantages and disadvantages with
respect to conventional thyristor here will talk about these advantages and
disadvantages.
Fig.1.5 GTO v-i characteristics.

14 Chapter One
The GTO has the following advantage over thyristor.
1- Elimination of commutating components in forced
commutation resulting in reduction in cost, weight and volume,
2- Reduction in acoustic and electromagnetic noise due to the
elimination of commutation chokes,
3- Faster turn OFF permitting high switching frequency,
4- Improved converters efficiency, and,
5- It has more di/dt rating at turn ON.
The thyristor has the following advantage over GTO.
1- ON state voltage drop and associated losses are higher in GTO
than thyristor,
2- Triggering gate current required for GTOs is more than those
of thyristor,
3- Latching and holding current is more in GTO than those of
thyristor,
4- Gate drive circuit loss is more than those of thyristor, and,
5- Its reverse voltage block capability is less than its forward
blocking capability.
Bi-Directional-Triode thyristor (TRIAC).
TRIAC are used for the control of power in AC circuits. A TRIAC is
equivalent of two reverse parallel-connected SCRs with one common
gate. Conduction can be achieved in either direction with an appropriate
gate current. A TRIAC is thus a bi-directional gate controlled thyristor
with three terminals. Fig.1.4 shows the schematic symbol of a TRIAC.
The terms anode and cathode are not applicable to TRIAC. Fig.1.6 shows
the i-v characteristics of the TRIAC.

Introduction 15
Fig.1.6 Operating characteristics of TRIAC.ele146
DIAC
DIAC is like a TRIAC without a gate terminal. DIAC conducts current
in both directions depending on the voltage connected to its terminals.
When the voltage between the two terminals greater than the break down
voltage, the DIAC conducts and the current goes in the direction from the
higher voltage point to the lower voltage one. The following figure shows
the layers construction, electric circuit symbol and the operating
characteristics of the DIAC. Fig.1.7 shows the DIAC construction and
electric symbol. Fig.1.8 shows a DIAC v-i characteristics.
The DIAC used in firing circuits of thyristors since its breakdown
voltage used to determine the firing angle of the thyristor.
Fig.1.7 DIAC construction and electric symbol.

16 Chapter One
Fig.1.8 DIAC v-i characteristics
1.9 Power Transistor
Power transistor has many applications now in power electronics and
become a better option than thyristor. Power transistor can switch on and
off very fast using gate signals which is the most important advantage
over thyristor. There are three famous types of power transistors used in
power electronics converters shown in the following items:
Bipolar Junction Transistor (BJT)
BJT has three terminals as shown in Fig.. These terminals are base,
collector, and, emitter each of them is connected to one of three
semiconductor materials layers. These three layers can be NPN or PNP.
Fig.1.9 shows the circuit symbol of NPN and PNP BJT transistor.
npn pnp
Fig.1.9 The electric symbol of npn and pnp transistors.

Introduction 17
Fig.1.10 shows the direction of currents in the NPN and PNP transistors.
It is clear that the emitter current direction takes the same direction as on
the electric symbol of BJT transistor and both gate and collector take the
opposite direction.
Fig.1.10 The currents of the NPN and PNP transistors.
When the transistor connected in DC circuit, the voltage BBV representing
a forward bias voltage and ccV representing a reverse bias for base to
collector circuit as shown in Fig.1.11 for NPN and PNP transistors.
Fig.1.11 Transistor connection to DC circuit.
The relation between the collector current and base current known as a
current gain of the transistor β as shown in ( )
B
C
I
I
=β
Current and voltage analysis of NPN transistors is shown if Fig.1.11. It is
clear from Fig.1.11 that:
BBBEBBR RIVVV b
*=−=
Then, the base current can be obtained as shown in the following
equation:
B
BEBB
B
R
VV
I
−
=

18 Chapter One
The voltage on CR resistor are:
CCR RIV C
*=
CCCCCE RIVV *−=
Fig.1.12 shows the collector characteristics of NPN transistor for
different base currents. This figure shows that four regions, saturation,
linear, break down, and, cut-off regions. The explanation of each region
in this figure is shown in the following points:
Increasing of CCV increases the voltage CEV gradually as shown in the
saturation region.
When CEV become more than 0.7 V, the base to collector junction
become reverse bias and the transistor moves to linear region. In linear
region CI approximately constant for the same amount of base current
when CEV increases.
When CEV become higher than the rated limits, the transistor goes to
break down region.
At zero base current, the transistor works in cut-off region and there is
only very small collector leakage current.
Fig.1.12 Collector characteristics of NPN transistor for different base currents.
1.10 Power MOSFET
The power MOSFET has two important advantages over than BJT,
First of them, is its need to very low operating gate current, the second of

Introduction 19
them, is its very high switching speed. So, it is used in the circuit that
requires high turning ON and OFF speed that may be greater than
100kHz. This switch is more expensive than any other switches have the
same ratings. The power MOSFET has three terminals source, drain and
gate. Fig.1.13 shows the electric symbol and static characteristics of the
power MOSFET.
Fig.1.13 The electric symbol and static characteristics of power MOSFET.
1.11 Insulated Gate Bipolar Transistor (IGBT)
IGBTs transistors introduce a performance same as BJT but it has the
advantage that its very high current density and it has higher switch speed
than BJT but still lower than MOSFET. The normal switching frequency
of the IGBT is about 40kHz. IGBT has three terminals collector, emitter,
and, gate.
Fig.1.14 shows the electric circuit symbol and operating characteristics of
the IGBT. IGBT used so much in PWM converters and in Adjustable
speed drives.
Fig.1.14 IGBT v-i transfer characteristics and circuit symbol:

20 Chapter One
1.12 Power Junction Field Effect Transistors
This device is also sometimes known as the static induction transistor
(SIT). It is effectively a JFET transistor with geometry changes to allow
the device to withstand high voltages and conduct high currents. The
current capability is achieved by paralleling up thousands of basic JFET
cells. The main problem with the power JFET is that it is a normally on
device. This is not good from a start-up viewpoint, since the device can
conduct until the control circuitry begins to operate. Some devices are
commercially available, but they have not found widespread usage.
1.13 Field Controlled Thyristor
This device is essentially a modification of the SIT. The drain of the
SIT is modified by changing it into an injecting contact. This is achieved
by making it a pn junction. The drain of the device now becomes the
anode, and the source of the SIT becomes the cathode. In operation the
device is very similar to the JFET, the main difference being quantitative
– the FCT can carry much larger currents for the same on-state voltage.
The injection of the minority carriers in the device means that there is
conductivity modulation and lower on-state resistance. The device also
blocks for reverse voltages due to the presence of the pn junction.
1.14 MOS-Controlled Thyristors
The MOS-controlled thyristor (MCT) is a relatively new device which
is available commercially. Unfortunately, despite a lot of hype at the time
of its introduction, it has not achieved its potential. This has been largely
due to fabrication problems with the device, which has resulted on low
yields. Fig.1.15 is an equivalent circuit of the device, and its circuit
symbol. From Fig.1.15 one can see that the device is turned on by the
ON-FET, and turned o. by the OFF-FET. The main current carrying
element of the device is the thyristor. To turn the device on a negative
voltage relative to the cathode of the device is applied to the gate of the
ON-FET. As a result this FET turns on, supplying current to the base of
the bottom transistor of the SCR. Consequently the SCR turns on. To turn
o. the device, a positive voltage is applied to the gate. This causes the
ON-FET to turn o., and the OFF-FET to turn on. The result is that the
base-emitter junction of the top transistor of the SCR is shorted, and
because vBE drops to zero. volt it turns o.. Consequently the regeneration
process that causes the SCR latching is interrupted and the device turns.

Introduction 21
The P-MCT is given this name because the cathode is connected to P
type material. One can also construct an N-MCT, where the cathode is
connected to N type material.
Fig.1.15 Schematic and circuit symbol for the P-MCT.

Chapter 2
Diode Circuits or Uncontrolled Rectifier
2.1 Introduction
The only way to turn on the diode is when its anode voltage becomes
higher than cathode voltage as explained in the previous chapter. So,
there is no control on the conduction time of the diode which is the main
disadvantage of the diode circuits. Despite of this disadvantage, the diode
circuits still in use due to it’s the simplicity, low price, ruggedness,
….etc.
Because of their ability to conduct current in one direction, diodes are
used in rectifier circuits. The definition of rectification process is “ the
process of converting the alternating voltages and currents to direct
currents and the device is known as rectifier” It is extensively used in
charging batteries; supply DC motors, electrochemical processes and
power supply sections of industrial components.
The most famous diode rectifiers have been analyzed in the following
sections. Circuits and waveforms drawn with the help of PSIM simulation
program [1].
There are two different types of uncontrolled rectifiers or diode
rectifiers, half wave and full wave rectifiers. Full-wave rectifiers has
better performance than half wave rectifiers. But the main advantage of
half wave rectifier is its need to less number of diodes than full wave
rectifiers. The main disadvantages of half wave rectifier are:
1- High ripple factor,
2- Low rectification efficiency,
3- Low transformer utilization factor, and,
4- DC saturation of transformer secondary winding.
2.2 Performance Parameters
In most rectifier applications, the power input is sine-wave voltage
provided by the electric utility that is converted to a DC voltage and AC
components. The AC components are undesirable and must be kept away
from the load. Filter circuits or any other harmonic reduction technique
should be installed between the electric utility and the rectifier and

Diode Circuits or Uncontrolled Rectifier 23
between the rectifier output and the load that filters out the undesired
component and allows useful components to go through. So, careful
analysis has to be done before building the rectifier. The analysis requires
define the following terms:
The average value of the output voltage, dcV ,
The average value of the output current, dcI ,
The rms value of the output voltage, rmsV ,
The rms value of the output current, rmsI
The output DC power, dcdcdc IVP *= (2.1)
The output AC power, rmsrmsac IVP *= (2.2)
The effeciency or rectification ratio is defiend as
ac
dc
P
P
=η (2.3)
The output voltage can be considered as being composed of two
components (1) the DC component and (2) the AC component or ripple.
The effective (rms) value of the AC component of output voltage is
defined as:-
22
dcrmsac VVV −= (2.4)
The form factor, which is the measure of the shape of output voltage, is
defiend as shown in equation (2.5). Form factor should be greater than or
equal to one. The shape of output voltage waveform is neare to be DC as
the form factor tends to unity.
dc
rms
V
V
FF = (2.5)
The ripple factor which is a measure of the ripple content, is defiend as
shown in (2.6). Ripple factor should be greater than or equal to zero. The
shape of output voltage waveform is neare to be DC as the ripple factor
tends to zero.
11 2
2
222
−=−=
−
== FF
V
V
V
VV
V
V
RF
dc
rms
dc
dcrms
dc
ac
(2.6)
The Transformer Utilization Factor (TUF) is defiend as:-
SS
dc
IV
P
TUF = (2.7)

24 Chapter Two
Where SV and SI are the rms voltage and rms current of the
transformer secondery respectively.
Total Harmonic Distortion (THD) measures the shape of supply
current or voltage. THD should be grearter than or equal to zero. The
shape of supply current or voltage waveform is near to be sinewave as
THD tends to be zero. THD of input current and voltage are defiend as
shown in (2.8.a) and (2.8.b) respectively.
12
1
2
2
1
2
1
2
−=
−
=
S
S
S
SS
i
I
I
I
II
THD (2.8.a)
12
1
2
2
1
2
1
2
−=
−
=
S
S
S
SS
v
V
V
V
VV
THD (2.8.b)
where 1SI and 1SV are the fundamental component of the input current
and voltage, SI and SV respectively.
Creast Factor CF, which is a measure of the peak input current IS(peak)
as compared to its rms value IS, is defiend as:-
S
peakS
I
I
CF
)(
= (2.9)
In general, power factor in non-sinusoidal circuits can be obtained as
following:
φcos
sVoltampereApparent
PowerR
===
SS IV
Peal
PF (2.10)
Where, φ is the angle between the current and voltage. Definition is
true irrespective for any sinusoidal waveform. But, in case of sinusoidal
voltage (at supply) but non-sinusoidal current, the power factor can be
calculated as the following:
Average power is obtained by combining in-phase voltage and current
components of the same frequency.
FaactorntDisplacemeFactorDistortion
I
I
IV
IV
IV
P
PF
S
S
SSSS
*cos
cos
1
111 ==== φ
φ (2.11)
Where 1φ is the angle between the fundamental component of current
and supply voltage.
Distortion Factor = 1 for sinusoidal operation and displacement factor is a
measure of displacement between ( )tv ω and ( )ti ω .

2.3 Single-Phase Half-Wave Diode Rectifier
Most of the power electronic applications operate at a relative high
voltage and in such cases; the voltage drop across the power diode tends
to be small with respect to this high voltage. It is quite often justifiable to
use the ideal diode model. An ideal diode has zero conduction drops
when it is forward-biased ("ON") and has zero current when it is reverse-
biased ("OFF"). The explanation and the analysis presented below are
based on the ideal diode model.
2.3.1 Single-Phase Half Wave Diode Rectifier With Resistive Load
Fig.2.1 shows a single-phase half-wave diode rectifier with pure
resistive load. Assuming sinusoidal voltage source, VS the diode beings
to conduct when its anode voltage is greater than its cathode voltage as a
result, the load current flows. So, the diode will be in “ON” state in
positive voltage half cycle and in “OFF” state in negative voltage half
cycle. Fig.2.2 shows various current and voltage waveforms of half wave
diode rectifier with resistive load. These waveforms show that both the
load voltage and current have high ripples. For this reason, single-phase
half-wave diode rectifier has little practical significance.
The average or DC output voltage can be obtained by considering the
waveforms shown in Fig.2.2 as following:
∫ ==
π
π
ωω
π
0
sin
2
1 m
mdc
V
tdtVV (2.12)
Where, mV is the maximum value of supply voltage.
Because the load is resistor, the average or DC component of load
current is:
R
V
R
V
I mdc
dc
π
== (2.13)
The root mean square (rms) value of a load voltage is defined as:
2
sin
2
1
0
22 m
mrms
V
tdtVV == ∫
π
ωω
π
(2.14)
Similarly, the root mean square (rms) value of a load current is defined
as:
R
V
R
V
I mrms
rms
2
== (2.15)

26 Chapter Two
It is clear that the rms value of the transformer secondary current, SI
is the same as that of the load and diode currents
Then
R
V
II m
DS
2
== (2.15)
Where, DI is the rms value of diode current.
Fig.2.1 Single-phase half-wave diode rectifier with resistive load.
Fig.2.2 Various waveforms for half wave diode rectifier with resistive load.

Example 1: The rectifier shown in Fig.2.1 has a pure resistive load of R
Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) TUF
(e) Peak inverse voltage (PIV) of diode D1 and (f) Crest factor.
Solution: From Fig.2.2, the average output voltage dcV is defiend as:
π
π
π
ωω
π
π
mm
mdc
VV
tdtVV =−−== ∫ ))0cos(cos(
2
)sin(
2
1
0
Then,
R
V
R
V
I mdc
dc
π
==
2
)sin(
2
1
0
2 m
mrms
V
tVV == ∫
π
ω
π
,
R
V
I m
rms
2
= and,
2
m
S
V
V =
The rms value of the transformer secondery current is the same as that of
the load:
R
V
I m
S
2
= Then, the efficiency or rectification ratio is:
rmsrms
dcdc
ac
dc
IV
IV
P
P
*
*
==η %53.40
2
*
2
*
==
R
VV
R
VV
mm
mm
ππ
(b) 57.1
2
2 ====
π
π
m
m
dc
rms
V
V
V
V
FF
(c) 211.1157.11 22
=−=−== FF
V
V
RF
dc
ac
(d) %6.28286.0
22
====
R
VV
R
VV
IV
P
TUF
mm
mm
SS
dc ππ
(e) It is clear from Fig2.2 that the PIV is mV .
(f) Creast Factor CF, 2
2/
/)(
===
RV
RV
I
I
CF
m
m
S
peakS

28 Chapter Two
2.3.2 Half Wave Diode Rectifier With R-L Load
In case of RL load as shown in Fig.2.3, The voltage source, SV is an
alternating sinusoidal voltage source. If ( )tVv ms ωsin= , sv is positive
when 0 < ω t < π, and sv is negative when π < ω t <2π. When sv starts
becoming positive, the diode starts conducting and the source keeps the
diode in conduction till ω t reaches π radians. At that instant defined by
ω t =π radians, the current through the circuit is not zero and there is
some energy stored in the inductor. The voltage across an inductor is
positive when the current through it is increasing and it becomes negative
when the current through it tends to fall. When the voltage across the
inductor is negative, it is in such a direction as to forward-bias the diode.
The polarity of voltage across the inductor is as shown in the waveforms
shown in Fig.2.4.
When sv changes from a positive to a negative value, the voltage
across the diode changes its direction and there is current through the load
at the instant ω t = π radians and the diode continues to conduct till the
energy stored in the inductor becomes zero. After that, the current tends
to flow in the reverse direction and the diode blocks conduction. The
entire applied voltage now appears across the diode as reverse bias
voltage.
An expression for the current through the diode can be obtained by
solving the deferential equation representing the circuit. It is assumed that
the current flows for 0 < ω t < β, where β > π ( β is called the conduction
angle). When the diode conducts, the driving function for the differential
equation is the sinusoidal function defining the source voltage. During the
period defined by β < ω t < 2π, the diode blocks current and acts as an
open switch. For this period, there is no equation defining the behavior of
the circuit.
For 0 < ω t < β, the following differential equation defines the circuit:
βωω ≤≤=+ ttViR
dt
di
L m 0),(sin* (2.17)
Divide the above equation by L we get:
βωω ≤≤=+ tt
L
V
i
L
R
dt
di m
0),(sin* (2.18)
The instantaneous value of the current through the load can be
obtained from the solution of the above equation as following:

⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+= ∫
∫∫−
Adtt
L
V
eeti m
dt
L
R
dt
L
R
ωsin*)( (2.19)
Where A is a constant.
Then;
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+= ∫
−
Adtt
L
V
eeti m
t
L
R
t
L
R
ωsin*)( (2.20)
By integrating (2.20) (see appendix) we get:
( )
t
L
R
m
AetLtR
LwR
V
ti
−
+−
+
= ωωω cossin)( 222
(2.21)
Fig.2.3 Half Wave Diode Rectifier With R-L Load
Fig.2.4 Various waveforms for Half wave diode rectifier with R-L load.

30 Chapter Two
Assume wLjRZ +=∠φ
Then 2222
LwRZ += ,
φcosZR = , φω sinZL = and
R
Lω
φ =tan
Substitute these values into (2.21) we get the following equation:
( )
t
L
R
m
Aett
Z
V
ti
−
+−= ωφωφ cossinsincos)(
Then, ( )
t
L
R
m
Aet
Z
V
ti
−
+−= φωsin)( (2.22)
The above equation can be written in the following form:
( ) ( ) φ
ω
ω
ω
φωφω tan
sinsin)(
t
m
t
L
R
m
Aet
Z
V
Aet
Z
V
ti
−−
+−=+−= (2.23)
The value of A can be obtained using the initial condition. Since the
diode starts conducting at ω t = 0 and the current starts building up from
zero, ( ) 00 =i (discontinuous conduction). The value of A is expressed by
the following equation:
( )φsin
Z
V
A m
=
Once the value of A is known, the expression for current is known. After
evaluating A, current can be evaluated at different values of tω .
( ) ( )
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
+−=
−
φ
ω
φφωω tan
sinsin)(
t
m
et
Z
V
ti (2.24)
Starting from ω t = π, as tω increases, the current would keep
decreasing. For some value of tω , say β, the current would be zero. If ω t
> β, the current would evaluate to a negative value. Since the diode
blocks current in the reverse direction, the diode stops conducting when
tω reaches β. The value of β can be obtained by substituting that
β
ω =
= wt
ti 0)( into (2.24) we get:
( ) ( ) 0sinsin)( tan
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
+−=
−
φ
β
φφββ e
Z
V
i m (2.25)
R
wL
Z
Φ

The value of β can be obtained from the above equation by using the
methods of numerical analysis. Then, an expression for the average
output voltage can be obtained. Since the average voltage across the
inductor has to be zero, the average voltage across the resistor and the
average voltage at the cathode of the diode to ground are the same. This
average value can be obtained as shown in (2.26). The rms output voltage
in this case is shown in equation (2.27).
)cos1(*
2
sin*
2
0
β
π
ωω
π
β
−== ∫ mm
dc
V
tdt
V
V (2.26)
)2sin(1(5.0*
2
)sin(*
2
1
0
2
ββ
π
ω
π
β
−+== ∫
Vm
dwttVV mrms (2.27)
2.3.3 Single-Phase Half-Wave Diode Rectifier With Free Wheeling Diode
Single-phase half-wave diode rectifier with free wheeling diode is
shown in Fig.2.5. This circuit differs from the circuit described above,
which had only diode D1. This circuit shown in Fig.2.5 has another
diode, marked D2. This diode is called the free-wheeling diode.
Let the source voltage sv be defined as ( )tVm ωsin which is positive
when πω << t0 radians and it is negative when π < ω t < 2π radians.
When sv is positive, diode D1 conducts and the output voltage, ov
become positive. This in turn leads to diode D2 being reverse-biased
during this period. During π < wt < 2π, the voltage ov would be negative
if diode D1 tends to conduct. This means that D2 would be forward-
biased and would conduct. When diode D2 conducts, the voltage ov
would be zero volts, assuming that the diode drop is negligible.
Additionally when diode D2 conducts, diode D1 remains reverse-biased,
because the voltage across it is sv which is negative.
Fig.2.5 Half wave diode rectifier with free wheeling diode.

32 Chapter Two
When the current through the inductor tends to fall (when the supply
voltage become negative), the voltage across the inductor become
negative and its voltage tends to forward bias diode D2 even when the
source voltage sv is positive, the inductor current would tend to fall if the
source voltage is less than the voltage drop across the load resistor.
During the negative half-cycle of source voltage, diode D1 blocks
conduction and diode D2 is forced to conduct. Since diode D2 allows the
inductor current circulate through L, R and D2, diode D2 is called the
free-wheeling diode because the current free-wheels through D2.
Fig.2.6 shows various voltage waveforms of diode rectifier with free-
wheeling diode. Fig.2.7 shows various current waveforms of diode
rectifier with free-wheeling diode.
It can be assumed that the load current flows all the time. In other
words, the load current is continuous. When diode D1 conducts, the
driving function for the differential equation is the sinusoidal function
defining the source voltage. During the period defined by π < ω t < 2π,
diode D1 blocks current and acts as an open switch. On the other hand,
diode D2 conducts during this period, the driving function can be set to
be zero volts. For 0 < ω t < π, the differential equation (2.18) applies. The
solution of this equation will be as obtained before in (2.20) or (2.23).
( ) ( )
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
+−=
−
φ
ω
φφωω tan
sinsin)(
t
m et
Z
V
ti πω << t0 (2.28)
For the negative half-cycle ( πωπ 2<< t ) of the source voltage D1 is
OFF and D2 is ON. Then the driving voltage is set to zero and the
following differential equation represents the circuit in this case.
πωπ 20* <<=+ tforiR
td
id
L (2.29)
The solution of (2.29) is given by the following equation:
φ
πω
ω tan
)(
−
−
=
t
eBti (2.30)
The constant B can be obtained from the boundary condition where
Bi =)(π is the starting value of the current in πωπ 2<< t and can be
obtained from equation (2.23) by substituting πω =t
Then, ( ) ( ) Be
Z
V
i m
=+−=
−
)sin(sin)( tanφ
π
φφππ

The above value of )(πi can be used as initial condition of equation
(2.30). Then the load current during πωπ 2<< t is shown in the
following equation.
( ) ( ) φ
πω
φ
π
φφπω tantan
sinsin)(
−
−−
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
+−=
t
m
ee
Z
V
ti for πωπ 2<< t (2.31)
Fig.2.6 Various voltage waveforms of diode rectifier with free-wheeling diode.
Fig.2.7 Various current waveforms of diode rectifier with free-wheeling diode.

34 Chapter Two
For the period πωπ 32 << t the value of )2( πi from (2.31) can be
used as initial condition for that period. The differential equation
representing this period is the same as equation (2.28) by replacing ω t by
πω 2−t and the solution is given by equation (2.32). This period
( πωπ 32 << t ) differ than the period π<< wt0 in the way to get the
constant A where in the πω << t0 the initial value was 0)0( =i but in
the case of πωπ 32 << t the initial condition will be )2( πi that given
from (2.31) and is shown in (2.33).
( ) φ
πω
φπωω tan
2
2sin)(
−
−
+−−=
t
m Aet
Z
V
ti for πωπ 32 << t (2.32)
The value of ( )π2i can be obtained from (2.31) and (2.32) as shown
in (2.33) and (2.34) respectively.
( ) ( ) φ
π
φ
π
φφππ tantan
sinsin)2(
−−
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
+−= ee
Z
V
i m (2.33)
( ) A
Z
V
i m
+−= φπ sin)2( (2.34)
By equating (2.33) and (2.34) the constant A in πωπ 32 << t can be
obtained from the following equation:
( ) ( )φπ sin2
Z
V
iA m
+= (2.35)
Then, the general solution for the period πωπ 32 << t is given by
equation (2.36):
( ) ( ) ( ) φ
πω
φπφπωω tan
2
sin22sin)(
−
−
⎟
⎠
⎞
⎜
⎝
⎛
++−−=
t
mm
e
Z
V
it
Z
V
ti πωπ 32 << t (2.36)
Where ( )π2i can be obtained from equation (2.33).
Example 2 A diode circuit shown in Fig.2.3 with R=10 Ω, L=20mH, and
VS=220 2 sin314t.
(a) Determine the expression for the current though the load in the
period πω 20 << t and determine the conduction angle β .
(b) If we connect free wheeling diode through the load as shown in
Fig.2.5 Determine the expression for the current though the load
in the period of πω 30 << t .

Solution: (a) For the period of πω << t0 , the expression of the load
current can be obtained from (2.24) as following:
.561.0
10
10*20*314
tantan
3
11
rad
R
L
===
−
−− ω
φ and 628343.0tan =φ
Ω=+=+= −
8084.11)10*20*314(10)( 23222
LRZ ω
( ) ( )
( )[ ]t
t
m
et
et
Z
V
ti
ω
φ
ω
ω
φφωω
5915.1
tan
*532.0561.0sin
8084.11
2220
sinsin)(
−
−
+−=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
+−=
( ) t
etti ω
ωω 5915.1
*0171.14561.0sin3479.26)( −
+−=
The value of β can be obtained from the above equation by substituting
for 0)( =βi . Then, ( ) β
β 5915.1
*0171.14561.0sin3479.260 −
+−= e
By using the numerical analysis we can get the value of β. The
simplest method is by using the simple iteration technique by assuming
( ) β
β 5915.1
*0171.14561.0sin3479.26 −
+−=Δ e and substitute different
values for β in the region πβπ 2<< till we get the minimum value of Δ
then the corresponding value of β is the required value. The narrow
intervals mean an accurate values of β . The following table shows the
relation between β and Δ:
β Δ
1.1 π 6.49518
1.12 π 4.87278
1.14 π 3.23186
1.16 π 1.57885
1.18 π -0.079808
1.2 π -1.73761
It is clear from the above table that πβ 18.1≅ rad. The current in
πβ 2<< wt will be zero due to the diode will block the negative current
to flow.
(b) In case of free-wheeling diode as shown in Fig.2.5, we have to divide
the operation of this circuit into three parts. The first one when

36 Chapter Two
πω << t0 (D1 “ON”, D2 “OFF”), the second case when πωπ 2<< t
(D1 “OFF” and D2 “ON”) and the last one when πωπ 32 << t (D1
“ON”, D2 “OFF”).
In the first part ( πω << t0 ) the expression for the load current
can be obtained as In case (a). Then:
( ) wt
etwti 5915.1
*0171.14561.0sin3479.26)( −
+−= ω for πω << t0
the current at πω =t is starting value for the current in the next part.
Then
( ) Aei 1124.14*0171.14561.0sin3479.26)( 5915.1
=+−= − π
ππ
In the second part πωπ 2<< t , the expression for the load current
can be obtained from (2.30) as following:
φ
πω
ω tan
)(
−
−
=
t
eBti
where AiB 1124.14)( == π
Then ( )πω
ω −−
= t
eti 5915.1
1124.14)( for ( πωπ 2<< t )
The current at πω 2=t is starting value for the current in the next part.
Then
Ai 095103.0)2( =π
In the last part ( πωπ 32 << t ) the expression for the load current
can be obtained from (2.36):
( ) ( ) ( ) φ
πω
φπφπωω tan
2
sin22sin)(
−
−
⎟
⎠
⎞
⎜
⎝
⎛
++−−=
t
mm e
Z
V
it
Z
V
ti
( ) ( ) ( )πω
ωω 25915.1
532.0*3479.26095103.08442.6sin3479.26)( −−
++−=∴ t
etti
( ) ( )πω
ωω 25915.1
1131.148442.6sin3479.26)( −−
+−=∴ t
etti for
( πωπ 32 << t )
2.4 Single-Phase Full-Wave Diode Rectifier
The full wave diode rectifier can be designed with a center-taped
transformer as shown in Fig.2.8, where each half of the transformer with
its associated diode acts as half wave rectifier or as a bridge diode
rectifier as shown in Fig. 2.12. The advantage and disadvantage of center-
tap diode rectifier is shown below:

Advantages
• The need for center-tapped transformer is eliminated,
• The output is twice that of the center tapped circuit for the same
secondary voltage, and,
• The peak inverse voltage is one half of the center-tap circuit.
Disadvantages
• It requires four diodes instead of two, in full wave circuit, and,
• There are always two diodes in series are conducting. Therefore,
total voltage drop in the internal resistance of the diodes and losses
are increased.
The following sections explain and analyze these rectifiers.
2.4.1 Center-Tap Diode Rectifier With Resistive Load
In the center tap full wave rectifier, current flows through the load in
the same direction for both half cycles of input AC voltage. The circuit
shown in Fig.2.8 has two diodes D1 and D2 and a center tapped
transformer. The diode D1 is forward bias “ON” and diode D2 is reverse
bias “OFF” in the positive half cycle of input voltage and current flows
from point a to point b. Whereas in the negative half cycle the diode D1
is reverse bias “OFF” and diode D2 is forward bias “ON” and again
current flows from point a to point b. Hence DC output is obtained across
the load.
Fig.2.8 Center-tap diode rectifier with resistive load.
In case of pure resistive load, Fig.2.9 shows various current and
voltage waveform for converter in Fig.2.8. The average and rms output
voltage and current can be obtained from the waveforms shown in Fig.2.9
as shown in the following:

38 Chapter Two
π
ωω
π
π
m
mdc
V
tdtVV
2
sin
1
0
== ∫ (2.36)
R
V
I m
dc
π
2
= (2.37)
( )
2
sin
1
0
2 m
mrms
V
tdtVV == ∫
π
ωω
π
(2.38)
R
V
I m
rms
2
= (2.39)
PIV of each diode = mV2 (2.40)
2
m
S
V
V = (2.41)
The rms value of the transformer secondery current is the same as that of
the diode:
R
V
II m
DS
2
== (2.41)
Fig.2.9 Various current and voltage waveforms for center-tap diode rectifier
with resistive load.

Example 3. The rectifier in Fig.2.8 has a purely resistive load of R
Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) TUF
(e) Peak inverse voltage (PIV) of diode D1 and(f) Crest factor of
transformer secondary current.
Solution:- The efficiency or rectification ratio is
%05.81
2
*
2
2
*
2
*
*
====
R
VV
R
VV
IV
IV
P
P
mm
mm
rmsrms
dcdc
ac
dc ππ
η
(b) 11.1
222
2 ====
π
π
m
m
dc
rms
V
V
V
V
FF
(c) 483.0111.11 22
=−=−== FF
V
V
RF
dc
ac
(d) 5732.0
22
2
22
2
===
R
VV
R
VV
IV
P
TUF
mm
mm
SS
dc ππ
(e) The PIV is mV2
(f) Creast Factor of secondary current, 2
2
)(
===
R
V
R
V
I
I
CF
m
m
S
peakS
2.4.2 Center-Tap Diode Rectifier With R-L Load
Center-tap full wave rectifier circuit with RL load is shown in Fig.2.10.
Various voltage and current waveforms for Fig.2.10 is shown in Fig.2.11.
An expression for load current can be obtained as shown below:
It is assumed that D1 conducts in positive half cycle of VS and D2
conducts in negative half cycle. So, the deferential equation defines the
circuit is shown in (2.43).
)sin(* tViR
td
id
L m ω=+ (2.43)
The solution of the above equation can be obtained as obtained before
in (2.24)

40 Chapter Two
Fig.2.10 Center-tap diode rectifier with R-L load
Fig.2.11 Various current and voltage waveform for Center-tap diode rectifier
with R-L load
( ) ( )
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
+−=
−
φ
ω
φφωω tan
sinsin)(
t
m et
Z
V
ti for πω << t0 (2.44)
In the second half cycle the same differential equation (2.43) and the
solution of this equation will be as obtained before in (2.22)

( ) φ
πω
φπωω tan
sin)(
−
−
+−−=
t
m
Aet
Z
V
ti (2.45)
The value of constant A can be obtained from initial condition. If we
assume that i(π)=i(2π)=i(3π)=……..=Io (2.46)
Then the value of oI can be obtained from (2.44) by letting πω =t
( ) ( )
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
+−==
−
φ
π
φφππ tan
sinsin)( e
Z
V
iI m
o (2.47)
Then use the value of oI as initial condition for equation (2.45). So we
can obtain the value of constant A as following:
( ) φ
ππ
φπππ tan
sin)(
−
−
+−−== Ae
Z
V
Ii m
o
Then; ( )φsin
Z
V
IA m
o += (2.48)
Substitute (2.48) into (2.45) we get:
( ) ( ) φ
πω
φφπωω tan
sinsin)(
−
−
⎟
⎠
⎞
⎜
⎝
⎛
++−−=
t
m
o
m e
Z
V
It
Z
V
ti , then,
( ) ( ) φ
πω
φ
πω
φφπωω tantan
sinsin)(
−
−
−
−
+
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+−−=
t
o
t
m eIet
Z
V
ti (for πωπ 2<< t ) (2.49)
In the next half cycle πωπ 32 << t the current will be same as
obtained in (2.49) but we have to take the time shift into account where
the new equation will be as shown in the following:
( ) ( ) φ
πω
φ
πω
φφπω tan
2
tan
2
sin2sin)(
−
−
−
−
+
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+−−=
t
o
t
m
eIewt
Z
V
ti (for πωπ 32 << t )(2.50)
2.4.3 Single-Phase Full Bridge Diode Rectifier With Resistive Load
Another alternative in single-phase full wave rectifier is by using four
diodes as shown in Fig.2.12 which known as a single-phase full bridge
diode rectifier. It is easy to see the operation of these four diodes. The
current flows through diodes D1 and D2 during the positive half cycle of
input voltage (D3 and D4 are “OFF”). During the negative one, diodes
D3 and D4 conduct (D1 and D2 are “OFF”).

42 Chapter Two
In positive half cycle the supply voltage forces diodes D1 and D2 to be
"ON". In same time it forces diodes D3 and D4 to be "OFF". So, the
current moves from positive point of the supply voltage across D1 to the
point a of the load then from point b to the negative marked point of the
supply voltage through diode D2. In the negative voltage half cycle, the
supply voltage forces the diodes D1 and D2 to be "OFF". In same time it
forces diodes D3 and D4 to be "ON". So, the current moves from
negative marked point of the supply voltage across D3 to the point a of
the load then from point b to the positive marked point of the supply
voltage through diode D4. So, it is clear that the load currents moves
from point a to point b in both positive and negative half cycles of supply
voltage. So, a DC output current can be obtained at the load in both
positive and negative halves cycles of the supply voltage. The complete
waveforms for this rectifier is shown in Fig.2.13
Fig.2.12 Single-phase full bridge diode rectifier.
Fig.2.13 Various current and voltage waveforms of Full bridge single-phase
diode rectifier.

Example 4 The rectifier shown in Fig.2.12 has a purely resistive load of
R=15 Ω and, VS=300 sin 314 t and unity transformer ratio. Determine (a)
The efficiency, (b) Form factor, (c) Ripple factor, (d) TUF, (e) The peak
inverse voltage, (PIV) of each diode, (f) Crest factor of input current, and,
(g) Input power factor.
Solution: 300=mV V
V
V
tdtVV m
mdc 956.190
2
sin
1
0
=== ∫ π
ωω
π
π
, A
R
V
I m
dc 7324.12
2
==
π
( ) V
V
tdtVV m
mrms 132.212
2
sin
1
2/1
0
2
==
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
= ∫
π
ωω
π
, A
R
V
I m
rms 142.14
2
==
(a) %06.81===
rmsrms
dcdc
ac
dc
IV
IV
P
P
η
(b) 11.1==
dc
rms
V
V
FF
(c) 482.011 2
2
222
=−=−=
−
== FF
V
V
V
VV
V
V
RF
dc
rms
dc
dcrms
dc
ac
(d) %81
142.14*132.212
7324.12*986.190
===
SS
dc
IV
P
TUF
(e) The PIV= mV =300V
(f) 414.1
142.14
15/300)(
===
S
peakS
I
I
CF
(g) Input power factor = 1
*Re 2
==
SS
rms
IV
RI
PowerApperant
Poweral
2.4.4 Full Bridge Single-phase Diode Rectifier with DC Load Current
The full bridge single-phase diode rectifier with DC load current is
shown in Fig.2.14. In this circuit the load current is pure DC and it is
assumed here that the source inductances is negligible. In this case, the
circuit works as explained before in resistive load but the current
waveform in the supply will be as shown in Fig.2.15.
The rms value of the input current is oS II =

44 Chapter Two
Fig.2.14 Full bridge single-phase diode rectifier with DC load current.
Fig.2.15 Various current and voltage waveforms for full bridge single-phase
diode rectifier with DC load current.
The supply current in case of pure DC load current is shown in
Fig.2.15, as we see it is odd function, then na coefficients of Fourier
series equal zero, 0=na , and
[ ]
[ ] .............,5,3,1
4
cos0cos
2
cos
2
sin*
2
0
0
==−=
−== ∫
nfor
n
I
n
n
I
tn
n
I
tdtnIb
oo
o
on
π
π
π
ω
π
ωω
π
π
π
(2.51)
Then from Fourier series concepts we can say:
)..........9sin
9
1
7sin
7
1
5sin
5
1
3sin
3
1
(sin*
4
)( +++++= ttttt
I
ti o
ωωωωω
π
(2.52)

%46
15
1
13
1
11
1
9
1
7
1
5
1
3
1
))((
2222222
=⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
=∴ tITHD s
or we can obtain ))(( tITHD s as the following:
From (2.52) we can obtain the value of is
π2
4
1
o
S
I
I =
%34.481
4
2
1
2
4
1))((
2
2
2
1
=−⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=−
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
=−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=∴
π
π
o
o
S
S
s
I
I
I
I
tITHD
Example 5 solve Example 4 if the load is 30 A pure DC
Solution: From example 4 Vdc= 190.986 V, Vrms=212.132 V
AIdc 30= and rmsI = 30 A
(a) %90===
rmsrms
dcdc
ac
dc
IV
IV
P
P
η
(b) 11.1==
dc
rms
V
V
FF
(c) 482.011 2
2
222
=−=−=
−
== FF
V
V
V
VV
V
V
RF
dc
rms
dc
dcrms
dc
ac
(d) %90
30*132.212
30*986.190
===
SS
dc
IV
P
TUF
(e) The PIV=Vm=300V
(f) 1
30
30)(
===
S
peakS
I
I
CF
(g) A
I
I o
S 01.27
2
30*4
2
4
1 ===
ππ
Input Power factor= =
PowerApperant
PoweralRe
Lag
I
I
IV
IV
S
S
SS
SS
9.01*
30
01.27cos*cos* 11 ====
φφ

46 Chapter Two
2.4.5 Effect Of LS On Current Commutation Of Single-Phase Diode
Bridge Rectifier.
Fig.2.15 Shows the single-phase diode bridge rectifier with source
inductance. Due to the value of LS the transitions of the AC side current
Si from a value of oI to oI− (or vice versa) will not be instantaneous.
The finite time interval required for such a transition is called
commutation time. And this process is called current commutation
process. Various voltage and current waveforms of single-phase diode
bridge rectifier with source inductance are shown in Fig.2.16.
Fig.2.15 Single-phase diode bridge rectifier with source inductance.
Fig.2.16 Various current and voltage waveforms for single-phase diode bridge
rectifier with source inductance.

Let us study the commutation time starts at t=10 ms as indicated in
Fig.2.16. At this time the supply voltage starts to be negative, so diodes
D1 and D2 have to switch OFF and diodes D3 and D4 have to switch ON
as explained in the previous case without source inductance. But due to
the source inductance it will prevent that to happen instantaneously. So, it
will take time tΔ to completely turn OFF D1 and D2 and to make D3 and
D4 carry the entire load current ( oI ). Also in the time tΔ the supply
current will change from oI to oI− which is very clear in Fig.2.16.
Fig.2.17 shows the equivalent circuit of the diode bridge at time tΔ .
Fig.2.17 The equivalent circuit of the diode bridge at commutation time tΔ .
From Fig.2.17 we can get the following equations
0=−
dt
di
LV S
sS (2.53)
Multiply the above equation by tdω then,
SsS diLtdV ωω = (2.54)
Integrate both sides of the above equation during the commutation
period ( tΔ sec or u rad.) we get the following:
SsS diLtdV ωω =
∫∫
−+
=
o
o
I
I
Ss
u
m diLtdtV ωωω
π
π
sin (2.55)
Then; ( )[ ] osm ILuV ωππ 2coscos −=+−
Then; ( )[ ] osm ILuV ω2cos1 −=+−

48 Chapter Two
Then; ( )
m
os
V
IL
u
ω2
1cos −=
Then; ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−= −
m
os
V
IL
u
ω2
1cos 1
(2.56)
And ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−==Δ −
m
os
V
ILu
t
ω
ωω
2
1cos
1 1
(2.57)
It is clear that the DC voltage reduction due to the source inductance is
the drop across the source inductance.
dt
di
Lv S
srd = (2.58)
Then oS
I
I
SS
u
rd ILdiLtdv
o
o
ωωω
π
π
2−== ∫∫
−+
(2.59)
∫
+u
rd tdv
π
π
ω is the reduction area in one commutation period tΔ . But we
have two commutation periods tΔ in one period of supply voltage. So the
total reduction per period is: oS
u
rd ILtdv ωω
π
π
42 −=∫
+
(2.60)
To obtain the average reduction in DC output voltage rdV due to
source inductance we have to divide the above equation by the period
time π2 . Then;
oS
oS
rd ILf
IL
V 4
2
4
−=
−
=
π
ω
(2.61)
The DC voltage with source inductance tacking into account can be
calculated as following:
os
m
rdceinducsourcewithoutdcactualdc IfL
V
VVV 4
2
tan
−=−=
π
(2.62)
To obtain the rms value and Fourier transform of the supply current it
is better to move the vertical axis to make the waveform odd or even this
will greatly simplfy the analysis. So, it is better to move the vertical axis
of supply current by 2/u as shown in Fig.2.18. Moveing the vertical axis
will not change the last results. If you did not bleave me keep going in the
analysis without moveing the axis.

Fig. 2.18 The old axis and new axis for supply currents.
Fig.2.19 shows a symple drawing for the supply current. This drawing
help us in getting the rms valuof the supply current. It is clear from the
waveform of supply current shown in Fig.2.19 that we obtain the rms
value for only a quarter of the waveform because all for quarter will be
the same when we squaret the waveform as shown in the following
equation:
]
2
[
2
2/
0
2
2/
2
2
∫ ∫+⎟
⎠
⎞
⎜
⎝
⎛
=
u
u
o
o
s tdItdt
u
I
I
π
ωωω
π
(2.63)
Then; ⎥⎦
⎤
⎢⎣
⎡
−=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−+=
32
2
2283
42 23
2
2
uIuu
u
I
I oo
s
π
π
π
π
(2.64)
2
u
−
oI
oI− 2
u
−π2
u
π
2
u
+π
2
2
u
−π
π2
u
sI
2
π
Fig.2.19 Supply current waveform

50 Chapter Two
To obtain the Fourier transform for the supply current waveform you
can go with the classic fourier technique. But there is a nice and easy
method to obtain Fourier transform of such complcated waveform known
as jump technique [ ]. In this technique we have to draw the wave form
and its drevatives till the last drivative values all zeros. Then record the
jump value and its place for each drivative in a table like the table shown
below. Then; substitute the table values in (2.65) as following:
2
u
−
oI
oI−
2
u
−π
u
Io2
u
Io2
−
sI′
2
u
π
2
u
+π
2
2
u
−π
π2
2
u
−
2
u
−π
2
u
π
2
u
+π
2
2
u
−π
u
sI
Fig.2.20 Supply current and its first derivative.
Table(2.1) Jumb value of supply current and its first derivative.
sJ
2
u
−
2
u
2
u
−π
2
u
+π
sI 0 0 0 0
sI′
u
Io2
u
Io2
−
u
Io2
−
u
Io2

It is an odd function, then 0== no aa
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
′−= ∑∑
==
m
s
ss
m
s
ssn tnJ
n
tnJ
n
b
11
sin
1
cos
1
ωω
π
(2.65)
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
++⎟
⎠
⎞
⎜
⎝
⎛
−−⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
−
−
=
2
sin
2
sin
2
sin
2
sin
2
*
11 u
n
u
n
u
n
u
n
u
I
nn
b o
n ππ
π
2
sin*
8
2
nu
un
I
b o
n
π
= (2.66)
2
sin*
8
1
u
u
I
b o
π
= (2.67)
Then;
2
sin*
2
8
1
u
u
I
I o
S
π
= (2.68)
( )
⎥⎦
⎤
⎢⎣
⎡
−
=
⎥⎦
⎤
⎢⎣
⎡
−
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎠
⎞
⎜
⎝
⎛
⎥⎦
⎤
⎢⎣
⎡
−
=⎟
⎠
⎞
⎜
⎝
⎛
=
32
sin2
32
2
cos
2
sin4
2
cos
32
2
2
sin*
2
8
2
cos*
2
1
u
u
u
u
u
uu
u
uI
u
u
I
u
I
I
pf
o
o
S
S
π
π
π
π
π
π
π
(2.69)
Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz,
source inductance mHXs 5= supply to feed 200 A pure DC load, find:
i. Average DC output voltage.
ii. Power factor.
iii. Determine the THD of the utility line current.
Solution: (i) From (2.62), VVm 155562*11000 ==
os
m
rdceinducsourcewithoutdcactualdc IfL
V
VVV 4
2
tan
−=−=
π
VV actualdc 9703200*005.0*50*4
15556*2
=−=
π
(ii) From (2.56) the commutation angle u can be obtained as following:

52 Chapter Two
.285.0
15556
200*005.0*50**2*2
1cos
2
1cos 11
rad
V
IL
u
m
os
=⎟
⎠
⎞
⎜
⎝
⎛
−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−= −− πω
The input power factor can be obtained from (2.69) as following
( ) ( ) 917.0
3
285.
2
285.0
285.0sin*2
32
sin*2
2
cos*1
=
⎥⎦
⎤
⎢⎣
⎡
−
=
⎥⎦
⎤
⎢⎣
⎡
−
=⎟
⎠
⎞
⎜
⎝
⎛
=
π
π
π
π
u
u
uu
I
I
pf
S
S
A
uI
I o
S 85.193
3
285.0
2
200*2
32
2 22
=⎥⎦
⎤
⎢⎣
⎡
−=⎥⎦
⎤
⎢⎣
⎡
−=
π
π
π
π
A
u
u
I
I o
S 46.179
2
285.0
sin*
285.0*2
200*8
2
sin*
2
8
1 =⎟
⎠
⎞
⎜
⎝
⎛
==
ππ
%84.401
46.179
85.193
1
22
1
=−⎟
⎠
⎞
⎜
⎝
⎛
=−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
S
S
i
I
I
THD
2.5 Three Phase Diode Rectifiers
2.5.1 Three-Phase Half Wave Rectifier
Fig.2.21 shows a half wave three-phase diode rectifier circuit with
delta star three-phase transformer. In this circuit, the diode with highest
potential with respect to the neutral of the transformer conducts. As the
potential of another diode becomes the highest, load current is transferred
to that diode, and the previously conduct diode is reverse biased “OFF
case”.
Fig.2.21 Half wave three-phase diode rectifier circuit with delta star three-phase
transformer.

For the rectifier shown in Fig.2.21 the load voltage, primary diode
currents and its FFT components are shown in Fig.2.22, Fig.2.23 and
Fig.2.24 respectively.
Fig.2.22 Secondary and load voltages of half wave three-phase diode rectifier.
Fig.2.23 Primary and diode currents.
6
π
6
5π

54 Chapter Two
Fig.2.24 FFT components of primary and diode currents.
By considering the interval from
6
π
to
6
5π
in the output voltage we can
calculate the average and rms output voltage and current as following:
m
m
mdc V
V
tdtVV 827.0
2
33
sin
2
3
6/5
6/
=== ∫ π
ωω
π
π
π
(2.70)
R
V
R
V
I mm
dc
*827.0
**2
33
==
π
(2.71)
( ) mmmrms VVtdtVV 8407.0
8
3*3
2
1
sin
2
3
6/5
6/
2
=+== ∫ π
ωω
π
π
π
(2.72)
R
V
I m
rms
8407.0
= (2.73)
Then the diode rms current is equal to secondery current and can be
obtaiend as following:
R
V
R
V
II mm
Sr 4854.0
3
08407
=== (2.74)
Note that the rms value of diode current has been obtained from the
rms value of load current divided by 3 because the diode current has
one third pulse of similar three pulses in load current.
ThePIV of the diodes is mLL VV 32 = (2.75)
Primary current
Diode current

Example 7 The rectifier in Fig.2.21 is operated from 460 V 50 Hz supply
at secondary side and the load resistance is R=20 Ω. If the source
inductance is negligible, determine (a) Rectification efficiency, (b) Form
factor (c) Ripple factor (d) Transformer utilization factor, (e) Peak
inverse voltage (PIV) of each diode and (f) Crest factor of input current.
Solution:
(a) VVVV mS 59.3752*58.265,58.265
3
460
====
m
m
dc V
V
V 827.0
2
33
==
π
,
R
V
R
V
I mm
dc
0827
2
33
==
π
mrms VV 8407.0=
R
V
I m
rms
8407.0
=
%767.96===
rmsrms
dcdc
ac
dc
IV
IV
P
P
η
(b) %657.101==
dc
rms
V
V
FF
(c) %28.1811 2
2
222
=−=−=
−
== FF
V
V
V
VV
V
V
RF
dc
rms
dc
dcrms
dc
ac
(d)
R
V
II m
rmsS
3
8407.0
3
1
==
%424.66
3
8407.0
*2/*3
/)827.0(
*3
2
===
R
V
V
RV
IV
P
TUF
m
m
m
SS
dc
(e) The PIV= 3 Vm=650.54V
(f) 06.2
3
8407.0
/)(
===
R
V
RV
I
I
CF
m
m
S
peakS

56 Chapter Two
2.5.2 Three-Phase Half Wave Rectifier With DC Load Current and
zero source inductance
In case of pure DC load current as shown in Fig.2.25, the diode current
and primary current are shown in Fig.2.26.
Fig.2.25 Three-phase half wave rectifier with dc load current
To calculate Fourier transform of the diode current of Fig.2.26, it is
better to move y axis to make the function as odd or even to cancel one
coefficient an or bn respectively. If we put Y-axis at point o
t 30=ω then
we can deal with the secondary current as even functions. Then, 0=nb of
secondary current. Values of na can be calculated as following:
32
1
3/
3/
0
o
o
I
tdIa ∫
−
==
π
π
ω
π
(2.76)
[ ]
harmonicstrepleanallfor
nfor
n
I
nfor
n
I
tn
n
I
dwttnIa
o
o
o
on
0
17,16,11,10,5,43*
,....14,13,8,7,2,13*
sin
cos*
1
3/
3/
3/
3/
=
=−=
==
=
=
−
−
∫
π
π
ω
π
ω
π
π
π
π
π
(2.77)
⎟
⎠
⎞
⎜
⎝
⎛
−−++−−++= ...8sin
8
1
7sin
7
1
5sin
5
1
4sin
4
1
2sin
2
1
sin
3
3
)( tttttt
II
tI OO
s ωωωωωω
π
(2.78)

Fig.2.26 Primary and secondary current waveforms and FFT components of
three-phase half wave rectifier with dc load current
Newaxis

58 Chapter Two
%24.1090924.11
9
*2
1
2
3
3/
1))((
2
2
2
1
==−=−
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎝
⎛
=−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
π
π
O
o
S
S
s
I
I
I
I
tITHD
It is clear that the primary current shown in Fig.2.26 is odd, then, an=0,
[ ]
harmonicstrepleanallfor
nfor
n
I
tn
n
I
tdtnIb
o
o
on
0
,....14,13,11,10,8,7,5,4,2,1
3
cos
2
sin*
2 3/2
0
3/2
0
=
==
−== ∫
π
ω
π
ωω
π
π
π
(2.79)
⎟
⎠
⎞
⎜
⎝
⎛
−−+++++= ...8sin
8
1
7sin
7
1
5sin
5
1
4sin
4
1
2sin
2
1
sin
3
)( tttttt
I
ti O
P ωωωωωω
π
(2.80)
The rms value of oP II
3
2
= (2.81)
%983.671
33
2
1
2
3
3
2
1))((
2
2
2
1
=−⎟
⎠
⎞
⎜
⎝
⎛
=−
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
=−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
π
π
O
o
P
P
P
I
I
I
I
tITHD (2.82)
Example 8 Solve example 7 if the load current is 100 A pure DC
Solution: (a) VVVV mS 59.3752*58.265,58.265
3
460
====
VV
V
V m
m
dc 613.310827.0
2
33
===
π
, AIdc 100=
VVV mrms 759.3158407.0 == , AIrms 100=
%37.98
100*759.315
100*613.310
====
rmsrms
dcdc
ac
dc
IV
IV
P
P
η
(b) %657.101==
dc
rms
V
V
FF
(c) %28.1811 2
2
222
=−=−=
−
== FF
V
V
V
VV
V
V
RF
dc
rms
dc
dcrms
dc
ac

(d) AII rmsS 735.57100*
3
1
3
1
===
%52.67
735.57*2/*3
100*613.310
*3
===
mSS
dc
VIV
P
TUF
(f) 732.1
735.57
100)(
===
S
peakS
I
I
CF
2.5.3 Three-Phase Half Wave Rectifier With Source Inductance
The source inductance in three-phase half wave diode rectifier Fig.2.27
will change the shape of the output voltage than the ideal case (without
source inductance) as shown in Fig.2.28. The DC component of the
output voltage is reduced due to the voltage drop on the source
inductance. To calculate this reduction we have to discuss Fig.2.27 with
reference to Fig.2.28. As we see in Fig.2.28 when the voltage bv is going
to be greater than the voltage av at time t (at the arrow in Fig.2.28) the
diode D1 will try to turn off, in the same time the diode D2 will try to
turn on but the source inductance will slow down this process and makes
it done in time tΔ (overlap time or commutation time). The overlap time
will take time tΔ to completely turn OFF D1 and to make D2 carry the
entire load current ( oI ). Also in the time tΔ the current in bL will change
from zero to oI and the current in aL will change from oI to zero. This is
very clear from Fig.2.28. Fig.2.29 shows the equivalent circuit of three
phase half wave diode bridge in commutation period tΔ .
Fig.2.27 Three-phase half wave rectifier with load and source inductance.

60 Chapter Two
Fig.2.28 Supply current and output voltage for three-phase half wave
rectifier with pure DC load and source inductance.
Fig.2.29 The equivalent circuit for three-phase half wave diode rectifier in
commutation period.
From Fig.2.29 we can get the following equations
01 =−− dc
D
aa V
dt
di
Lv (2.83)
02 =−− dc
D
bb V
dt
di
Lv (2.84)
subtract (2.84) from(2.83) we get:

012 =⎟
⎠
⎞
⎜
⎝
⎛
−+−
dt
di
dt
di
Lvv DD
ba
Multiply the above equation by tdω the following equation can be
obtained: ( ) ( ) 012 =−+− DDba didiLtdvv ωω
substitute the voltage waveforms of av and bv into the above equation
we get: ( ) ( )21
3
2
sinsin DDmm didiLtdtVtV −=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
−− ωω
π
ωω
Then; ( )21
6
sin3 DDm didiLtdtV −=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+ ωω
π
ω
Integrating both parts of the above equation we get the following:
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
−=⎟
⎠
⎞
⎜
⎝
⎛
+ ∫∫∫
+
o
o
I
D
I
D
u
m didiLtdtV
0
2
0
1
6
5
6
5
6
sin3 ωω
π
ω
π
π
Then; om LIuV ω
ππππ
2
66
5
cos
66
5
cos3 −=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
++−⎟
⎠
⎞
⎜
⎝
⎛
+
Then; ( ) ( )( ) om LIuV ωππ 2coscos3 −=+−
Then; ( )( ) om LIuV ω2cos13 −=+−
Then; ( )
m
o
V
LI
u
3
2
cos1
ω
=−
Then; ( )
m
o
V
LI
u
3
2
1cos
ω
−=
Then ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−= −
m
o
V
LI
u
3
2
1cos 1 ω
(2.85)
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−==Δ −
m
o
V
LIu
t
3
2
1cos
1 1 ω
ωω
(2.86)
equal to the drop across the source inductance. Then;
dt
di
Lv D
rd =

62 Chapter Two
Then, o
I
D
u
rd LIdiLtdv
o
ωωω
π
π
== ∫∫
+
0
6
5
6
5
(2.87)
∫
+u
rd tdv
6
5
6
5
π
π
ω is the reduction area in one commutation period tΔ . But,
we have three commutation periods, tΔ in one period. So, the total
reduction per period is:
o
u
rd LItdv ωω
π
π
3*3
6
5
6
5
=∫
+
source inductance we have to divide the total reduction per period by π2
as following:
o
o
rd ILf
LI
V 3
2
3
==
π
ω
(2.88)
Then, the DC component of output voltage due to source inductance is:
o
ceinduc
source
without
dcActualdc ILfVV 3
tan
−= (2.89)
o
m
Actualdc ILf
V
V 3
2
33
−=
π
(2.90)
Example 9 Three-phase half-wave diode rectifier connected to 66 kV, 50
Hz , 5mH supply to feed a DC load with 500 A DC, fined the average DC
output voltage.
Solution: Vvm 538892*
3
66000
=⎟
⎠
⎞
⎜
⎝
⎛
=
(i) o
ceinduc
source
without
dcActualdc ILfVV 3
tan
−=
VILf
V
V o
m
Actualdc 44190500*005.0*50*3
2
53889*3*3
3
2
33
=−=−=
ππ

2.5 Three-Phase Full Wave Diode Rectifier
The three phase bridge rectifier is very common in high power
applications and is shown in Fig.2.30. It can work with or without
transformer and gives six-pulse ripples on the output voltage. The diodes
are numbered in order of conduction sequences and each one conduct for
120 degrees. These conduction sequence for diodes is 12, 23, 34, 45, 56,
and, 61. The pair of diodes which are connected between that pair of
supply lines having the highest amount of instantaneous line to line
voltage will conduct. Also, we can say that, the highest positive voltage
of any phase the upper diode connected to that phase conduct and the
highest negative voltage of any phase the lower diode connected to that
phase conduct.
2.5.1 Three-Phase Full Wave Rectifier With Resistive Load
In the circuit of Fig.2.30, the AC side inductance LS is neglected and
the load current is pure resistance. Fig.2.31 shows complete waveforms
for phase and line to line input voltages and output DC load voltages.
Fig.2.32 shows diode currents and Fig.2.33 shows the secondary and
primary currents and PIV of D1. Fig.2.34 shows Fourier Transform
components of output DC voltage, diode current secondary current and
Primary current respectively.
For the rectifier shown in Fig.2.30 the waveforms is as shown in
Fig.2.31. The average output voltage is :-
LLm
LLm
mdc VV
VV
tdtVV 3505.1654.1
2333
sin3
3
3/2
3/
===== ∫ ππ
ωω
π
π
π
(2.91)
R
V
R
V
R
V
R
V
I LLLLmm
dc
3505.123654.133
====
ππ
(2.92)
( ) LLmmmrms VVVtdtVV 3516.16554.1
4
3*9
2
3
sin3
3
3/2
3/
2
==+== ∫ π
ωω
π
π
π
(2.93)
R
V
I m
rms
6554.1
= (2.94)
Then the diode rms current is
R
V
R
V
I mm
r 9667.0
3
6554.1
== (2.95)
R
V
I m
S 29667.0= (2.96)

64 Chapter Two
1 3 5
4 6 2
b
c
IL
VL
Is
Ip
a
Fig.2.30 Three-phase full wave diode bridge rectifier.
Fig.2.31 shows complete waveforms for phase and line to line input voltages
and output DC load voltages.

Fig.2.32 Diode currents.
Fig.2.33 Secondary and primary currents and PIV of D1.

66 Chapter Two
Fig.2.34 Fourier Transform components of output DC voltage, diode current
secondary current and Primary current respectively of three-phase full wave
diode bridge rectifier.
Example 10 The rectifier shown in Fig.2.30 is operated from 460 V 50
Hz supply and the load resistance is R=20 Ω. If the source inductance is
negligible, determine (a) The efficiency, (b) Form factor (c) Ripple factor
(d) Transformer utilization factor, (e) Peak inverse voltage (PIV) of each
diode and (f) Crest factor of input current.
Solution: (a) VVVV mS 59.3752*58.265,58.265
3
460
====
VV
V
V m
m
dc 226.621654.1
33
===
π
,
A
R
V
R
V
I mm
dc 0613.31
654.133
===
π
VVVV mmrms 752.6216554.1
4
3*9
2
3
==+=
π
,
A
R
V
I m
rms 0876.31
6554.1
==

%83.99===
rmsrms
dcdc
ac
dc
IV
IV
P
P
η
(b) %08.100==
dc
rms
V
V
FF
(c) %411 2
2
222
=−=−=
−
== FF
V
V
V
VV
V
V
RF
dc
rms
dc
dcrms
dc
ac
(d)
R
V
R
V
II mm
rmsS 352.1
6554.1
*8165.0
3
2
===
%42.95
352.1*2/*3
/)654.1(
*3
2
===
R
V
V
RV
IV
P
TUF
m
m
m
SS
dc
(f) 281.1
352.1
/3)(
===
R
V
RV
I
I
CF
m
m
S
peakS
2.5.2 Three-Phase Full Wave Rectifier With DC Load Current
The supply current in case of pure DC load current is shown in
Fig.2.35. Fast Fourier Transform of Secondary and primary currents
respectively is shown in Fig2.36.
As we see it is odd function, then an=0, and
[ ]
....,.........15,14,12,10,9,8,6,4,3,2,0
.....),........3(
13
2
),3(
11
2
)3(
7
2
),3(
5
2
,3
2
cos
2
sin*
2
1311
751
6/5
6/
6/5
6/
==
==
−=−==
−=
= ∫
nforb
I
b
I
b
I
b
I
b
I
b
tn
n
I
tdtnIb
n
oo
ooo
o
on
ππ
πππ
ω
π
ωω
π
π
π
π
π
(2.97)
⎟
⎠
⎞
⎜
⎝
⎛
++−−= ttttvt
I
tI o
s ωωωωω
π
13sin
13
1
11sin
11
1
7sin
7
1
5sin
5
1
sin
32
)( (2.98)

68 Chapter Two
%31
25
1
23
1
19
1
17
1
13
1
11
1
7
1
5
1
))((
22222222
=
⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
=tITHD s
Also ))(( tITHD s can be obtained as following:
oS II
3
2
= , oS II
π
3*2
1 =
%01.311
/3*2
3/2
1))(( 2
2
1
=−=−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
πS
S
s
I
I
tITHD
Fig.2.35 The D1 and D2 currents, secondary and primary currents.

Fig2.36 Fast Fourier Transform of Secondary and primary currents respectively.
For the primary current if we move the t=0 to be as shown in Fig.2.28,
then the function will be odd then, 0=na , and
....,.........15,14,12,10,9,8,6,4,3,2,0
......,.........13,11,7,5,1
3*2
3
2
cos
3
coscos1
2
sin*sin*2sin*
2
1
1
3/2
1
3/2
3/
1
3/
0
1
==
==
⎟
⎠
⎞
⎜
⎝
⎛
−+−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
++= ∫∫∫
nforb
nfor
n
I
b
nnn
n
I
tdtnItdtnItdtnIb
n
n
n
π
ππ
π
π
ωωωωωω
π
π
π
π
π
π
(2.99)
⎟
⎠
⎞
⎜
⎝
⎛
++++= ttttt
I
tIP ωωωωω
π
13sin
13
1
11sin
11
1
7sin
7
1
5sin
5
1
sin
3*2
)( 1 (2.100)
%30
25
1
23
1
19
1
17
1
13
1
11
1
7
1
5
1
))((
22222222
=
⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
=tITHD P
Power Factor =
S
S
S
S
I
I
I
I 11
)0cos(* =

70 Chapter Two
2.5.4 Three-Phase Full Wave Diode Rectifier With Source Inductance
The source inductance in three-phase diode bridge rectifier Fig.2.37
will change the shape of the output voltage than the ideal case (without
source inductance) as shown in Fig.2.31. The DC component of the
output voltage is reduced. Fig.2.38 shows The output DC voltage of
three-phase full wave rectifier with source inductance.
Fig.2.37 Three-phase full wave rectifier with source inductance
Fig.2.38 The output DC voltage of three-phase full wave rectifier with source
inductance
Let us study the commutation time starts at t=5ms as shown in
Fig.2.39. At this time cV starts to be more negative than bV so diode D6
has to switch OFF and D2 has to switch ON. But due to the source
inductance will prevent that to happen instantaneously. So it will take
time tΔ to completely turn OFF D6 and to make D2 carry all the load

current ( oI ). Also in the time tΔ the current in bL will change from oI
to zero and the current in cL will change from zero to oI . This is very
clear from Fig.2.39. The equivalent circuit of the three phase diode bridge
at commutation time tΔ at mst 5= is shown in Fig.2.40 and Fig.2.41.
Fig.2.39 Waveforms represent the commutation period at time t=5ms.
Fig.2.40 The equivalent circuit of the three phase diode bridge at commutation
time tΔ at mst 5=

72 Chapter Two
Fig.2.41 Simple circuit of the equivalent circuit of the three phase diode bridge
at commutation time tΔ at mst 5=
From Fig.2.41 we can get the following defferntial equations:
061 =−−−− b
D
bdc
D
aa V
dt
di
LV
dt
di
LV (2.101)
021 =−−−− c
D
cdc
D
aa V
dt
di
LV
dt
di
LV (2.102)
Note that, during the time tΔ , 1Di is constant so 01 =
dt
diD , substitute this
value in (2.101) and (2.102) we get the following differential equations:
dc
D
bba V
dt
di
LVV =−− 6 (2.103)
dc
D
cca V
dt
di
LVV =−− 2 (2.104)
By equating the left hand side of equation (2.103) and (2.104) we get the
following differential equation:
dt
di
LVV
dt
di
LVV D
cca
D
bba
26
−−=−− (2.105)
026 =−+−
dt
di
L
dt
di
LVV D
c
D
bcb (2.106)
The above equation can be written in the following manner:
( ) 026 =−+− DcDbcb diLdiLdtVV (2.107)
( ) 026 =−+− DcDbcb diLdiLtdVV ωωω (2.108)
Integrate the above equation during the time tΔ with the help of
Fig.2.39 we can get the limits of integration as shown in the following:
( ) 0
0
2
0
6
2/
2/
=−+− ∫∫∫
+ o
o
I
Dc
I
Db
u
cb diLdiLtdVV ωωω
π
π

( ) 0
3
2
sin
3
2
sin
2/
2/
=−−+⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+−⎟
⎠
⎞
⎜
⎝
⎛
−∫
+
ocob
u
mm ILILtdtVtV ωωω
π
ω
π
ω
π
π
assume Scb LLL ==
oS
u
m ILttV ω
π
ω
π
ω
π
π
2
3
2
cos
3
2
cos
2/
2/
=⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
++⎟
⎠
⎞
⎜
⎝
⎛
−−
+
oS
m
IL
uuV
ω
ππππππππ
2
3
2
2
cos
3
2
2
cos
3
2
2
cos
3
2
2
cos
=
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+−⎟
⎠
⎞
⎜
⎝
⎛
−+⎟
⎠
⎞
⎜
⎝
⎛
+++⎟
⎠
⎞
⎜
⎝
⎛
−+−
oSm ILuuV ω
ππππ
2
6
7
cos
6
cos
6
7
cos
6
cos =⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛ −
+⎟
⎠
⎞
⎜
⎝
⎛
++⎟
⎠
⎞
⎜
⎝
⎛
−−
( ) ( ) ( ) ( )
m
oS
V
IL
uuuu
ω
ππππ
2
2
3
2
3
6
7
sinsin
6
7
coscos
6
sinsin
6
coscos
=
⎥
⎦
⎤
⎢
⎣
⎡
++⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
−
( ) ( ) ( ) ( )
m
oS
V
IL
uuuu
ω2
3sin5.0cos
2
3
sin5.0cos
2
3
=⎥
⎦
⎤
⎢
⎣
⎡
++−−−
( )[ ]
m
oS
V
IL
u
ω2
cos13 =−
( )
LL
oS
LL
o
m
o
V
IL
V
LI
V
LI
u
ωωω 2
1
2
2
1
3
2
1cos −=−=−=
⎥
⎦
⎤
⎢
⎣
⎡
−= −
LL
oS
V
IL
u
ω2
1cos 1
(2.109)
⎥
⎦
⎤
⎢
⎣
⎡
−==Δ −
LL
oS
V
ILu
t
ω
ωω
2
1cos
1 1
(2.110)
the drop across the source inductance.
dt
di
Lv D
Srd = (2.111)
Multiply (2.111) by tdω and integrate both sides of the resultant
equation we get:

74 Chapter Two
oS
I
D
u
rd ILLditdv
o
ωωω
π
π
== ∫∫
+
0
2
2
(2.112)
∫
+u
rd tdv
2
2
π
π
ω is the reduction area in one commutation period tΔ . But we
have six commutation periods tΔ in one period so the total reduction per
period is:
oS
u
rd ILtdv ωω
π
π
66
2
2
=∫
+
(2.113)
source inductance we have to divide by the period time π2 . Then,
o
o
rd fLI
LI
V 6
2
6
==
π
ω
(2.114)
The DC voltage without source inductance tacking into account can be
calculated as following:
dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan
−=−= (2.115)
Fig.2.42 shows the utility line current with some detailes to help us to
calculate its rms value easly.
u
oI
oI− 3
2π
sI
u+
3
2π
26
2 u
+
π
Fig.2.42 The utility line current

⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
+⎟
⎠
⎞
⎜
⎝
⎛
= ∫ ∫
+
u
u
u
d
o
s tdItdt
u
I
I
0
23
2
2
2
π
ωωω
π ⎥
⎦
⎤
⎢
⎣
⎡
−++= u
u
u
u
Io
233
12 3
2
2
π
π
Then ⎥⎦
⎤
⎢⎣
⎡
−=
63
2 2
uI
I o
S
π
π
(2.116)
Fig.2.43 shows the utility line currents and its first derivative that help
us to obtain the Fourier transform of supply current easily. From Fig.2.43
we can fill Table(2.2) as explained before when we study Table (2.1).
26
u
−
π
u
oI
oI−
26
5 u
−
π
26
7 u
−
π 26
11 u
−
π
sI
u
u
Io
u
Io−
sI′
26
u
−
π
26
5 u
−
π
26
7 u
−
π
26
11 u
−
π
Fig.2.43 The utility line currents and its first derivative.
Table(2.2) Jumb value of supply current and its first derivative.
sJ
26
u
−
π
26
u
+
π
26
5 u
−
π
26
5 u
+
π
26
7 u
−
π
26
7 u
+
π
26
11 u
−
π
26
11 u
+
π
sI 0 0 0 0 0 0 0 0
sI′
u
Io
u
Io−
u
Io−
u
Io
u
Io−
u
Io
u
Io
u
Io−

76 Chapter Two
It is an odd function, then 0== no aa
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
′−= ∑∑
==
m
s
ss
m
s
ssn tnJ
n
tnJ
n
b
11
sin
1
cos
1
ωω
π
(2.117)
⎥
⎦
⎤
⎟
⎠
⎞
⎟
⎠
⎞
⎜
⎝
⎛
+−⎟
⎠
⎞
⎜
⎝
⎛
−+⎟
⎠
⎞
⎜
⎝
⎛
++⎟
⎠
⎞
⎜
⎝
⎛
−−
⎢
⎣
⎡
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
++⎟
⎠
⎞
⎜
⎝
⎛
−−⎟
⎠
⎞
⎜
⎝
⎛
+−⎟
⎠
⎞
⎜
⎝
⎛
−
−
=
26
11
sin
26
11
sin
26
7
sin
26
7
sin
26
5
sin
26
5
sin
26
sin
26
sin*
11
u
n
u
n
u
n
u
n
u
n
u
n
u
n
u
n
u
I
nn
b o
n
ππππ
ππππ
π
⎥⎦
⎤
⎢⎣
⎡
+−−=
6
11
cos
6
7
cos
6
5
cos
6
cos
2
sin*
2
2
ππππ
π
nnnnnu
un
I
b o
n (2.118)
Then, the utility line current can be obtained as in (2.119).
( ) ( ) ( ) ( )
( ) ( ) ⎥
⎦
⎤
++−−⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+
⎢
⎣
⎡
+⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
=
t
u
t
u
t
u
t
u
t
u
u
ti
ωω
ωωω
π
ω
13sin
2
13
sin
13
1
11sin
2
11
sin
11
1
7sin
2
7
sin
7
1
5sin
2
5
sin
5
1
sin
2
sin
34
22
22
(2.119)
Then; ⎟
⎠
⎞
⎜
⎝
⎛
=
2
sin
62
1
u
u
I
I o
S
π
2.120)
The power factor can be calculated from the following equation:
⎟
⎠
⎞
⎜
⎝
⎛
⎥⎦
⎤
⎢⎣
⎡
−
⎟
⎠
⎞
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
=
2
cos
63
2
2
sin
62
2
cos
2
1 u
uI
u
u
I
u
I
I
pf
o
o
S
S
π
π
π
Then;
( )
⎥⎦
⎤
⎢⎣
⎡
−
=
63
sin*3
u
u
u
pf
π
π
(2.121)
Example 11 Three phase diode bridge rectifier connected to tree phase
33kV, 50 Hz supply has 8 mH source inductance to feed 300A pure DC
load current Find;
(i) commutation time and commutation angle.
(ii) DC output voltage.
(iii) Power factor.
(iv) Total harmonic distortion of line current.

Solution: (i) By substituting for 50**2 πω = , AId 300= , HL 008.0= ,
VVLL 33000= in (2.109), then o
radu 61.14.2549.0 ==
Then, ms
u
t 811.0==Δ
ω
.
(ii) The the actual DC voltage can be obtained from (2.115) as following:
dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan
−=−=
VVdcactual 43830300*008.*50*633000*35.1 =−=
(iii) the power factor can be obtained from (2.121) then
( ) ( ) 9644.0
6
2549.0
3
*2549.0
2549.0sin3
63
sin*3
=
⎥⎦
⎤
⎢⎣
⎡
−
=
⎥⎦
⎤
⎢⎣
⎡
−
=
π
π
π
π
u
u
u
pf Lagging
(iv) The rms value of supply current can be obtained from (2.116)as
following
A
uI
I d
s 929.239
6
2549.0
3
*
300*2
63
2 22
=⎟
⎠
⎞
⎜
⎝
⎛
−=⎥⎦
⎤
⎢⎣
⎡
−=
π
π
π
π
The rms value of fundamental component of supply current can be
obtained from (2.120) as following:
A
u
u
I
I o
S 28.233
2
2549.0
sin*
2*2549.0*
300*34
32*
2
sin
2
34
1 =⎟
⎠
⎞
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
=
ππ
9644.0
2
2549.0
cos*
929.239
28.233
2
cos*1
=⎟
⎠
⎞
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
=
u
I
I
pf
s
S Lagging.
%05.241
28.233
929.239
1
22
1
=−⎟
⎠
⎞
⎜
⎝
⎛
=−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
S
S
i
I
I
THD
2.7 Multi-pulse Diode Rectifier
Twelve-pulse bridge connection is the most widely used in high
number of pulses operation. Twelve-pulse technique is using in most
HVDC schemes and in very large variable speed drives for DC and AC
motors as well as in renewable energy system. An example of twelve-
pulse bridge is shown in Fig.2.33. In fact any combination such as this
which gives a 30o
-phase shift will form a twelve-pulse converter. In this
kind of converters, each converter will generate all kind of harmonics

78 Chapter Two
described above but some will cancel, being equal in amplitude but 180o
out of phase. This happened to 5th
and 7th
harmonics along with some of
higher order components. An analysis of the waveform shows that the AC
line current can be described by (2.83).
( ) ( ) ( ) ( ) ( )⎟
⎠
⎞
⎜
⎝
⎛
+−+−= tttttIti dP ωωωωω
π
25cos
25
1
23cos
23
1
13cos
13
1
5cos
11
1
cos
32
)( (2.83)
%5.13
35
1
35
1
25
1
23
1
13
1
11
1
))((
222222
=
⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
=tITHD P
As shown in (10) the THDi is about 13.5%. The waveform of utility
line current is shown in Fig.2.34. Higher pulse number like 18-pulse or
24-pulse reduce the THD more and more but its applications very rare. In
all kind of higher pulse number the converter needs special transformer.
Sometimes the transformers required are complex, expensive and it will
not be ready available from manufacturer. It is more economic to connect
the small WTG to utility grid without isolation transformer. The main
idea here is to use a six-pulse bridge directly to electric utility without
transformer. But the THD must be lower than the IEEE-519 1992 limits.
2N :1
32 N :1
Vd
a
c
b
a1
b1
c1
a2
b2
c2
Fig.2.33 Twelve-pulse converter arrangement

(a) Utility input current. (b) FFT components of utility current.
Fig.2.34 Simulation results of 12.pulse system.

80 Chapter Two
Problems
1- Single phase half-wave diode rectifier is connected to 220 V, 50 Hz
supply to feed Ω5 pure resistor. Draw load voltage and current and
diode voltage drop waveforms along with supply voltage. Then,
calculate (a) The rectfication effeciency. (b) Ripple factor of load
voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse
Voltage (PIV) of the diode. (e) Crest factor of supply current.
2- The load of the rectifier shown in problem 1 is become Ω5 pure
resistor and 10 mH inductor. Draw the resistor, inductor voltage
drops, and, load current along with supply voltage. Then, find an
expression for the load current and calculate the conduction angle,
β . Then, calculate the DC and rms value of load voltage.
3- In the rectifier shown in the following figure assume VVS 220= ,
50Hz, mHL 10= and VEd 170= . Calculate and plot the current an
the diode voltage drop along with supply voltage, sv .
sv
diodev
+ -
Lv i
dE
+ -
+
-
4- Assume there is a freewheeling diode is connected in shunt with the
load of the rectifier shown in problem 2. Calculate the load current
during two periods of supply voltage. Then, draw the inductor,
resistor, load voltages and diode currents along with supply voltage.
5- The voltage v across a load and the current i into the positive
polarity terminal are as follows:
( ) ( ) ( ) ( )tVtVtVVtv d ωωωω 3cos2sin2cos2 311 +++=
( ) ( ) ( )φωωω −++= tItIIti d 3cos2cos2 31
Calculate the following:
(a) The average power supplied to the load.
(b) The rms value of ( )tv and ( )ti .
(c) The power factor at which the load is operating.

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