Powerelectronicsnote 121011031400-phpapp02

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 I2R 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 I2R 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 tnbtnaa

tf ωω (1.1)

8 Chapter One

where ∫=π

ωωπ

2

0

)(cos)(1 tdtntfan (1.2)

and ∫=π

ωωπ

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

njnnn e

baF ϕ

2

22 += (1.5)

Where n

nn a

b−= −1tanϕ (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

21

2*100*100

s

nnsn

s

ssi I

I

III

THD∑≠=

−= (1.7)

1

2

1

21

2*100*100

s

nnsn

s

ssv V

V

VVV

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

nn

MLI

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

CII

=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

BEBBB R

VVI −=

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

dcPP

=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:-

22dcrmsac 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

rmsVVFF = (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 22

222−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac (2.6)

The Transformer Utilization Factor (TUF) is defiend as:-

SS

dcIV

PTUF = (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.

121

2

21

21

2−=

−=

S

S

S

SSi

II

IIITHD (2.8.a)

121

2

21

21

2−=

−=

S

S

S

SSv

VV

VVVTHD (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

II

CF )(= (2.9)

In general, power factor in non-sinusoidal circuits can be obtained as following:

φcossVoltampereApparent

PowerR===

SS IVPealPF (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.

FaactorntDisplacemeFactorDistortionII

IVIV

IVPPF

S

S

SSSS*coscos

1111 ==== φφ (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

πωω

π0

sin21 m

mdcVtdtVV (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:

RV

RVI mdc

dc π== (2.13)

The root mean square (rms) value of a load voltage is defined as:

2sin

21

0

22 mmrms

VtdtVV == ∫π

ωωπ

(2.14)

Similarly, the root mean square (rms) value of a load current is defined as:

RV

RVI 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

VII mDS 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:

ππ

πωω

π

πmm

mdcVVtdtVV =−−== ∫ ))0cos(cos(

2)sin(

21

0

Then, R

VR

VI mdcdc π

==

2)sin(

21

0

2 mmrms

VtVV == ∫π

ωπ

, R

VI mrms 2

= and, 2m

SVV =

The rms value of the transformer secondery current is the same as that of

the load: R

VI mS 2

= Then, the efficiency or rectification ratio is:

rmsrms

dcdc

ac

dcIVIV

PP

**

==η %53.40

2*

2

*==

RVV

RVV

mm

mmππ

(b) 57.12

2 ====π

πm

m

dc

rmsV

V

VVFF

(c) 211.1157.11 22 =−=−== FFVVRF

dc

ac

(d) %6.28286.0

22

====

RVV

RVV

IVPTUF

mm

mm

SS

dc ππ

(e) It is clear from Fig2.2 that the PIV is mV .

(f) Creast Factor CF, 22//)( ===

RVRV

II

CFm

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 ≤≤=+ ttViRdtdiL m 0),(sin* (2.17)

Divide the above equation by L we get:

βωω ≤≤=+ ttL

ViLR

dtdi 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

LVeeti mdt

LRdt

LR

ωsin*)( (2.19)

Where A is a constant.

Then; ⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+= ∫

−Adtt

LV

eeti mtLRt

LR

ωsin*)( (2.20)

By integrating (2.20) (see appendix) we get:

( )t

LR

m AetLtRLwR

Vti−

+−+

= 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 RLωφ =tan

Substitute these values into (2.21) we get the following equation:

( )t

LR

m AettZ

Vti−

+−= ωφωφ cossinsincos)(

Then, ( )t

LR

m AetZ

Vti−

+−= φωsin)( (2.22)

The above equation can be written in the following form:

( ) ( ) φω

ωω φωφω tansinsin)(

tm

tL

Rm Aet

ZVAet

ZVti

−−+−=+−= (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:

( )φsinZ

VA 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 tansinsin)(t

m etZ

Vti (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

βω == wtti 0)( into (2.24) we get:

( ) ( ) 0sinsin)( tan =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−=

−φ

β

φφββ eZ

Vi m (2.25)

R

wLZ

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

βπ

ωωπ

β

−== ∫ mmdc

VtdtVV (2.26)

)2sin(1(5.0*2

)sin(*21

0

2 ββπ

ωπ

β

−+== ∫VmdwttVV 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


m etZ

Vti 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* <<=+ tforiRtdidL (2.29)

The solution of (2.29) is given by the following equation:

φπω

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, ( ) ( ) BeZ

Vi 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.

( ) ( ) φπω

φπ

φφπω tantansinsin)(−

−−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−=

tm eeZ

Vti 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).

( ) φπω

φπωω tan2

2sin)(−

−+−−=

tm AetZ

Vti 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.

( ) ( ) φπ

φπ

φφππ tantansinsin)2(−−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−= ee

ZVi m (2.33)

( ) AZ

Vi 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 sin2Z

ViA m+= (2.35)

Then, the general solution for the period πωπ 32 << t is given by equation (2.36):

( ) ( ) ( ) φπω

φπφπωω tan2

sin22sin)(−

−⎟⎠⎞

⎜⎝⎛ ++−−=

tmm eZ

VitZ

Vti 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:

.561.010

10*20*314tantan3

11 radRL

===−

−− ωφ and 628343.0tan =φ

Ω=+=+= − 8084.11)10*20*314(10)( 23222 LRZ ω

( ) ( )

( )[ ]t

tm

et

etZ

Vti

ω

φω

ω

φφωω

5915.1

tan

*532.0561.0sin8084.11

2220

sinsin)(

−

−

+−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−=

( ) tetti ωωω 5915.1*0171.14561.0sin3479.26)( −+−= The value of β can be obtained from the above equation by substituting for 0)( =βi . Then, ( ) ββ 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

( ) ββ 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:

( ) wtetwti 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 =+−= − πππ In the second part πωπ 2<< t , the expression for the load current

can be obtained from (2.30) as following:

φπω

ω tan)(−

−=

t

eBti where AiB 1124.14)( == π

Then ( )πωω −−= teti 5915.11124.14)( 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):

( ) ( ) ( ) φπω

φπφπωω tan2

sin22sin)(−

−⎟⎠⎞

⎜⎝⎛ ++−−=

tmm eZ

VitZ

Vti

( ) ( ) ( )πωωω 25915.1532.0*3479.26095103.08442.6sin3479.26)( −−++−=∴ tetti

( ) ( )πωωω 25915.11131.148442.6sin3479.26)( −−+−=∴ tetti 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

πωω

π

πm

mdcVtdtVV 2sin1

0

== ∫ (2.36)

RVI m

dc π2

= (2.37)

( )2

sin1

0

2 mmrms

VtdtVV == ∫π

ωωπ

(2.38)

RVI m

rms 2= (2.39)

PIV of each diode = mV2 (2.40)

2m

SVV = (2.41)

The rms value of the transformer secondery current is the same as that of the diode:

RVII 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

**

====

RVV

RVV

IVIV

PP

mm

mm

rmsrms

dcdc

ac

dc ππη

(b) 11.1222

2 ====π

πm

m

dc

rmsV

V

VVFF

(c) 483.0111.11 22 =−=−== FFVVRF

dc

ac

(d) 5732.0

222

22

2===

RVV

RVV

IVPTUF

mm

mm

SS

dc ππ

(e) The PIV is mV2

(f) Creast Factor of secondary current, 2

2

)( ===

RVR

V

II

CFm

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(* tViRtdidL 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


m etZ

Vti 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)


( ) φπω

φπωω tansin)(−

−+−−=

tm AetZ

Vti (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 tansinsin)( eZ

ViI mo (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:

( ) φππ

φπππ tansin)(−

−+−−== Ae

ZVIi m

o

Then; ( )φsinZ

VIA mo += (2.48)

Substitute (2.48) into (2.45) we get:

( ) ( ) φπω

φφπωω tansinsin)(−

−⎟⎠⎞

⎜⎝⎛ ++−−=

tm

om e

ZVIt

ZVti , then,

( ) ( ) φπω

φπω

φφπωω tantansinsin)(−

−−

−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+−−=

t

o

tm eIetZ

Vti (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:

( ) ( ) φπω

φπω

φφπω tan2

tan2

sin2sin)(−

−−

−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+−−=

t

o

tm eIewtZ

Vti (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

VVtdtVV mmdc 956.1902sin1

0

=== ∫ πωω

π

π

, AR

VI mdc 7324.122

==π

( ) VVtdtVV mmrms 132.212

2sin1

2/1

0

2 ==⎥⎥⎦

⎤

⎢⎢⎣

⎡= ∫

π

ωωπ

, AR

VI m

rms 142.142

==

(a) %06.81===rmsrms

dcdc

ac

dcIVIV

PPη

(b) 11.1==dc

rmsVVFF

(c) 482.011 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

(d) %81142.14*132.212

7324.12*986.190===

SS

dcIV

PTUF

(e) The PIV= mV =300V

(f) 414.1142.14

15/300)( ===S

peakS

II

CF

(g) Input power factor = 1*Re 2==

SS

rmsIV

RIPowerApperant

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

[ ]

[ ] .............,5,3,14

cos0cos2

cos2

sin*20

0

==−=

−== ∫

nforn

In

nI

tnn

ItdtnIb

oo

oon

ππ

π

ωπ

ωωπ

ππ

(2.51)

Then from Fourier series concepts we can say: )..........9sin

917sin

715sin

513sin

31(sin*4)( +++++= tttttIti o ωωωωω

π(2.52)


%46151

131

111

91

71

51

31))((

2222222=⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=∴ tITHD s

or we can obtain ))(( tITHD s as the following:

From (2.52) we can obtain the value of is π2

41

oS

II =

%34.4814

21

24

1))((2

2

2

1=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=−

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=−⎟⎟⎠

⎞⎜⎜⎝

⎛=∴

π

πo

o

S

Ss I

III

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

dcIVIV

PPη

(b) 11.1==dc

rmsVVFF

(c) 482.011 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

(d) %9030*132.21230*986.190

===SS

dcIV

PTUF

(e) The PIV=Vm=300V

(f) 13030)( ===

S

peakS

II

CF

(g) AII oS 01.27

230*4

24

1 ===ππ

Input Power factor= =PowerApperant

PoweralRe

LagI

IIV

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

diLV SsS (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

ISs

u

m diLtdtV ωωωπ

π

sin (2.55)

Then; ( )[ ] osm ILuV ωππ 2coscos −=+− Then; ( )[ ] osm ILuV ω2cos1 −=+−

48 Chapter Two

Then; ( )m

osV

ILu ω21cos −=

Then; ⎟⎟⎠

⎞⎜⎜⎝

⎛−= −

m

osV

ILu ω21cos 1 (2.56)

And ⎟⎟⎠

⎞⎜⎜⎝

⎛−==Δ −

m

osV

ILut ωωω

21cos1 1 (2.57)

It is clear that the DC voltage reduction due to the source inductance is the drop across the source inductance.

dtdiLv S

srd = (2.58)

Then oS

I

ISS

u

rd ILdiLtdvo

o

ωωωπ

π

2−== ∫∫−+

(2.59)

∫+u

rd tdvπ

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 ωωπ

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;

oSoS

rd ILfILV 42

4−=

−=

πω (2.61)

The DC voltage with source inductance tacking into account can be calculated as following:

osm

rdceinducsourcewithoutdcactualdc IfLVVVV 42tan −=−=

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/

22

∫ ∫+⎟⎠⎞

⎜⎝⎛=

u

uo

os tdItdt

uI

I

π

ωωωπ

(2.63)

Then; ⎥⎦⎤

⎢⎣⎡ −=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−+=

322

228342 23

2

2 uIuuu

II oos

ππ

ππ

(2.64)

2u

−

oI

oI− 2u

−π2u

π

2u

+π

22 u

−π

π2

usI

2π

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:

2u

−

oI

oI−2u

−π

uIo2

uIo2

−

sI ′

2u

π

2u

+π

22 u

−π

π2

2u

−

2u

−π

2u

π

2u

+π

22 u

−π

usI

Fig.2.20 Supply current and its first derivative.

Table(2.1) Jumb value of supply current and its first derivative.

sJ 2u

− 2u

2u

−π 2u

+π

sI 0 0 0 0

sI ′ uIo2

uIo2

− uIo2

− uIo2


It is an odd function, then 0== no aa

⎥⎥⎦

⎤

⎢⎢⎣

⎡′−= ∑∑

==

m

sss

m

sssn tnJ

ntnJ

nb

11sin1cos1 ωω

π (2.65)

⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−

−=

2sin

2sin

2sin

2sin2*11 unununun

uI

nnb o

n πππ

2sin*8

2nu

unIb o

nπ

= (2.66)

2sin*8

1u

uIb o

π= (2.67)

Then; 2

sin*28

1u

uII o

S π= (2.68)

( )

⎥⎦⎤

⎢⎣⎡ −

=

⎥⎦⎤

⎢⎣⎡ −

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

⎟⎠⎞

⎜⎝⎛

⎥⎦⎤

⎢⎣⎡ −

=⎟⎠⎞

⎜⎝⎛=

32

sin2

32

2cos

2sin4

2cos

322

2sin*

28

2cos*

21

uu

uuu

uu

u

uI

uu

Iu

IIpf

o

o

S

S

ππππ

ππ

π

(2.69)

Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance mHX s 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 ==

osm

rdceinducsourcewithoutdcactualdc IfLVVVV 42tan −=−=

π

VV actualdc 9703200*005.0*50*415556*2=−=

π

(ii) From (2.56) the commutation angle u can be obtained as following:

52 Chapter Two

.285.015556

200*005.0*50**2*21cos21cos 11 radV

ILum

os =⎟⎠⎞

⎜⎝⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−= −− πω

The input power factor can be obtained from (2.69) as following ( ) ( ) 917.0

3285.

2285.0

285.0sin*2

32

sin*22

cos*1 =

⎥⎦⎤

⎢⎣⎡ −

=

⎥⎦⎤

⎢⎣⎡ −

=⎟⎠⎞

⎜⎝⎛=

ππππ uu

uuIIpf

S

S

AuII oS 85.193

3285.0

2200*2

322 22

=⎥⎦⎤

⎢⎣⎡ −=⎥⎦

⎤⎢⎣⎡ −=

ππ

ππ

Auu

II oS 46.179

2285.0sin*

285.0*2200*8

2sin*

28

1 =⎟⎠⎞

⎜⎝⎛==

ππ

%84.40146.17985.1931

22

1=−⎟

⎠⎞

⎜⎝⎛=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

S

Si I

ITHD

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.

6π

65π

54 Chapter Two

Fig.2.24 FFT components of primary and diode currents.

By considering the interval from6π to

65π in the output voltage we can

calculate the average and rms output voltage and current as following:

mm

mdc VVtdtVV 827.0233sin

23 6/5

6/

=== ∫ πωω

π

π

π

(2.70)

RV

RVI mm

dc*827.0

**233

==π

(2.71)

( ) mmmrms VVtdtVV 8407.08

3*321sin

23 6/5

6/

2 =+== ∫ πωω

π

π

π

(2.72)

RVI m

rms8407.0

= (2.73)

Then the diode rms current is equal to secondery current and can be obtaiend as following:

RV

RVII mm

Sr 4854.03

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*58.265,58.2653

460====

mm

dc VVV 827.0233

==π

,

RV

RVI mm

dc0827

233

==π

mrms VV 8407.0=

RVI m

rms8407.0

=

%767.96===rmsrms

dcdc

ac

dcIVIV

PPη

(b) %657.101==dc

rmsVVFF

(c) %28.1811 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

(d) RVII m

rmsS 38407.0

31

==

%424.66

38407.0*2/*3

/)827.0(*3

2===

RVV

RVIV

PTUFm

m

m

SS

dc

(e) The PIV= 3 Vm=650.54V

(f) 06.2

38407.0

/)( ===

RV

RVI

ICF

mm

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 ot 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:

321 3/

3/0

oo

ItdIa ∫−

==π

π

ωπ

(2.76)

[ ]

harmonicstrepleanallfor

nfornI

nfornI

tnnI

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

ωπ

ωπ

ππ

π

π

(2.77)

⎟⎠⎞

⎜⎝⎛ −−++−−++= ...8sin

817sin

715sin

514sin

412sin

21sin

33

)( ttttttII

tI OOs 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

New

axi

s

58 Chapter Two

%24.1090924.119

*21

23

3/1))((2

2

2

1==−=−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=−⎟⎟⎠

⎞⎜⎜⎝

⎛=

π

πO

o

S

Ss I

IIItITHD

It is clear that the primary current shown in Fig.2.26 is odd, then, an=0,

[ ]

harmonicstrepleanallfor

nfornI

tnnItdtnIb

o

oon

0

,....14,13,11,10,8,7,5,4,2,13

cos2sin*2 3/20

3/2

0

=

==

−== ∫

π

ωπ

ωωπ

ππ

(2.79)

⎟⎠⎞

⎜⎝⎛ −−+++++= ...8sin

817sin

715sin

514sin

412sin

21sin3)( ttttttIti O

P ωωωωωωπ

(2.80)

The rms value of oP II32

= (2.81)

%983.67133

21

2332

1))((2

2

2

1=−⎟

⎠⎞

⎜⎝⎛=−

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=−⎟⎟⎠

⎞⎜⎜⎝

⎛=

π

πO

o

P

PP I

I

IItITHD (2.82)

Example 8 Solve example 7 if the load current is 100 A pure DC

Solution: (a) VVVV mS 59.3752*58.265,58.2653

460====

VVVV mm

dc 613.310827.0233

===π

, AIdc 100=

VVV mrms 759.3158407.0 == , AIrms 100=

%37.98100*759.315100*613.310

====rmsrms

dcdc

ac

dcIVIV

PPη

(b) %657.101==dc

rmsVVFF

(c) %28.1811 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac


(d) AII rmsS 735.57100*31

31

===

%52.67735.57*2/*3

100*613.310*3

===mSS

dcVIV

PTUF


(f) 732.1735.57

100)( ===S

peakS

II

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 =−− dcD

aa Vdt

diLv (2.83)

02 =−− dcD

bb Vdt

diLv (2.84)

subtract (2.84) from(2.83) we get:


012 =⎟⎠⎞

⎜⎝⎛ −+−

dtdi

dtdiLvv 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: ( ) ( )2132sinsin DDmm didiLtdtVtV −=⎟

⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −− ωωπωω

Then; ( )216sin3 DDm didiLtdtV −=⎟

⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ + ωωπω

Integrating both parts of the above equation we get the following:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=⎟

⎠⎞

⎜⎝⎛ + ∫∫∫

+o

o

I

DI

D

u

m didiLtdtV0

2

0

1

65

65 6

sin3 ωωπω

π

π

Then; om LIuV ωππππ 266

5cos66

5cos3 −=⎟⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ++−⎟

⎠⎞

⎜⎝⎛ +

Then; ( ) ( )( ) om LIuV ωππ 2coscos3 −=+− Then; ( )( ) om LIuV ω2cos13 −=+−

Then; ( )m

oVLIu

32cos1 ω

=−

Then; ( )m

oVLIu

321cos ω

−=

Then ⎟⎟⎠

⎞⎜⎜⎝

⎛−= −

m

oVLIu

321cos 1 ω (2.85)

⎟⎟⎠

⎞⎜⎜⎝

⎛−==Δ −

m

oVLIut

321cos1 1 ω

ωω (2.86)

It is clear that the DC voltage reduction due to the source inductance is equal to the drop across the source inductance. Then;

dtdiLv D

rd =

62 Chapter Two

Then, o

I

D

u

rd LIdiLtdvo

ωωω

π

π

== ∫∫+

0

65

65

(2.87)

∫+u

rd tdv6

5

65

π

πω 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

π3*3

65

65

=∫+

To obtain the average reduction in DC output voltage rdV due to source inductance we have to divide the total reduction per period by π2 as following:

oo

rd ILfLIV 32

3==

πω (2.88)

Then, the DC component of output voltage due to source inductance is: o

ceinducsourcewithout

dcActualdc ILfVV 3tan

−= (2.89)

om

Actualdc ILfVV 3233

−=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) oceinduc

sourcewithout

dcActualdc ILfVV 3tan

−=

VILfVV om

Actualdc 44190500*005.0*50*32

53889*3*33233

=−=−=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 :-

LLmLLm

mdc VVVVtdtVV 3505.1654.12333sin33 3/2

3/

===== ∫ ππωω

π

π

π

(2.91)

RV

RV

RV

RVI LLLLmm

dc3505.123654.133

====ππ

(2.92)

( ) LLmmmrms VVVtdtVV 3516.16554.14

3*923sin33 3/2

3/

2==+== ∫ π

ωωπ

π

π

(2.93)

RVI m

rms6554.1

= (2.94)

Then the diode rms current is

RV

RVI mm

r 9667.03

6554.1== (2.95)

RVI m

S 29667.0= (2.96)

64 Chapter Two

1 3 5

4 6 2

b

c

IL

VLIs

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*58.265,58.2653

460====

VVVV mm

dc 226.621654.133===

π,

AR

VRVI mm

dc 0613.31654.133===

π

VVVV mmrms 752.6216554.14

3*923

==+=π

,

AR

VI mrms 0876.316554.1

==


%83.99===rmsrms

dcdc

ac

dcIVIV

PPη

(b) %08.100==dc

rmsVVFF

(c) %411 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

(d) RV

RVII mm

rmsS 352.16554.1*8165.032

===

%42.95352.1*2/*3

/)654.1(*3

2===

RVV

RVIV

PTUFm

m

m

SS

dc


(f) 281.1352.1

/3)( ===

RV

RVI

ICF

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(132),3(

112

)3(7

2),3(5

2,32

cos2

sin*2

1311

751

6/56/

6/5

6/

==

==

−=−==

−=

= ∫

nforb

IbIb

IbIbIb

tnn

I

tdtnIb

n

oo

ooo

o

on

ππ

πππ

ωπ

ωωπ

ππ

π

π

(2.97)

⎟⎠⎞

⎜⎝⎛ ++−−= tttt vtItI o

s ωωωωωπ

13sin13111sin

1117sin

715sin

51sin32)( (2.98)

68 Chapter Two

%31251

231

191

171

131

111

71

51))((

22222222

=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=tITHD s

Also ))(( tITHD s can be obtained as following:

oS II32

= , oS IIπ

3*21 =

%01.311/3*23/21))(( 2

2

1=−=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

πS

Ss I

ItITHD

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

......,.........13,11,7,5,13*23

2cos3

coscos12

sin*sin*2sin*2

1

1

3/21

3/2

3/1

3/

01

==

==

⎟⎠⎞

⎜⎝⎛ −+−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛++= ∫∫∫

nforb

nforn

Ib

nnnnI

tdtnItdtnItdtnIb

n

n

n

π

ππππ

ωωωωωωπ

π

π

π

π

π

(2.99)

⎟⎠⎞

⎜⎝⎛ ++++= tttttItIP ωωωωω

π13sin

13111sin

1117sin

715sin

51sin3*2)( 1 (2.100)

%30251

231

191

171

131

111

71

51))((

22222222

=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=tITHD P

Power Factor =S

S

S

SII

II 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 =−−−− bD

bdcD

aa Vdt

diLVdt

diLV (2.101)

021 =−−−− cD

cdcD

aa Vdt

diLVdt

diLV (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:

dcD

bba Vdt

diLVV =−− 6 (2.103)

dcD

cca Vdt

diLVV =−− 2 (2.104)

By equating the left hand side of equation (2.103) and (2.104) we get the following differential equation:

dtdiLVV

dtdiLVV D

ccaD

bba26 −−=−− (2.105)

026 =−+−dt

diLdt

diLVV Dc

Dbcb (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:

( ) 00

2

0

6

2/

2/

=−+− ∫∫∫+ o

o

I

DcI

Db

u

cb diLdiLtdVV ωωωπ

π


( ) 03

2sin3

2sin2/

2/

=−−+⎟⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −∫

+

ocob

u

mm ILILtdtVtV ωωωπωπωπ

π

assume Scb LLL ==

oS

u

m ILttV ωπωπωπ

π2

32cos

32cos

2/

2/=⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−

+

oS

m

IL

uuV

ω

ππππππππ

23

22

cos3

22

cos3

22

cos3

22

cos

=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ +++⎟

⎠⎞

⎜⎝⎛ −+−

oSm ILuuV ωππππ 26

7cos6

cos6

7cos6

cos =⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ −

+⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−

( ) ( ) ( ) ( )

m

oSV

IL

uuuu

ω

ππππ

2

23

23

67sinsin

67coscos

6sinsin

6coscos

=

⎥⎦

⎤⎢⎣

⎡++⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−

( ) ( ) ( ) ( )m

oSV

ILuuuu ω23sin5.0cos23sin5.0cos

23

=⎥⎦

⎤⎢⎣

⎡++−−−

( )[ ]m

oSV

ILu ω2cos13 =−

( )LL

oS

LL

o

m

oV

ILVLI

VLIu ωωω 21

221

321cos −=−=−=

⎥⎦

⎤⎢⎣

⎡−= −

LL

oSV

ILu ω21cos 1 (2.109)

⎥⎦

⎤⎢⎣

⎡−==Δ −

LL

oSV

ILut ωωω

21cos1 1 (2.110)

It is clear that the DC voltage reduction due to the source inductance is the drop across the source inductance.

dtdiLv 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 ILLditdvo

ωωω

π

π== ∫∫

+

0

2

2

(2.112)

∫+u

rd tdv2

2

π

πω 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

π66

2

2

=∫+

(2.113)

To obtain the average reduction in DC output voltage rdV due to source inductance we have to divide by the period time π2 . Then,

oo

rd fLILIV 62

6==

πω (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− 32π

sI

u+3

2π

262 u

+π

Fig.2.42 The utility line current


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+⎟⎠⎞

⎜⎝⎛= ∫ ∫

+u

u

ud

os tdItdt

uII

0

232

22π

ωωωπ ⎥⎦

⎤⎢⎣⎡ −++= uuu

uIo

23312 3

2

2 ππ

Then ⎥⎦⎤

⎢⎣⎡ −=

632 2 uII o

Sπ

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).

26u

−π

u

oI

oI−26

5 u−

π

267 u

−π 26

11 u−

π

sI

u

uIo

uIo−

sI ′

26u

−π

265 u

−π

267 u

−π

2611 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

26u

−π

26u

+π

265 u

−π

265 u

+π

267 u

−π

267 u

+π

2611 u

−π

2611 u

+π

sI 0 0 0 0 0 0 0 0

sI ′ uIo

uIo−

uIo−

uIo

uIo−

uIo

uIo

uIo−

76 Chapter Two


⎥⎥⎦

⎤

⎢⎢⎣

⎡′−= ∑∑

==

m

sss

m

sssn tnJ

ntnJ

nb

11sin1cos1 ωω

π (2.117)

⎥⎦

⎤⎟⎠

⎞⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−

⎢⎣

⎡⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −

−=

2611sin

2611sin

267sin

267sin

265sin

265sin

26sin

26sin*11

unununun

ununununuI

nnb o

n

ππππ

πππππ

⎥⎦⎤

⎢⎣⎡ +−−=

611cos

67cos

65cos

6cos

2sin*2

2ππππ

πnnnnnu

unIb o

n (2.118)

Then, the utility line current can be obtained as in (2.119).

( ) ( ) ( ) ( )

( ) ( ) ⎥⎦

⎤++−−⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+

⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

tutu

tututuu

ti

ωω

ωωωπ

ω

13sin2

13sin13

111sin2

11sin11

1

7sin2

7sin715sin

25sin

51sin

2sin34

22

22 (2.119)

Then; ⎟⎠⎞

⎜⎝⎛=

2sin62

1u

uII o

S π 2.120)

The power factor can be calculated from the following equation:

⎟⎠⎞

⎜⎝⎛

⎥⎦⎤

⎢⎣⎡ −

⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛=

2cos

632

2sin62

2cos

21 u

uI

uuI

uIIpf

o

o

S

S

ππ

π

Then; ( )

⎥⎦⎤

⎢⎣⎡ −

=

63

sin*3uu

upfππ

(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 oradu 61.14.2549.0 ==

Then, msut 811.0==Δω

.

(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

62549.0

3*2549.0

2549.0sin3

63

sin*3=

⎥⎦⎤

⎢⎣⎡ −

=

⎥⎦⎤

⎢⎣⎡ −

=ππππ uu

upf Lagging

(iv) The rms value of supply current can be obtained from (2.116)as following

AuII ds 929.239

62549.0

3*300*2

632 22

=⎟⎠⎞

⎜⎝⎛ −=⎥⎦

⎤⎢⎣⎡ −=

ππ

ππ

The rms value of fundamental component of supply current can be obtained from (2.120) as following:

Auu

II oS 28.233

22549.0sin*

2*2549.0*300*3432*

2sin

234

1 =⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

ππ

9644.02

2549.0cos*929.23928.233

2cos*1 =⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

uIIpf

s

S Lagging.

%05.24128.233

929.239122

1=−⎟

⎠⎞

⎜⎝⎛=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

S

Si I

ITHD

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

π25cos

25123cos

23113cos

1315cos

111cos32)( (2.83)

%5.13351

351

251

231

131

111))((

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

Vda

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.


6- Center tap diode rectifier is connected to 220 V, 50 Hz supply via unity turns ratio center-tap transformer to feed Ω5 resistor load. Draw load voltage and currents and diode currents 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.

7- Single phase diode bridge rectifier is connected to 220 V, 50 Hz supply to feed Ω5 resistor. Draw the load voltage, diodes currents and 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.

8- If the load of rectifier shown in problem 7 is changed to be Ω5 resistor in series with 10mH inductor. Calculate and draw the load current during the first two periods of supply voltages waveform.

9- Solve problem 8 if there is a freewheeling diode is connected in shunt with the load.

10- If the load of problem 7 is changed to be 45 A pure DC. Draw diode diodes currents and supply currents 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. (f) input power factor.

11- Single phase diode bridge rectifier is connected to 220V ,50Hz supply. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor, and, input power factor and THD of the voltage at the point of common coupling.

12- Three-phase half-wave diode rectifier is connected to 380 V, 50Hz supply via 380/460 V delta/way transformer to feed the load with 45 A DC current. Assuming ideal transformer and zero source inductance. Then, draw the output voltage, secondary and primary currents along with supply voltage. Then, calculate (a) Rectfication effeciency. (b) Crest factor of secondary current. (c) Transformer Utilization Factor (TUF). (d) THD of primary current. (e) Input power factor.

82 Chapter Two

13- Solve problem 12 if the supply has source inductance of 4 mH. 14- Three-phase full bridge diode rectifier is connected to 380V, 50Hz

supply to feed Ω10 resistor. Draw the output voltage, diode currents and supply current of phase a. 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.

15- Solve problem 14 if the load is 45A pure DC current. Then find THD of supply current and input power factor.

16- If the supply connected to the rectifier shown in problem 14 has a 5 mH source inductance and the load is 45 A DC. Find, average DC voltage, and THD of input current.

17- Single phase diode bridge rectifier is connected to square waveform with amplitude of 200V, 50 Hz. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor.

18- In the single-phase rectifier circuit of the following figure, 1=SL mH and VVd 160= . The input voltage sv has the pulse waveform shown in the following figure. Plot si and di waveforms and find the average value of dI .

+

-SV

dVSi

di

Hzf 50=

o60 o60 o60o120

o120o120

V200tω

Chapter 3 Thyristor Converters or Controlled

Converters 3.1 Introduction The controlled rectifier circuit is divided into three main circuits:-

(1) Power Circuit This is the circuit contains voltage source, load and switches as diodes, thyristors or IGBTs.

(2) Control Circuit This circuit is the circuit, which contains the logic of the firing of switches that may, contains amplifiers, logic gates and sensors.

(3) Triggering circuit This circuit lies between the control circuit and power thyristors. Sometimes this circuit called switch drivers circuit. This circuit contains buffers, opt coupler or pulse transformers. The main purpose of this circuit is to separate between the power circuit and control circuit.

The thyristor is normally switched on by applying a pulse to its gate. The forward drop voltage is so small with respect to the power circuit voltage, which can be neglected. When the anode voltage is greater than the cathode voltage and there is positive pulse applied to its gate, the thyristor will turn on. The thyristor can be naturally turned off if the voltage of its anode becomes less than its cathode voltage or it can be turned off by using commutation circuit. If the voltage of its anode is become positive again with respect to its cathode voltage the thyristor will not turn on again until gets a triggering pulse to its gate.

The method of switching off the thyristor is known as Thyristor commutation. The thyristor can be turned off by reducing its forward current below its holding current or by applying a reverse voltage across it. The commutation of thyristor is classified into two types:-

1- Natural Commutation If the input voltage is AC, the thyristor current passes through a natural zero, and a reverse voltage appear across the thyristor, which in turn automatically turned off the device due to the natural behavior of AC voltage source. This is known as natural commutation or line commutation. This type of commutation is applied in AC voltage

SCR Rectifier or Controlled Rectifier 81

controller rectifiers and cycloconverters. In case of DC circuits, this technique does not work as the DC current is unidirectional and does not change its direction. Thus the reverse polarity voltage does not appear across the thyristor. The following technique work with DC circuits: 2- Forced Commutation In DC thyristor circuits, if the input voltage is DC, the forward current of the thyristor is forced to zero by an additional circuit called commutation circuit to turn off the thyristor. This technique is called forced commutation. Normally this method for turning off the thyristor is applied in choppers.

There are many thyristor circuits we can not present all of them. In the following items we are going to present and analyze the most famous thyristor circuits. By studying the following circuits you will be able to analyze any other circuit.

3.2 Half Wave Single Phase Controlled Rectifier 3.2.1 Half Wave Single Phase Controlled Rectifier With Resistive Load The circuit with single SCR is similar to the single diode circuit, the difference being that an SCR is used in place of the diode. Most of the power electronic applications operate at a relative high voltage and in such cases; the voltage drop across the SCR tends to be small. It is quite often justifiable to assume that the conduction drop across the SCR is zero when the circuit is analyzed. It is also justifiable to assume that the current through the SCR is zero when it is not conducting. It is known that the SCR can block conduction in either direction. The explanation and the analysis presented below are based on the ideal SCR model. All simulation carried out by using PSIM computer simulation program [ ].

A circuit with a single SCR and resistive load is shown in Fig.3.1. The source vs is an alternating sinusoidal source. If ( )tVv ms ωsin= , vs is positive when 0 < ω t < π, and vs is negative when π < ω t <2π. When vs starts become positive, the SCR is forward-biased but remains in the blocking state till it is triggered. If the SCR is triggered at ω t = α, then α is called the firing angle. When the SCR is triggered in the forward-bias state, it starts conducting and the positive source keeps the SCR in conduction till ω t reaches π radians. At that instant, the current through the circuit is zero. After that the current tends to flow in the reverse

82 Chapter Three

direction and the SCR blocks conduction. The entire applied voltage now appears across the SCR. Various voltages and currents waveforms of the half-wave controlled rectifier with resistive load are shown in Fig.3.2 for α=40o. FFT components for load voltage and current of half wave single phase controlled rectifier with resistive load at α=40o are shown in Fig.3.3. It is clear from Fig.3.3 that the supply current containes DC component and all other harmonic components which makes the supply current highly distorted. For this reason, this converter does not have acceptable practical applecations.

Fig.3.1 Half wave single phase controlled rectifier with resistive load.

Fig.3.2 Various voltages and currents waveforms for half wave single-phase

controlled rectifier with resistive load at α=40o.


Fig.3.3 FFT components for load voltage and current of half wave single phase

controlled rectifier with resistive load at α =40o.

The average voltage, dcV , across the resistive load can be obtained by considering the waveform shown in Fig.3.2.

)cos1(2

))cos(cos(2

)sin(21 α

παπ

πωω

π

π

α

+=+−== ∫ mmmdc

VVtdtVV (3.1)

The maximum output voltage and can be acheaved when 0=α which is the same as diode case which obtained before in (2.12).

πm

dmVV = (3.2)

The normalized output voltage is the DC voltage devideded by maximum DC voltage, dmV which can be obtained as shown in equation (3.3).

)cos1(5.0 α+==dm

dcn V

VV (3.3

The rms value of the output voltage is shown in the following equation:-

( ) ⎟⎠⎞

⎜⎝⎛ +−== ∫ 2

)2sin(12

)sin(21 2 ααπ

πωω

π

π

α

mmrms

VtdtVV (3.3)

The rms value of the transformer secondery current is the same as that of the load:

RVI rms

s = (3.4)

84 Chapter Three

Example 1 In the rectifier shown in Fig.3.1 it has a load of R=15 Ω and, Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain an average output voltage of 70% of the maximum possible output voltage, calculate:- (a) The firing angle, α, (b) The efficiency, (c) Ripple factor (d) Transformer utilization factor, (e) Peak inverse voltage (PIV) of the thyristor and (f) The crest factor of input current. Solution:

(a) Vdm is the maximum output voltage and can be acheaved when 0=α , The normalized output voltage is shown in equation (3.3) which is

required to be 70%. Then,

7.0)cos1(5.0 =+== αdm

dcn V

VV . Then, α=66.42o =1.15925 rad.

(b) 220=mV V

VVVV mdmdc 02.49*7.0*7.0 ===

π, A

RVI dc

dc 268.315

02.49===

⎟⎠⎞

⎜⎝⎛ +−=

2)2sin(1

2ααπ

πm

rmsVV ,

at α=66.42o, Vrms=95.1217 V. Then, Irms=95.1217/15=6.34145 A

VVV mS 56.155

2==

The rms value of the transformer secondery current is: AII rmsS 34145.6==

Then, the rectification efficiency is:

%56.2634145.6*121.95268.3*02.49**

==

==rmsrms

dcdc

ac

dcIVIV

PP

η

(b) 94.12202.49

121.95====

π

dc

rmsVVFF

(c) 6624.1194.11 22 =−=−== FFVVRF

dc

ac

(d) 1624.034145.6*56.155268.3*02.49

===SS

dcIV

PTUF


(e) The PIV is mV (f) Creast factor of input current CF is as following:

313.234145.66667.14

34145.6)( ==== R

V

II

CFm

S

peakS

3.2.2 Half Wave Single Phase Controlled Rectifier With RL Load A circuit with single SCR and RL load is shown in Fig.3.4. The source vs is an alternating sinusoidal source. If vs = Vm sin (ω t), vs is positive when 0 < ω t < π, and vs is negative when π <ω t <2π. When vs starts become positive, the SCR is forward-biased but remains in the blocking state till it is triggered. If the SCR is triggered when ω t = α, then it starts conducting and the positive source keeps the SCR in conduction till ω t reaches π radians. At that instant, the current through the circuit is not zero and there is some energy stored in the inductor at ω t = π radians. The voltage across the inductor is positive when the current through it is increasing and it becomes negative when the current through the inductor tends to fall. When the voltage across the inductor is negative, it is in such a direction as to forward bias the SCR. There is current through the load at the instant ω t = π radians and the SCR 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 SCR blocks conduction. Fig.3.5 shows the output voltage, resistor, inductor voltages and thyristor voltage drop waveforms.

Fig.3.4 Half wave single phase controlled rectifier with RL load.

86 Chapter Three

Fig.3.5 Various voltages and currents waveforms for half wave single phase

controlled rectifier with RL load.

It is assumed that the current flows for α < ω t < β, where πβπ >>2 . When the SCR conducts, the driving function for the differential equation is the sinusoidal function defining the source voltage. Outside this period, the SCR blocks current and acts as an open switch and the current through the load and SCR is zero at this period there is no differential equation representing the circuit. For α < ω t < β , equation (3.5) applies.

βωαω ≤≤=+ ttViRdtdiL m ),(sin* (3.5)

Divide the above equation by L we get the following equation:

βωω ≤≤=+ ttL

ViLR

dtdi m 0),(sin* (3.6)

The instantaneous value of the current through the load can be obtained from the solution of the above equation as following:

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+= ∫

∫∫−Adtt

LVeeti mdt

LRdt

LR

ωω sin*)(

Then, ⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+= ∫

−Adtt

LVeeti mt

LRt

LR

ωω sin*)(


Where, A is constant. By integrating the above equation we get:

( )t

LR

m AetLtRLwR

Vti−

+−+

= ωωωω cossin)( 222 (3.7)

Assume LjRZ ωφ +=∠ . Then, 2222 LRZ ω+= ,

φcosZR = , φω sinZL = and RLωφ =tan

Substitute that in (3.7) we get the following equation:

( ) φω

ωφωφω tancossinsincos)(t

m AettZ

Vti−

+−=

Then, ( ) φω

φωω tansin)(t

m AetZ

Vti−

+−= βωα << t (3.8)

The value of A can be obtained using the initial condition. Since the diode starts conducting at ω t = α and the current starts building up from zero, then, i(α) = 0. Then, the value of A is expressed by the following equation:

)sin( φα −−=A (3.9) Once the value of A is known, the expression for current is known.

After evaluating A, current can be evaluated at different values of ω t.

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−−=

−−

RLt

m etZ

Vti /)sin()sin()( ωαω

φαφωω βωα << t (3.10)

When the firing angle α and the extinction angle β are known, the average and rms output voltage at the cathode of the SCR can be evaluated. We know that, i=0 when ω t=β substitute this condition in equation (3.10) gives equation (3.11) which used to determine β. Once the value of A, α and the extinction angle β are known, the average and rms output voltage at the cathode of the SCR can be evaluated as shown in equation (3.12) and (3.13) respectively.

RLe /)(

)sin()sin( ωαβ

φαφβ−

−−=− (3.11)

)cos(cos*2

sin*2

βαπ

ωωπ

β

α

−== ∫ mmdc

VtdtVV (3.12)

R

Z

Φ

Lω

88 Chapter Three

The rms load voltage is:

( ) ( ) )2cos2(cos21*

2sin*

21 2 αβαβ

πωω

π

β

α

−−−== ∫ mmrms

VtdtVV (3.13)

The average load current can be obtained as shown in equation (3.14) by dividing the average load voltage by the load resistance, since the average voltage across the inductor is zero.

)cos(cos*2

βαπ

−=R

VI mdc (3.14)

Example 2 A thyristor circuit shown in Fig.3.4 with R=10 Ω, L=20mH, and VS=220 sin314t and the firing angle α is 30o. Determine the expression for the current though the load. Also determine the rectification efficiency of this rectifier. Solution:

Ω=+= − 81.11)10*20*314(10 232Z , .561.0tan 1 RadR

wL== −φ

.5236.06180

*30 Rad===ππα

From equation (3.10) the current α < ω t < β can be obtained as

follows:-

( )( )5236.0628.0

/

0374.0)561.0sin(*6283.18

)561.05236.0sin()561.0sin(81.11

220)(

−−

−−

+−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−−=

t

RLt

et

etti

ω

ωαω

ω

ωω

The excitation angle β can be obtained from equation (3.11) as: )5236.0(628.0)056105236sin()561.0sin( −−−=− ββ e

Then, by using try and error technique which explained in Chapter 2 we can get the value of β , β = 3.70766 Rad. The DC voltage can be obtained from equation (3.12)

VVdc 9.59)70766.3cos5236.0(cos*2220

=−=π

Then, AR

VI dcdc 99.5

109.59

===

Similarly, Vrms can be obtained from equation (3.13)

( ) VVrms 384.111)05236*2cos70766.3*2(cos215236.070766.3*

2220

=−−−=π


AR

VI rmsrms 1384.11

10384.111

===

%92.28384.111

9.59

**

2

2

2

2====

rms

dc

rmsrms

dcdc

VV

IVIV

η

3.2.3 Half Wave Single Phase Controlled Rectifier With Free Wheeling Diode The circuit shown in Fig.3.6 differs than the circuit described in Fig.3.4, which had only a single SCR. This circuit has a freewheeling diode in addition, marked D1. The voltage source vs is an alternating sinusoidal source. If vs = Vm sin (ω t), vs is positive when 0 < ω t < π, and it is negative when π < ω t <2π. When vs starts become positive, the SCR is forward-biased but remains in the blocking state till it is triggered. When the SCR is triggered in the forward-bias state, it starts conducting and the positive source keeps the SCR in conduction till ω t reaches π radians. At that instant, the current through the circuit is not zero and there is some energy stored in the inductor at ω t = π radians. In the absence of the free wheeling diode, the inductor would keep the SCR in conduction for part of the negative cycle till the energy stored in it is discharged. But, when a free wheeling diode is present as shown in the circuit shown in Fig.3.7 the supply voltage will force D1 to turn on because its anode voltage become more positive than its cathode voltage and the current has a path that offers almost zero resistance. Hence the inductor discharges its energy during π < ω t < (2π + α) (Assuming continuous conduction mode) through the load. When there is a free wheeling diode, the current through the load tends to be continuous, at least under ideal conditions. When the diode conducts, the SCR remains reverse-biased, because the voltage vs is negative.

Fig.3.6 Half wave single phase controlled rectifier with RL load and freewheeling diode.

90 Chapter Three

Fig.3.7 Various voltages and currents waveforms for half wave single phase

controlled rectifier with RL load and freewheeling diode.


An expression for the current through the load can be obtained as shown below:

When the SCR 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 + α), the SCR blocks current and acts as an open switch. On the other hand, the free wheeling diode conducts during this period, and the driving function can be set to be zero volts. For α < ω t < π , equation (3.15) applies whereas equation (3.16) applies for the rest of the cycle. As in the previous cases, the solution is obtained in two parts.

πωαω ≤≤=+ ttViRdtdiL m ),(sin* (3.15)

απωπ +≤≤=+ 2,0* tiRdtdiL (3.16)

The solution of (3.15) is the same as obtained before in (3.8) which is shown in (3.17).

RLt

m eAtZ

Vti /*)sin(*)( ωαω

φω−

−+−= πωα << t (3.17)

The difference in the solution of (3.17) than (3.8) is how the constant A is evaluated? In the circuit without free-wheeling diode ( ) 0=αi , since the current starts build up from zero when the SCR is triggered during the positive half cycle. Assuming in the steady state the load is continuous and periodical which means that:

( ) ( ) ( )απαπα +=+= niii 22 (3.18) The solution of (3.16) can be obtained as following:

απωπφπ

+2<<=

−−

teBwtiwttan

)(

*)( (3.19) At πω =t then, ( )πiB = (3.20) To obtain B, we have to obtain ( )πi from (3.17) by letting tω equal

π . Then,

φαπ

φπ tan*)sin(*)(−

−+= eA

ZVi m (3.22)

Substitute (3.21) into (3.19) we get the current during ( )απωπ +<< 2t as following:

92 Chapter Three

απωπφω φπω

φαπ

+2<<⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+=

−−−

−teeA

ZVti

tm tan

)(tan **)sin(*)( (3.22)

απωπφω φαω

φπω

+2<<+=∴

−−

−−

teAeZ

Vtitt

m tantan)(

**)sin(*)( (3.23)

From the above equation we can obtain ( )απ +2i as following:

απωπφαπ φπ

φπα

+2<<+=+−

+−

teAeZ

Vi m tan2

tan **)sin(*)2( (3.24)

From (3.17) we can obtain ( )αι by letting αω =t as following:

( ) AZ

Vi m +−= )sin(* φαα (3.25)

Substitute (3.24) and (3.25) into (3.18) we get the following:

φπ

φπα

φφα tan2

tan **)sin(*)sin(*−

+−+=+− eAe

ZVA

ZV mm

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

+−−)sin()sin(*1 tantan

2

φαφ φπα

φπ

eZ

VeA m

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

=∴−

+−

φπ

φπα

φαφ

tan2

tan

1

)sin()sin(

*

e

e

ZVA m (3.26)

Then, the load current in the period of ( )πωαπ 122 +<<+ ntn where .......,3,2,1,0=n can be obtained from substituting (3.26) into (3.17).

Also, the load current in the period of ( ) ( ) απωπ ++<<+ 1212 ntn .......,3,2,1,0=n can be obtained from substituting (3.26) into (3.23).

Example 3 A thyristor circuit shown in Fig.3.6 with R=10 Ω, L=100 mH, and VS=220 sin314t V, and the firing angle α is 30o. Determine the expression for the current though the load. Also determine the rectification efficiency of this converter.


Solution: Ω=+= − 9539.32)10*100*314(10 232Z

.2625.1tan 1 RadR

wL== −φ ,

.5236.06180

*30 Rad===ππα

From (3.26) we can obtain A , A=7.48858 From (3.17) we can obtain ( )ti ω in the period of

( )πωαπ 122 +<<+ ntn where .......,3,2,1,0=n Then, )5236.0(*31845.0*48858.7)2625.1sin(*676.6)( −−+−= tetti ωωω

61457.9)( =πi , 99246.2)( =αi During the period of ( ) ( ) απωπ ++<<+ 1212 ntn where

.......,3,2,1,0=n we can obtain the solution from (3.18) where ( )πiB = .

Then, ,*61457.9)( )(31845.0 πωω −−= teti

VVVdwtwtVV mmmdc 3372.65)cos1(

2))cos(cos(

2)sin(

21

=+=+−== ∫ απ

απππ

π

α

The rms value of the output voltage is shown in the following equation:-

( ) VVtdtVV mmrms 402.108)

2)2sin((1

2)sin(

21 2/12/1

2 =⎥⎦⎤

⎢⎣⎡ +−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡= ∫

ααππ

ωωπ

π

α

( )

( ) Atde

tdetI

t

tdc

2546.4*61457.9

*48858.7)2625.1sin(*676.621

2)(31845.0

)5236.0(*31845.0

=⎥⎥⎦

⎤+

⎢⎢⎣

⎡+−=

∫

∫+

−−

−−

απ

π

πω

π

α

ω

ω

ωωπ

( )

( ) Atde

tdetI

t

trms

43288.5*61457.9

*48858.7)2625.1sin(*676.621

2/12 2)(31845.0

2)5236.0(*31845.0

=⎥⎥⎦

⎤

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎢⎢⎣

⎡++−=

∫

∫

+−−

−−

απ

π

πω

π

α

ω

ω

ωωπ

94 Chapter Three

%2.4743288.5*402.1082546.4*3372.65

**

==

==rmsrms

dcdc

ac

dcIVIV

PP

η

3.3 Single-Phase Full Wave Controlled Rectifier 3.3.1 Single-Phase Center Tap Controlled Rectifier With Resistive Load Center tap controlled rectifier is shown in Fig.3.8. When the upper half of the transformer secondary is positive and thyristor T1 is triggered, T1 will conduct and the current flows through the load from point a to point b. When the lower half of the transformer secondary is positive and thyristor T2 is triggered, T2 will conduct and the current flows through the load from point a to point b. So, each half of input wave a unidirectional voltage (from a to b ) is applied across the load. Various voltages and currents waveforms for center tap controlled rectifier with resistive load are shown in Fig.3.9 and Fig.3.10.

Fig.3.8 Center tap controlled rectifier with resistive load.

a b


Fig.3.9 The output voltgae and thyristor T1 reverse voltage wavforms along

with the supply voltage wavform.

Fig.3.10 Load current and thyristors currents for Center tap controlled rectifier

with resistive load.

96 Chapter Three

The average voltage, Vdc, across the resistive load is given by:

)cos1())cos(cos()sin(1 απ

αππ

ωωπ

π

α

+=−−== ∫ mmmdc

VVtdtVV (3.27)

Vdm is the maximum output voltage and can be acheaved when α=0 in the above equation. The normalized output voltage is:

)cos1(5.0 α+==dm

dcn V

VV (3.28)

From the wavfrom of the output voltage shown in Fug.3.9 the rms output voltage can be obtained as following:

( )2

)2sin(2

)sin(1 2 ααππ

ωωπ

π

α

+−== ∫ mmrms

VtdtVV (3.29)

Example 4 The rectifier shown in Fig.3.8 has load of R=15 Ω and, Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain an average output voltage of 70 % of the maximum possible output voltage, calculate:- (a) The delay angle α, (b) The efficiency, (c) The ripple factor (d) The transformer utilization factor, (e) The peak inverse voltage (PIV) of the thyristor and (f) The crest factor of input current. (g) Input power factor. Solution :

(a) Vdm is the maximum output voltage and can be acheaved when α=0, the normalized output voltage is shown in equation (3.28) which is required to be 70%. Then:

7.0)cos1(5.0 =+== αdm

dcn V

VV , then, α=66.42o

(b) 220=mV , then, VVVV mdmdc 04.982*7.0*7.0 ===

π

AR

VI dcdc 536.6

1504.98

===

2)2sin(

2ααπ

π+−= m

rmsVV

at α=66.42o Vrms=134.638 V Then, Irms=134.638/15=8.976 A


VVV mS 56.155

2==

The rms value of the transformer secondery current is:

AI

I rmsS 347.6

2==

Then, The rectification efficiency is:

%04.53976.8*638.134

536.6*04.98**

==

==rmsrms

dcdc

ac

dcIVIV

PP

η

(c) 3733.104.98638.134

===dc

rmsVVFF and,

9413.013733.11 22 =−=−== FFVVRF

dc

ac

(d) 32479.034145.6*56.155*2

536.6*04.982

===SS

dcIV

PTUF

(e) The PIV is 2 Vm

(f) Creast Factor CF, 313.234145.66667.14

34145.6)( ==== R

V

II

CFm

S

peakS

3.3.2 Single-Phase Fully Controlled Rectifier Bridge With Resistive Load This section describes the operation of a single-phase fully-controlled bridge rectifier circuit with resistive load. The operation of this circuit can be understood more easily when the load is pure resistance. The main purpose of the fully controlled bridge rectifier circuit is to provide a variable DC voltage from an AC source.

The circuit of a single-phase fully controlled bridge rectifier circuit is shown in Fig.3.11. The circuit has four SCRs. For this circuit, vs is a sinusoidal voltage source. When the supply voltage is positive, SCRs T1 and T2 triggered then current flows from vs through SCR T1, load resistor R (from up to down), SCR T2 and back into the source. In the next half-cycle, the other pair of SCRs T3 and T4 conducts when get pulse on their

98 Chapter Three

gates. Then current flows from vs through SCR T3, load resistor R (from up to down), SCR T4 and back into the source. Even though the direction of current through the source alternates from one half-cycle to the other half-cycle, the current through the load remains unidirectional (from up to down).

Fig.3.11 Single-phase fully controlled rectifier bridge with resistive load.

Fig.3.12 Various voltages and currents waveforms for converter shown in

Fig.3.11 with resistive load.


Fig.3.13 FFT components of the output voltage and supply current for converter

shown in Fig.3.11. The main purpose of this circuit is to provide a controllable DC output

voltage, which is brought about by varying the firing angle, α. Let vs = Vm sin ω t, with 0 < ω t < 360o. If ω t = 30o when T1 and T2 are triggered, then the firing angle α is said to be 30o. In this instance, the other pair is triggered when ω t = 30+180=210o. When vs changes from positive to negative value, the current through the load becomes zero at the instant ω t = π radians, since the load is purely resistive. After that there is no current flow till the other is triggered. The conduction through the load is discontinuous.

The average value of the output voltage is obtained as follows.:- Let the supply voltage be vs = Vm*Sin (ω t), where ω t varies from 0 to

2π radians. Since the output waveform repeats itself every half-cycle, the average output voltage is expressed as a function of α, as shown in equation (3.27).

( )[ ] )cos1()cos(cos)sin(1 απ

αππ

ωωπ

π

α

+=−−−== ∫ mmmdc

VVtdtVV (3.27)

Vdm is the maximum output voltage and can be acheaved when α=0, The normalized output voltage is:

)cos1(5.0 α+==dm

dcn V

VV (3.28)

The rms value of output voltage is obtained as shown in equation (3.29).

( )2

)2sin(2

)sin(1 2 ααππ

ωωπ

π

α

+−== ∫ mmrms

VtdtVV (3.29)

100 Chapter Three

Example 5 The rectifier shown in Fig.3.11 has load of R=15 Ω and, Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain an average output voltage of 70% of the maximum possible output voltage, calculate:- (a) The delay angle α, (b) The efficiency, (c) Ripple factor of output voltage(d) The transformer utilization factor, (e) The peak inverse voltage (PIV) of one thyristor and (f) The crest factor of input current. Solution:

(a) Vdm is the maximum output voltage and can be acheaved when α=0, The normalized output voltage is shown in equation (3.31) which is

required to be 70%.

Then, 7.0)cos1(5.0 =+== αdm

dcn V

VV , then, α=66.42o

(b) 220=mV , then, VV

VV mdmdc 04.98

2*7.0*7.0 ===

π

AR

VI dcdc 536.6

1504.98

===

2)2sin(

2ααπ

π+−= m

rmsVV

At α=66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A

VVV mS 56.155

2==

The rms value of the transformer secondery current is: AII rmsS 976.8==

Then, The rectification efficiency is

%04.53976.8*638.134

536.6*04.98**

==

==rmsrms

dcdc

ac

dcIVIV

PP

η

(c) 3733.104.98638.134

===dc

rmsVVFF

9413.013733.11 22 =−=−== FFVVRF

dc

ac


(d) 4589.0976.8*56.155536.6*04.98

===SS

dcIV

PTUF

(e) The PIV is Vm

(f) Creast Factor CF, 634.1976.86667.14

976.8)( ==== R

V

II

CFm

S

peakS

3.3.3 Full Wave Fully Controlled Rectifier With RL Load In Continuous Conduction Mode

The circuit of a single-phase fully controlled bridge rectifier circuit is shown in Fig.3.14. The main purpose of this circuit is to provide a variable DC output voltage, which is brought about by varying the firing angle. The circuit has four SCRs. For this circuit, vs is a sinusoidal voltage source. When it is positive, the SCRs T1 and T2 triggered then current flows from +ve point of voltage source, vs through SCR T1, load inductor L, load resistor R (from up to down), SCR T2 and back into the –ve point of voltage source. In the next half-cycle the current flows from -ve point of voltage source, vs through SCR T3, load resistor R, load inductor L (from up to down), SCR T4 and back into the +ve point of voltage source. Even though the direction of current through the source alternates from one half-cycle to the other half-cycle, the current through the load remains unidirectional (from up to down). Fig.3.15 shows various voltages and currents waveforms for the converter shown in Fig.3.14. Fig.3.16 shows the FFT components of load voltage and supply current.

Fig.3.14 Full wave fully controlled rectifier with RL load.

102 Chapter Three

Fig.3.15 Various voltages and currents waveforms for the converter shown in

Fig.3.14 in continuous conduction mode.

Fig.3.16 FFT components of load voltage and supply current in continuous

conduction mode.

Let vs = Vm sin ω t, with 0 < ω t < 360o. If ω t = 30o when T1 and T2 are triggered, then the firing angle is said to be 30o. In this instance the other pair is triggered when ω t= 210o. When vs changes from a positive to a negative value, the current through the load does not fall to zero value at the instant ω t = π radians, since the load contains an inductor


and the SCRs continue to conduct, with the inductor acting as a source. When the current through an inductor is falling, the voltage across it changes sign compared with the sign that occurs when its current is rising. When the current through the inductor is falling, its voltage is such that the inductor delivers power to the load resistor, feeds back some power to the AC source under certain conditions and keeps the SCRs in conduction forward-biased. If the firing angle is less than the load angle, the energy stored in the inductor is sufficient to maintain conduction till the next pair of SCRs is triggered. When the firing angle is greater than the load angle, the current through the load becomes zero and the conduction through the load becomes discontinuous. Usually the description of this circuit is based on the assumption that the load inductance is sufficiently large to keep the load current continuous and ripple-free.

Since the output waveform repeats itself every half-cycle, the average output voltage is expressed in equation (3.33) as a function of α, the firing angle. The maximum average output voltage occurs at a firing angle of 0o as shown in equation (3.34). The rms value of output voltage is obtained as shown in equation (3.35).

απ

ωωπ

απ

α

cos2)sin(1 mmdc

VtdtVV == ∫+

(3.33)

The normalized output voltage is αcos==dm

dcn V

VV (3.34)

The rms value of output voltage is obtained as shown in equation (3.35).

( )2

)2cos(1(2

)sin(1 2 mmmrms

Vtdt

VtdtVV =−== ∫∫

++ απ

α

απ

α

ωωπ

ωωπ

(3.35)

Example 6 The rectifier shown in Fig.3.14 has pure DC load current of 50 A and, Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain an average output voltage of 70% of the maximum possible output voltage, calculate:- (a) The delay angle α, (b) The efficiency, (c) Ripple factor (d) The transformer utilization factor, (e) The peak inverse voltage (PIV) of the thyristor and (f) Crest factor of input current. (g) Input displacement factor.

104 Chapter Three

Solution: (a) Vdm is the maximum output voltage and can be acheaved when α=0. The normalized output voltage is shown in equation (3.30) which is

required to be 70%. Then, 7.0cos === αdm

dcn V

VV , then, α=45.5731o= 0.7954

(b) 220=mV

VVVV mdmdc 04.982*7.0*7.0 ===

π,

2m

rmsVV =

At α=45.5731o Vrms=155.563 V. Then, Irms=50 A

VVV mS 56.155

2==

The rms value of the transformer secondery current is: AII rmsS 50== Then, The rectification efficiency is

%02.6350*563.155

50*04.98**

====rmsrms

dcdc

ac

dcIVIV

PP

η

(c) 587.104.98563.155

===dc

rmsVVFF

23195.113733.11 22 =−=−== FFVVRF

dc

ac

(d) 4589.050*56.15550*04.98

===SS

dcIV

PTUF

(e) The PIV is Vm

(f) Creast Factor CF, 1)( ==S

peakS

II

CF

3.3.4 Full Wave Fully Controlled Rectifier With R-L Load In discontinuous Conduction Mode The converter circuit of Fig.3.14 discussed before was assumed to operate in continuous conduction mode (i.e. the load angle is bigger than the firing angle). Sometimes the converter shown in Fig.3.14 can work in discontinue mode where the load current falls to zero every half cycle and before the next thyristor in sequence is fired as shown in Fig.3.17. The equation during the conduction can be given as shown in equation (3.36). Which can be solved for i as shown in equation (3.37).


Fig.3.17 Load, resistor and inductor voltages waveforms along with supply

voltage waveforms of the converter shown in Fig.3.14 in case of discontinuous conduction mode.

Fig.3.18 Supply current waveform along with supply voltage waveforms of the

converter shown in Fig.3.14 in case of discontinuous conduction mode.

106 Chapter Three

Fig.3.18 FFT components of supply current along with supply voltage of the

converter shown in Fig.3.14 in discontinuous conduction mode.

During the period of βωα ≤≤ t the following differential equation can be obtained:

βωαω ≤≤=+ ttViRdtdiL m ),(sin* (3.35)

The solution of the above equation is given as in the following:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−−=

−−

φαω

φαφωω tan)(

)sin()sin()(t

m etZ

Vti (3.36)

Where 2221 )(tan LRZandRL ωωφ +== −

But i=0 when ω t=β substitute this condition in equation (3.36) gives equation (3.37) which used to determine β. Once the value of A, α and the extinction angle β are known, the average and rms output voltage at the cathode of the SCR can be evaluated as shown in equation (3.38) and (3.39) respectively.

φαβ

φαφβ tan)(

)sin()sin(−

−−=− e (3.37)

)cos(cos*sin* βαπ

ωωπ

β

α

−== ∫ mmdc

VtdtVV (3.38)

The rms load voltage is:


( ) ( ) )2cos2(cos21*

2sin*1 2 αβαβ

πωω

π

β

α

−−−== ∫ mmrms

VtdtVV (3.39)

The average load current can be obtained as shown in equation (3.40) by dividing the average load voltage by the load resistance, since the average voltage across the inductor is zero.

)cos(cos* βαπ

−=R

VI m

dc (3.40)

3.3.5 Single Phase Full Wave Fully Controlled Rectifier With Source Inductance: Full wave fully controlled rectifier with source inductance is shown in Fig.3.19. The presence of source inductance changes the way the circuit operates during commutation time. Let vs = Vm sin wt, with 0 < ω t < 360o. Let the load inductance be large enough to maintain a steady current through the load. Let firing angle α be 30o. Let SCRs T3 and T4 be in conduction before ω t < 30o. When T1 and T2 are triggered at ω t = 30o, there is current through the source inductance, flowing in the direction opposite to that marked in the circuit diagram and hence commutation of current from T3 and T4 to T1 and T2 would not occur instantaneously. The source current changes from dcI− to dcI due to the whole of the source voltage being applied across the source inductance. When T1 is triggered with T3 in conduction, the current through T1 would rise from zero to dcI and the current through T3 would fall from

dcI to zero. Similar process occurs with the SCRs T2 and T4. During this period, the current through T2 would rise from zero to dcI and, the current through T4 would fall from dcI to zero.

Fig.3.19 single phase full wave fully controlled rectifier with source inductance

108 Chapter Three

Various voltages and currents waveforms of converter shown in Fig.3.19 are shown in Fig.3.20 and Fig.3.21. You can observe how the currents through the devices and the line current change during commutation overlap.

Fig.3.20 Output voltage, thyristors current along with supply voltage waveform

of a single phase full wave fully controlled rectifier with source inductance.

Fig.3.21 Output voltage, supply current along with supply voltage waveform of

a single phase full wave fully controlled rectifier with source inductance.


Let us study the commutation period starts at ut +<< αωα . When T1 and T2 triggered, then T1 and T2 switched on and T3 and T4 try to switch off. If this happens, the current in the source inductance has to change its direction. But source inductance prevents that to happen instantaneously. So, it will take time tΔ to completely turn off T3 and T4 and to make T1 and T2 carry the entire load current oI which is very clear from Fig.3.20. Also, in the same time ( tΔ ) the supply current changes from oI− to oI which is very clear from Fig.3.21. Fig.3.22 shows the equivalent circuit of the single phase full-wave controlled rectifier during that commutation period. From Fig.3.22 We can get the differential equation representing the circuit during the commutation time as shown in (3.41).

Fig.3.22 The equivalent circuit of single phase fully controlled rectifier

during the commutation period.

0=−dtdiLv s

ss (3.41)

Multiply the above equation by tω then, 0sin =− ssm diLtdtV ωωω

Integrate the above equation during the commutation period we get the following:

∫∫−

+

=o

o

I

Iss

u

m diLtdtV ωωωα

α

sin

Then, ( )[ ] osm ILuV ωαα 2coscos =+− . Then,

( )m

osV

ILu ωαα

2coscos −=+

110 Chapter Three

Then ( ) αωα −⎥⎦

⎤⎢⎣

⎡−= −

m

os

VILu 2coscos 1 (3.42)

Then ( )⎭⎬⎫

⎩⎨⎧

−⎥⎦

⎤⎢⎣

⎡−−==Δ − αωα

ωω m

os

VILut 2coscos1 1 (3.43)

It is clear that the DC voltage reduction due to the source inductance, rdv is the drop across the source inductor. Then,

dtdiLv s

srd = (3.44)

Then, os

I

Iss

u

rd ILdiLtdtvo

o

ωωωα

α

2== ∫∫−

+

(3.45)

∫+u

rd tdtvα

α

ω is the reduction area in one commutation period. But we

have two commutation periods in one period of supply voltage waveform. So, the total reduction per period is shown in (3.46):

os

u

rd ILtdtV ωωα

α

42 =∫+

(3.46)

To obtain the average reduction in DC output voltage rdV due to source inductance we have to divide the above equation by the period of supply voltage waveform, π2 . Then,

osos

rd IfLILV 42

4==

πω (3.47)

The DC voltage with source inductance taking into account can be calculated as following:

osm

rdceinducsourcewithoutdcactualdc IfLVVVV 4cos2tan −=−= α

π(3.48)

The rms value of supply current is the same as obtained before in single phase full bridge diode rectifier in (2.64)

⎥⎦⎤

⎢⎣⎡ −=

322 2 uII o

sπ

π (3.49)


The Fourier transform of supply current is the same as obtained for single phase full bridge diode rectifier in (2.66) and the fundamental component of supply current 1sI is shown in (2.68) as following:

2sin*

28

1u

uII o

S π= (3.50)

The power factor of this rectifier is shown in the following:

⎟⎠⎞

⎜⎝⎛ +=

2cos. 1 u

IIfp

s

s α (3.51)

3.3.5 Single-Phase Half Controlled Bridge Rectifier (Semi Bridge Converter)

A half controlled single-phase bridge is shown in Fig.3.23. This rectifier uses two SCRs and two diodes in addition to freewheeling diode. When the source voltage is in its positive half cycle and thyristor T2 is triggered and it will conduct with diode D1. When the supply voltage is going to negative and thyristor T4 is triggered and it will conduct with diode D3. This circuit cannot work without freewheeling diode in case of inductive load. The inductive energy in the load would freewheel through the diode D1 and Thyristor T4 or diode D3 and thyristor T2 even if there is no gate signal. If there is freewheeling diode is connected across the load of a bridge converter, it will remove the negative voltage in the load voltage. In this converter the diode D1 and Thyristor T2 work in positive half cycle of the supply voltage and D3 and T4 can work in the negative half cycle. As an example at 60o firing angle as shown in Fig.3.24, then the thyristor T2 triggers at 60o conducts till ω t =π. At ω t=π the voltage across the freewheeling diode will start to be forward then the stored energy in the load will take the freewheeling diode as a pass to circulate the stored reactive power, In this case D1 and T2 will turned off. The circuit will still like that till the thyristor T4 gets its triggering pulse at angle 240o. In this case, the voltage across the freewheeling diode will reverse and diode D3 and thyristor T4 will conduct till ω t=360o. When ω t=360o the voltage across the freewheeling diode will be forward again and the load will freewheel the stored reactive power in the load through the freewheeling diode and the diode D3 and T4 will turned off and cycle will repeated again. The average load voltage is shown in shown in the following equation:-

112 Chapter Three

Fig.3.23 Single-phase half controlled bridge rectifier (semi bridge converter).

( )[ ] )cos1()cos(cos)sin(1 απ

αππ

ωωπ

π

α

+=−−−== ∫ mmmdc

VVtdtVV (3.52)

dmV is the maximum output voltage and can be acheaved when α=0,

The normalized output voltage is: )cos1(5.0 α+==dm

dcn V

VV (3.53)

The rms value of output voltage is obtained as shown in the following equation:-

( )2

)2sin(2

)sin(1 2 ααππ

ωωπ

π

α

+−== ∫ mmrms

VtdtVV (3.54)

Fig.3.24 Various voltages and currents waveforms for the converter shown in Fig.3.23.


3.3.6 Inverter Mode Of Operation The thyristor converters can also operate in an inverter mode, where dV has a negative value, and hence the power flows from the do side to the ac side. The easiest way to understand the inverter mode of operation is to assume that the DC side of the converter can be replaced by a current source of a constant amplitude dI , as shown in Fig.3.25. For a delay angle a greater than 90° but less than 180°, the voltage and current waveforms are shown in Fig.3.26. The average value of dv is negative, given by (3.48), where 90° < α < 180°. Therefore, the average power

ddd IVP *= is negative, that is, it flows from the DC to the AC side. On

the AC side, 11 cosφSsac IVP = is also negative because o90>φ .

Fig.3.25 Single phase SCR inverter.

Fig.3.26 Waveform output from single phase inverter assuming DC load current.

114 Chapter Three

There are several points worth noting here. This inverter mode of operation is possible since there is a source of energy on the DC side. On the ac side, the ac voltage source facilitates the commutation of current from one pair of thyristors to another. The power flows into this AC source.

Generally, the DC current source is not a realistic DC side representation of systems where such a mode of operation may be encountered. Fig.3.27 shows a voltage source dE on the DC side that may represent a battery, a photovoltaic source, or a DC voltage produced by a wind-electric system. It may also be encountered in a four-quadrant DC motor supplied by a back-to-back connected thyristor converter.

An assumption of a very large value of dL allows us to assume di to be a constant DC, and hence the waveforms of Fig.3.28 also apply to the circuit of Fig.3.27. Since the average voltage across dL is zero,

dsdodd ILVVE ωπ

α 2cos −== (3.55)

The equation is exact if the current is constant at dI ; otherwise, a value of di at αω =t should be used in (3.55) instead of dI . Fig.3.28 shows that for a given value of α , for example, 1α , the intersection of the do source voltage 1dd EE = , and the converter characteristic at 1α , determines the do current 1dI , and hence the power flow 1dP .

During the inverter mode, the voltage waveform across one of the thyristors is shown in Fig.3.29. An extinction angle γ is defined to be as shown in (3.56) during which the voltage across the thyristor is negative and beyond which it becomes positive. The extinction time interval

ωγγ /=t should be greater than the thyristor turn-off time qτ Otherwise, the thyristor will prematurely begin to conduct, resulting in the failure of current to commutate from one thyristor pair to the other, an abnormal operation that can result in large destructive currents.

( )u+−= αγ 180 (3.56)


Fig.3.27 SCR inverter with a DC voltage source.

Fig.3.28 dV versus dI in SCR inverter with a DC voltage source.

Fig.3.29 Voltage across a thyristor in the inverter mode.

116 Chapter Three

Inverter startup For startup of the inverter in Fig.3.25, the delay angle α is initially

made sufficiently large (e.g.,165o) so that di is discontinuous as shown in Fig.3.30. Then, α is decreased by the controller such that the desired dI and dP are obtained.

Fig.3.30 Waveforms of single phase SCR inverter at startup.

3.5 Three Phase Half Wave Controlled Rectifier 3.5.1 Three Phase Half Wave Controlled Rectifier With Resistive Load Fig.3.31 shows the circuit of a three-phase half wave controlled rectifier, the control circuit of this rectifier has to ensure that the three gate pulses for three thyristor are displaced 120o relative to each other’s. Each thyristor will conduct for 120o. A thyristor can be fired to conduct when its anode voltage is positive with respect to its cathode voltage. The maximum output voltage occurred when α=0 which is the same as diode case. This rectfier has continuous load current and voltage in case of α ≤ 30. However, the load voltage and current will be discontinuous in case of α > 30.

Fig.3.31 Three phase half wave controlled rectifier with resistive load.


In case of α ≤ 30, various voltages and currents of the converter shown in Fig.3.31 are shown in Fig.3.32. Fig.3.33 shows FFT components of load voltage, secondary current and primary current. As we see the load voltage contains high third harmonics and all other triplex harmonics. Also secondary current contains DC component, which saturate the transformer core. The saturation of the transformer core is the main drawback of this system. Also the primary current is highly distorted but without a DC component. The average output voltage and current are shown in equation (3.57) and (3.58) respectively. The rms output voltage and current are shown in equation (3.59) and (3.60) respectively.

ααπ

ααπ

ωωπ

απ

απ

cos675.0cos23

cos827.0cos233sin

23 6/5

6/

LLLL

mm

mdc

VV

VVtdtVV

==

=== ∫+

+ (3.57)

ααπ

cos*827.0

cos**2

33R

VR

VI mm

dc == (3.58)

( ) απ

ωωπ

απ

απ

2cos8

3613sin

23 6/5

6/

2 +== ∫+

+mmrms VtdtVV (2.57)

R

VI

m

rms

απ

2cos8

3613 +

= (3.60)

Then the thyristor rms current is equal to secondery current and can be obtaiend as follows:

R

VI

IIm

rmsSr

2/1

2cos8

361

3

⎟⎟⎠

⎞⎜⎜⎝

⎛+

===

απ

(3.61)

The PIV of the diodes is mLL VV 32 = (3.62)

118 Chapter Three

Fig.3.32 Voltages and currents waveforms for rectifier shown in Fig.3.31 at α ≤ 30.


Fig.3.33 FFT components of load voltage, secondary current and supply current

for the converter shown in Fig.3.31 for α ≤ 30.

In case of α > 30, various voltages and currents of the rectifier shown in Fig.3.31 are shown in Fig.3.34. Fig.3.35 shows FFT components of load voltage, secondary current and primary current. As we can see the load voltage and current equal zero in some regions (i.e. discontinuous load current). The average output voltage and current are shown in equation (3.63) and (3.64) respectively. The rms output voltage and current are shown in equation (3.65) and (3.66) respectively. The average output voltage is :-

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++=⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++== ∫

+

απαππ

ωωπ

π

απ6

cos14775.06

cos12

3sin23

6/m

mmdc VVtdtVV (3.63)

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++= απ

π 6cos1

23

RVI m

dc (3.64)

( ) )23/sin(81

42453sin

23

6/

2 απππ

αωωπ

π

απ

++−== ∫+

mmrms VtdtVV (2.63)

)23/sin(81

42453

απππ

α++−=

RVI m

rms (3.66)

Then the diode rms current can be obtaiend as follows:

)23/sin(81

4245

3απ

ππα

++−===RVIII mrms

Sr (3.67)

The PIV of the diodes is mLL VV 32 = (3.68)

120 Chapter Three

Fig.3.34 Various voltages and currents waveforms for converter shown in

Fig.3.22 for α > 30.



for the converter shown in Fig.3.22 for α > 30.

Example 7 Three-phase half-wave controlled rectfier is connected to 380 V three phase supply via delta-way 380/460V transformer. The load of the rectfier is pure resistance of 5Ω . The delay angle o25=α . Calculate: The rectfication effeciency (b) Transformer Utilization Factor (TUF) (c) Crest Factor FC of the input current (d) PIV of thyristors

Solution: From (3.57) the DC value of the output voltage can be obtained as following:

VVV LLdc 5.28125cos46023cos

23

===π

απ

Then; AR

VI dcdc 3.56

55.281

===

From (3.59) we can calculate rmsV as following:

απ

απ

2cos8

361*22cos

83

613 +=+= LLmrms VVV

Then, ( ) VVrms 8.29825*2cos8

361*460*2 =+=

π

122 Chapter Three

Then AR

VI rmsrms 76.59

58.298

===

Then, the rectfication effeciency can be calculated as following:

%75.88100* ==rmsrms

dcdcIVIV

η

The rms value of the secondary current can be calculated as following:

AII rmsS 5.34

376.59

3===

%66.57100*5.34*460*3

3.56*5.281*3

===sLL

dcdcIV

IVTUF

( ) ( ) AVR

VI LLmpeakS 12.75

5460*3/2

5*3/2

, ====

177.25.3412.75, ===

S

peakSF I

IC

VVPIV LL 54.650460*22 === Example 8 Solve the previous example (evample 7) if the firing angle

o60=α Slution: From (3.63) the DC value of the output voltage can be obtained as following:

VVV mdc 33.179

36cos1

2

460*323

6cos1

23

=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++

⎟⎟⎠

⎞⎜⎜⎝

⎛

=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++=

πππ

αππ

Then; AR

VI dcdc 87.35

533.179

===

From (3.65) we can calculate rmsV as following:

V

VV mrms

230)3/23/sin(81

43/

245*460*2

)23/sin(81

42453

=++−=

++−=

ππππ

π

απππ

α

Then AR

VI rmsrms 46

5230

===

Then, the rectfication effeciency can be calculated as following


%79.60100* ==rmsrms

dcdcIVIV

η

The rms value of the secondary current can be calculated as following:

AII rmsS 56.26

346

3===

%4.30100*56.26*460*3

87.35*33.179*3

===sLL

dcdcIV

IVTUF

( ) ( ) AVR

VI LLmpeakS 12.75

5460*3/2

5*3/2

, ====

83.256.2612.75, ===

S

peakSF I

IC

VVPIV LL 54.650460*22 === 3.5 Three Phase Half Wave Controlled Rectifier With DC Load Current

The Three Phase Half Wave Controlled Rectifier With DC Load Current is shown in Fig.3.36, the load voltage will reverse its direction only if α > 30. However if α < 30 the load voltage will be positive all the time. Then in case of α > 30 the load voltage will be negative till the next thyristor in the sequence gets triggering pulse. Also each thyristor will conduct for 120o if the load current is continuous as shown in Fig.3.37. Fig.3.38 shows the FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.36 for α > 30 and pure DC current load.

Fig.3.36 Three Phase Half Wave Controlled Rectifier With DC Load Current

124 Chapter Three


Fig.3.36 for α > 30 and pure DC current load.


for the converter shown in Fig.3.36 for α > 30 and pure DC current load.

t=0


As explained before the secondary current of transformer contains DC component. Also the source current is highly distorted which make this system has less practical significance. The THD of the supply current can be obtained by the aid of Fourier analysis as shown in the following:-

If we move y-axis of supply current to be as shown in Fig.3.33, then the waveform can be represented as odd function. So, an=0 and bn can be obtained as the following:-

⎟⎠⎞

⎜⎝⎛ −== ∫ 3

2cos12)sin(2 3/2

0

ππ

ωωπ

π nn

ItdtnIb dcdcn for n=1,2,3,4,…

Then, 23*

2n

Ib dc

n π= for n=1,2,4,5,7,8,10,….. (3.69)

And 0=nb For n=3,6,9,12,….. (3.70) Then the source current waveform can be expressed as the following

equation

⎥⎦⎤

⎢⎣⎡ +++++= ......7sin

715sin

514sin

412sin

21sin3)( tttttIti dc

p ωωωωωπ

ω (3.71)

The resultant waveform shown in equation (3.61) agrees with the result from simulation (Fig.3.38). The THD of source current can be obtained by two different methods. The first method is shown below:-

21

21

2

p

pp

I

IITHD

−= (3.72)

Where, dcp II *32

= (3.73)

The rms of the fundamental component of supply current can be obtained from equation (3.71) and it will be as shown in equation (3.74)

π23

1dc

pII = (3.74)

Substitute equations (3.73) and (3.74) into equation (3.72), then,

%68

29

29

32

22

22

2

=−

=dc

dcdc

I

IITHD

π

π (3.75)

126 Chapter Three

Another method to determine the THD of supply current is shown in the following:- Substitute from equation (3.71) into equation (3.72) we get the following equation:-

%68....141

131

111

101

81

71

51

41

21 222222222

≅+⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=THD (3.76)

The supply current THD is very high and it is not acceptable by any electric utility system. In case of full wave three-phase converter, the THD in supply current becomes much better than half wave (THD=35%) but still this value of THD is not acceptable.

Example 9 Three phase half wave controlled rectfier is connected to 380 V three phase supply via delta-way 380/460V transformer. The load of the rectfier draws 100 A pure DC current. The delay angle, o30=α . Calculate:

(a) THD of primary current. (b) Input power factor. Solution: The voltage ratio of delta-way transformer is 380/460V. Then,

the peak value of primary current is A05.121380460*100 = . Then,

AI rmsP 84.9832*05.121, == .

1PI can be obtained from equation (372) where

AII dcP 74.81

205.121*3

23

1 ===ππ

.

Then, ( ) %98.67100*174.8184.98100*1

22

1

, =−⎟⎠⎞

⎜⎝⎛=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

P

rmsPI I

ITHD

P

The input power factor can be calculated as following:

LaggingI

IfPrmsP

P 414.066

cos*84.9874.81

6cos*.

,

1 =⎟⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ +=

πππα

3.6 Three Phase Half Wave Controlled Rectifier With Free Wheeling Diode

The circuit of three-phase half wave controlled rectifier with free wheeling diode is shown in Fig.3.39. Various voltages and currents waveforms of this converter are shown in Fig.3.40. FFT components of


load voltage, secondary current and supply current for the converter shown in Fig.3.39 for α > 30 and RL load is shown in Fig.3.41. In case of firing angle α less than 30o, the output voltage and current will be the same as the converter without freewheeling diode, because of the output voltage remains positive all the time. However, for firing angle α greater than 30o, the freewheeling diode eliminates the negative voltage by bypassing the current during this period. The freewheeling diode makes the output voltage less distorted and ensures continuous load current. Fig.3.40 shows various voltages and currents waveforms of the converter shown in Fig.3.39. The average and rms load voltage is shown below:-

The average output voltage is :-

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++=⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++== ∫

+

απαππ

ωωπ

π

απ6

cos14775.06

cos12

3sin

23

6/m

mmdc V

VtdtVV (3.77)

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++= απ

π 6cos1

23

RVI m

dc (3.78)

( ) )23/sin(81

42453sin

23

6/

2 απππ

αωωπ

π

απ

++−== ∫+

mmrms VtdtVV

(3.79)

)23/sin(81

42453

απππ

α++−=

RV

I mrms (3.80)

Fig.3.39 Three-phase half wave controlled rectifier with free wheeling diode.

128 Chapter Three


Fig.3.36 for α > 30 with RL load and freewheeling diode.


for the converter shown in Fig.3.36 for α > 30 and RL load.


3.7 Three Phase Full Wave Fully Controlled Rectifier Bridge 3.7.1 Three Phase Full Wave Fully Controlled Rectifier With Resistive Load

Three-phase full wave controlled rectifier shown in Fig.3.42. As we can see in this figure the thyristors has labels T1, T2,……,T6. The label of each thyristor is chosen to be identical to triggering sequence where thyristors are triggered in the sequence of T1, T2,……,T6 which is clear from the thyristors currents shown in Fig.3.43.

Fig.3.42 Three-phase full wave controlled rectifier.

Fig.3.43 Thyristors currents of three-phase full wave controlled rectifier.

130 Chapter Three

The operation of the circuit explained here depending on the understanding of the reader the three phase diode bridge rectifier. The Three-phase voltages vary with time as shown in the following equations:

)120(sin)120(sin

)(sin

+=−=

=

tVvtVvtVv

mc

mb

ma

ωωω

It can be seen from Fig.3.44 that the voltage av is the highest positive voltage of the three phase voltage when tω is in the range

ot 15030 << ω . So, the thyristor T1 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44 the firing angle 40=α as an example. So, T1 takes a pulse at ot 70403030 =+=+= αω as shown in Fig.3.44. Also, it is clear

from Fig3.38 that thyristor T1 or any other thyristor remains on for o120 .

Fig.3.44 Phase voltages and thyristors currents of three-phase full wave

controlled rectifier at o40=α .

It can be seen from Fig.3.44 that the voltage bv is the highest positive voltage of the three phase voltage when tω is in the range of

ot 270150 << ω . So, the thyristor T3 is forward bias during this period


and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle 40=α as an example. So, T3 takes a pulse at ot 19040150150 =+=+= αω .

It can be seen from Fig.3.44 that the voltage cv is the highest positive voltage of the three phase voltage when tω is in the range

ot 390270 << ω . So, the thyristor T5 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle 40=α as an example. So, T3 takes a pulse at ot 310270 =+= αω .

It can be seen from Fig.3.44 that the voltage av is the highest negative voltage of the three phase voltage when tω is in the range

ot 330210 << ω . So, the thyristor T4 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.338, the firing angle 40=α as an example. So, T4 takes a pulse at ot 25040210210 =+=+= αω .

It can be seen from Fig.3.44 that the voltage bv is the highest negative voltage of the three phase voltage when tω is in the range

ot 450330 << ω or ot 90330 << ω in the next period of supply voltage waveform. So, the thyristor T6 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle 40=α as an example. So, T6 takes a pulse at ot 370330 =+= αω .

It can be seen from Fig.3.44 that the voltage cv is the highest negative voltage of the three phase voltage when tω is in the range

ot 21090 << ω . So, the thyristor T2 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle 40=α as an example. So, T2 takes a pulse at ot 13090 =+= αω .

From the above explanation we can conclude that there is two thyristor in conduction at any time during the period of supply voltage. It is also clear that the two thyristors in conduction one in the upper half (T1, T3, or, T5) which become forward bias at highest positive voltage connected to its anode and another one in the lower half (T2, T4, or, T6)

132 Chapter Three

which become forward bias at highest negative voltage connected to its cathode. So the load is connected at any time between the highest positive phase voltage and the highest negative phase voltage. So, the load voltage equal the highest line to line voltage at any time which is clear from Fig.3.45. The following table summarizes the above explanation.

Period, range of wt SCR Pair in conduction α + 30o to α + 90o T1 and T6 α + 90o to α + 150o T1 and T2 α + 150o to α + 210o T2 and T3 α + 210o to α + 270o T3 and T4 α + 270o to α + 330o T4 and T5 α + 330o to α + 360o and α + 0o to α + 30o T5 and T6

Fig.3.45 Output voltage along with three phase line to line voltages of rectifier

in Fig.3.42 at o40=α .

The line current waveform is very easy to obtain it by applying kerchief's current law at the terminals of any phase. As an example

41 TTa III −= which is clear from Fig.3.42. The input current of this rectifier for ( )60,40 ≤= αα is shown in Fig.3.46. Fast Fourier transform (FFT) of output voltage and supply current are shown in Fig.3.47.


Fig.3.46 The input current of this rectifier of rectifier in Fig.3.42 at

( )60,40 ≤= αα .

Fig.3.46 FFT components of output voltage and supply current of rectifier in

Fig.3.42 at ( )60,40 ≤= αα .

134 Chapter Three

Analysis of this three-phase controlled rectifier is in many ways similar to the analysis of single-phase bridge controlled rectifier circuit. The average output voltage, the rms output voltage, the ripple content in output voltage, the total rms line current, the fundamental rms current, THD in line current, the displacement power factor and the apparent power factor are to be determined. In this section, the analysis is carried out assuming that the load is pure resistance.

απ

ωπωπ

απ

απ

cos33)6

sin(33 2/

6/

mmdc

VtdtVV =+= ∫+

+

(3.81)

The maximum average output voltage for delay angle α=0 is

πm

dmVV 33

= (3.82)

The normalized average output voltage is as shown in (3.83)

αcos==dm

dcn V

VV (3.83)

The rms value of the output voltage is found from the following equation:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟

⎠⎞

⎜⎝⎛ += ∫

+

+

απ

ωπωπ

απ

απ

2cos4

33213)

6sin(33 2/

6/

2

mmrms VtdtVV (3.84)

In the converter shown in Fig.3.42 the output voltage will be continuous only and only if o60≤α . If o60>α the output voltage, and phase current will be as shown in Fig.3.47.

Fig.3.47 Output voltage along with three phase line to line voltages of rectifier

in Fig.3.42 at o75=α .


The average and rms values of output voltage is shown in the following equation:

( )[ ]αππ

ωπωπ

π

απ

++=+= ∫+

3/cos133)6

sin(33 6/5

6/

mmdc

VtdtVV (3.85)

The maximum average output voltage for delay angle α=0 is

πm

dmVV 33

= (3.86)

The normalized average output voltage is

( )[ ]απ ++== 3/cos1dm

dcn V

VV (3.87)

The rms value of the output voltage is found from the following equation:

⎟⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +−−=

⎟⎠⎞

⎜⎝⎛ += ∫

+

62cos2

4313

)6

sin(33 6/5

6/

2

πααπ

ωπωπ

π

απ

m

mrms

V

tdtVV (3.88)

Example 10 Three-phase full-wave controlled rectifier is connected to 380 V, 50 Hz supply to feed a load of 10Ω pure resistance. If it is required to get 400 V DC output voltage, calculate the following: (a) The firing angle, α (b) The rectfication effeciency (c) The crest factor of input current. (d) PIV of the thyristors. Solution: From (3.81) the average voltage is :

VVV mdc 400cos

380*32*33

cos33=== α

πα

π.

Then o79.38=α , AR

VI dcdc 40

10400

===

From (3.84) the rms value of the output voltage is:

( )⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+= 79.38*2cos

433

21*380*

32*32cos

433

213

πα

πmrms VV

Then, VVrms 412.412=

136 Chapter Three

Then, AR

VI rmsrms 24.41

10412.412

===

Then, %07.94100*24.41*4.412

40*400100***

===rmsrms

dcdcIVIV

η

The crest factor of input current, rmss

peakSF I

IC

,

,=

( ) AR

tI peakS 11.53

103079.3830sin380*26

sin380*2, =

++=

⎟⎠⎞

⎜⎝⎛ +

=

πω

AII rmsrmss 67.3324.41*32*

32

, ===

Then, 577.167.3311.53

,

, ===rmss

peakSF I

IC

The PIV= 3 Vm=537.4V

Example 11 Solve the previous example if the required dc voltage is 150V. Solution: From (3.81) the average voltage is :

VVV mdc 150cos

380*32*33

cos33=== α

πα

π. Then, o73=α

It is not acceptable result because the above equation valid only for 60≤α . Then we have to use the (3.85) to get dcV as following:

( )[ ] VVdc 1503/cos1380*

32*33

=++== αππ

. Then, o05.75=α

Then AR

VI dcdc 15

10150

===

From (3.88) the rms value of the output voltage is:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +−−=

3005.75*2cos180

*05.75*2431*380*

32*3

62cos2

4313

ππ

πααπmrms VV


Then, VVrms 075.198=

Then, AR

VI rmsrms 8075.19

10075.198

===

Then, %35.57100*81.19*075.198

15*150100***

===rmsrms

dcdcIVIVη

The crest factor of input current, rmss

peakSF I

IC

,

,=

( ) AR

tI peakS 97.37

103005.7530sin380*26

sin380*2, =

++=

⎟⎠⎞

⎜⎝⎛ +

=

πω

AII rmsrmss 1728.168075.19*32*

32

, ===

Then, 348.21728.16

97.37

,

, ===rmss

peakSF I

IC

The PIV= 3 Vm=537.4V 3.7.1 Three Phase Full Wave Fully Controlled Rectifier With pure DC Load Current Three-phase full wave-fully controlled rectifier with pure DC load current is shown in Fig.3.48. Fig.3.49 shows various currents and voltage of the converter shown in Fig.3.48 when the delay angle is less than 60o. As we see in Fig.3.49, the load voltage is only positive and there is no negative period in the output waveform. Fig.3.50 shows FFT components of output voltage of rectifier shown in Fig.3.48 for o60<α .

Fig.3.48 Three phase full wave fully controlled rectifier with pure dc load current

138 Chapter Three

Fig.3.49 Output voltage and supply current waveforms along with three phase line voltages for the rectifier shown in Fig.3.48 for α < 60o with pure DC current load.

Fig.3.50 FFT components of SCR, secondary, primary currents respectively of

rectifier shown in Fig.3.48.


In case of the firing angle is greater than o60 , the output voltage contains negaive portion as shown in Fig.3.51. Fig.3.52 shows FFT components of output voltage of rectifier shown in Fig.3.48 for o60>α . The average and rms voltage is the same as in equations (3.81) and (3.84) respectively. The line current of this rectifier is the same as line current of three-phase full-wave diode bridge rectifier typically except the phase shift between the phase voltage and phase current is zero in case of diode bridge but it is α in case of three-phase full-wave controlled rectifier with pure DC load current as shown in Fig.3.53. So, the input power factor of three-phase full-wave diode bridge rectifier with pure DC load current is:

αcos1

s

sIIrPowerFacto = (3.89)

Fig.3.51 Output voltage and supply current waveforms along with three phase

line voltages for the rectifier shown in Fig.3.48 for α > 60o with pure DC current load.

140 Chapter Three

Fig.3.52 FFT components of SCR, secondary, primary currents respectively of

rectifier shown in Fig.3.48 for α > 60o.

Fig.3.53 Phase a voltage, current and fundamental components of phase a of three phase full bridge fully controlled rectifier with pure DC current load and 60>α .

In case of three-phase full-wave controlled rectifier with pure DC load and source inductance the waveform of output voltage and line current and their FFT components are shown in Fig.3.54 and Fig.3.55 respectively. The output voltage reduction due to the source inductance is the same as obtained before in Three-phase diode bridge rectifier. But, the commutation time will differ than the commutation time obtained in case of Three-phase diode bridge rectifier. It is left to the reader to determine the commutation angle u in case of three-phase full-wave


diode bridge rectifier with pure DC load and source inductance. The Fourier transform of line current and THD will be the same as obtained before in Three-phase diode bridge rectifier with pure DC load and source inductance which explained in the previous chapter.

Fig.3.54 Output voltage and supply current of rectifier shown n Fig.3.48 with

pure DC load and source inductance the waveforms.

Fig.3.55 FFT components of output voltage of rectifier shown in Fig.3.48 for α

> 60o and there is a source inductance.

142 Chapter Three

Let us study the commutation time shown in Fig.3.56n. At this time cV starts to be more negative than bV so T2 becomes forward bias and it is ready to switch as soon as it gets a pulse on its gate. Thyristor T2 gets a pulse after that by α as shown in Fig.3.56n, so, T6 has to switch OFF and T2 has to switch ON. But due to the source inductance will prevent that to happen instantaneously. So it will take time redut =Δ sec to completely turn OFF T6 and to make T2 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.3.56n. The equivalent circuit of the three phase full wave controlled rectifier at commutation time tΔ is shown in Fig.3.57n and Fig.3.58n.

α u

oI

oI

oI

oI

aVcV bV

2/π

Fig.3.56n Waveforms represent the one commutation period.


Fig.3.57n The equivalent circuit of the three phase controlled rectifier at

commutation time tΔ shown in Fig.3.56n.

From Fig.3.57n we can get the following defferntial equations:

061 =−−−− bT

bdcT

aa Vdt

diLVdt

diLV (3.90)

021 =−−−− cT

cdcT

aa Vdt

diLVdt

diLV (3.91)

Note that, during the time tΔ , 1Ti is constant so 01 =dt

diT , substitute this

value in (3.90) and (3.91) we get the following differential equations:

dcT

bba Vdt

diLVV =−− 6 (3.92)

dcT

cca Vdt

diLVV =−− 2 (3.93)

By equating the left hand side of equation (3.92) and (3.93) we get the following differential equation:

dtdiLVV

dtdiLVV T

ccaT

bba26 −−=−− (3.94)

026 =−+−dt

diLdt

diLVV Tc

Tbcb (3.95)

The above equation can be written in the following manner: ( ) 026 =−+− TcTbcb diLdiLdtVV (3.96) ( ) 026 =−+− TcTbcb diLdiLtdVV ωωω (3.97)

Integrate the above equation during the time tΔ with the help of Fig.3.56n we can get the limits of integration as shown in the following:

144 Chapter Three

( ) 00

2

0

6

2/

2/

=−+− ∫∫∫++

+

o

o

I

TcI

Tb

u

cb diLdiLtdVV ωωωαπ

απ

( ) 03

2sin3

2sin2/

2/

=−−+⎟⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −∫

++

+ocob

u

mm ILILtdtVtV ωωωπωπωαπ

απ

assume Scb LLL ==

oS

u

m ILttV ωπωπωαπ

απ2

32cos

32cos

2/

2/=⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−

++

+

oS

m

IL

uuV

ω

παππαππαππαπ

23

22

cos3

22

cos3

22

cos3

22

cos

=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++−⎟

⎠⎞

⎜⎝⎛ −++⎟

⎠⎞

⎜⎝⎛ ++++⎟

⎠⎞

⎜⎝⎛ −++−

oSm ILuuV ωπαπαπαπα 26

7cos6

cos6

7cos6

cos =⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ +++⎟

⎠⎞

⎜⎝⎛ −+−

( ) ( ) ( ) ( )

m

oSV

IL

uuuu

ωπαπαπαπα

παπαπαπα

26

7sinsin6

7coscos6

sinsin6

coscos

67sinsin

67coscos

6sinsin

6coscos

=⎥⎦⎤+−++

⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+−⎟

⎠⎞

⎜⎝⎛++⎟

⎠⎞

⎜⎝⎛+−⎟

⎠⎞

⎜⎝⎛+−

( ) ( ) ( ) ( )

m

oSV

IL

uuuu

ωαααα

αααα

2sin5.0cos23sin5.0cos

23

sin5.0cos23sin5.0cos

23

=⎥⎦

⎤−+

⎢⎣

⎡+++−+−+−

( )[ ]m

oSV

ILu ωαα

2coscos3 =+−

( ) ( )LL

oS

LL

o

m

oV

ILVLI

VLIu ωωω

αα2

22

32coscos ===+− (3.98)

( ) αωα −⎥⎦

⎤⎢⎣

⎡−= −

LL

oSV

ILu 2coscos 1 (3.99)

( )⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎥⎦

⎤⎢⎣

⎡−==Δ − α

ωα

ωω LL

oSV

ILut 2coscos1 1 (3.100)


It is clear that the DC voltage reduction due to the source inductance is

the drop across the source inductance. dt

diLv TSrd = (3.101)

Multiply (3.101) by tdω and integrate both sides of the resultant equation we get:

oS

I

D

u

rd ILLditdvo

ωωω

απ

απ

== ∫∫++

+ 0

2

2

(3.102)

∫++

+

u

rd tdv

απ

απ

ω2

2

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

απ

απ

662

2

=∫++

+

(3.103)

To obtain the average reduction in DC output voltage rdV due to source inductance we have to divide by the period time π2 . Then,

oo

rd fLILIV 62

6==

πω (3.104)

The DC voltage without source inductance tacking into account can be calculated as following:

osLLrdceinducsourcewithoutdcactualdc IfLVVVV 6cos23tan −=−= α

π(3.105)

Fig.3.58n shows the utility line current with some detailes to help us to calculate its rms value easly.

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+⎟⎠⎞

⎜⎝⎛= ∫ ∫

+u

u

ud

os tdItdt

uII

0

232

22

π

ωωωπ ⎥⎦

⎤⎢⎣⎡ −++= uuu

uIo

23312 3

2

2 ππ

146 Chapter Three

Then ⎥⎦⎤

⎢⎣⎡ −=

632 2 uI

I oS

ππ

(3.106)

u

oI

oI− 32π

sI

u+3

2π

262 u

+π

Fig.3.58n The utility line current

Fig.3.59 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(3.1) as explained before when we study Table (2.1).

26u

−π

u

oI

oI−26

5 u−

π

267 u

−π 26

11 u−

π

sI

u

uIo

uIo−

sI′

26u

−π

265 u

−π

267 u

−π

2611 u

−π

Fig.3.59 The utility line currents and its first derivative.


Table(3.1) Jumb value of supply current and its first derivative. sJ

26u

−π

26u

+π

265 u

−π

265 u

+π

267 u

−π

267 u

+π

2611 u

−π

2611 u

+π

sI 0 0 0 0 0 0 0 0

sI ′

uIo

uIo−

uIo−

uIo

uIo−

uIo

uIo

uIo−


⎥⎥⎦

⎤

⎢⎢⎣

⎡′−= ∑∑

==

m

sss

m

sssn tnJ

ntnJ

nb

11sin1cos1 ωω

π (3.107)

⎥⎦

⎤⎟⎠

⎞⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−

⎢⎣

⎡⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −

−=

2611sin

2611sin

267sin

267sin

265sin

265sin

26sin

26sin*11

unununun

ununununuI

nnb o

n

ππππ

πππππ

⎥⎦⎤

⎢⎣⎡ +−−=

611cos

67cos

65cos

6cos

2sin*2

2ππππ

πnnnnnu

unIb o

n (3.108)

Then, the utility line current can be obtained as in (3.109).

( ) ( ) ( ) ( )

( ) ( ) ⎥⎦

⎤++−−⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+

⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

tutu

tututuu

ti

ωω

ωωωπ

ω

13sin2

13sin13

111sin2

11sin11

1

7sin2

7sin715sin

25sin

51sin

2sin34

22

22 (3.109)

Then; ⎟⎠⎞

⎜⎝⎛=

2sin

621

uuI

I oS π

(3.110)

The power factor can be calculated from the following equation:

⎟⎠⎞

⎜⎝⎛ +

⎥⎦⎤

⎢⎣⎡ −

⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛=

2cos

632

2sin62

2cos

21 u

uI

uuI

uIIpf

o

o

S

S απ

π

π

Then; ⎟⎠⎞

⎜⎝⎛ +

⎥⎦⎤

⎢⎣⎡ −

⎟⎠⎞

⎜⎝⎛

=2

cos

63

2sin*32

uuu

u

pf αππ

(3.111)

148 Chapter Three

Note, if we approximate the source current to be trapezoidal as shown in Fig.3.58n, the displacement power factor will be as shown in (3.111) is

⎟⎠⎞

⎜⎝⎛ +

2cos uα . Another expression for the displacement power factor, by

equating the AC side and DC side powers [ ] as shown in the following derivation: From (3.98) and (3.105) we can get the following equation:

( )( )uVVV LLLLdc +−−= ααπ

απ

coscos23cos23

( )( )[ ]uVV LLdc +−−=∴ αααπ

coscoscos22

23

( )[ ]uVV LLdc ++=∴ ααπ

coscos2

23 (3.112)

Then the DC power output from the rectifier is odcdc IVP = . Then,

( )[ ]uIVP oLLdc ++= ααπ

coscos*2

23 (3.113)

On the AC side, the AC power is: 11 cos3 φSLLac IVP = (3.114) Substitute from (3.110) into (3.114) we get the following equation:

11 cos2

sin26cos2

sin2

343 φπ

φπ

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

uu

IVuu

IVP oLLoLLac (3.115)

By equating (3.113) and (3.115) we get the following: ( )[ ]

⎟⎠⎞

⎜⎝⎛

++=

2sin*4

coscoscos 1 uuu ααφ (2.116)

The source inductance reduces the magnitudes of the harmonic currents. Fig.3.60a through d show the effects of SL (and hence of u) on various harmonics for various values of α , where dI is a constant dc. The harmonic currents are normalized by 1I I, with 0=SL , which is

given by (2.98) where os IIπ6

1 = in this case. Normally, the DC-side

current is not a constant DC. Typical and idealized harmonics are shown in Table 3.2.


Fig.3.60 Normalized harmonic current in the presence of SL [ ].

Table 3.2 Typical and idealized harmonics.

3.7.2 Inverter Mode of Operation Once again, to understand the inverter mode of operation, we will assume that the do side of the converter can be represented by a current source of a constant amplitude dI , as shown in Fig.3.61. For a delay angle a greater than 90° but less than 180°, the voltage and current

150 Chapter Three

waveforms are shown in Fig.3.62a. The average value of dV is negative according to (3.81). On the ac side, the negative power implies that the phase angle 1φ , between sv and si , is greater than 90°, as shown in Fig.3.62b.

Fig.3.61 Three phase SCR inverter with a DC current.

Fig.3.62 Waveforms in the inverter shown in Fig.3.56.

In a practical circuit shown in Fig.3.63, the operating point for a given

dE and α can be obtained from the characteristics shown in Fig.3.64.


Similar to the discussion in connection with single-phase converters, the extinction angle ( )uo −−= αγ 180 must be greater than the thyristor turn-off interval qtω in the waveforms of Fig.3.54, where 5v is the voltage across thyristor 5.

Fig.3.63 Three phase SCR inverter with a DC voltage source.

Fig.3.64 dV versus dI of Three phase SCR inverter with a DC voltage source. Inverter Startup As discussed for start up of a single-phase inverter, the delay angle α in the three-phase inverter of Fig.3.63 is initially made sufficiently large (e.g., 165°) so that id is discontinuous. Then, α is decreased by the controller such that the desired dI and dP are obtained.

152 Chapter Three

Problems 1- Single phase half-wave controlled rectifier is connected to 220 V,

50Hz supply to feed Ω10 resistor. If the firing angle o30=α draw output voltage and drop voltage across the thyristor along with the supply voltage. Then, calculate, (a) The rectfication effeciency. (b) Ripple factor. (c) Peak Inverse Voltage (PIV) of the thyristor. (d) The crest factor FC of input current.

2- Single phase half-wave controlled rectfier is connected to 220 V, 50Hz supply to feed Ω5 resistor in series with mH10 inductor if the firing angle o30=α .

(a) Determine an expression for the current through the load in the first two periods of supply current, then fiend the DC and rms value of output voltage.

(b) Draw the waveforms of load, resistor, inductor voltages and load current.


4- single phase full-wave fully controlled rectifier is connected to 220V, 50 Hz supply to feed Ω5 resistor, if the firing angle

o40=α . Draw the load voltage and current, diode currents and supply current. Then, calculate (a) The rectfication effeciency. (b) Peak Inverse Voltage (PIV) of the thyristor. (c) Crest factor of supply current.

5- In the problem 4, if there is a 5mH inductor is connected in series with the Ω5 resistor. Draw waveforms of output voltage and current, resistor and inductor voltages, diode currents, supply currents. Then, find an expression of load current, DC and rms values of output voltages.

6- Solve problem 5 if the load is connected with freewheeling diode. 7- Single phase full wave fully controlled rectifier is connected to

220V, 50 Hz supply to feed the load with 47 A pure dc current. The firing angle o40=α . Draw the load voltage, thyristor, and load currents. Then, calculate (a) the rectfication effeciency. (b) Ripple factor of output voltage. (c) Crest factor of supply current. (d) Use Fourier series to fiend an expression for supply current. (e) THD of supply current. (f) Input power factor.

8- Solve problem 7 if the supply has a 3 mH source inductance.


9- Single phase full-wave semi-controlled rectifier is connected to 220 V, 50Hz supply to feed Ω5 resistor in series with 5 mH inductor, the load is connected in shunt with freewheeling diode. Draw the load voltage and current, resistor voltage and inductor voltage diodes and thyristor currents. Then, calculate dcV and

rmsV of the load voltages. If the freewheeling diode is removed, explain what will happen?

10- The single-phase full wave controlled converter is supplying a DC load of 1 kW with pure DC current. A 1.5-kVA-isolation transformer with a source-side voltage rating of 120 V at 50 Hz is used. It has a total leakage reactance of 8% based on its ratings. The ac source voltage of nominally 120 V is in the range of -10% and +5%. Then, Calculate the minimum transformer turns ratio if the DC load voltage is to be regulated at a constant value of 100 V. What is the value of a when VS = 120 V + 5%.

11- In the single-phase inverter of, SV = 120 V at 50 Hz, SL = 1.2 mH,

dL = 20 mH, dE = 88 V, and the delay angle α = 135°. Using PSIM, obtain sv , si , dv , and di waveforms in steady state.

12- In the inverter of Problem 12, vary the delay angle α from a value of 165° down to 120° and plot di versus α . Obtain the delay angle bα , below which di becomes continuous. How does the slope of the characteristic in this range depend on SL ?

13- In the three-phase fully controlled rectifier is connected to 460 V at 50 Hz and mHLs 1= . Calculate the commutation angle u if the load draws pure DC current at VVdc 515= and dcP = 500 kW.

14- In Problem 13 compute the peak inverse voltage and the average and the rms values of the current through each thyirstor in terms of

LLV and oI . 15- Consider the three-phase, half-controlled converter shown in the

following figure. Calculate the value of the delay angle α for which dmdc VV 5.0= . Draw dv waveform and identify the devices that conduct during various intervals. Obtain the DPF, PF, and %THD in the input line current and compare results with a full-bridge converter operating at dmdc VV 5.0= . Assume SL .

154 Chapter Three

16- Repeat Problem 15 by assuming that diode fD is not present in

the converter. 17- The three-phase converter of Fig.3.48 is supplying a DC load of 12

kW. A Y- Y connected isolation transformer has a per-phase rating of 5 kVA and an AC source-side voltage rating of 120 V at 50 Hz. It has a total per-phase leakage reactance of 8% based on its ratings. The ac source voltage of nominally 208 V (line to line) is in the range of -10% and +5%. Assume the load current is pure DC, calculate the minimum transformer turns ratio if the DC load voltage is to be regulated at a constant value of 300 V. What is the value of α when LLV = 208 V +5%.

18- In the three-phase inverter of Fig.3.63, LLV = 460 V at 60 Hz, E = 550 V, and SL = 0.5 mH. Assume the DC-side current is pure DC, Calculate α and γ if the power flow is 55 kW.

Chapter 4

SWITCH-MODE dc-ac INVERTERS: dc SINUSOIDAL ac

4.1 Introduction Switch-mode dc-to-ac inverters are used in ac motor drives and uninterruptible ac power supplies where the objective is to produce a sinusoidal ac output whose magnitude and frequency can both be controlled. As an example, consider an ac motor drive, shown in Fig.4.1 in a block diagram form. The dc voltage is obtained by rectifying and filtering the line voltage, most often by the diode rectifier circuits. In an ac motor load, the voltage at its terminals is desired to be sinusoidal and adjustable in its magnitude and frequency. This is accomplished by means of the switch-mode dc-to-ac inverter of Fig.4.1, which accepts a dc voltage as the input and produces the desired ac voltage input.

To be precise, the switch-mode inverter in Fig.4.1 is a converter through which the power flow is reversible. However, most of the time the power flow is from the dc side to the motor on the ac side, requiring an inverter mode of operation. Therefore, these switch-mode converters are often referred to as switch-mode inverters.

To slow down the ac motor in Fig.4.1, the kinetic energy associated with the inertia of the motor and its load is recovered and the ac motor acts as a generator. During the so-called braking of the motor, the power flows from the ac side to the dc side of the switch-mode converter and it operates in a rectifier mode. The energy recovered during the braking of the ac motor can be dissipated in a resistor, which can be switched in parallel with the dc bus capacitor for this purpose in Fig.4.1. However, in applications where this braking is performed frequently, a better alternative is regenerative braking where the energy recovered from the motor load inertia is fed back to the utility grid, as shown in the system of Fig.4.2. This requires that the converter connecting the drive to the utility grid be a two-quadrant converter with a reversible dc current, which can operate as a rectifier during the motoring mode of the ac motor and as an inverter during the braking of the motor. Such a reversible-current two-quadrant converter can be realized by two back-to-back connected line-frequency thyristor converters or by means of a switch-mode converter as shown in Fig.4.2. There are other reasons for using such a switch-mode

SWITCH-MODE dc-ac 153

rectifier (called a rectifier because, most of the time, the power flows from the ac line input to the dc bus) to interface the drive with the utility system.

Fig.4.1 Switch mode inverter in ac motor drive.

Fig.4.2 Switch-mode converters for motoring and regenerative braking in ac

motor drive.

In this chapter, we will discuss inverters with single-phase and three-phase ac outputs. The input to switch-mode inverters will be assumed to be a dc voltage source, as was assumed in the block diagrams of Fig.4.1 and Fig.4.2. Such inverters are referred to as voltage source inverters (VSIs). The other types of inverters, now used only for very high power ac motor drives, are the current source inverters (CSIs), where the dc input to the inverter is a dc current source. Because of their limited applications, the CSIs are not discussed.

The VSIs can be further divided into the following three general categories: 1. Pulse-width-modulated inverters. In these inverters, the input dc voltage is essentially constant in magnitude, such as in the circuit of Fig.4.1, where a diode rectifier is used to rectify the line voltage. Therefore, the inverter must control the magnitude and the frequency of the ac output voltages. This is achieved by PWM of the inverter switches and hence such inverters are called PWM inverters. There are various

154 Chapter Three

schemes to pulse-width modulate the inverter switches in order to shape the output ac voltages to be as close to a sine wave as possible. Out of these various PWM schemes, a scheme called the sinusoidal PWM will be discussed in detail, and some of the other PWM techniques will be described in a separate section at the end of this chapter. 2. Square-wave inverters. In these inverters, the input dc voltage is controlled in order to control the magnitude of the output ac voltage, and therefore the inverter has to control only the frequency of the output voltage. The output ac voltage has a waveform similar to a square wave, and hence these inverters are called square-wave inverters. 3. Single-phase inverters with voltage cancellation. In case of inverters with single-phase output, it is possible to control the magnitude and the frequency of the inverter output voltage, even though the input to the inverter is a constant dc voltage and the inverter switches are not pulse-width modulated (and hence the output voltage wave-shape is like a square wave). Therefore, these inverters combine the characteristics of the previous two inverters. It should be noted that the voltage cancellation technique works only with single-phase inverters and not with three-phase inverters. 4.2 BASIC CONCEPTS OF SWITCH-MODE INVERTERS In this section, we will consider the requirements on the switch-mode inverters. For simplicity, let us consider a single-phase inverter, which is shown in block diagram form in Fig.4.3a, where the output voltage of the inverter is filtered so that ov can be assumed to be sinusoidal. Since the inverter supplies an inductive load such as an ac motor, oi will lag ov , as shown in Fig.4.3b. The output waveforms of Fig.4.3b show that during interval 1, ov and oi are both positive, whereas during interval 3, ov and

oi are both negative. Therefore, during intervals 1 and 3, the instantaneous power flow ooo Ivp *= is from the dc side to the ac side, corresponding to an inverter mode of operation. In contrast, ov and oi are of opposite signs during intervals 2 and 4, and therefore op flows from the ac side to the dc side of the inverter, corresponding to a rectifier mode of operation. Therefore, the switch-mode inverter of Fig.4.3a must be capable of operating in all four quadrants of the oo vi − plane, as shown in Fig.4.3c during each cycle of the ac output. Such a four-quadrant


inverter is reversible and ov can be of either polarity independent of the direction of oi . Therefore, the full-bridge converter meets the switch-mode inverter requirements. Only one of the two legs of the full-bridge converter, for example leg A, is shown in Fig.4.4. All the dc-to-ac inverter topologies described in this chapter are derived from the one-leg converter of Fig.4.4. For ease of explanation, it will be assumed that in the inverter of Fig.4.4, the midpoint "o" of the dc input voltage is available, although in most inverters it is not needed and also not available.

Fig.4.3 Single-Phase switch-mode inverter.

Fig.4.4 One-leg switch-mode inverter.

156 Chapter Three

To understand the dc-to-ac inverter characteristics of the one-leg inverter of Fig.4.4, we will first assume that the input dc voltage dV is constant and that the inverter switches are pulse-width modulated to shape and control the output voltage. Later on, it will be shown that the square-wave switching is a special case of the PWM switching scheme.

4.2.1 PULSE-WIDTH-MODULATED SWITCHING SCHEME In inverter circuits, we would like the inverter output to be sinusoidal

with magnitude and frequency controllable. In order to produce a sinusoidal output voltage waveform at a desired frequency, a sinusoidal control signal at the desired frequency is compared with a triangular waveform, as shown in Fig.4.5a. The frequency of the triangular waveform establishes the inverter switching frequency and is generally kept constant along with its amplitude triV .

Before discussing the PWM behavior, it is necessary to define a few terms. The triangular waveform triV in Fig.4.5a is at a switching frequency sf which establishes the frequency with which the inverter switches are switched ( sf is also called the carrier frequency). The control signal controlv is used to modulate the switch duty ratio and has a frequency 1f , which is the desired fundamental frequency of the inverter voltage output ( 1f is also called the modulating frequency), recognizing that the inverter output voltage ' will not be a perfect sine wave and will contain voltage components at harmonic frequencies of 1f . The amplitude modulation ratio am is defined as

tri

controla V

Vm ˆ

ˆ= (4.1)

where controlV is the peak amplitude of the control signal. The amplitude

triV of the triangular signal is generally kept constant. The frequency modulation ratio fm is defined as:

1ffm s

f = (4.2)

In the inverter of Fig.4.4b, the switches +AT and −AT are controlled based on the comparison of controlv and triv and the following output voltage results, independent of the direction of oi :


tricontrol vv > onisTA+ , dAo Vv21

= (4.3)

tricontrol vv < onisTA− , dAo Vv21

−=

Fig.4.5 Pulse width modulation.

Since the two switches are never off simultaneously, the output voltage Aov fluctuates between two values ( dV

21 and dV

21

− ). Voltage Aov and

its fundamental frequency component (dashed curve) are shown in Fig.4.5b, which are drawn for 15=fm and am = 0.8.

158 Chapter Three

The harmonic spectrum of Aov under the conditions indicated in Figs.4.5a and Fig.4.5b is shown in Fig.4.5c, where the normalized

harmonic voltages ( ) dhAo VV21/ˆ having significant amplitudes are plotted.

This plot (for 0.1≤am ) shows three items of importance: 1. The peak amplitude of the fundamental-frequency component

( )1ˆAoV is am times 2/dV . This can be explained by first considering a

constant controlv , as shown in Fig.4.6a. This results in an output

waveform Aov . From the PWM in a full-bridge dc-dc converter, it can be noted that the average output voltage (or more specifically, the output voltage averaged over one switching time period ss fT /1= ) Aov depends on the ratio of controlv to triV for a given dV :

tricontrold

tri

controlAo VvV

VvV ˆ

2ˆ ≤= (4.4)

Let us assume (though this assumption is not necessary) that controlv varies very little during a switching time period, that is, fm is large, as shown in Fig.4.6b. Therefore, assuming controlv to be constant over a switching time period, Eq. (4.4) indicates how the "instantaneous average" value of Aov (averaged over one switching time period sT ) varies from one switching time period to the next. This "instantaneous average" is the same as the fundamental-frequency component of Aov . The foregoing argument shows why controlv is chosen to be sinusoidal to provide a sinusoidal output voltage with fewer harmonics. Let the control voltage vary sinusoidally at the frequency πω 2/11 =f , which is the desired (or the fundamental) frequency of the inverter output:

tVv controlcontrol 1sinˆ ω= where tricontrol VV ˆˆ ≤ (4.5)

Using Eqs. (4.4) and (4.5) and the foregoing arguments, which show that the fundamental-frequency component ( )1Aov varies sinusoidally and in phase with controlv as a function of time, results in

( ) 12

sin2

sinˆˆ

111 ≤== ad

ad

tri

controlAo mforVtmVt

VVv ωω (4.6)


Therefore, ( ) 12

ˆ1 ≤= a

daAo mforVmV (4.7)

which shows that in a sinusoidal PWM, the amplitude of the fundamental-frequency component of the output voltage varies linearly with am (provided 1≤am .0). Therefore, the range of am from 0 to 1 is referred to as the linear range.

Fig.4.6 Sinusoidal PWM.

2. The harmonics in the inverter output voltage waveform appear as sidebands, centered around the switching frequency and its multiples, that is, around harmonics fm , fm2 , fm3 , and so on. This general pattern holds true for all values of am in the range 0 to 1. For a frequency modulation ratio 9≤fm (which is always the case, except in very high power ratings), the harmonic amplitudes are almost independent of fm , though mf defines the frequencies at which they occur. Theoretically, the frequencies at which voltage harmonics occur can be indicated as: ( ) 1fkjmf fh ±= that is, the harmonic order h corresponds to the kth sideband of j times the frequency modulation ratio fm :

( ) kmjh f ±= (4.8) where the fundamental frequency corresponds to 1=h . For odd values of j, the harmonics exist only for even values of k. For even values of j, the harmonics exist only for odd values of k.

160 Chapter Three

In Table 8-1, the normalized harmonics ( ) dhAo VV21/ˆ are tabulated as a

function of the amplitude modulation ratio ma, assuming 9≥fm . Only those with significant amplitudes up to j = 4 in Eq.4.8 are shown.

It will be useful later on to recognize that in the inverter circuit of Fig.4.4

dAoAN Vvv21

+= (4.9)

Therefore, the harmonic voltage components in ANv and Aov are the same: ( ) ( )hAohAN VV ˆˆ = (4.10) Table 1 shows that Eq.7 is followed almost exactly and the amplitude of the fundamental component in the output voltage varies linearly with am .

3. The harmonic fm should be an odd integer. Choosing fm as an odd integer results in an odd symmetry as well as a half-wave symmetry with the time origin shown in Fig.4.5b, which is plotted for 15=fm . Therefore, only odd harmonics are present and the even harmonics disappear from the waveform of Aov . Moreover, only the coefficients of the sine series in the Fourier analysis are finite; those for the cosine series are zero. The harmonic spectrum is plotted in Fig.4.5c. Table 1 Generalized Harmonics of Aov , for a Large fm


Example 1 In the circuit of Fig.4.4, VVd 300= , 8.0=am , 39=fm , and the fundamental frequency is 47 Hz. Calculate the rms values of the fundamental-frequency voltage and some of the dominant harmonics in

Aov using Table 1. Solution: From Table 1, the rms voltage at any value of h is given as:

(4.11) Therefore, from Table 1 the rms voltages are as follows:

Now we discuss the selection of the switching frequency and the

frequency modulation ratio fm . Because of the relative ease in filtering harmonic voltages at high frequencies, it is desirable to use as high a switching frequency as possible, except for one significant drawback: Switching losses in the inverter switches increase proportionally with the switching frequency sf . Therefore, in most applications, the switching frequency is selected to be either less than 6 kHz or greater than 20 kHz to be above the audible range. If the optimum switching frequency (based on the overall system performance) turns out to be somewhere in the 6-20-kHz range, then the disadvantages of increasing it to 20 kHz are often outweighed by the advantage of no audible noise with sf of 20 kHz or greater. Therefore, in 50- or 60-Hz type applications, such as ac motor drives (where the fundamental frequency of the inverter output may be required to be as high as 200 Hz), the frequency modulation ratio fm may be 9 or even less for switching frequencies of less than 2 kHz. On the other hand, fm will be larger than 100 for switching frequencies higher than 20 kHz. The desirable relationships between the triangular waveform signal and the control voltage signal are dictated by how large

fm is. In the discussion here, fm = 21 is treated as the borderline

162 Chapter Three

between large and small, though its selection is somewhat arbitrary. Here, it is assumed that the amplitude modulation ratio am is less than 1.

4.2.1.1 Small ( )21≤ff mm 1. Synchronous PWM. For small values of fm , the triangular

waveform signal and the control signal should be synchronized to each other (synchronous PWM) as shown in Fig.4.5a. This synchronous PWM requires that fm be an integer. The reason for using the synchronous PWM is that the asynchronous PWM (where fm is not an integer) results in subharmonics (of the fundamental frequency) that are very undesirable in most applications. This implies that the triangular waveform frequency varies with the desired inverter frequency (e.g., if the inverter output frequency and hence the frequency of controlv , is 65.42 Hz and fm = 15, the triangular wave frequency should be exactly 15 x 65.42 = 981.3 Hz).

2. fm should be an odd integer. As discussed previously, fm should be an odd integer except in single-phase inverters with PWM unipolar voltage switching, to be discussed in the following sections.

4.2.1.2 Large ( )21>ff mm The amplitudes of subharmonics due to asynchronous PWM are small

at large values of fm . Therefore, at large values of fm , the asynchronous PWM can be used where the frequency of the triangular waveform is kept constant, whereas the frequency of controlv varies, resulting in noninteger values of fm (so long as they are large). However, if the inverter is supplying a load such as an ac motor, the subharmonics at zero or close to zero frequency, even though small in amplitude, will result in large currents that will be highly undesirable. Therefore, the asynchronous PWM should be avoided.

4.2.1.3 Overmodulation 0.1>am In the previous discussion, it was assumed that 0.1≤am ,

corresponding to a sinusoidal PWM in the linear range. Therefore, the amplitude of the fundamental-frequency voltage varies linearly with am , as derived in Eq.(4.7). In this range of 0.1≤am , PWM pushes the harmonics into a high-frequency range around the switching frequency


and its multiples. In spite of this desirable feature of a sinusoidal PWM in the linear range, one of the drawbacks is that the maximum available amplitude of the fundamental-frequency component is not as high as we wish. This is a natural consequence of the notches in the output voltage waveform of Fig.4.5b.

To increase further the amplitude of the fundamental-frequency component in the output voltage, ma is increased beyond 1.0, resulting in what is called overmodulation. Overmodulation causes the output voltage to contain many more harmonics in the sidebands as compared with the linear range (with 0.1≤am ), as shown in Fig.4.7. The harmonics with dominant amplitudes in the linear range may not be dominant during overmodulation. More significantly, with overmodulation, the amplitude of the fundamental-frequency component does not vary linearly with the amplitude modulation ratio am . Figure 4.8 shows the normalized peak

amplitude of the fundamental-frequency component ( ) dhAo VV21/ˆ as a

function of the amplitude modulation ratio am . Even at reasonably large

values of fm , ( ) dhAo VV21/ˆ depends on fm in the overmodulation

region. This is contrary to the linear range ( 1≤am .0) where ( ) dhAo VV21/ˆ

varies linearly with am , almost independent of mf (provided fm > 9). With overmodulation regardless of the value of fm , it is recommended that a synchronous PWM operation be used, thus meeting the requirements indicated previously for a small value of fm .

Fig.4.7 Harmonics due to overmodulation, drawn for 5.2=am and 15=fm .

164 Chapter Three

Fig.4.8 Voltage control by varying am .

The overmodulation region is avoided in uninterruptible power supplies because of a stringent requirement on minimizing the distortion in the output voltage. In induction motor drives, overmodulation is normally used.

For sufficiently large values of am , the inverter voltage waveform degenerates from a pulse-width-modulated waveform into a square wave, which is discussed in detail in the next section. From Fig.4.8 and the discussion of square-wave switching to be presented in the next section, it can be concluded that in the overmodulation region with 1>am

( )2

4ˆ2 1

dAo

d VVVπ

<< (4.12)

4.2.2 SQUARE-WAVE SWITCHING SCHEME In the square-wave switching scheme, each switch of the inverter leg

of Fig.4.4 is on for one half-cycle (180°) of the desired output frequency. This results in an output voltage waveform as shown in Fig.4.9a. From Fourier analysis, the peak values of the fundamental-frequency and


harmonic components in the inverter output waveform can be obtained for a given input dV as:

( ) ⎟⎠⎞

⎜⎝⎛==

2273.1

24ˆ

1dd

AoVVV

π (4.13)

and ( ) ( )h

VV Ao

hAo1

ˆˆ = (4.14)

where the harmonic order h takes on only odd values, as shown in Fig.4.9b. It should be noted that the square-wave switching is also a special case of the sinusoidal PWM switching when am becomes so large that the control voltage waveform intersects with the triangular waveform in Fig.4.5a only at the zero crossing of controlv . Therefore, the output voltage is independent of am in the square-wave region, as shown in Fig.4.8.

One of the advantages of the square-wave operation is that each inverter switch changes its state only twice per cycle, which is important at very high power levels where the solid-state switches generally have slower turn-on and turn-off speeds. One of the serious disadvantages of square-wave switching is that the inverter is not capable of regulating the output voltage magnitude. Therefore, the dc input voltage dV to the inverter must be adjusted in order to control the magnitude of the inverter output voltage.

Fig.4.9 Square wave switching.

4.3 SINGLE PHASE IVERTERS 4.3.1 HALF-BRIDGE INVERTERS (SINGLE PHASE)

Figure 4.10 shows the half-bridge inverter. Here, two equal capacitors are connected in series across the dc input and their junction is at a

166 Chapter Three

midpotential, with a voltage dV21 , across each capacitor. Sufficiently

large capacitances should be used such that it is reasonable to assume that the potential at point o remains essentially constant with respect to the negative dc bus N. Therefore, this circuit configuration is identical to the basic one-leg inverter discussed in detail earlier, and Aoo vv = .

Assuming PWM switching, we find that the output voltage waveform will be exactly as in Fig.4.5b. It should be noted that regardless of the switch states, the current between the two capacitors C+ and C- (which have equal and very large values) divides equally. When T+ is on, either T+ or D+ conducts depending on the direction of the output current, and io splits equally between the two capacitors. Similarly, when the switch −T is in its on state, either −T or −D conducts depending on the direction

of oi , and oi splits equally between the two capacitors. Therefore, the capacitors C+ and C_ are "effectively" connected in parallel in the path of oi . This also explains why the junction o in Fig.4.10 stays at

midpotential. Since oi must flow through the parallel combination of C+ and C_, oi

in steady state cannot have a dc component. Therefore, these capacitors act as dc blocking capacitors, thus eliminating the problem of transformer saturation from the primary side, if a transformer is used at the output to provide electrical isolation. Since the current in the primary winding of such a transformer would not be forced to zero with each switching, the transformer leakage inductance energy does not present a problem to the switches.

In a half-bridge inverter, the peak voltage and current ratings of the switches are as follows:

dT VV = (4.15) and peakoT iI ,= (4.16) 4.3.2 FULL-BRIDGE INVERTERS (SINGLE PHASE)

A full-bridge inverter is shown in Fig.4.11. This inverter consists of two one-leg inverters of the type discussed in Section 4-2 and is preferred over other arrangements in higher power ratings. With the same dc input voltage, the maximum output voltage of the full-bridge inverter is twice that of the half-bridge inverter. This implies that for the same power, the output current and the switch currents are one-half of those for a


half-bridge inverter. At high power levels, this is a distinct advantage, since it requires less paralleling of devices.

Fig.4.10 Half-bridge inverter.

Fig.4.11 Single-phase full-bridge inverter.

4.3.2.1 PWM with Bipolar Voltage Switching

The diagonally opposite switches (TA+, TB-) and (TA-, TB+) from the two legs in Fig.4.11 are switched as switch pairs 1 and 2, respectively. With this type of PWM switching, the output voltage waveform of leg A is identical to the output of the basic one-leg inverter, which is determined in the same manner by comparison of controlv and triv in Fig.4.12a. The output of inverter leg B is negative of the leg A output; for example,

when TA+ is on and Aov is equal to dV21

+ is also on and dBo Vv21

−= .

Therefore: ( ) ( )tvtv AoBo −= (4.17)

and ( ) ( ) ( ) ( )tvtvtvtv AoBoAoo 2=−= (4.18) The ov waveform is shown in Fig.4.12b. The analysis carried out in Section 4.2 for the basic one-leg inverter completely applies to this type of PWM switching. Therefore, the peak of the fundamental-frequency

168 Chapter Three

component in the output voltage ( )1oV can be obtained from Eqs. (4.7), (4.12), and (4.18) as:

( )0.1ˆ1 ≤= adao mVmV (4.19)

and ( )0.14ˆ1 ><< adod mVVV

π (4.20).

Fig4.12 PWM with bipolar voltage switching.

In Fig.4.12b, we observe that the output voltage ov switches between

dV− and dV+ voltage levels. That is the reason why this type of switching is called a PWM with bipolar voltage switching. The amplitudes of harmonics in the output voltage can be obtained by using Table 1, as illustrated by the following example. Example 2 In the full-bridge converter circuit of Fig.4.11, VVd 300= ,

am =0.8, 39=fm , and the fundamental frequency is 47 Hz. Calculate the rms values of the fundamental-frequency voltage and some of the dominant harmonics in the output voltage ov if a PWM bipolar voltage-switching scheme is used. Solution: From Eq.(4.18), the harmonics in ov can be obtained by multiplying the harmonics in Table 1 and Example 1 by a factor of 2. Therefore from Eq. (4.11), the rms voltage at any harmonic h is given as

( ) ( ) ( ) ( )2/

ˆ13.212

2/

ˆ*

22/

ˆ*

2*2*

21

d

hAo

d

hAod

d

hAodho V

VVVV

VVVV === (4.21)


Therefore, the rms voltages are as follows: Fundamental: 1oV = 212.13 x 0. 8 = 169.7 V at 47 Hz ( )37oV = 212.13 x 0.22 = 46.67 V at 1739 Hz ( )39oV = 212.13 x 0.818 = 173.52 V at 1833 Hz - ( )41oV = 212.13 x 0.22 = 46.67 V at 1927 Hz ( )77oV = 212.13 x 0.314 = 66.60 V at 3619 Hz ( )79oV = 212.13 x 0.314 = 66.60 V at 3713 Hz etc. dc-Side Current di It is informative to look at the dc-side current di in the PWM biopolar voltage-switching scheme.

For simplicity, fictitious L-C high-frequency filters will be used at the dc side as well as at the ac side, as shown in Fig.4.13. The switching frequency is assumed to be very high, approaching infinity. Therefore, to filter out the high-switching-frequency components in ov and di , the filter components L and C required in both ac and dc-side filters approach zero. This implies that the energy stored in the filters is negligible. Since the converter itself has no energy storage elements, the instantaneous power input must equal the instantaneous power output.

Fig.4.13 Inverter with "fictitious" filters.

Having made these assumptions, ov in Fig.4.13 is a pure sine wave at the fundamental output frequency 1ω ,

tVvv ooo 11 sin2 ω== (4.22) If the load is as shown in Fig.4.13, where oe is a sine wave at frequency

1ω , then the output current would also be sinusoidal and would lag ov for an inductive load such as an ac motor:

( )φω −= tIi oo 1sin2 (4.23) where φ is the angle by which oi lags ov .

170 Chapter Three

On the dc side, the L-C filter will filter the high-switching-frequency components in di and *

di would only consist of the low-frequency and dc components. Assuming that no energy is stored in the filters,

( ) ( ) ( ) ( )φωω −== tItVtitvtiV oooodd 11* sin2sin2 (4.24)

Therefore ( ) ( ) 21* 2coscos dd

d

oo

d

ood iIt

VIV

VIVti +=−−= φωφ (4.25)

( ) ( )φω −−= tIIti ddd 12* 2cos2 (4.26)

where φcosd

ood V

IVI = (4.27)

and d

ood V

IVI2

12 = (4.28)

Equation (4.26) for *di shows that it consists of a dc component dI ,

which is responsible for the power transfer from dV on the dc side of the

inverter to the ac side. Also, *di contains a sinusoidal component at twice

the fundamental frequency. The inverter input current di consists of *di

and the high-frequency components due to inverter switchings, as shown in Fig.4.14.

Fig.4.14 The dc-side current in a single-phase inverter with PWM bipolar

voltage switching. In practical systems, the previous assumption of a constant dc voltage

as the input to the inverter is not entirely valid. Normally, this dc voltage is obtained by rectifying the ac utility line voltage. A large capacitor is used across the rectifier output terminals to filter the dc voltage. The


ripple in the capacitor voltage, which is also the dc input voltage to the inverter, is due to two reasons: (1) The rectification of the line voltage to produce dc does not result in a pure dc, dealing with the line-frequency rectifiers. (2) As shown earlier by Eq.(4.26), the current drawn by a single-phase inverter from the dc side is not a constant dc but has a second harmonic component (of the fundamental frequency at the inverter output) in addition to the high switching-frequency components. The second harmonic current component results in a ripple in the capacitor voltage, although the voltage ripple due to the high switching frequencies is essentially negligible.

4.3.2.2 PWM with Unipolar Voltage Switching In PWM with unipolar voltage switching, the switches in the two legs of the full-bridge inverter of Fig.4.11 are not switched simultaneously, as in the previous PWM scheme. Here, the legs A and B of the full-bridge inverter are controlled separately by comparing triv with controlv and

controlv− , respectively. As shown in Fig.4.15a, the comparison of controlv with the triangular waveform results in the following logic signals to control the switches in leg A:

dANAtricontrol VVandonTvv => + (4.29) 0=< − ANAtricontrol VandonTvv

The output voltage of inverter leg A with respect to the negative dc bus N is shown in Fig.4.15b. For controlling the leg B switches, controlv− is compared with the same triangular waveform, which yields the following:

dBNBtricontrol VVandonTvv =>− + (4.30) 0=<− − BNBtricontrol VandonTvv

Because of the feedback diodes in antiparallel with the switches, the foregoing voltages given by Eqs.(4.29) and (4.30) are independent of the direction of the output current oi .

The waveforms of Fig.4.15 show that there are four combinations of switch on-states and the corresponding voltage levels:

(4.31)

172 Chapter Three

Fig.4.15 PWM with unipolar voltage switching (single phase).

We notice that when both the upper switches are on, the output voltage is zero. The output current circulates in a loop through +AT and

+BD or +AD and +BT depending on the direction of oi . During this interval, the input current di is zero. A similar condition occurs when both bottom switches −AT and −BT are on.

In this type of PWM scheme, when a switching occurs, the output voltage changes between zero and dV+ or between zero and dV− voltage levels. For this reason, this type of PWM scheme is called PWM with a unipolar voltage switching, as opposed to the PWM with bipolar (between dV+ and dV− ) voltage-switching scheme described earlier.


This scheme has the advantage of "effectively" doubling the switching frequency as far as the output harmonics are concerned, compared to the bipolar voltage switching scheme. Also, the voltage jumps in the output voltage at each switching are reduced to dV as compared to dV2 in the previous scheme.

The advantage of "effectively" doubling the switching frequency appears in the harmonic spectrum of the output voltage waveform, where the lowest harmonics (in the idealized circuit) appear as sidebands of twice the switching frequency. It is easy to understand this if we choose the frequency modulation ratio fm to be even ( fm should be odd for PWM with bipolar voltage switching) in a single-phase inverter. The voltage waveforms ANv and BNv are displaced by 180° of the fundamental frequency 1f with respect to each other. Therefore, the harmonic components at the switching frequency in ANv and BNv have

the same phase ( 0.180 ==− fo

BNAN mφφ , since the waveforms are 180° displaced and fm is assumed to be even). This results in the cancellation of the harmonic component at the switching frequency in the output voltage BNANo vvv −= . In addition, the sidebands of the switching-frequency harmonics disappear. In a similar manner, the other dominant harmonic at twice the switching frequency cancels out, while its sidebands dc not. Here also

( )0.1ˆ1 ≤= adao mVmV (4.32)

and ( )0.14ˆ 1 ><< adod mVVVπ

(4.33)

Example 3 In Example 2, suppose that a PWM with unipolar voltage switching scheme is used, with 38=fm . Calculate the rms values of the fundamental frequency voltage and some of the dominant harmonics in the output voltage. Solution: Based on the discussion of unipolar voltage switching, the harmonic order h can be written as

( ) kmjh f ±= 2 (4.34) where the harmonics exist as sidebands around fm2 and the multiples of

fm2 . Since h is odd, k in Eq.(34) attains only odd values. From Example 2,

174 Chapter Three

( ) ( )2/

13.212d

hAoho V

VV = (4.35)

Using Eq.(35) and Table 1, we find that the rms voltages are as follows: At fundamental or 47 Hz: 1oV = 0.8 x 212.13 = 169.7 V At h = fm2 - 1 = 75 or 3525 Hz: ( )75oV = 0.314 x 212.13 = 66.60 V At h = fm2 + 1 = 77 or 3619 Hz: ( )77oV = 0.314 x 212.13 = 66.60 V etc. Comparison of the unipolar voltage switching with the bipolar voltage switching of Example 2 shows that, in both cases, the fundamental-frequency voltages are the same for equal am However, with unipolar voltage switching, the dominant harmonic voltages centered around fm disappear, thus resulting in a significantly lower harmonic content.

dc-Side Current di . Under conditions similar to those in the circuit of Fig.4.13 for the PWM with bipolar voltage switching, Fig.4.16 shows the dc-side current di for the PWM unipolar voltage-switching scheme, where 14=fm (instead of 15=fm for the bipolar voltage switching).

By comparing Figs.4.14 and 4.16, it is clear that using PWM with unipolar voltage switching results in a smaller ripple in the current on the dc side of the inverter.

Fig.4.16 The dc-side current in a single-phase inverter with PWM unipolar

voltage switching.

4.3.2.3 Square-Wave Operation The full-bridge inverter can also be operated in a square-wave mode.

Both types of PWM discussed earlier degenerate into the same square-wave mode of operation, where the switches ( +AT , −BT ) and ( +BT , −AT ) are operated as two pairs with a duty ratio of 0.5.


As is the case in the square-wave mode of operation, the output voltage magnitude given below is regulated by controlling the input dc voltage:

do VVπ4ˆ 1 = (4.36)

4.3.2.4 Output Control by Voltage Cancellation This type of control is feasible only in a single-phase, full-bridge

inverter circuit. It is based on the combination of square-wave switching and PWM with a unipolar voltage switching. In the circuit of Fig.4.17a, the switches in the two inverter legs are controlled separately (similar to PWM unipolar voltage switching). But all switches have a duty ratio of 0.5, similar to a square-wave control. This results in waveforms for ANv and BNv shown in Fig.4.17b, where the waveform overlap angle a can be controlled. During this overlap interval, the output voltage is zero as a consequence of either both top switches or both bottom switches being on. With 0=α , the output waveform is similar to a square-wave inverter with the maximum possible fundamental output magnitude.

Fig.17 Full-bridge- single-phase inverter control by voltage cancellation: (a)

power circuit: (b) waveforms; (c) normalized fundamental and harmonic voltage output and total harmonic distortion as a function of α .

176 Chapter Three

It is easier to derive the fundamental and the harmonic frequency

components of the output voltage in terms of αβ2190 −= o , as is shown

in Fig.4.17b:

(4.37)

where αβ2190 −= o and h is an odd integer.

Fig.4.17c shows the variation in the fundamental-frequency component as well as the harmonic voltages as a function of α . These are normalized with respect to the fundamental-frequency component for the square-wave (α = 0) operation. The total harmonic distortion, which is the ratio of the rms value of the harmonic distortion to the rms value of the fundamental-frequency component, is also plotted as a function of α . Because of a large distortion, the curves are shown as dashed for large values of α . 4.3.2.5 Switch Utilization in Full-Bridge Inverters

Similar to a half-bridge inverter, if a transformer is utilized at the output of a full-bridge inverter, the transformer leakage inductance does not present a problem to the switches.

Independent of the type of control and the switching scheme used, the peak switch voltage and current ratings required in a full-bridge inverter are as follows:

dT VV = (4.38) and peakoT iI ,= (4.39) 4.3.2.6 Ripple in the Single-Phase Inverter Output

The ripple in a repetitive waveform refers to the difference between the instantaneous values of the waveform and its fundamental-frequency component.


Fig.4.18a shows a single-phase switch-mode inverter. It is assumed to be supplying an induction motor load, which is shown by means of a simplified equivalent circuit with a counter electromotive force (emf) oe . Since ( )teo is sinusoidal, only the sinusoidal (fundamental-frequency) components of the inverter output voltage and current are responsible for the real power transfer to the load.

We can separate the fundamental-frequency and the ripple components in ov and oi by applying the principle of superposition to the linear circuit of Fig.4.18a. Let rippleoo vvv += 1 and rippleoo iii += 1 . Figs.4.18b, c show the circuits at the fundamental frequency and at the ripple frequency, respectively, where the ripple frequency component consists of sub-components at various harmonic frequencies.

Therefore, in a phasor form (with the fundamental frequency components designated by subscript 1) as shown in Fig.4.18d,

1111 ooLoo LIjEVEV ω+=+= (4.40)

Fig.4.18 Single-phase inverter: (a) circuit; (6) fundamental- frequency

components; (c) ripple frequency components: (d) fundamental-frequency phasor diagram.

Since the superposition principle is valid here, all the ripple in v, appears across L, where ( ) 1ooripple vvtv −= (4.41) The output current ripple can be calculated as

( ) ( )∫ +=t

rippleripple kdvL

ti0

1 ζζ (4.42)

178 Chapter Three

where k is a constant and ζ is a variable of integration. With a properly selected time origin t = 0, the constant k in Eq.(4.42)

will be zero. Therefore, Eqs.(4.41) and (4.42) show that the current ripple is independent of the power being transferred to the load.

As an example, Fig.4.19a shows the ripple current for a square-wave inverter output. Fig.4.19b shows the ripple current in a PWM bipolar voltage switching. In drawing Figs.4.19a and 4.19b, the fundamental-frequency components in the inverter output voltages are kept equal in magnitude (this requires a higher value of dV in the PWM inverter). The PWM inverter results in a substantially smaller peak ripple current compared to the square-wave inverter. This shows the advantage of pushing the harmonics in the inverter output voltage to as high frequencies as feasible, thereby reducing the losses in the load by reducing the output current harmonics. This is achieved by using higher inverter switching frequencies, which would result in more frequent switching and hence higher switching losses in the inverter. Therefore, from the viewpoint of the overall system energy efficiency, a compromise must be made in selecting the inverter switching frequency.

Fig. 4.19 Ripple in the inverter output: (a) square-wave switching; (6) PWM

bipolar voltage switching.


4.3.3 PUSH-PULL INVERTERS Fig.4.20 shows a push-pull inverter circuit. It requires a transformer

with a center tapped primary. We will initially assume that the output current oi flows continuously. With this assumption, when the switch 1T is on (and 2T is off), 1T would conduct for a positive value of oi , and 1D would conduct for a negative value of oi . Therefore, regardless of the direction of oi , nVv do /= , where n is the transformer turns ratio between the primary half and the secondary windings, as shown in Fig.4.20. Similarly, when 2T is on (and 1T is off), nVv do /−= . A push-pull inverter can be operated in a PWM or a square-wave mode and the waveforms are identical to those in Figs.4.5 and 4.12 for half-bridge and full-bridge inverters. The output voltage in Fig.4.20 equals:

( )0.1ˆ1 ≤= a

dao m

nVmV (4.43)

and ( )0.14ˆ 1 ><< ad

od m

nVV

nV

π (4.44)

In a push-pull inverter, the peak switch voltage and current ratings are niIVV peakoTdT /2 ,== (4.45)

Fig.4.20 Push-Pull inverter (single phase).

The main advantage of the push-pull circuit is that no more than one switch in series conducts at any instant of time. This can be important if the dc input to the converter is from a low-voltage source, such as a battery, where the voltage drops across more than one switch in series would result in a significant reduction in energy efficiency. Also, the control drives for the two switches have a common ground. It is, however, difficult to avoid the dc saturation of the transformer in a push-pull inverter.

180 Chapter Three

The output current, which is the secondary current of the transformer, is a slowly varying current at the fundamental output frequency. It can be assumed to be a constant during a switching interval. When a switching occurs, the current shifts from one half to the other half of the primary winding. This requires very good magnetic coupling between these two half-windings in order to reduce the energy associated with the leakage inductance of the two primary windings. This energy will be dissipated in the switches or in snubber circuits used to protect the switches. This is a general phenomenon associated with all converters (or inverters) with isolation where the current in one of the windings is forced to go to zero with every switching. This phenomenon is very important in the design of such converters.

In a pulse-width-modulated push-pull inverter for producing sinusoidal output (unlike those used in switch-mode dc power supplies), the transformer must be designed for the fundamental output frequency. The number of turns will therefore be high compared to a transformer designed to operate at the switching frequency in a switch-mode dc power supply. This will result in a high transformer leakage inductance, which is proportional to the square of the number of turns, provided all other dimensions are kept constant. This makes it difficult to operate a sine-wave-modulated PWM push-pull inverter at switching frequencies higher than approximately 1 kHz.

4.3.4 SWITCH UTILIZATION IN SINGLE-PHASE INVERTERS Since the intent in this section is to compare the utilization of switches

in various single-phase inverters, the circuit conditions are idealized. We will assume that max,dV is the highest value of the input voltage, which establishes the switch voltage ratings. In the PWM mode, the input remains constant at max,dV In the square-wave mode, the input voltage is decreased below max,dV to decrease the output voltage from its maximum value. Regardless of the PWM or the square-wave mode of operation, we assume that there is enough inductance associated with the output load to yield a purely sinusoidal current (an idealized condition indeed for a square-wave output) with an rms value of max,oI at the maximum load.

If the output current is assumed to be purely sinusoidal, the inverter rms volt-ampere output at the fundamental frequency equals max,1 oo IV at the maximum rated output, where the subscript 1 designates the


fundamental-frequency component of the inverter output. With TV and

TI as the peak voltage and current ratings of a switch, the combined utilization of all the switches in an inverter can be defined as

Switch utilization ratio = TT

oo

IqVIV max,1 (4.46)

where q is the number of switches in an inverter. To compare the utilization of switches in various single-phase

inverters, we will initially compare them for a square-wave mode of operation at the maximum rated output. (The maximum switch utilization occurs at max,dd VV = ).

In practice, the switch utilization ratio would be much smaller than

0.16 for the following reasons: (1) switch ratings are chosen conservatively to provide safety margins; (2) in determining the switch current rating in a PWM inverter, one would have to take into account the variations in the input dc voltage available; and (3) the ripple in the output current would influence the switch current rating. Moreover, the

182 Chapter Three

inverter may be required to supply a short-term overload. Thus, the switch utilization ratio, in practice, would be substantially less than the 0.16 calculated.

At the lower output volt-amperes compared to the maximum rated output, the switch utilization decreases linearly. It should be noted that using a PWM switching with mp !~ 1.0, this ratio would be smaller by a factor of ( ) am4/π as compared to the square-wave switching:

Maximum switch utilization ratio = 0.1,81

421

≤= aaa mmmππ

) (4.54)

Therefore, the theoretical maximum switch utilization ratio in a PWM switching is only 0.125 at 1=am , as compared with 0.16 in square –wave inverter.

Example 4 In a single-phase full-bridge PWM inverter, dV varies in a range of 295-325 V. The output voltage is required to be constant at 200 V (rms), and the maximum load current (assumed to be sinusoidal) is 10 A (rms). Calculate the combined switch utilization ratio (under these idealized conditions, not accounting for any overcurrent capabilities). Solution: In this inverter

VVV madT 325, ==

14.1410*22 === oT II 4. == switchesofnoq

The maximum output volt-ampere (fundamental frequency) is VAIV oo 200010*200max,1 == (4.55)

Therefore, from Eq.(4.46)

Switch utilization ratio = 11.014.14*325*4

2000max,1 ==TT

oo

IqVIV

4.4 THREE-PHASE INVERTERS In applications such as uninterruptible ac power supplies and ac motor

drives, three-phase inverters are commonly used to supply three-phase loads. It is possible to supply a three-phase load by means of three separate single-phase inverters, where each inverter produces an output displaced by 120° (of the fundamental frequency) with respect to each other. Though this arrangement may be preferable under certain conditions, it requires either a three-phase output transformer or separate access to each of the three phases of the load. In practice, such access is generally not available. Moreover, it requires 12 switches.


The most frequently used three-phase inverter circuit consists of three legs, one for each phase, as shown in Fig.4.21. Each inverter leg is similar to the one used for describing the basic one-leg inverter in Section 4.2. Therefore, the output of each leg, for example ANv , (with respect to the negative dc bus), depends only on dV and the switch status; the output voltage is independent of the output load current since one of the two switches in a leg is always on at any instant. Here, we again ignore the blanking time required in practical circuits by assuming the switches to be ideal. Therefore, the inverter output voltage is independent of the direction of the load current.

Fig.4.21 Three-phase inverter.

4.4.1 PWM IN THREE-PHASE VOLTAGE SOURCE INVERTERS Similar to the single-phase inverters, the objective in

pulse-width-modulated three-phase inverters is to shape and control the three-phase output voltages in magnitude and frequency with an essentially constant input voltage dV . To obtain balanced three-phase output voltages in a three-phase PWM inverter, the same triangular voltage waveform is compared with three sinusoidal control voltages that are 120° out of phase, as shown in Fig.4.22a (which is drawn for

15=fm ). It should also be noted from Fig.4.22b that an identical amount of

average dc component is present in the output voltages ANv and BNv , which are measured with respect to the negative dc bus. These dc components are canceled out in the line-to-line voltages, for example in

ABv shown in Fig.4.22b. This is similar to what happens in a single-phase full-bridge inverter utilizing a PWM switching.

In the three-phase inverters, only the harmonics in the line-to-line voltages are of concern. The harmonics in the output of any one of the

184 Chapter Three

legs, for example ABv in Fig.4.22b, are identical to the harmonics in Aov in Fig.4.5, where only the odd harmonics exist as sidebands, centered around fm and its multiples, provided mf is odd. Only considering the harmonic at fm (the same applies to its odd multiples), the phase difference between the mf harmonic in ANv and BNv is (120 fm )°. This phase difference will be equivalent to zero (a multiple of 360°) if fm is odd and a multiple of 3. As a consequence, the harmonic at fm is suppressed in the line-to-line voltage ABv . The same argument applies in the suppression of harmonics at the odd multiples of fm if fm is chosen to be an odd multiple of 3 (where the reason for choosing fm to be an odd multiple of 3 is to keep fm odd and, hence, eliminate even harmonics). Thus, some of the dominant harmonics in the one-leg inverter can be eliminated from the line-to-line voltage of a three-phase inverter. PWM considerations are summarized as follows: 1. For low values of fm , to eliminate the even harmonics, a synchronized PWM should be used and mf should be an odd integer. Moreover, mf should be a multiple of 3 to cancel out the most dominant harmonics in the line-to-line voltage. 2. For large values of fm , the comments in Section 4.2.1.2 for a single-phase PWM apply. 3. During overmodulation ( am > 1.0), regardless of the value of fm , the conditions pertinent to a small fm should be observed. 4.4.1.1 Linear Modulation ( 0.1≤am ) In the linear region ( 0.1≤am ), the fundamental-frequency component in the output voltage varies linearly with the amplitude modulation ratio am . From Figs.4.5b and 4.22b, the peak value of the fundamental-frequency component in one of the inverter legs is

( )2

ˆ1

daAN

VmV = (4.56)

Therefore, the line-to-line rms voltage at the fundamental frequency, due to 120° phase displacement between phase voltages, can be written as


(4.57)

Fig.4.22 Three-phase PWM waveforms and harmonic spectrum.

186 Chapter Three

The harmonic components of the line-to-line output voltages can be calculated in a similar manner from Table 1, recognizing that some of the harmonics are canceled out in the line-to-line voltages. These rms harmonic voltages are listed in Table 2.

4.4.1.2 Overmodulation ( 0.1>am ) In PWM overmodulation, the peak of the control voltages are allowed

to exceed the peak of the triangular waveform. Unlike the linear region, in this mode of operation the fundamental-frequency voltage magnitude does not increase proportionally with ma. This is shown in Fig.4.23, where the rms value of the fundamental-frequency line-to-line voltage

1LLV is plotted as a function of am . Similar to a single-phase PWM, for sufficiently large values of ma, the PWM degenerates into a square-wave inverter waveform. This results in the maximum value of 1LLV equal to 0.78 dV as explained in the next section.

In the overmodulation region compared to the region with 0.1≤am , more sideband harmonics appear centered around the frequencies of harmonics mf and its multiples. However, the dominant harmonics may not have as large an amplitude as with 0.1≤am . Therefore, the power loss in the load due to the harmonic frequencies may not be as high in the overmodulation region as the presence of additional sideband harmonics would suggest. Depending on the nature of the load and on the switching frequency, the losses due to these harmonics in overmodulation may be even less than those in the linear region of the PWM.

Table 2 Generalized Harmonics of LLv for a large and odd fm that is a multiple of 3.


Fig.4.23 Three-phase inverter ( ) drmsLL VV /1 as a function of am .

4.4.2 SQUARE-WAVE OPERATION IN THREE-PHASE INVERTERS If the input dc voltage dV is controllable, the inverter in Fig.4.24a can be operated in a square-wave mode. Also for sufficiently large values of am , PWM degenerates into square-wave operation and the voltage waveforms are shown in Fig.4.24b. Here, each switch is on for 180° (i.e., its duty ratio is 50%). Therefore, at any instant of time, three switches are on.

Fig.4.24 Square-wave inverter (three phase).

188 Chapter Three

In the square-wave mode of operation, the inverter itself cannot control the magnitude of the output ac voltages. Therefore, the dc input voltage must be controlled in order to control the output in magnitude. Here, the fundamental-frequency line-to-line rms voltage component in the output can be obtained from Eq. (13) for the basic one-leg inverter operating in a square-wave mode:

(4.58) The line-to-line output voltage waveform does not depend on the load

and contains harmonics (6n ± 1; n = 1, 2, . . .), whose amplitudes decrease inversely proportional to their harmonic order, as shown in Fig.4.24c:

( ) dhLL Vh

V 78.0= (4.59)

where ( ),.......3,2,11 =±= nnh It should be noted that it is not possible to control the output

magnitude in a three-phase, square-wave inverter by means of voltage cancellation as described in Section 4.3.2.4.

4.4.3 SWITCH UTILIZATION IN THREE-PHASE INVERTERS We will assume that max,dV is the maximum input voltage that remains constant during PWM and is decreased below this level to control the output voltage magnitude in a square-wave mode. We will also assume that there is sufficient inductance associated with the load to yield a pure sinusoidal output current with an rms value of max.oI (both in the PWM and the square-wave mode) at maximum loading. Therefore, each switch would have the following peak ratings:

max,dT VV = (4.60)

and max,2 oT II = (4.61) If 1LLV is the rms value of the fundamental-frequency line-to-line

voltage component, the three-phase output volt-amperes (rms) at the fundamental frequency at the rated output is

( ) max,3 13 OLLphase IVVA =− (4.62) Therefore, the total switch utilization ratio of all six switches combined is


(4.63) In the PWM linear region ( )0.1≤am using Eq.(4.57) and noting that the maximum switch utilization occurs at max,dd VV =

(4.64) In the square-wave mode, this ratio is 16.02/1 ≅π compared to a

maximum of 0.125 for a PWM linear region with 0.1=am . In practice, the same derating in the switch utilization ratio applies as

discussed in Section 4.3.4 for single-phase inverters. Comparing Eqs.(4.54) and (4.64), we observe that the maximum

switch utilization ratio is the same in a three-phase, three-leg inverter as in a single-phase inverter. In other words, using the switches with identical ratings, a three-phase inverter with 50% increase in the number of switches results in a 50% increase in the output volt-ampere, compared to a single-phase inverter.

4.4.4 RIPPLE IN THE INVERTER OUTPUT Figure 4.25a shows a three-phase, three-leg, voltage source, switch-

mode inverter in a block diagram form. It is assumed to be supplying a three-phase ac motor load. Each phase of the load is shown by means of its simplified equivalent circuit with respect to the load neutral n. The induced back ( ) ( )tete BA , , and ( )tec are assumed to be sinusoidal.

Fig.4.25 Three-phase inverter: (a) circuit diagram; (b) phasor diagram

(fundamental frequency).

190 Chapter Three

Under balanced operating conditions, it is possible to express the inverter phase output voltages ANv , and so on (with respect to the load neutral n), in terms of the inverter output voltages with respect to the negative dc bus N:

( )CBAkvvv nNkNkn ,,=−= (4.65) Each phase voltage can be written as

( )CBAkedtdiLv kn

kkn ,,=+= (4.66)

In a three-phase, three-wire load 0=++ CBA iii (4.67a)

and ( ) 0=++ CBA iiidtd (4.67b)

Similarly, under balanced operating conditions, the three back-emfs are a balanced three-phase set of voltages, and therefore

0=++ CBA eee (4.68) From the foregoing equations, the following condition for the inverter voltages can be written:

0=++ CnBnAn vvv (4.69) Using Eqs. (4.65) through (4.69),

( )CNBNANnN vvvv ++=31 (4.70)

Substituting nNv from Eq.(4.70) into Eq.(4.65), we can write the phase-to-neutral voltage for phase A as

( )CNBNANAn vvvv +−=31

32 (4.71)

Similar equations can be written for phase B and C voltages. Similar to the discussion in Section 4.3.2.6 for the ripple in the

single-phase inverter output, only the fundamental-frequency components of the phase voltage 1AnV and the output current 1Ai are responsible for the real power transformer since the back-emf ( )teA is assumed to be sinusoidal and the load resistance is neglected. Therefore, in a phasor form as shown in Fig.4.25b

111 AAAn ILjEV ω+= (4.72) By using the principle of superposition, all the ripple in Anv appears

across the load inductance L. Using Eq.(4.71), the waveform for the


phase-to-load-neutral voltage AnV is shown in Figs.4.26a and 4.26b for square-wave and PWM operations, respectively. Both inverters have identical magnitudes of the fundamental-frequency voltage component

1AnV , which requires a higher dV in the PWM operation. The voltage ripple ( )1AnAnripple vvv −= is the ripple in the phase-to-neutral voltage. Assuming identical loads in these two cases, the output current ripple is obtained by using Eq.(4.42) and plotted in Fig.4.26. This current ripple is independent of the power being transferred, that is, the current ripple would be the same so long as for a given load inductance L, the ripple in the inverter output voltage remains constant in magnitude and frequency. This comparison indicates that for large values of fm , the current ripple in the PWM inverter will be significantly lower compared to a square-wave inverter.

Fig.4.26 Phase-to-load-neutral variables of a three-phase inverter: (a) square

wave: (b) PWM.

4. 5 RECTIFIER MODE OF OPERATION

192 Chapter Three

Fig.4.37 Operation modes: (a) circuit; (b) inverter mode; (c) rectifier mode:

(d) constant AI .

AnV can be varied. For a balanced operation, the control voltages for phases B and C are equal in magnitude, but ± 120° displaced with respect to the control voltage of phase A.


4.6 PROGRAMMABLE PWM CONVERTERS This technique combines the square wave switching and PWM to

control the fundamental output voltage as well as to eliminate the designated harmonics from the output voltage. This technique provides the facility to adjust the output voltage and simultaneous optimization of an objective function. The types of objective functions that technique can optimize are:-

1- Selective harmonic elimination. 2- Minimum THD. 3- Reduced acoustic noise. 4- Minimum losses. 5- Minimum torque pulsations.

The main advantages of programmable PWM technique are: 1- High quality output voltage 2- About 50% reduction in switching frequency compared to

conventional sine PWM. 3- Suitable for higher voltage and high power inverter systems, where

switching frequency is a limitation. 4- Higher voltage gain due to over modulation. 5- Reduced size of dc link filter components. 6- Selective elimination of lower order harmonics guarantee

avoidance of resonance with external line filtering networks.

The voltage Aov , of an inverter leg, normalized by dV21 is plotted in

Fig.4.34a, where six notches are introduced in the otherwise square-wave output, to control the magnitude of the fundamental voltage and to eliminate fifth and seventh harmonics. On a half-cycle basis, each notch provides one degree of freedom, that is, having three notches per half-cycle provides control of fundamental and elimination of two harmonics (in this case fifth and seventh).

Figure 4.38 shows that the output waveform has odd half-wave symmetry (sometimes it is referred to as odd quarter-wave symmetry). Therefore, only odd harmonics (coefficients of sine series) will be present. Since in a three-phase inverter (consisting of three such inverter legs), the third harmonic and its multiples are canceled out in the output, these harmonics need not be eliminated from the output of the inverter leg by means of waveform notching.

194 Chapter Three

A careful examination shows that the switching frequency of a switch in Fig.4.38 is seven times the switching frequency associated with a square-wave operation.

In a square-wave operation, the fundamental-frequency voltage component is:

( )273.14

2/

ˆ1 ==

πd

AoVV

(4.73)

Because of the notches to eliminate 5th and 7th harmonics, the maximum available fundamental amplitude is reduced. It can be shown that:

( )188.1

2/

ˆmax,1 =

d

Ao

VV

(4.74)

Fig.4.38 Programable harmonic elemination of fifth and seventh harmonics.

It is clear that the waveform in Fig.4.38 is odd function. So,

⎥⎥⎦

⎤

⎢⎢⎣

⎡= ∑

=

m

ssn tnJ

nb

1cos1 ω

π (4.75)

Where m is the number of jumps in the waveform. Jumps of the waveform shown in the figure is for only quarter of the waveform (because of similarity). The jumps are tabulated in the following table.

sJ 1J 2J 3J 4J Time 0 1α 2α 3α Value 2 -2 2 -2

Then, [ ]321 cos2cos2cos20cos22 αααπ

nnnnn

bn −+−= (4.76)

Where 2321πααα <<<


The above equation has three variables 321 ,, ααα and so we need three equation to obtain them. The First equation can be obtained by assigning a specific value to the amplitude of the fundamental component

1b . Another two equations can be obtained by equating 75 , bandb by zero to eliminate fifth and seventh harmonics. So, the following equations can be obtained.

[ ]3211 coscoscos14 αααπ

−+−=b (4.77)

[ ] 05cos5cos5cos143215 =−+−= ααα

πb (4.78)

[ ] 07cos7cos7cos143217 =−+−= ααα

πb (4.79)

The above equation can be rearranged to be in the following form:

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡ −

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

11

41

7cos7coscos5cos5cos5cos

coscoscos 1

321

321

321bπ

ααααααααα

(4.80)

These equations are nonlinear having multiple solution depending the value of 1b . Computer programs help us in solving the above equations. The required values of 321 ,, ααα and are plotted in Fig.4.39 as a function of the normalized fundamental in the output voltage.

Fig.4.39 The required values of 321 ,, ααα and .

196 Chapter Three

To allow control over the fundamental output and to eliminate the fifth-, seventh-, eleventh-, and the thirteenth-order harmonics, five notches per half-cycle would be needed. In that case, each switch would have 11 times the switching frequency compared with a square-wave operation.

Example Eliminate fifth and seventh harmonics from square waveform with no control on the fundamental amplitude:

Solution: It is clear that we have only two conditions which are 05 =b and 07 =b . So, we have only two notches per half cycle as shown in the following figure.

1α 2α 2απ − 1απ −π

1απ +

2απ +

22 απ − 12 απ −π2

2/d

AoVV

t1ω

Notch1Notch2

So we need only two variables 1α and 2α which can be obtained from

the following equations:

[ ]21 coscos14 ααπ

nnn

bn +−= (4.81)

[ ] 05cos5cos14215 =+−= αα

πb (4.82)

[ ] 07cos7cos14217 =+−= αα

πb (4.83)

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡−−

11

7cos7cos5cos5cos

21

21αααα

(4.84)

By solving the above equation we can get the value of 1α and 2α as following:

o8111.121 =α and o8458.242 =α The following table shows the absolute value of each harmonics after

eliminating fifth and seventh harmonics.


Harmonic order nb for square wave

πnbn

4= nb after eliminating 5th and 7th

[ ]21 coscos14 ααπ

nnn

bn +−=

1 1.27324 1.1871 3 0.244413 0.20511 5 0.25468 0.0 7 0.18189 0.0 9 0.14147 0.0995 11 0.11575 0.21227 13 0.09794 0.27141 15 0.08488 0.25067 17 0.07490 0.16881 19 0.06701 0.07183 21 0.06063 0.00407 THD 22.669% 26.201%

4.7 LOW COST PWM CONVERTER FOR UTILITY INTERFACE. As discussed in regular three phase PWM inverter, the existing

configuration uses six switches as shown in Fig.4.21. The low cost PWM inverter (Four switch inverter) uses four semiconductor switches. The reduction in number of switches reduces switching losses, system cost and enhances reliability of the system.

Fig.4.40 shows the proposed converter with four switches (Four Switch Topology, FST) By comparing Fig.4.40 and Fig.4.21, it is clear that, by using one additional capacitor one can replace two switches and the system will perform the same function. It is apparent that the cost and reliability are two major advantages of the proposed converter. The cost reduction can be accomplished by reducing the number of switches and the complexity of control system. The proposed converter current regulated with good power quality characterization.

ab

c

2dV

2dV

S1S3

S2S4

Fig.4.40 Four switch converter.

198 Chapter Three

• System Analysis For the proposed converter Fig.4.40, the switching requirements can

be stated as follows. Let the input three-phase generated voltages are:

)150(*3

)270(*3

)30(*3

+=

+=

+=

tSinVV

tSinVV

tSinVV

imca

imbc

imab

ω

ω

ω

(4.85)

The line voltages at the generator terminals can be expressed as follows:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

2

2*43

21

d

d

cb

abV

V

SSSS

VV

(4.86)

Where S1 , S2 , S3 and S4 are the switching functions of switches 1,2,3 and 4 respectively. Vd is the DC-link voltage. But, 12 1 SS −= and 34 1 SS −= (4.87)

Then, ⎟⎠⎞

⎜⎝⎛−=

⎟⎠⎞

⎜⎝⎛−=

2)12(

2)12(

3

1

dcb

dab

VSV

andVSV (4.88)

Then from (4.85), (4.86), (4.87) and (4.88) we get the following equation:

( )

( )

( )

( )90sin35.0

90sin35.0

30sin35.0

30sin35.0

14

13

12

11

+−=

++=

+−=

++=

tVVS

tVVS

tVVS

tVVS

d

m

d

m

d

m

d

m

ω

ω

ω

ω

(4.89) Then, the shift angle for switching signal of leg ‘a’ is 30o and for leg ‘c’

is 90o. Then, odaab

VmV 30

22∠= (4.90)


odabc

VmV 270

22∠= (4.91)

odaca

VmV 150

221 ∠= (4.92)

Then, 22

daLL

VmV = (4.93)

From the above equations it is clear that the DC voltage must be at least twice the maximum of input line-to-line voltage to avoid the input current distortion.

The main disadvantage of four switch converter is it needs for higher dc voltage to give the same line-to-line voltage as the six switch inverter which is clear from comparing the following equations:

22d

aLLV

mV = (four switch converter) (4.94)

223 d

aLLV

mV = (six switch converter) (4.95)

200 Chapter Three

PROBLEMS SINGLE PHASE 1- In a single-phase full-bridge PWM inverter, the input dc voltage

varies in a range of 295-325 V. Because of the low distortion required in the output ov , 0.1≤am

(a) What is the highest 1oV , that can be obtained and stamped on its nameplate as its voltage rating?

(b) Its nameplate volt-ampere rating is specified as 2000 VA, that is, VAIV oo 2000max,1max,1 = , where oi is assumed to be sinusoidal.

Calculate the combined switch utilization ratio when the inverter is supplying its rated volt-amperes.

2- Consider the problem of ripple in the output current of a single-phase full-bridge inverter. Assume 1oV = 220 V at a frequency of 47 Hz and the type of load is as shown in Fig.4.18a with L = 100 mH. If the inverter is operating in a square-wave mode, calculate the peak value of the ripple current.

3- Repeat Problem 2 with the inverter operating in a sinusoidal PWM mode, with fm = 21 and am = 0.8. Assume a bipolar voltage switching.

4- Repeat Problem 2 but assume that the output voltage is controlled by voltage cancellation and dV has the same value as required in the PWM inverter of Problem 3.

5- Calculate and compare the peak values of the ripple currents in Problems 2 through 4.

THREE-PHASE 6- Consider the problem of ripple in the output current of a three-

phase square-wave inverter. Assume ( ) 2201 =LLV V at a frequency of 52 Hz and the type of load is as shown in Fig.4.25a with L = 100 mH. Calculate the peak ripple current defined in Fig.4.26a.

7- Repeat Problem 6 if the inverter of Problem 6 is operating in a synchronous PWM mode with 39=fm and 8.0=am . Calculate the peak ripple current defined in Fig.4.26b.

8- In the three-phase, square-wave inverter of Fig.4.24a, consider the load to be balanced and purely resistive with a load-neutral n. Draw the


steady-state +DAAAn iuv ,, , and di waveforms, where +DAi is the current through +DA .

9- Repeat Problem 8 by assuming that the toad is purely inductive, where the load resistance, though finite, can be neglected.

Chapter 5 dc MOTOR DRIVES

5.1 INTRODUCTION

Traditionally, dc motor drives have been used for speed and position control applications. In the past few years, the use of ac motor servo drives in these applications is increasing. In spite of that, in applications where an extremely low maintenance is not required, dc drives continue to be used because of their low initial cost and excellent drive performance.

5.2 EQUIVALENT CIRCUIT OF dc MOTORS In a dc motor, the field flux fφ is established by the stator, either by

means of permanent magnets as shown in Fig.5-1a, where fφ stays constant, or by means of a field winding as shown in Fig.5-1b, where the field current If controls fφ . If the magnetic saturation in the flux path can be neglected, then:

fff Ik=φ (5.1) where fk is a field constant of proportionality.

The rotor carries in its slots the so-called armature winding, which handles the electrical power. This is in contrast to most ac motors, where the power-handling winding is on the stator for ease of handling the larger amount of power. However, the armature winding in a dc machine has to be on the rotor to provide a "mechanical" rectification of voltages and currents (which alternate direction as the conductors rotate from the influence

Fig.5-1 A dc motor (a) permanent-magnet motor (b) dc motor with a field

winding.

dc MOTOR DRIVES 201

of one stator pole to the next) in the armature-winding conductors, thus producing a dc voltage and a dc current at the terminals of the armature winding. The armature winding, in fact, is a continuous winding, without any beginning or end, and it is connected to the commutator segments. These commutator segments, usually made up of copper, are insulated from each other and rotate with the shaft. At least one pair of stationary carbon brushes is used to make contact between the commutator segments (and, hence, the armature conductors), and the stationary terminals of the armature winding that supply the dc voltage and current.

In a dc motor, the electromagnetic torque is produced by the interaction of the field flux fφ and the armature current ai :

affem ikT φ= (5.2) where fk is the torque constant of the motor. In the armature circuit, a back-emf is produced by the rotation of armature conductors at a speed wm in the presence of a field flux fφ

mfea ke ωφ= (5.3) where ek is the voltage constant of the motor.

In SI units, fk and ek are numerically equal, which can be shown by equating the electrical power aaie and the mechanical power emmTω . The electrical power is calculated as

amfeaae ikieP ωφ== (using Eq. 5-3) (5.4) and the mechanical power as:

amffemmm IkTP ωφω == (using Eq. 5-2) (5.5) In steady state,

me PP = (5.6) Therefore, from the foregoing equations

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡.sec/.. radWb

VkWbA

Nmk et (5.7)

In practice, a controllable voltage source tv is applied to the armature terminals to establish ai . Therefore, the current ai in the armature circuit is determined by tv , the induced back-emf ae , the armature-winding resistance aR , and the armature-winding inductance aL:

202 Chapter Five

dtdi

iRev aaaaat ++= (5.8)

Equation 5-8 is illustrated by an equivalent circuit in Fig.5-2.

Fig.5.2 A DC motor equivalent circuit.

The interaction of emT with the load torque, determines how the motor speed builds up:

( )tTBdt

dJT WLmm

em ++= ωω (5.9)

where J and B are the total equivalent inertia and damping, respectively, of the motor load combination and WLT is the equivalent working torque of the load.

Seldom are dc machines used as generators. However, they act as generators while braking, where their speed is being reduced. Therefore, it is important to consider dc machines in their generator mode of operation. In order to consider braking, we will assume that the flux fφ is kept constant and the motor is initially driving a load at a speed of mω . To reduce the motor speed, if tv is reduced below ae in Fig.5.2, then the current ai will reverse in direction. The electromagnetic torque emT given by Eq. 5-2 now reverses in direction and the kinetic energy associated with the motor load inertia is converted into electrical energy by the dc machine, which now acts as a generator. This energy must be somehow absorbed by the source of v, or dissipated in a resistor.

During the braking operation, the polarity of ea does not change, since the direction of rotation has not changed. Eqn.(5.3) still determines the magnitude of the induced emf. As the rotor slows down, ae decreases in

dc MOTOR DRIVES 203

magnitude (assuming that fφ is constant). Ultimately, the generation stops when the rotor comes to a standstill and all the inertial energy is extracted. If the terminal-voltage polarity is also reversed, the direction of rotation of the motor will reverse. Therefore, a dc motor can be operated in either direction and its electromagnetic torque can be reversed for braking, as shown by the four quadrants of the torque-speed in Fig.5-3.

Fig.5.3 Four-quadrant operation of a dc motor.

5.3 PERMANENT-MAGNET dc MOTORS Often in small dc motors, permanent magnets on the stator as shown in

Fig. 5-1a produce a constant field flux fφ . In steady state, assuming a constant field flux ( fφ , Eqs. 5.2, 5.3 and 5.8 result in

aTem IkT = (5.10)

mEa kE ω= (5.11)

aaat IREV += (5.12) where ffT kk φ= and feE KK φ= . Equations 5-10 through 5-12

correspond to the equivalent circuit of Fig.5-4a. From the above equations, it is possible to obtain the steady-state speed wm as a function of emT for a given tV :

⎟⎟⎠

⎞⎜⎜⎝

⎛−= em

T

at

Em T

kR

Vk1ω (5.13)

The plot of this equation in Fig. 5-4b shows that as the torque is increased, the torque-speed characteristic at a given tV is essentially vertical, except for the droop to the voltage drop IaRa across the armature-winding resistance. This droop in speed is quite small in integral horsepower dc motors but may be substantial in small servo motors. More importantly, however, the torque-speed characteristics can be shifted horizontally

204 Chapter Five

Figure 5.4 Permanent-magnet dc motor: (a) equivalent circuit: (b) torque-speed

characteristics: 12345 ttttt VVVVV >>>> where 4tV is the rated voltage; (c) continuous torque-speed capability.

In Fig.5.4b by controlling the applied terminal voltage Vt. Therefore, the speed of a load with an arbitrary torque-speed characteristic can be controlled by controlling tv in a permanent-magnet dc motor with a constant fφ .

In a continuous steady state, the armature current aI should not exceed its rated value, and therefore, the torque should not exceed the rated torque. Therefore, the characteristics beyond the rated torque are shown as dashed in Fig. 5-4b. Similarly, the characteristic beyond the rated speed is shown as dashed, because increasing the speed beyond the rated speed would require the terminal voltage tV to exceed its rated value, which is not desirable. This is a limitation of permanent-magnet dc motors, where the maximum speed is limited to the rated speed of the motor. The torque capability as a function of speed is plotted in Fig. 5-4c. It shows the steady-state operating limits of the torque and current; it is possible to significantly exceed current and torque limits on a short-term basis. Figure 5.4c also shows the terminal voltage required as a function of speed and the corresponding aE.

5.4 dc MOTORS WITH A SEPARATELY EXCITED FIELD WINDING Permanent-magnet dc motors are limited to ratings of a few horsepower and also have a maximum speed limitation. These limitations can be overcome if (~f is produced by means of a field winding on the stator, which is supplied' by a dc current 1 f, as shown in Fig.5.1b. To offer the most flexibility in controlling the dc motor, the field winding is excited by a separately controlled dc source fv , as shown in Fig. 5-5a. As

dc MOTOR DRIVES 205

indicated by Eq. 5-1, the steady-state value of fφ is controlled by

fff RVI /= , where fR is the resistance of the field winding. Since (~f is controllable, Eq. 5-13 can be written as follows:

(5.14) recognizing that feE kk φ= and ffkk φ= . Equation (5.14) shows that in a dc motor with a separately excited field winding, both tV and fφ can be controlled to yield the desired torque and speed. As a general practice, to maximize the motor torque capability, fφ (hence fI ) is kept at its rated value for speeds less than the rated speed. With fφ at its rated value, the relationships are the same as given by Eqs.(5.10) through (5.13) of a permanent-magnet dc motor. Therefore, the torque-speed characteristics are also the same as those for a permanent-magnet dc motor that were shown in Fig. 5-4b. With fφ constant and equal to its rated value, the motor torque-speed capability is as shown in Fig. 5-5b, where this region of constant fφ is often called the constant-torque region. The required terminal voltage tV in this region increases linearly from approximately zero to its rated value as the speed increases from zero to its rated value. The voltage tV and the corresponding aE are shown in Fig. 5-5b.

To obtain speeds beyond its rated value, tV is kept constant at its rated value and fφ is decreased by decreasing fI . Since aI is not allowed to exceed its rated value on a continuous basis, the torque capability declines, since fφ is reduced in Eq. (5.2). In this so-called field-weakening region, the maximum power aa IE (equal to mm Tω ) into the motor is not allowed to exceed its rated value on a continuous basis. This region, also called the constant-power region, is shown in Fig. 5-5b, where emT declines with mω and tV , aE , and aI stay constant at their rated values. It should be emphasized that Fig.5.5b is the plot of the maximum continuous capability of the motor in steady state. Any operating point within the regions shown is, of course, permissible. In the field-weakening region, the speed may be exceeded by 50-100% of its rated value, depending on the motor specifications.

206 Chapter Five

Figure 5-5 Separately excited dc motor: (a) equivalent circuit; (6) continuous

torque-speed capability.

5.5 EFFECT OF ARMATURE CURRENT WAVEFORM In dc motor drives, the output voltage of the power electronic converter contains an ac ripple voltage superimposed on the desired dc voltage. Ripple in the terminal voltage can lead to a ripple in the armature current with the following consequences that must be recognized: the form factor and torque pulsations.

5.5.1 FORM FACTOR The form factor for the dc motor armature current is defined as

(5.15) The form factor will be unity only if ai is a pure dc. The more ai

deviates from a pure dc, the higher will be the value of the form factor. The power input to the motor (and hence the power output) varies proportionally with the average value of ai , whereas the losses in the

resistance of the armature winding depend on 2aI (rms). Therefore, the

higher the form factor of the armature current, the higher the losses in the motor (i.e., higher heating) and, hence, the lower the motor efficiency.

Moreover, a form factor much higher than unity implies a much larger value of the peak armature current compared to its average value, which may result in excessive arcing in the commutator and brushes. To avoid serious damage to the motor that is caused by large peak currents, the motor may have to be derated (i.e., the maximum power or torque would have to be kept well below its rating) to keep the motor temperature from exceeding its specified limit and to protect the commutator and brushes.

dc MOTOR DRIVES 207

Therefore, it is desirable to improve the form factor of the armature current as much as possible. 5.5.2 TORQUE PULSATIONS Since the instantaneous electromagnetic torque ( )tTem developed by the motor is proportional to the instantaneous armature current ( )tia , a ripple in ai results in a ripple in the torque and hence in speed if the inertia is not large. This is another reason to minimize the ripple in the armature current. It should be noted that a high-frequency torque ripple will result in smaller speed fluctuations, as compared with a low-frequency torque ripple of the same magnitude.

5.6 dc SERVO DRIVES In servo applications, the speed and accuracy of response is important.

In spite of the increasing popularity of ac servo drives, dc servo drives are still widely used. If it were not for the disadvantages of having a commutator and brushes, the dc motors would be ideally suited for servo drives. The reason is that the instantaneous torque emT in Eq. 5-2 can be controlled linearly by controlling the armature current ai of the motor.

5.6.1 TRANSFER FUNCTION MODEL FOR SMALL-SIGNAL DYNAMIC PERFORMANCE

Figure 5-6 shows a dc motor operating in a closed loop to deliver controlled speed or controlled position. To design the proper controller that will result in high performance (high speed of response, low steady-state error, and high degree of stability), it is important to know the transfer function of the motor. It is then combined with the transfer function of the rest of the system in order to determine the dynamic response of the drive for changes in the desired speed and position or for a change in load. As we will explain later on, the linear model is valid only for small changes where the motor current is not limited by the converter supplying the motor.

208 Chapter Five

Figure 5-6 Closed-loop position/speed dc servo drive.

For analyzing small-signal dynamic performance of the motor-load combination around a steady-state operating point, the following equations can be written in terms of small deviations around their steady-state values:

If we take Laplace transform of these equations, where Laplace variables represent only the small-signal Δ values in Eqs.5.16 through 5.19,

(5.20) These equations for the motor-load combination can be represented by

transfer function blocks, as shown in Fig.5.7. The inputs to the motor-load combination in Fig. 5.7 are the armature terminal voltage ( )sVt and the load torque ( )sTWL . Applying one input at a time by setting the other input to zero, the superposition principle yields (note that this is a linearized system)

(5.21) This equation results in two closed-loop transfer functions:

(5.22)

(5.16)

(5.17)

(5.18)

(5.19)

dc MOTOR DRIVES 209

(5.23)

Fig.5.7 Block diagram representation of the motor and load (without ay feedback). As a simplification to gain better insight into the dc motor behavior,

the friction term, which is usually small, will be neglected by setting B = 0 in Eq. 5.22. Moreover, considering just the motor without the load, J in Eq. 5.22 is then the motor inertia mJ . Therefore

(5.24) We will define the following constants:

(5.25)

(5.26) Using mτ and eτ in the expression for ( )sG1 yields

(5.27) Since in general em ττ >> , it is a reasonable approximation to replace

msτ by ( )ems ττ + in the foregoing expression. Therefore

(5.28) The physical significance of the electrical and the mechanical time

constants of the motor should also be understood. The electrical time constant eτ , determines how quickly the armature current builds up, as shown in Fig.5-8, in response to a step change tvΔ in the terminal voltage, where the rotor speed is assumed to be constant.

210 Chapter Five

Figure 5-8 Electrical time constant Te; speed cam is assumed to be constant. The mechanical time constant mτ determines how quickly the speed

builds up in response to a step change tvΔ in the terminal voltage, provided that the electrical time constant eτ is assumed to be negligible and, hence, the armature current can change instantaneously. Neglecting

eτ in Eq. 5-28, the change in speed from the steady-state condition can be obtained as

(5.29) recognizing that ( ) svsV tt /Δ= . From Eq. (5.29)

(5.30) where mτ is the mechanical time constant with which the speed changes in response to a step change in the terminal voltage, as shown in Fig.5.9a. The corresponding change in the armature current is plotted in Fig.5.9b. Note that if the motor current is limited by the converter during large transients, the torque produced by the motor is simply max,aT Ik .

5.6.2 POWER ELECTRONIC CONVERTER Based on the previous discussion, a power electronic converter supplying a dc motor should have the following capabilities: 1- The converter should allow both its output voltage and current to

reverse in order to yield a four-quadrant operation as shown in Fig.5.3. 2- The converter should be able to operate in a current-controlled mode

by holding the current at its maximum acceptable value during fast acceleration and deceleration. The dynamic current limit is generally

dc MOTOR DRIVES 211

several times higher than the continuous steady-state current rating of the motor.

3- For accurate control of position, the average voltage output of the converter should vary linearly with its control input, independent of the load on the motor. This item is further discussed in Section 5-6-5.

4- The converter should produce an armature current with a good form factor and should minimize the fluctuations in torque and speed of the motor.

5- The converter output should respond as quickly as possible to its control input, thus allowing the converter to be represented essentially by a constant gain without a dead time in the overall servo drive transfer function model.

Figure 5-9Mechanical time constant mτ ; load torque is assumed to be constant.

A linear power amplifier satisfies all the requirements listed above. However, because of its low energy efficiency, this choice is limited to a very low power range. Therefore, the choice must be made between switch-mode dc-dc converters or the line-frequency-controlled converters. Here, only the switch-mode dc-dc converters are described. Drives with line-frequency converters can be analyzed in the same manner.

A full-bridge switch-mode dc-dc converter produces a four-quadrant controllable dc output. This full-bridge dc-dc converter (also called an H-

212 Chapter Five

bridge). The overall system is shown in Fig. 5-10, where the line-frequency ac input is rectified into dc by means of a diode rectifier of the type and filtered by means of a filter capacitor. An energy dissipation circuit is included to prevent the filter capacitor voltage from becoming large in case of braking of the dc motor. All four switches in the converter of Fig. 5-10 are switched during each cycle of the switching frequency. This results in a true four-quadrant operation with a continuous-current conduction, where both tV and aI can smoothly reverse, independent of each other. Ignoring the effect of blanking time, the average voltage output of the converter varies linearly with the input control voltage controlv , independent of the load:

controlct vkV = (5.31) where ck is the gain of the converter. Either a PWM bipolar voltages witching scheme or a PWM unipolar voltage-switching scheme can be used. Thus, the converter in Fig.5.6 can be replaced by an amplifier gain ck given by Eq. 5.31.

Figure 5-10 A dc motor servo drive; four-quadrant operation.

5.6.3 RIPPLE IN THE ARMATURE CURRENT ai Te current through a PWM full-bridge dc-dc converter supplying a dc motor load flows continuously even at small values of aI . However, it is important to consider the peak-to-peak ripple in the armature current because of its impact on the torque pulsations and heating of the motor. Moreover, a larger current ripple requires a larger peak current rating of the converter switches.

In the system of Fig. 5-10 under a steady-state operating condition, the instantaneous speed wm can be assumed to be constant if there is sufficient inertia, and therefore ( ) aa Ete = . The terminal voltage and the

dc MOTOR DRIVES 213

armature current can be expressed in terms of their dc and the ripple components as ( ) ( )tvVtv rtt += (5.32) ( ) ( )tiIti raa += (5.33)

where ( )tvr and ( )tir are the ripple components in tv and ai , respectively. Therefore, in the armature circuit, from Eq. 5.8,

( ) ( )[ ] ( )dt

tdiLtiIREtvV raraAart +++=+ (5.34)

Where aaat IREV += (5.35)

And ( ) ( ) ( )dt

tdiLtiRtv rarar += (5.36)

Assuming that the ripple current is primarily determined by the armature inductance aL and aR has a negligible effect, from Eq. 5-36

( ) ( )dt

tdiLtv rar ≅ (5.37)

The additional heating in the motor is approximately 2ra IR where rI is

the rms value of the ripple current ri . The ripple voltage is maximum when the average output voltage in a

dc is zero and all switches operate at equal duty ratios. Applying these results to the dc motor drive, Fig. 5-11a shows the voltage ripple ( )tvr and the resulting ripple current ( )tir using Eq. 5.37. From these waveforms, the maximum peak-to-peak ripple can be calculated as:

( )sa

dPP fL

VI

2max =Δ − (5.38)

where dV is the input dc voltage to the full-bridge converter.

214 Chapter Five

Figure 5-11 Ripple ri in the armature current: (a) PWM bipolar voltage switching, 0=tV ; (b) PWM unipolar voltage switching, dt VV 2/1= .

The ripple voltage for a PWM unipolar voltage switching is shown to be maximum when the average output voltage is dV2/1 . Applying this result to a dc motor drive, Fig.5.11b shows ( )tir waveform, where

( )sa

dPP fL

VI

8max =Δ − (5.39)

Equations 5-38 and 5-39 show that the maximum peak-to-peak ripple current is inversely proportional to aL and sf . Therefore, careful consideration must be given to the selection of sf and aL , where aL can be increased by adding an external inductor in the series with the motor armature. 5.6.4 CONTROL OF SERVO DRIVES

A servo system where the speed error directly controls the power electronic converter is shown in Fig. 5-12a. The current-limiting circuit comes into operation only when the drive current tries to exceed an acceptable limit max,aI during fast accelerations and decelerations. During these intervals, the output of the speed regulator is suppressed and the current is held at its limit until the speed and position approach their desired values.

To improve the dynamic response in high-performance servo drives, an internal current loop is used as shown in Fig.5-12b, where the armature current and, hence, the torque are controlled. The current control is accomplished by comparing the actual measured armature current ai with its reference value *

ai produced by the speed regulator. The current ai is inherently controlled from exceeding the current rating

of the drive by limiting the reference current *ai to max,aI .

The armature current provided by the dc-dc converter in Fig. 5-12b can be controlled in a similar manner as the current-regulated modulation in a dc-to-ac inverter. The only difference is that the reference current in steady state in a dc-dc converter is a dc rather than a sinusoidal waveform. Either a variable-frequency tolerance band control, or a fixed frequency control can be used for current control.

5.6.5 NONLINEARITY DUE TO BLANKING TIME

dc MOTOR DRIVES 215

In a practical full-bridge dc-dc converter, where the possibility of a short circuit across the input dc bus exists, a blanking time is introduced between the instant at which a switch turns off and the instant at which the other switch in the same leg turns on. The effect of the blanking time on the output of dc-to-ac full bridge PWM inverters. That analysis is also valid for PWM full-bridge dc-dc converters for dc servo drives. The output voltage of the converter is proportional to the motor speed com and the output current ai is proportional to the torque emT , Fig. 5.13. If at an arbitrary speed mω , the torque and, hence, ai are to be reversed, there is a dead zone in controlv as shown in Fig. 5-13, during which ai and emT remain small. The effect of this nonlinearity due to blanking time on the performance of the servo system is minimized by means of the current-controlled mode of operation discussed in the block diagram of Fig. 5-12b, where an internal current loop directly controls ai .

Fig.5.12 Control of servo drives: (a) no internal current-control loop; (b)

internal current-control loop.

216 Chapter Five

Fig.5.13 Effect of blanking time.

5.6.6 SELECTION OF SERVO DRIVE PARAMETERS Based on the foregoing discussion, the effects of armature inductance

aL , switching frequency sf , blanking time ct , and switching times t, of the solid state devices in the dc-dc converter can be summarized as follows: 1- The ripple in the armature current, which causes torque ripple and

additional armature heating, is proportional to sa fL / . 2- The dead zone in the transfer function of the converter, which degrades

the servo performance, is proportional to Δtf s . 3- Switching losses in the converter are proportional to cstf .

All these factors need to be considered simultaneously in the selection of the appropriate motor and the power electronic converter.

5.7 ADJUSTABLE-SPEED dc DRIVES Unlike servo drives, the response time to speed and torque commands is not as critical in adjustable-speed drives. Therefore, either switch-mode dc-dc converters as discussed for servo drives or the line-frequency controlled converters can be used for speed control.

5.7.1 SWITCH-MODE dc-dc CONVERTER If a four-quadrant operation is needed and a switch-mode converter is utilized, then the full-bridge converter shown in Fig. 5.10 is used.

If the speed does not have to reverse but braking is needed, then the two-quadrant converter shown in Fig. 5-14a can be used. It consists of two switches, where one of the switches is on at any time, to keep the output voltage independent of the direction of ai . The armature current

dc MOTOR DRIVES 217

can reverse, and a negative value of aI corresponds to the braking mode of operation, where the power flows from the dc motor to dV . The output voltage tV can be controlled in magnitude, but it always remains unipolar. Since ai can flow in both directions, unlike in the single-switch step-down and step-up dc-dc converters, ai in the circuit of Fig. 5-14a will not become discontinuous. For a single-quadrant operation where the speed remains unidirectional and braking is not required, the step-down converter shown in Fig. 5-14b can be used. 5.7.2 LINE-FREQUENCY CONTROLLED CONVERTERS In many adjustable-speed dc drives, especially in large power ratings, it may be economical to utilize a line-frequency controlled converter of the type discussed in Chapter 6. Two of these converters are repeated in Fig. 5-15 for single-phase and three-phase ac inputs. The output of these line-frequency converters, also called the phase-controlled converters, contains an ac ripple that is a multiple of the 60-Hz line frequency. Because of this low frequency ripple, an inductance in series with the motor armature may be required to keep the ripple in ai low, to minimize its effect on armature heating and the ripple in torque and speed.

Fig. 5.14 (a) Two-quadrant operation; (b) single-quadrant operation.

A disadvantage of the line-frequency converters is the longer dead time in responding to the changes in the speed control signal, compared to high-frequency switch-mode dc-dc converters. Once a thyristor or a pair of thyristors is triggered on in the circuits of Fig.5.15, the delay angle α that controls the converter output voltage applied to the motor terminals cannot be increased for a portion of the 50-Hz cycle. This may

218 Chapter Five

not be a problem in adjustable-speed drives where the response time to speed and torque commands is not too critical. But it clearly shows the limitation of line-frequency converters in servo drive applications.

The current through these line-frequency controlled converters is unidirectional, but the output voltage can reverse polarity. The two-quadrant operation with the reversible voltage is not suited for dc motor braking, which requires the voltage to be unidirectional but the current to be reversible. Therefore, if regenerative braking is required, two back-to-back connected thyristor converters can be used, as shown in Fig. 5-16a. This, in fact, gives a capability to operate in all four quadrants, as depicted in Fig. 5-16b.

An alternative to using two converters is to use one phase-controlled converter together with two pairs of contactors, as shown in Fig. 5-16c. When the machine is to be operated as a motor, the contactors 1M and

2M are closed. During braking when the motor speed is to be reduced rapidly, since the direction of rotation remains the same, Ea is of the same polarity as in the motoring mode. Therefore, to let the converter go into an inverter mode, contactors 1M and 2M are opened and 1R and 2R are closed. It should be noted that tie contactors switch at zero current when the current through them is brought to zero by the converter.

Fig.5.15 Line-frequency-controlled converters for dc motor drives: (a) single-phase

input, (b) three-phase input.

dc MOTOR DRIVES 219

Fig.5.16 Line-frequency-controlled converters for four-quadrant operation: (a) back-to-back converters for four-quadrant operation (without circulating current); (b) converter operation

modes; (c) contactors for four-quadrant operation.

5.7.3 EFFECT OF DISCONTINUOUS ARMATURE CURRENT In line-frequency phase-controlled converters and single-quadrant

step-down switch mode dc-dc converters, the output current can become discontinuous at light loads on the motor. For a fixed control voltage

controlv or the delay angle α , the discontinuous current causes the output voltage to go up. This voltage rise causes the motor speed to increase at low values of aI (which correspond to low torque load), as shown generically by Fig.5.17. With a continuously flowing ai , the drop in speed at higher torques is due to the voltage drop aa IR across the armature resistance; additional drop in speed occurs in the phase-controlled converter-driven motors due to commutation voltage drops across the ac-side inductance sL , which approximately equal ( ) aS IL πω /2 in single-phase converters and ( ) aS IL πω /3 in three-phase converters. These effects results in poor speed regulation under an open-loop operation.

220 Chapter Five

Fig.5.17 Effects of discontinuous ai on mω .

5.7.4 CONTROL OF ADJUSTABLE-SPEED DRIVES The type of control used depends on the drive requirements. An open-

loop control is shown in Fig.5.18 where the speed command *ω is generated by comparing the drive output with its desired value (which, e.g., may be temperature in case of a capacity modulated heat pump). A

dtd / limiter allows the speed command to change slowly, thus preventing the rotor current from exceeding its rating. The slope of the

dtd / limiter can be adjusted to match the motor-load inertia. The current limner in such drives may be just a protective measure, whereby if the measured current exceeds its rated value, the controller shuts the drive off. A manual restart may be required. As discussed in Section 5-6, a closed-loop control can also be implemented. 5.7.5 FIELD WEAKENING IN ADJUSTABLE-SPEED dc MOTOR DRIVES

In a dc motor with a separately excited field winding, the drive can be operated at higher than the rated speed of the motor by reducing the field flux fφ . Since many adjustable speed drives, especially at higher power ratings, employ a motor with a wound field, this capability can be exploited by controlling the field current and fφ . The simple line frequency phase-controlled converter shown in Fig. 5-15 is normally used to control fI through the field winding, where the current is controlled in magnitude but always flows in only one direction. If a converter topology consisting of only thyristors (such as in Fig. 5-15) is chosen, where the converter output voltage is reversible, the field current can be decreased rapidly.

dc MOTOR DRIVES 221

Figure 5-18 Open-loop speed control.

5.7.6 POWER FACTOR OF THE LINE CURRENT IN ADJUSTABLE-SPEED DRIVES

The motor operation at its torque limit is shown in Fig. 5-19a in the constant-torque region below the rated speed and in the field-weakening region above the rated speed. In a switch-mode drive, which consists of a diode rectifier bridge and a PWM dc-dc converter, the fundamental-frequency component 1sI of the line current as a function of speed is shown in Fig. 5-196. Figure 5-19c shows 1sI for a line-frequency phase controlled thyristor drive. Assuming the load torque to be constant, ISO decreases with decreasing speed in a switch-mode drive. Therefore, the switch-mode drive results in a good displacement power factor. On the other hand, in a phase-controlled thyristor drive, 1sI remains essentially constant as speed decreases, thus resulting in a very poor displacement power factor at low speeds.

Both the diode rectifiers and the phase-controlled rectifiers draw line currents that consist of large harmonics in addition to the fundamental. These harmonics cause the power factor of operation to be poor in both types of drives. The circuits described in Chapter 18 can be used to remedy the harmonics problem in the switch-mode drives, thus resulting in a high power factor of operation.

222 Chapter Five

Fig.5.19 Line current in adjustable-speed dc drives: (a) drive capability; (b) switch-mode converter drive; (c) line-frequency thyristor converter drive.

dc MOTOR DRIVES 223

PROBLEMS 1- Consider a permanent-magnet dc servo motor with the following parameters:

224 Chapter Five

Chapter 6

A.C. Voltage Regulators

6.1 Single-phase control of load voltage using thyristor switching A. Resistive load Assumptions : a. Triggering circuit provides pulse train to gate thyristor any point on

wave 0-180o.

b. Ideal supply, i.e. zero Z, voltage remains sinusoidal in spite of non-sinusoidal pulses of current drawn from supply .

c. Thyristors are ideal, i.e. no dissipation when conducting, no voltage drop when conducting , infinite Z when off.

||

||,

,0

2,

,

,

,0

2,

,

sin

sin

sin

απα

π

ππ

απα

απα

π

ππ

απα

ω

ω

ω

+

+

+

+

+=

+=

=

tEe

tEe

tEe

mT

mL

m

224 Chapter Six

6.2 Harmonics analysis of the load voltage (and current) The load voltage Le can be expressed as :-

∑∑∞

=

∞

=++=++=

1

0

1

0 )(sin2

sincos2

)(n

nn

nnL tnCa

tnbtnaa

te ψωωωω

where, ∫ ==π

ωωπ

2

00 valueAverage)(

21

2tdtea

L

For fundamental component :-

point.datumthe

andlfundamentathebetweenanglentdisplacemetan

componentlfundamentaofvaluepeak

sin)(1,cos)(1

1

111

21

211

2

012

01

==

=+=

==

−

∫∫

ba

bac

tdttebtdttea LL

ψ

ωωωπ

ωωωπ

ππ

For the n-th harmonics,

∫∫ ==ππ

ωωωπ

ωωωπ

2

0

2

0sin)(1,cos)(1 tdtntebtdtntea LnLn

For the fundamental of the load voltage,

ααπαψ

ααπαπ

ααππ

απ

ωωωπ

π

2sin)(212costantan

]2sin)(2[)12(cos2

)2sin)(2(2

)12(cos2

cos)(1

1

1

111

221

1

2

01

+−−

==

+−+−=

+−=

−==

−−

∫

ba

Ec

Eb

Etdttea

m

m

mL

RMS Load Voltage

r.m.s. value of tdteEe LLL ωωπ

π∫==

2

02 )(

21

]2sin)(2[21

2

]2sin)(2[4

)sin(21 22,

,22

ααππ

ααππ

ωωπ

ππ

παα

+−=

+−== ∫ +

mL

mmL

EE

EtdtEE

A.C. Voltage Regulators 225

with resistive load, the instantaneous load current is :- ππαπα

ω 2,,|sin+==

RtE

Rei mL

L

The r.m.s. load current

]2sin)(2[21]2sin)(2[

21

2ααπ

πααπ

π+−=+−=

RE

RE

I mL

6.3 Power Dissipation

]2sin)(2[2

valuesr.m.s.theare&where

powerAverage

product.ampvoltousinstantaneofvalueAverage21

2

22

2

0

ααππ

ππ

+−=

==

=

−== ∫

RE

EIR

ERI

dtieP

LLL

L

L

In terms of harmonic components:-

.......)(1.......)( 25

23

21

25

23

21 +++=+++= LLL EEE

RIIIRP

In terms of fundamental components: 11 cosψIEP =

where1

11

11 cos,1

2currentlfundamentar.m.s.

cb

Rc

I === ψ

6.4 Power factor in non-sinusoidal circuits

In general, IE

PPF ==sVoltampereApparent

PowerAverage

.supplyatcurrentr.m.s.

supply.atvoltager.m.s.==

IE

Definition is true irrespective for any waveform and frequency. Let e and i be periodic in ,2π

∫∫

∫==

=

ππ

π

ωπ

ωπ

ωπ

2

022

02

2

0

21;

21

21

tdiItdeE

tdieP

The usual aim is to obtain unity PF at the supply.

226 Chapter Six

To obtain unity P.F., certain conditions must be observed w.r.t. e(ω t) and i(ωt):- (i) e(ω t) and i(ωt) must be of the same frequency. (ii) e(ω t) and i(ωt) must be of the same wave shape (whatever the

wave shape) (iii) e(ω t) and i(ωt) must be in time-phase at every instant of the cycle.

Power Factor in systems with sinusoidal voltage (at supply) but non-sinusoidal current

)(wti is periodic in π2 but is non-sinusoidal. Average power is obtained by combining in-phase voltage and current components of the same frequency.

FactorntDisplacemexFactorDistortion

coscoscos

1111

11

=

===

=

ψψψ

II

IEIE

IEPPF

IEP

Distortion Factor = 1 for sinusoidal operation Displacement factor is a measure of displacement between e(ω t) and i(ωt). Displacement Factor =1 for sinusoidal resistive operation. Calculation of PF

]2sin)(2[21

]2sin)(2[21

]2sin)(2[2 2

2

2ααπ

πααπ

π

ααππ

+−=+−

+−

==

REE

RRE

IERIPF

R-L load

RLLRZ ωφω 1222 tan, −=+=


Steady-state at 0=α (sinusoidal operation)

Applying gate signal at ωt = α, where α > φ The start of conduction is delayed until .αω =t Subsequent to triggering, let the instantaneous current )( ti ω consists of hypothetical steady-state components )( tiss ω and transient component )( titrans ω ,

)()()( tititi transss ωωω +=∴ Now the αω =t , the instantaneous steady-state component has the value,

)sin()( φαα −=Z

Ei mss

But at ,αω =t total current 0)( =ti ω . At ωt = α, iss(α) = -itrans(α)

)sin()( φαω −−=∴Z

Eti mtrans

Subsequent to α The transient component decay exponentially from it’s instantaneous -

)sin( φα −Z

Em by the constant .R

LZ =

τφαωt

mtrans e

ZEti

−−

−= )sin()(

For ωt > α, )(

)sin()(αω

ωφαω−−

−−=t

LR

mtrans e

ZEti

Complete solution for first cycle

228 Chapter Six

παπ

απωω

ααω

ω

παπωωππ

απα

φαφα

φαφωω

2)()(

0)(2,,

,,0

)sin()sin(

)sin()[sin()(

+−−−−−

−−+−−+

−+−−

++−=

tL

Rxt

tR

xtt

Rxxm

e

etZ

Eti

l

Extinction Angle For the current in the interval xt ≤≤ωα

)()sin()sin()(

αωωφαφωω

−−−−−=

tL

Rmm e

ZEt

ZEti

But at 0)(, == tit ωαω , hence

)()sin()sin(0

αωφαφ

−−−−−=

xL

Rmm e

ZEx

ZE


But φω

cotand0 =≠L

RZ

Em )cot()sin()sin(0 αφαφ −−−−−=∴ xex

If φα and are known, x can be calculated. However, this is a transcendental equation (i.e. cannot be solved explicitly and no way of obtaining ),( φαfx = ). Method of solution is by iteration, e.g. If φ = 600, α = 1200 = 2π / 3, 578.0cot =φ

0)3

2(578.000 222)60120sin()60sin( =⇒−=−−−

xexx π

Rough Approximation :- Δ−+= φ0180x

where LlargeandRsmallfor20~15LRfor15~10

LsmallandRlargefor10~50000

00

ωω

ω

===

=Δ

For the previous example,

0000

0000

000

2202060180

2251560180

2015,60

=−+=

=−+=∴

−=Δ=

xor

x

φ

Load voltage

]2sin2sin)(2[2

]2cos2[cos2

sin)(

1

1

2,,,,0

ααπ

απ

ωω ππαπα

+−−=

−=

= −+

xxEb

xEa

tEte

m

m

xxmL

230 Chapter Six

A.C. voltage control using Integral cycle switching

N = no. of conducting cycles. T = Total ‘ON’+’OFF’ cycles, representing a control period. Analytical Properties.

Power, P ∝ TN ; r.m.s. load voltage, EL ∝

TN

Power Factor = TN

TN

REE

TN

RE

EIP

==

2

6.6 Advantages and Disadvantages of Integral cycle Control Advantages 1. Avoids the radio frequency interference created by phase - angle

switching . 2. Able to switch the loads with a large thermal time const . Disadvantages 1. Produces lamp flicker with incandescent lighting loads. 2. Inconsistent flashing with discharge lighting. 3. Not suitable for motor control (because of interruption of motor

current) 4. Total supply Distortion is greater than for symmetrical phase

control.

1

Problems Of Chapter 2 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.

2

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.

3

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.

6- Center tap diode rectifier is connected to 220 V, 50 Hz supply via

unity turns ratio center-tap transformer to feed Ω5 resistor load. Draw load voltage and currents and diode currents 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.

4

7- Single phase diode bridge rectifier is connected to 220 V, 50 Hz

supply to feed Ω5 resistor. Draw the load voltage, diodes currents and 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.

5

8- If the load of rectifier shown in problem 7 is changed to be Ω5 resistor in series with 10mH inductor. Calculate and draw the load current during the first two periods of supply voltages waveform.

9- Solve problem 8 if there is a freewheeling diode is connected in

shunt with the load.

10- If the load of problem 7 is changed to be 45 A pure DC. Draw diode

diodes currents and supply currents 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. (f) input power factor.

6

11- Single phase diode bridge rectifier is connected to 220V ,50Hz

supply. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor, and, input power factor and THD of the voltage at the point of common coupling.

7

12- Three-phase half-wave diode rectifier is connected to 380 V, 50Hz supply via 380/460 V delta/way transformer to feed the load with 45 A DC current. Assuming ideal transformer and zero source inductance. Then, draw the output voltage, secondary and primary currents along with supply voltage. Then, calculate (a) Rectfication effeciency. (b) Crest factor of secondary current. (c) Transformer Utilization Factor (TUF). (d) THD of primary current. (e) Input power factor.

13- Solve problem 12 if the supply has source inductance of 4 mH.

8

14- Three-phase full bridge diode rectifier is connected to 380V, 50Hz supply to feed Ω10 resistor. Draw the output voltage, diode currents and supply current of phase a. 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.

15- Solve problem 14 if the load is 45A pure DC current. Then find

THD of supply current and input power factor.

9

16- If the supply connected to the rectifier shown in problem 14 has a 5

mH source inductance and the load is 45 A DC. Find, average DC voltage, and THD of input current.

10

17- Single phase diode bridge rectifier is connected to square waveform with amplitude of 200V, 50 Hz. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor.

18- In the single-phase rectifier circuit of the following figure, 1=SL

mH and VVd 160= . The input voltage sv has the pulse waveform shown in the following figure. Plot si and di waveforms and find the average value of dI .

+

-SV

dVSi

di

Hzf 50=

o60 o60 o60o120

o120o120

V200tω

11

Problems Of Chapter 3 1- Single phase half-wave controlled rectifier is connected to 220 V,

50Hz supply to feed Ω10 resistor. If the firing angle o30=α draw output voltage and drop voltage across the thyristor along with the supply voltage. Then, calculate, (a) The rectfication effeciency. (b) Ripple factor. (c) Peak Inverse Voltage (PIV) of the thyristor. (d) The crest factor FC of input current.

2- Single phase half-wave controlled rectifier is connected to 220 V,

50Hz supply to feed Ω5 resistor in series with mH10 inductor if the firing angle o30=α .

(a) Determine an expression for the current through the load in the first two periods of supply current, then fiend the DC and rms value of output voltage.

(b) Draw the waveforms of load, resistor, inductor voltages and load current.

12


4- single phase full-wave fully controlled rectifier is connected to 220V,

50 Hz supply to feed Ω5 resistor, if the firing angle o40=α . Draw the load voltage and current, thyristor currents and supply current. Then, calculate (a) The rectfication effeciency. (b) Peak Inverse Voltage (PIV) of the thyristor. (c) Crest factor of supply current.

13

5- In the problem 4, if there is a 5mH inductor is connected in series with the Ω5 resistor. Draw waveforms of output voltage and current, resistor and inductor voltages, thyristor currents, supply currents. Then, find an expression of load current, DC and rms values of output voltages.

6- Solve problem 5 if the load is connected with freewheeling diode.

14

7- Single phase full wave fully controlled rectifier is connected to 220V, 50 Hz supply to feed the load with 47 A pure dc current. The firing angle o40=α . Draw the load voltage, thyristor, and load currents. Then, calculate (a) the rectfication effeciency. (b) Ripple factor of output voltage. (c) Crest factor of supply current. (d) Use Fourier series to fiend an expression for supply current. (e) THD of supply current. (f) Input power factor.

8- Solve problem 7 if the supply has a 3 mH source inductance.

15

9- Single phase full-wave semi-controlled rectifier is connected to 220 V, 50Hz supply to feed Ω5 resistor in series with 5 mH inductor, the load is connected in shunt with freewheeling diode. Draw the load voltage and current, resistor voltage and inductor voltage diodes and thyristor currents. Then, calculate dcV and rmsV of the load voltages. If the freewheeling diode is removed, explain what will happen?

10- The single-phase full wave controlled converter is supplying a DC

load of 1 kW with pure DC current. A 1.5-kVA-isolation transformer with a source-side voltage rating of 120 V at 50 Hz is used. It has a total leakage reactance of 8% based on its ratings. The ac source voltage of nominally 120 V is in the range of -10% and +5%. Then, Calculate the minimum transformer turns ratio if the DC load voltage is to be regulated at a constant value of 100 V. What is the value of a when VS = 120 V + 5%.

16

11- In the single-phase inverter of, SV = 120 V at 50 Hz, SL = 1.2 mH,

dL = 20 mH, dE = 88 V, and the delay angle α = 135°. Using PSIM, obtain sv , si , dv , and di waveforms in steady state.

12- In the inverter of Problem 12, vary the delay angle α from a value of 165° down to 120° and plot di versus α . Obtain the delay angle bα , below which di becomes continuous. How does the slope of the characteristic in this range depend on SL ?

13- In the three-phase fully controlled rectifier is connected to 460 V at 50 Hz and mHLs 1= . Calculate the commutation angle u if the load draws pure DC current at VVdc 515= and dcP = 500 kW.

14- In Problem 13 compute the peak inverse voltage and the average and the rms values of the current through each thyristor in terms of LLV and oI .

15- Consider the three-phase, half-controlled converter shown in the following figure. Calculate the value of the delay angle α for which

dmdc VV 5.0= . Draw dv waveform and identify the devices that conduct during various intervals. Obtain the DPF, PF, and %THD in the input line current and compare results with a full-bridge converter operating at dmdc VV 5.0= . Assume SL .

17

16- Repeat Problem 15 by assuming that diode fD is not present in the

converter.

17- The three-phase converter of Fig.3.48 is supplying a DC load of 12

kW. A Y- Y connected isolation transformer has a per-phase rating of 5 kVA and an AC source-side voltage rating of 120 V at 50 Hz. It has a total per-phase leakage reactance of 8% based on its ratings. The ac source voltage of nominally 208 V (line to line) is in the range of -10% and +5%. Assume the load current is pure DC, calculate the minimum transformer turns ratio if the DC load voltage is to be regulated at a constant value of 300 V. What is the value of α when

LLV = 208 V +5%. 18- In the three-phase inverter of Fig.3.63, LLV = 460 V at 60 Hz, E = 550

V, and SL = 0.5 mH. Assume the DC-side current is pure DC, Calculate α and γ if the power flow is 55 kW.

18

Problems Of Chapter 4 1- In a single-phase full-bridge PWM inverter, the input dc

voltage varies in a range of 295-325 V. Because of the low distortion required in the output ov , 0.1≤am

(a) What is the highest 1oV , that can be obtained and stamped on its nameplate as its voltage rating?

(b) Its nameplate volt-ampere rating is specified as 2000 VA, that is, VAIV oo 2000max,1max,1 = , where oi is assumed to be sinusoidal. Calculate the combined switch utilization ratio when the inverter is supplying its rated volt-amperes.

19

2- Consider the problem of ripple in the output current of a single-phase full-bridge inverter. Assume 1oV = 220 V at a frequency of 47 Hz and the type of load is as shown in Fig.18a with L = 100 mH. If the inverter is operating in a square-wave mode, calculate the peak value of the ripple current.

20

3- Repeat Problem 2 with the inverter operating in a sinusoidal PWM mode, with fm = 21 and am = 0.8. Assume a bipolar voltage switching.

4- Repeat Problem 2 but assume that the output voltage is controlled by voltage cancellation and dV has the same value as required in the PWM inverter of Problem 3.

21

5- Calculate and compare the peak values of the ripple currents in Problems 2 through 4.

22

THREE-PHASE 6- Consider the problem of ripple in the output current of a three-

phase square-wave inverter. Assume ( ) 2201 =LLV V at a frequency of 52 Hz and the type of load is as shown in Fig.25a with L = 100 mH. Calculate the peak ripple current defined in Fig.26a.

23

7- Repeat Problem 6 if the inverter of Problem 6 is operating in a synchronous PWM mode with 39=fm and 8.0=am . Calculate the peak ripple current defined in Fig.26b.

24

8- In the three-phase, square-wave inverter of Fig.24a, consider the load to be balanced and purely resistive with a load-neutral n. Draw the steady-state +DAAAn iuv ,, , and di waveforms, where +DAi is the current through +DA .

25

9- Repeat Problem 8 by assuming that the toad is purely inductive, where the load resistance, though finite, can be neglected.

26

Problems Of Chapter 5 1- Consider a permanent-magnet dc servo motor with the following parameters:

33

Power Electronics (2) Assignment

Powerelectronicsnote 121011031400-phpapp02

Documents

Transcript of Powerelectronicsnote 121011031400-phpapp02