113

Chapter 11

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 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

114Chapter 3 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 WheelingDiode

98

3.7 Three Phase Full Wave Fully Controlled Rectifier 100

Chapter 4 112

4-1 Introduction 112

4-2 Determination Of Fourier Coefficients 113

4-3 Determination Of Fourier Coefficients Without Integration 119

Chapter 1

1.1. Definition Of Power ElectronicsPower electronics refers to control and conversion of electrical power bypower semiconductor devices wherein these devices operate as switches.Advent of silicon-controlled rectifiers, abbreviated as SCRs, led to thedevelopment of a new area of application called the power electronics.Once the SCRs were available, the application area spread to many fieldssuch as drives, power supplies, aviation electronics, high frequencyinverters and power electronics originated.

Power electronics has applications that span the whole field ofelectrical power systems, with the power range of these applicationsextending from a few VA/Watts to several MVA / MW."Electronic power converter" is the term that is used to refer to a powerelectronic circuit that converts voltage and current from one form toanother. 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 DCvoltage to another DC voltage, andCycloconverter and cycloinverter converting an AC voltageto another AC voltage.

In addition, SCRs and other power semiconductor devices are used asstatic switches.

1.2 RectificationRectifiers can be classified as uncontrolled and controlled rectifiers, andthe controlled rectifiers can be further divided into semi-controlled andfully controlled rectifiers. Uncontrolled rectifier circuits are built withdiodes, and fully controlled rectifier circuits are built with SCRs. Bothdiodes and SCRs are used in semi-controlled rectifier circuits.

There are several rectifier configurations. The most famous rectifierconfigurations are listed below.

Single-phase semi-controlled bridge rectifier,Single-phase fully-controlled bridge rectifier,Three-phase three-pulse, star-connected rectifier,

2

Double three-phase, three-pulse star-connected rectifiers withinter-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-connectedand 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 currentapplications, a pair of three-phase, three-pulse rectifiers interconnected byan 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 specificapplication like electroplating

1.3 DC-To-AC ConversionThe converter that changes a DC voltage to an alternating voltage, AC iscalled an inverter. Earlier inverters were built with SCRs. Since thecircuitry required turning the SCR off tends to be complex, other powersemiconductor devices such as bipolar junction transistors, powerMOSFETs, insulated gate bipolar transistors (IGBT) and MOS-controlledthyristors (MCTs) are used nowadays. Currently only the inverters with ahigh power rating, such as 500 kW or higher, are likely to be built witheither SCRs or gate turn-off thyristors (GTOs). There are many invertercircuits 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 ConversionWhen the SCR came into use, a DC-to-DC converter circuit was called achopper. Nowadays, an SCR is rarely used in a DC-to-DC converter.

3

Either a power BJT or a power MOSFET is normally used in such aconverter and this converter is called a switch-mode power supply. Aswitch-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 chopperare:

DC drive,Battery charger, and,DC power supply.

1.5 AC-TO-AC ConversionA cycloconverter or a Matrix converter converts an AC voltage, such asthe mains supply, to another AC voltage. The amplitude and thefrequency of input voltage to a cycloconverter tend to be fixed values,whereas both the amplitude and the frequency of output voltage of acycloconverter tend to be variable specially in Adjustable Speed Drives(ASD). A typical application of a cycloconverter is to use it forcontrolling the speed of an AC traction motor and most of thesecycloconverters have a high power output, of the order a few megawattsand SCRs are used in these circuits. In contrast, low cost, low powercycloconverters for low power AC motors are also in use and many ofthese circuit tend to use triacs in place of SCRs. Unlike an SCR whichconducts in only one direction, a triac is capable of conducting in eitherdirection and like an SCR, it is also a three terminal device. It may benoted that the use of a cycloconverter is not as common as that of aninverter and a cycloinverter is rarely used because of its complexity andits high cost.

1.6 Additional Insights Into Power ElectronicsThere are several striking features of power electronics, the foremostamong them being the extensive use of inductors and capacitors. In manyapplications of power electronics, an inductor may carry a high current ata high frequency. The implications of operating an inductor in thismanner are quite a few, such as necessitating the use of litz wire in placeof single-stranded or multi-stranded copper wire at frequencies above 50

4

kHz, using a proper core to limit the losses in the core, and shielding theinductor properly so that the fringing that occurs at the air-gaps in themagnetic path does not lead to electromagnetic interference. Usually thecapacitors used in a power electronic application are also stressed. It istypical for a capacitor to be operated at a high frequency with currentsurges passing through it periodically. This means that the current ratingof the capacitor at the operating frequency should be checked before itsuse. In addition, it may be preferable if the capacitor has self-healingproperty. Hence an inductor or a capacitor has to be selected or designedwith care, taking into account the operating conditions, before its use in apower electronic circuit.In many power electronic circuits, diodes play a crucial role. A normalpower diode is usually designed to be operated at 400 Hz or less. Many ofthe inverter and switch-mode power supply circuits operate at a muchhigher frequency and these circuits need diodes that turn ON and OFFfast. In addition, it is also desired that the turning-off process of a diodeshould not create undesirable electrical transients in the circuit. Sincethere are several types of diodes available, selection of a proper diode isvery 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 DCresponse or a sinusoidal response. Most of the power electronic circuitshave 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 polesof the network. Since the responses are non-sinusoidal, harmonic analysisis often necessary. In order to obtain the time response, it may benecessary to resort to the use of a computer program.Power electronics is a subject of interdisciplinary nature. To design andbuild control circuitry of a power electronic application, one needsknowledge of several areas, which are listed below.

Design of analogue and digital electronic circuits, to build thecontrol circuitry.Microcontrollers and digital signal processors for use insophisticated applications.Many power electronic circuits have an electrical machine astheir load. In AC variable speed drive, it may be a reluctance

5

motor, an induction motor or a synchronous motor. In a DCvariable speed drive, it is usually a DC shunt motor.In a circuit such as an inverter, a transformer may be connectedat its output and the transformer may have to operate with anonsinusoidal waveform at its input.A pulse transformer with a ferrite core is used commonly totransfer the gate signal to the power semiconductor device. Aferrite-cored transformer with a relatively higher power outputis also used in an application such as a high frequency inverter.Many power electronic systems are operated with negativefeedback. A linear controller such as a PI controller is used inrelatively simple applications, whereas a controller based ondigital or state-variable feedback techniques is used in moresophisticated applications.Computer simulation is often necessary to optimize the designof a power electronic system. In order to simulate, knowledgeof software package such as MATLAB, Pspice, Orcad,…..etc.and the know-how to model nonlinear systems may benecessary.

The study of power electronics is an exciting and a challengingexperience. The scope for applying power electronics is growing at a fastpace. New devices keep coming into the market, sustaining developmentwork in power electronics.

1.7 HarmonicsThe invention of the semiconductor controlled rectifier (SCR or

thyristor) in the 1950s led to increase of development new typeconverters, all of which are nonlinear. The major part of power systemloads is in the form of nonlinear loads too much harmonics are injected tothe power system. It is caused by the interaction of distorting customerloads with the impedance of supply network. Also, the increase ofconnecting renewable energy systems with electric utilities injects toomuch harmonics to the power system.

There are a number of electric devices that have nonlinear operatingcharacteristics, and when it used in power distribution circuits it willcreate and generate nonlinear currents and voltages. Because of periodicnon-linearity can best be analyzed using the Fourier transform, thesenonlinear currents and voltages have been generally referred to as

6

“Harmonics”. Also, the harmonics can be defined as a sinusoidalcomponent of a periodic waves or quality having frequencies that are anintegral multiple of the fundamental frequency.

Among the devices that can generate nonlinear currents transformersand induction machines (Because of magnetic core saturation) and powerelectronics assemblies.

The electric utilities recognized the importance of harmonics as earlyas the 1930’s such behavior is viewed as a potentially growing concern inmodern power distribution network.

1.7.1 Harmonics Effects on Power System ComponentsThere are many bad effects of harmonics on the power system

components. These bad effects can derated the power system componentor it may destroy some devices in sever cases [Lee]. The following is theharmonic effects on power system components.In Transformers and Reactors

The eddy current losses increase in proportion to the square of theload 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 andline capacitor.

In CapacitorsThe life expectancy decreases due to increased dielectric lossesthat cause additional heating, reactive power increases due toharmonic voltages, and,Over voltage can occur and resonance may occur resulting inharmonic magnification.

In CablesAdditional heating occurs in cables due to harmonic currentsbecause of skin and proximity effects which are function offrequency, and,The I2R losses increase.

In SwitchgearChanging the rate of rise of transient recovery voltage, and,Affects the operation of the blowout.

In RelaysAffects the time delay characteristics, and,

7

False tripping may occurs.In Motors

Stator and rotor I2R losses increase due to the flow of harmoniccurrents,In the case of induction motors with skewed rotors the fluxchanges in both the stator and rotor and high frequency canproduce substantial iron losses, and,Positive sequence harmonics develop shaft torque that aid shaftrotation; negative sequence harmonics have opposite effect.

In GeneratorsRotor and stator heating ,Production of pulsating or oscillating torques, and,Acoustic noise.

In Electronic EquipmentUnstable operation of firing circuits based on zero voltagecrossing,Erroneous operation in measuring equipment, and,Malfunction of computers allied equipment due to the presence ofac supply harmonics.

1.7.2 Harmonic StandardsIt should be clear from the above that there are serious effects on the

power system components. Harmonics standards and limits evolved togive a standard level of harmonics can be injected to the power systemfrom any power system component. The first standard (EN50006) byEuropean Committee for Electro-technical Standardization (CENELEE)that was developed by 14th European committee. Many otherstandardizations were done and are listed in IEC61000-3-4, 1998 [1].

The IEEE standard 519-1992 [2] is a recommended practice for powerfactor correction and harmonic impact limitation for static powerconverters. It is convenient to employ a set of analysis tools known asFourier transform in the analysis of the distorted waveforms. In general, anon-sinusoidal waveform f(t) repeating with an angular frequency canbe expressed as in the following equation.

1

0 )sin()cos(2

)(n

nn tnbtnaa

tf (1.1)

8

where2

0

)(cos)(1

tdtntfan (1.2)

and2

0

)(sin)(1

tdtntfbn (1.3)

Each frequency component n has the following value)(sin)(cos)( tnbtnatf nnn (1.4)

fn(t) can be represented as a phasor in terms of its rms value as shown inthe following equation

njnnn e

baF

2

22

(1.5)

Wheren

nn a

b1tan (1.6)

The amount of distortion in the voltage or current waveform isqualified by means of an Total Harmonic Distortion (THD). The THD incurrent and voltage are given as shown in (1.7) and (1.8) respectively.

1

2

1

21

2

*100*100s

nnsn

s

ssi I

I

I

IITHD (1.7)

1

2

1

21

2

*100*100s

nnsn

s

ssv V

V

V

VVTHD (1.8)

Where THDi & THDv The Total Harmonic Distortion in the currentand voltage waveforms

Current and voltage limitations included in the update IEE 519 1992are shown in Table(1.1) and Table(1.2) respectively [2].

Table (1.1) IEEE 519-1992 current distortion limits for general distributionsystems (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 5.020<50 7.0 3.5 2.5 1.0 0.5 8.050<100 10.0 4.5 4.0 1.5 0.7 12.0100<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

9

Where; TDD (Total Demand Distortion) =2

2100

nn

MLI

I,

Where MLI is the maximum fundamental demand load current (15 or30min 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 commoncoupling (PCC).

Table (1.2) Voltage distortion limitsBus voltage at PCC Individual voltage distortion (%) vTHD (%)

69 kV and blow 3.0 5.069.001 kV through 161kV 1.5 2.5161.001kV and above 1 1.5

1.8 Semiconductors Switch typesAt this point it is beneficial to review the current state of semiconductordevices used for high power applications. This is required because theoperation of many power electronic circuits is intimately tied to thebehavior of various devices.

1.8.1 DiodesA sketch of a PN junction diode characteristic is drawn in Fig.1.1. Theicon used to represent the diode is drawn in the upper left corner of thefigure, together with the polarity markings used in describing thecharacteristics. The icon 'arrow' itself suggests an intrinsic polarityreflecting the inherent nonlinearity of the diode characteristic.

Fig.1.1 shows the i-v characteristics of the silicon diode andgermanium diode. As shown in the figure the diode characteristics havebeen divided into three ranges of operation for purposes of description.Diodes operate in the forward- and reverse-bias ranges. Forward bias is arange of 'easy' conduction, i.e., after a small threshold voltage level ( »0.7 volts for silicon) is reached a small voltage change produces a largecurrent 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 thereverse applied voltage exceeds the maximum limit that the diode canwithstand. At this range the diode destroyed.

10

Fig.1.1 The diode iv characteristics

On the other hand if the polarity of the voltage is reversed the currentflows 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 appearsacross its terminals. However, the voltage drop is about 0.7 V for silicondiodes and 0.3 V for germanium diodes, so it can be neglected in mostelectronic circuits because this voltage drop is small with respect to othercircuit voltages. So, a perfect diode behaves like normally closed switchwhen it is forward bias (as soon as its anode voltage is slightly positivethan cathode voltage) and open switch when it is in reverse biased (assoon as its cathode voltage is slightly positive than anode voltage). Thereare two important characteristics have to be taken into account inchoosing diode. These two characteristics are:

Peak Inverse voltage (PIV): Is the maximum voltage that a diodecan withstand only so much voltage before it breaks down. So ifthe PIV is exceeded than the PIV rated for the diode, then thediode will conduct in both forward and reverse bias and the diodewill be immediately destroyed.Maximum Average Current: Is the average current that the diodecan carry.

It is convenient for simplicity in discussion and quite useful in makingestimates of circuit behavior ( rather good estimates if done with care andunderstanding) to linearize the diode characteristics as indicated inFig.1.2. Instead of a very small reverse-bias current the idealized modelapproximates this current as zero. ( The practical measure of theappropriateness of this approximation is whether the small reverse biascurrent causes negligible voltage drops in the circuit in which the diode isembedded. If so the value of the reverse-bias current really does not enterinto calculations significantly and can be ignored.) Furthermore the zero

11

current approximation is extended into forward-bias right up to the kneeof the curve. Exactly what voltage to cite as the knee voltage is somewhatarguable, although usually the particular value used is not very important.

1.8.2 ThyristorThe thyristor is the most important type of the power semiconductordevices. They are used in very large scale in power electronic circuits.The thyristor are known also as Silicon Controlled Rectifier (SCR). Thethyristor 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 schematicdiagram of this device and its symbolic representation. The thyristor hasthree terminals, anode A, cathode K and gate G as shown in Fig.1.2.Theanode and cathode are connected to main power circuit. The gate terminalis connected to control circuit to carry low current in the direction fromgate 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 devicei.e. anode is positive with respect to cathode, junction J1 and J3 areforward bias while J2 remains reverse biased, and therefore the anodecurrent is so small leakage current. If the forward voltage reaches acritical limit, called forward break over voltage, the thyristor switchesinto high conduction, thus forward biasing junction J2 to turn thyristorON in this case the thyristor will break down. The forward voltage dropthen falls to very low value (1 to 2 Volts). The thyristor can be switchedto on state by injecting a current into the central p type layer via the gateterminal. The injection of the gate current provides additional holes in the

12

central p layer, reducing the forward breakover voltage. If the anodecurrent falls below a critical limit, called the holding current IH thethyristor turns to its forward state.

If the reverse voltage is applied across the thyristor i.e. the anode isnegative with respect to cathode, the outer junction J1 and J3 are reversebiased and the central junction J2 is forward biased. Therefore only asmall leakage current flows. If the reverse voltage is increased, then at thecritical breakdown level known as reverse breakdown voltage, anavalanche will occur at J1 and J3 and the current will increase sharply. Ifthis current is not limited to safe value, it will destroy the thyristor.

The gate current is applied at the instant turn on is desired. Thethyristor turn on provided at higher anode voltage than cathode. Afterturn on with IA reaches a value known as latching current, the thyristorcontinuous to conduct even after gate signal has been removed. Henceonly pulse of gate current is required to turn the Thyrstor ON.

Fig.1.3 Thyristor v-i characteristics1.8.3 Thyristor types:There is many types of thyristors all of them has three terminals butdiffers only in how they can turn ON and OFF. The most famous types ofthyristors 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 inFig.1.4.

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 bya pulse of negative gate current. The gate current therefore controls bothON state and OFF state operation of the device. GTO v-i characteristics isshown in Fig.1.5. The GTO has many advantages and disadvantages withrespect to conventional thyristor here will talk about these advantages anddisadvantages.

Fig.1.5 GTO v-i characteristics.

14

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 isequivalent of two reverse parallel-connected SCRs with one commongate. Conduction can be achieved in either direction with an appropriategate current. A TRIAC is thus a bi-directional gate controlled thyristorwith 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 showsthe i-v characteristics of the TRIAC.

15

Fig.1.6 Operating characteristics of TRIAC.ele146

DIACDIAC 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 downvoltage, the DIAC conducts and the current goes in the direction from thehigher voltage point to the lower voltage one. The following figure showsthe layers construction, electric circuit symbol and the operatingcharacteristics of the DIAC. Fig.1.7 shows the DIAC construction andelectric symbol. Fig.1.8 shows a DIAC v-i characteristics.

The DIAC used in firing circuits of thyristors since its breakdownvoltage used to determine the firing angle of the thyristor.

Fig.1.7 DIAC construction and electric symbol.

16

Fig.1.8 DIAC v-i characteristics

1.9 Power TransistorPower transistor has many applications now in power electronics andbecome a better option than thyristor. Power transistor can switch on andoff very fast using gate signals which is the most important advantageover thyristor. There are three famous types of power transistors used inpower 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 threesemiconductor materials layers. These three layers can be NPN or PNP.Fig.1.9 shows the circuit symbol of NPN and PNP BJT transistor.

npn pnpFig.1.9 The electric symbol of npn and pnp transistors.

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 onthe electric symbol of BJT transistor and both gate and collector take theopposite direction.

Fig.1.10 The currents of the NPN and PNP transistors.

When the transistor connected in DC circuit, the voltage BBV representinga forward bias voltage and ccV representing a reverse bias for base tocollector 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 acurrent gain of the transistor as shown in ( )

B

C

I

I

Current and voltage analysis of NPN transistors is shown if Fig.1.11. It isclear from Fig.1.11 that:

BBBEBBR RIVVVb

*

Then, the base current can be obtained as shown in the followingequation:

B

BEBBB R

VVI

18

The voltage on CR resistor are:

CCR RIVC

*

CCCCCE RIVV *Fig.1.12 shows the collector characteristics of NPN transistor fordifferent base currents. This figure shows that four regions, saturation,linear, break down, and, cut-off regions. The explanation of each regionin 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 linearregion CI approximately constant for the same amount of base currentwhen CEV increases.

When CEV become higher than the rated limits, the transistor goes tobreak down region.At zero base current, the transistor works in cut-off region and there isonly very small collector leakage current.

Fig.1.12 Collector characteristics of NPN transistor for different base currents.

1.10 Power MOSFETThe power MOSFET has two important advantages over than BJT,

First of them, is its need to very low operating gate current, the second of

19

them, is its very high switching speed. So, it is used in the circuit thatrequires high turning ON and OFF speed that may be greater than100kHz. This switch is more expensive than any other switches have thesame ratings. The power MOSFET has three terminals source, drain andgate. Fig.1.13 shows the electric symbol and static characteristics of thepower 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 speedthan BJT but still lower than MOSFET. The normal switching frequencyof the IGBT is about 40kHz. IGBT has three terminals collector, emitter,and, gate.Fig.1.14 shows the electric circuit symbol and operating characteristics ofthe IGBT. IGBT used so much in PWM converters and in Adjustablespeed drives.

Fig.1.14 IGBT v-i transfer characteristics and circuit symbol:

20

1.12 Power Junction Field Effect TransistorsThis device is also sometimes known as the static induction transistor

(SIT). It is effectively a JFET transistor with geometry changes to allowthe device to withstand high voltages and conduct high currents. Thecurrent capability is achieved by paralleling up thousands of basic JFETcells. The main problem with the power JFET is that it is a normally ondevice. This is not good from a start-up viewpoint, since the device canconduct until the control circuitry begins to operate. Some devices arecommercially available, but they have not found widespread usage.

1.13 Field Controlled ThyristorThis device is essentially a modification of the SIT. The drain of the

SIT is modified by changing it into an injecting contact. This is achievedby making it a pn junction. The drain of the device now becomes theanode, and the source of the SIT becomes the cathode. In operation thedevice 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 isconductivity modulation and lower on-state resistance. The device alsoblocks for reverse voltages due to the presence of the pn junction.

1.14 MOS-Controlled ThyristorsThe MOS-controlled thyristor (MCT) is a relatively new device which

is available commercially. Unfortunately, despite a lot of hype at the timeof its introduction, it has not achieved its potential. This has been largelydue to fabrication problems with the device, which has resulted on lowyields. Fig.1.15 is an equivalent circuit of the device, and its circuitsymbol. From Fig.1.15 one can see that the device is turned on by theON-FET, and turned o. by the OFF-FET. The main current carryingelement of the device is the thyristor. To turn the device on a negativevoltage relative to the cathode of the device is applied to the gate of theON-FET. As a result this FET turns on, supplying current to the base ofthe bottom transistor of the SCR. Consequently the SCR turns on. To turno. the device, a positive voltage is applied to the gate. This causes theON-FET to turn o., and the OFF-FET to turn on. The result is that thebase-emitter junction of the top transistor of the SCR is shorted, andbecause vBE drops to zero. volt it turns o.. Consequently the regenerationprocess that causes the SCR latching is interrupted and the device turns.

21

The P-MCT is given this name because the cathode is connected to Ptype material. One can also construct an N-MCT, where the cathode isconnected to N type material.

Fig.1.15 Schematic and circuit symbol for the P-MCT.

Chapter 2

2.1 IntroductionThe only way to turn on the diode is when its anode voltage becomeshigher than cathode voltage as explained in the previous chapter. So,there is no control on the conduction time of the diode which is the maindisadvantage of the diode circuits. Despite of this disadvantage, the diodecircuits still in use due to it’s the simplicity, low price, ruggedness,….etc.

Because of their ability to conduct current in one direction, diodes areused in rectifier circuits. The definition of rectification process is “ theprocess of converting the alternating voltages and currents to directcurrents and the device is known as rectifier” It is extensively used incharging batteries; supply DC motors, electrochemical processes andpower supply sections of industrial components.

The most famous diode rectifiers have been analyzed in the followingsections. Circuits and waveforms drawn with the help of PSIM simulationprogram [1].

There are two different types of uncontrolled rectifiers or dioderectifiers, half wave and full wave rectifiers. Full-wave rectifiers hasbetter performance than half wave rectifiers. But the main advantage ofhalf wave rectifier is its need to less number of diodes than full waverectifiers. 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 ParametersIn most rectifier applications, the power input is sine-wave voltage

provided by the electric utility that is converted to a DC voltage and ACcomponents. The AC components are undesirable and must be kept awayfrom the load. Filter circuits or any other harmonic reduction techniqueshould be installed between the electric utility and the rectifier and

23

between the rectifier output and the load that filters out the undesiredcomponent and allows useful components to go through. So, carefulanalysis has to be done before building the rectifier. The analysis requiresdefine 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, rmsIThe output DC power, dcdcdc IVP * (2.1)

The output AC power, rmsrmsac IVP * (2.2)

The effeciency or rectification ratio is defiend asac

dc

P

P (2.3)

The output voltage can be considered as being composed of twocomponents (1) the DC component and (2) the AC component or ripple.The effective (rms) value of the AC component of output voltage isdefined as:-

22dcrmsac VVV (2.4)

The form factor, which is the measure of the shape of output voltage, isdefiend as shown in equation (2.5). Form factor should be greater than orequal to one. The shape of output voltage waveform is neare to be DC asthe form factor tends to unity.

dc

rms

V

VFF (2.5)

The ripple factor which is a measure of the ripple content, is defiend asshown in (2.6). Ripple factor should be greater than or equal to zero. Theshape of output voltage waveform is neare to be DC as the ripple factortends to zero.

11 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac (2.6)

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

SS

dc

IV

PTUF (2.7)

24

Where SV and SI are the rms voltage and rms current of thetransformer secondery respectively.

Total Harmonic Distortion (THD) measures the shape of supplycurrent or voltage. THD should be grearter than or equal to zero. Theshape of supply current or voltage waveform is near to be sinewave asTHD tends to be zero. THD of input current and voltage are defiend asshown in (2.8.a) and (2.8.b) respectively.

121

2

21

21

2

S

S

S

SSi

I

I

I

IITHD (2.8.a)

121

2

21

21

2

S

S

S

SSv

V

V

V

VVTHD (2.8.b)

where 1SI and 1SV are the fundamental component of the input current

and voltage, SI and SV respectively.

Creast Factor CF, which is a measure of the peak input current IS(peak)

as compared to its rms value IS, is defiend as:-

S

peakS

I

ICF )( (2.9)

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

cossVoltampereApparent

PowerR

SS IV

PealPF (2.10)

Where, is the angle between the current and voltage. Definition istrue irrespective for any sinusoidal waveform. But, in case of sinusoidalvoltage (at supply) but non-sinusoidal current, the power factor can becalculated as the following:

Average power is obtained by combining in-phase voltage and currentcomponents of the same frequency.

FaactorntDisplacemeFactorDistortionI

I

IV

IV

IV

PPF

S

S

SSSS*cos

cos1

111 (2.11)

Where 1 is the angle between the fundamental component of currentand supply voltage.Distortion Factor = 1 for sinusoidal operation and displacement factor is ameasure of displacement between tv and ti .

25

2.3 Single-Phase Half-Wave Diode RectifierMost of the power electronic applications operate at a relative high

voltage and in such cases; the voltage drop across the power diode tendsto be small with respect to this high voltage. It is quite often justifiable touse the ideal diode model. An ideal diode has zero conduction dropswhen it is forward-biased ("ON") and has zero current when it is reverse-biased ("OFF"). The explanation and the analysis presented below arebased 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 pureresistive load. Assuming sinusoidal voltage source, VS the diode beingsto conduct when its anode voltage is greater than its cathode voltage as aresult, the load current flows. So, the diode will be in “ON” state inpositive voltage half cycle and in “OFF” state in negative voltage halfcycle. Fig.2.2 shows various current and voltage waveforms of half wavediode rectifier with resistive load. These waveforms show that both theload voltage and current have high ripples. For this reason, single-phasehalf-wave diode rectifier has little practical significance.

The average or DC output voltage can be obtained by considering thewaveforms shown in Fig.2.2 as following:

0

sin2

1 mmdc

VtdtVV (2.12)

Where, mV is the maximum value of supply voltage.Because the load is resistor, the average or DC component of load

current is:

R

V

R

VI mdc

dc (2.13)

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

2sin

21

0

22 mmrms

VtdtVV (2.14)

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

R

V

R

VI mrms

rms 2 (2.15)

26

It is clear that the rms value of the transformer secondary current, SIis the same as that of the load and diode currents

ThenR

VII m

DS 2 (2.15)

Where, DI is the rms value of diode current.

Fig.2.1 Single-phase half-wave diode rectifier with resistive load.

Fig.2.2 Various waveforms for half wave diode rectifier with resistive load.

27

Example 1: The rectifier shown in Fig.2.1 has a pure resistive load of RDetermine (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:

mmmdc

VVtdtVV ))0cos(cos(

2)sin(

2

1

0

Then,R

V

R

VI mdc

dc

2)sin(

2

1

0

2 mmrms

VtVV ,

R

VI m

rms 2 and,

2m

SV

V

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

the load:R

VI m

S 2 Then, the efficiency or rectification ratio is:

rmsrms

dcdc

ac

dc

IV

IV

P

P

*

*%53.40

2*

2

*

R

VVR

VV

mm

mm

(b) 57.12

2m

m

dc

rmsV

V

V

VFF

(c) 211.1157.11 22FFV

VRF

dc

ac

(d) %6.28286.0

22 R

VVR

VV

IV

PTUF

mm

mm

SS

dc

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

(f) Creast Factor CF, 22/

/)(

RV

RV

I

ICF

m

m

S

peakS

28

2.3.2 Half Wave Diode Rectifier With R-L LoadIn case of RL load as shown in Fig.2.3, The voltage source, SV is an

alternating sinusoidal voltage source. If tVv ms sin , sv is positive

when 0 < t < , and sv is negative when < t <2 . When sv startsbecoming positive, the diode starts conducting and the source keeps thediode in conduction till t reaches radians. At that instant defined by

t = radians, the current through the circuit is not zero and there issome energy stored in the inductor. The voltage across an inductor ispositive when the current through it is increasing and it becomes negativewhen the current through it tends to fall. When the voltage across theinductor 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 waveformsshown in Fig.2.4.

When sv changes from a positive to a negative value, the voltageacross the diode changes its direction and there is current through the loadat the instant t = radians and the diode continues to conduct till theenergy stored in the inductor becomes zero. After that, the current tendsto flow in the reverse direction and the diode blocks conduction. Theentire applied voltage now appears across the diode as reverse biasvoltage.

An expression for the current through the diode can be obtained bysolving the deferential equation representing the circuit. It is assumed thatthe current flows for 0 < t < , where > is called the conductionangle). When the diode conducts, the driving function for the differentialequation is the sinusoidal function defining the source voltage. During theperiod defined by < t < 2 , the diode blocks current and acts as anopen switch. For this period, there is no equation defining the behavior ofthe circuit.For 0 < t < , the following differential equation defines the circuit:

ttViRdt

diL m 0),(sin* (2.17)

Divide the above equation by L we get:

ttL

Vi

L

R

dt

di m 0),(sin* (2.18)

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

29

AdttL

Veeti m

dtL

Rdt

L

R

sin*)( (2.19)

Where A is a constant.

Then; AdttL

Veeti m

tL

Rt

L

R

sin*)( (2.20)

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

tL

Rm AetLtR

LwR

Vti 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

Assume wLjRZ

Then 2222 LwRZ ,

cosZR , sinZL andR

Ltan

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

tL

Rm AettZ

Vti cossinsincos)(

Then,t

L

Rm AetZ

Vti sin)( (2.22)

The above equation can be written in the following form:

tansinsin)(

t

mt

L

R

m AetZ

VAet

Z

Vti (2.23)

The value of A can be obtained using the initial condition. Since thediode starts conducting at t = 0 and the current starts building up fromzero, 00i (discontinuous conduction). The value of A is expressed bythe following equation:

sinZ

VA m

Once the value of A is known, the expression for current is known. Afterevaluating A, current can be evaluated at different values of t .

tansinsin)(

t

m etZ

Vti (2.24)

Starting from t = , as t increases, the current would keepdecreasing. For some value of t , say , the current would be zero. If t> , the current would evaluate to a negative value. Since the diodeblocks current in the reverse direction, the diode stops conducting when

t reaches . The value of can be obtained by substituting that

wtti 0)( into (2.24) we get:

0sinsin)( taneZ

Vi m (2.25)

R

wLZ

31

The value of can be obtained from the above equation by using themethods of numerical analysis. Then, an expression for the averageoutput voltage can be obtained. Since the average voltage across theinductor has to be zero, the average voltage across the resistor and theaverage voltage at the cathode of the diode to ground are the same. Thisaverage value can be obtained as shown in (2.26). The rms output voltagein this case is shown in equation (2.27).

)cos1(*2

sin*2

0

mmdc

Vtdt

VV (2.26)

)2sin(1(5.0*2

)sin(*2

1

0

2 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 isshown 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 anotherdiode, marked D2. This diode is called the free-wheeling diode.

Let the source voltage sv be defined as tVm sin which is positive

when t0 radians and it is negative when < t < 2 radians.When sv is positive, diode D1 conducts and the output voltage, ov

become positive. This in turn leads to diode D2 being reverse-biasedduring this period. During < wt < 2 the voltage ov would be negativeif 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

When the current through the inductor tends to fall (when the supplyvoltage become negative), the voltage across the inductor becomenegative and its voltage tends to forward bias diode D2 even when thesource voltage sv is positive, the inductor current would tend to fall if thesource voltage is less than the voltage drop across the load resistor.

During the negative half-cycle of source voltage, diode D1 blocksconduction and diode D2 is forced to conduct. Since diode D2 allows theinductor current circulate through L, R and D2, diode D2 is called thefree-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 dioderectifier with free-wheeling diode.

It can be assumed that the load current flows all the time. In otherwords, the load current is continuous. When diode D1 conducts, thedriving function for the differential equation is the sinusoidal functiondefining the source voltage. During the period defined by < t < 2 ,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 tobe zero volts. For 0 < t < , the differential equation (2.18) applies. Thesolution of this equation will be as obtained before in (2.20) or (2.23).

tansinsin)(

t

m etZ

Vti t0 (2.28)

For the negative half-cycle ( 2t ) of the source voltage D1 isOFF and D2 is ON. Then the driving voltage is set to zero and thefollowing differential equation represents the circuit in this case.

20* tforiRtd

idL (2.29)

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

tan)(

t

eBti (2.30)The constant B can be obtained from the boundary condition where

Bi )( is the starting value of the current in 2t and can beobtained from equation (2.23) by substituting t

Then, BeZ

Vi m )sin(sin)( tan

33

The above value of )(i can be used as initial condition of equation(2.30). Then the load current during 2t is shown in thefollowing equation.

tantansinsin)(

t

m eeZ

Vti for 2t (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

For the period 32 t the value of )2(i from (2.31) can beused as initial condition for that period. The differential equationrepresenting this period is the same as equation (2.28) by replacing t by

2t and the solution is given by equation (2.32). This period( 32 t ) differ than the period wt0 in the way to get theconstant A where in the t0 the initial value was 0)0(i but inthe case of 32 t the initial condition will be )2(i that givenfrom (2.31) and is shown in (2.33).

tan

2

2sin)(

t

m AetZ

Vti for 32 t (2.32)

The value of 2i can be obtained from (2.31) and (2.32) as shownin (2.33) and (2.34) respectively.

tantansinsin)2( eeZ

Vi m (2.33)

AZ

Vi m sin)2( (2.34)

By equating (2.33) and (2.34) the constant A in 32 t can beobtained from the following equation:

sin2Z

ViA m (2.35)

Then, the general solution for the period 32 t is given byequation (2.36):

tan2

sin22sin)(

t

mm eZ

Vit

Z

Vti 32 t (2.36)

Where 2i can be obtained from equation (2.33).

Example 2 A diode circuit shown in Fig.2.3 with R=10 , L=20mH, andVS=220 2 sin314t.

(a) Determine the expression for the current though the load in theperiod 20 t and determine the conduction angle .

(b) If we connect free wheeling diode through the load as shown inFig.2.5 Determine the expression for the current though the loadin the period of 30 t .

35

Solution: (a) For the period of t0 , the expression of the loadcurrent can be obtained from (2.24) as following:

.561.010

10*20*314tantan

311 rad

R

L and 628343.0tan

8084.11)10*20*314(10)( 23222 LRZ

t

t

m

et

etZ

Vti

5915.1

tan

*532.0561.0sin8084.11

2220

sinsin)(

tetti 5915.1*0171.14561.0sin3479.26)(

The value of can be obtained from the above equation by substituting

for 0)(i . Then, 5915.1*0171.14561.0sin3479.260 e

By using the numerical analysis we can get the value of . Thesimplest method is by using the simple iteration technique by assuming

5915.1*0171.14561.0sin3479.26 e and substitute different

values for in the region 2 till we get the minimum value ofthen the corresponding value of is the required value. The narrowintervals mean an accurate values of . The following table shows therelation between and :

1.1 6.495181.12 4.872781.14 3.231861.16 1.57885

1.18 -0.079808

1.2 -1.73761It is clear from the above table that 18.1 rad. The current in

2wt will be zero due to the diode will block the negative currentto flow.(b) In case of free-wheeling diode as shown in Fig.2.5, we have to dividethe operation of this circuit into three parts. The first one when

36

t0 (D1 “ON”, D2 “OFF”), the second case when 2t(D1 “OFF” and D2 “ON”) and the last one when 32 t (D1

“ON”, D2 “OFF”).In the first part ( t0 ) the expression for the load currentcan be obtained as In case (a). Then:

wtetwti 5915.1*0171.14561.0sin3479.26)( for t0

the current at t is starting value for the current in the next part.Then

Aei 1124.14*0171.14561.0sin3479.26)( 5915.1

In the second part 2t , the expression for the load currentcan be obtained from (2.30) as following:

tan)(

t

eBti

where AiB 1124.14)(

Then teti 5915.11124.14)( for ( 2t )

The current at 2t is starting value for the current in the next part.Then

Ai 095103.0)2(In the last part ( 32 t ) the expression for the load currentcan be obtained from (2.36):

tan

2

sin22sin)(

t

mm eZ

Vit

Z

Vti

25915.1532.0*3479.26095103.08442.6sin3479.26)( tetti

25915.11131.148442.6sin3479.26)( tetti for( 32 t )

2.4 Single-Phase Full-Wave Diode RectifierThe full wave diode rectifier can be designed with a center-tapedtransformer as shown in Fig.2.8, where each half of the transformer withits associated diode acts as half wave rectifier or as a bridge dioderectifier as shown in Fig. 2.12. The advantage and disadvantage of center-tap diode rectifier is shown below:

37

AdvantagesThe need for center-tapped transformer is eliminated,The output is twice that of the center tapped circuit for the samesecondary voltage, and,The peak inverse voltage is one half of the center-tap circuit.

DisadvantagesIt 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 lossesare increased.

The following sections explain and analyze these rectifiers.

2.4.1 Center-Tap Diode Rectifier With Resistive LoadIn the center tap full wave rectifier, current flows through the load in

the same direction for both half cycles of input AC voltage. The circuitshown in Fig.2.8 has two diodes D1 and D2 and a center tappedtransformer. The diode D1 is forward bias “ON” and diode D2 is reversebias “OFF” in the positive half cycle of input voltage and current flowsfrom point a to point b. Whereas in the negative half cycle the diode D1is reverse bias “OFF” and diode D2 is forward bias “ON” and againcurrent flows from point a to point b. Hence DC output is obtained acrossthe load.

Fig.2.8 Center-tap diode rectifier with resistive load.

In case of pure resistive load, Fig.2.9 shows various current andvoltage waveform for converter in Fig.2.8. The average and rms outputvoltage and current can be obtained from the waveforms shown in Fig.2.9as shown in the following:

38

mmdc

VtdtVV

2sin

1

0

(2.36)

R

VI m

dc2

(2.37)

2sin

1

0

2 mmrms

VtdtVV (2.38)

R

VI m

rms2

(2.39)

PIV of each diode = mV2 (2.40)

2m

SV

V (2.41)

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

R

VII m

DS 2 (2.41)

Fig.2.9 Various current and voltage waveforms for center-tap diode rectifierwith resistive load.

39

Example 3. The rectifier in Fig.2.8 has a purely resistive load of RDetermine (a) The efficiency, (b) Form factor (c) Ripple factor (d) TUF(e) Peak inverse voltage (PIV) of diode D1 and(f) Crest factor oftransformer secondary current.Solution:- The efficiency or rectification ratio is

%05.81

2*

2

2*

2

*

*

R

VVR

VV

IV

IV

P

P

mm

mm

rmsrms

dcdc

ac

dc

(b) 11.1222

2m

m

dc

rmsV

V

V

VFF

(c) 483.0111.11 22FFV

VRF

dc

ac

(d) 5732.0

222

22

2R

VVR

VV

IV

PTUF

mm

mm

SS

dc

(e) The PIV is mV2

(f) Creast Factor of secondary current, 2

2

)(

R

VR

V

I

ICF

m

m

S

peakS

2.4.2 Center-Tap Diode Rectifier With R-L LoadCenter-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 D2conducts in negative half cycle. So, the deferential equation defines thecircuit is shown in (2.43).

)sin(* tViRtd

idL m (2.43)

The solution of the above equation can be obtained as obtained beforein (2.24)

40

Fig.2.10 Center-tap diode rectifier with R-L load

Fig.2.11 Various current and voltage waveform for Center-tap diode rectifierwith R-L load

tansinsin)(

t

m etZ

Vti for t0 (2.44)

In the second half cycle the same differential equation (2.43) and thesolution of this equation will be as obtained before in (2.22)

41

tansin)(

t

m AetZ

Vti (2.45)

The value of constant A can be obtained from initial condition. If weassume that i( )=i(2 )=i(3 )=……..=Io (2.46)Then the value of oI can be obtained from (2.44) by letting t

tansinsin)( eZ

ViI m

o (2.47)

Then use the value of oI as initial condition for equation (2.45). So wecan obtain the value of constant A as following:

tansin)( AeZ

VIi m

o

Then; sinZ

VIA m

o (2.48)

Substitute (2.48) into (2.45) we get:

tansinsin)(

t

mo

m eZ

VIt

Z

Vti , then,

tantansinsin)(

t

o

t

m eIetZ

Vti (for 2t ) (2.49)

In the next half cycle 32 t the current will be same asobtained in (2.49) but we have to take the time shift into account wherethe new equation will be as shown in the following:

tan

2

tan

2

sin2sin)(

t

o

t

m eIewtZ

Vti (for 32 t )(2.50)

2.4.3 Single-Phase Full Bridge Diode Rectifier With Resistive LoadAnother 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 bridgediode rectifier. It is easy to see the operation of these four diodes. Thecurrent flows through diodes D1 and D2 during the positive half cycle ofinput voltage (D3 and D4 are “OFF”). During the negative one, diodesD3 and D4 conduct (D1 and D2 are “OFF”).

42

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, thecurrent moves from positive point of the supply voltage across D1 to thepoint a of the load then from point b to the negative marked point of thesupply voltage through diode D2. In the negative voltage half cycle, thesupply voltage forces the diodes D1 and D2 to be "OFF". In same time itforces diodes D3 and D4 to be "ON". So, the current moves fromnegative marked point of the supply voltage across D3 to the point a ofthe load then from point b to the positive marked point of the supplyvoltage through diode D4. So, it is clear that the load currents movesfrom point a to point b in both positive and negative half cycles of supplyvoltage. So, a DC output current can be obtained at the load in bothpositive and negative halves cycles of the supply voltage. The completewaveforms 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-phasediode rectifier.

43

Example 4 The rectifier shown in Fig.2.12 has a purely resistive load ofR=15 and, VS=300 sin 314 t and unity transformer ratio. Determine (a)The efficiency, (b) Form factor, (c) Ripple factor, (d) TUF, (e) The peakinverse voltage, (PIV) of each diode, (f) Crest factor of input current, and,(g) Input power factor.Solution: 300mV V

VV

tdtVV mmdc 956.190

2sin

1

0

, AR

VI m

dc 7324.122

VV

tdtVV mmrms 132.212

2sin

12/1

0

2 , AR

VI m

rms 142.142

(a) %06.81rmsrms

dcdc

ac

dc

IV

IV

P

P

(b) 11.1dc

rms

V

VFF

(c) 482.011 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac

(d) %81142.14*132.212

7324.12*986.190

SS

dc

IV

PTUF

(e) The PIV= mV =300V

(f) 414.1142.14

15/300)(

S

peakS

I

ICF

(g) Input power factor = 1*Re 2

SS

rms

IV

RI

PowerApperant

Poweral

2.4.4 Full Bridge Single-phase Diode Rectifier with DC Load CurrentThe 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 isassumed here that the source inductances is negligible. In this case, thecircuit works as explained before in resistive load but the currentwaveform in the supply will be as shown in Fig.2.15.The rms value of the input current is oS II

44

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-phasediode rectifier with DC load current.

The supply current in case of pure DC load current is shown inFig.2.15, as we see it is odd function, then na coefficients of Fourier

series equal zero, 0na , and

.............,5,3,14

cos0cos2

cos2

sin*2

00

nforn

In

n

I

tnn

ItdtnIb

oo

oon

(2.51)

Then from Fourier series concepts we can say:

)..........9sin91

7sin71

5sin51

3sin31

(sin*4

)( tttttI

ti o (2.52)

45

%4615

1

13

1

11

1

9

1

7

1

5

1

3

1))((

2222222

tITHD s

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

From (2.52) we can obtain the value of is2

41

oS

II

%34.4814

21

2

41))((

2

2

2

1 o

o

S

Ss I

I

I

ItITHD

Example 5 solve Example 4 if the load is 30 A pure DCSolution: From example 4 Vdc= 190.986 V, Vrms=212.132 V

AIdc 30 and rmsI = 30 A

(a) %90rmsrms

dcdc

ac

dc

IV

IV

P

P

(b) 11.1dc

rms

V

VFF

(c) 482.011 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac

(d) %9030*132.212

30*986.190

SS

dc

IV

PTUF

(e) The PIV=Vm=300V

(f) 13030)(

S

peakS

I

ICF

(g) AI

I oS 01.27

2

30*4

2

41

Input Power factor=PowerApperant

PoweralRe

LagI

I

IV

IV

S

S

SS

SS 9.01*30

01.27cos*cos* 11

46

2.4.5 Effect Of LS On Current Commutation Of Single-Phase DiodeBridge Rectifier.

Fig.2.15 Shows the single-phase diode bridge rectifier with sourceinductance. 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 calledcommutation time. And this process is called current commutationprocess. Various voltage and current waveforms of single-phase diodebridge 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 bridgerectifier with source inductance.

47

Let us study the commutation time starts at t=10 ms as indicated inFig.2.16. At this time the supply voltage starts to be negative, so diodesD1 and D2 have to switch OFF and diodes D3 and D4 have to switch ONas explained in the previous case without source inductance. But due tothe source inductance it will prevent that to happen instantaneously. So, itwill take time t to completely turn OFF D1 and D2 and to make D3 andD4 carry the entire load current ( oI ). Also in the time t the supplycurrent 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 t .

Fig.2.17 The equivalent circuit of the diode bridge at commutation time t .

From Fig.2.17 we can get the following equations

0dt

diLV S

sS (2.53)

Multiply the above equation by td then,

SsS diLtdV (2.54)Integrate both sides of the above equation during the commutation

period ( t sec or u rad.) we get the following:

SsS diLtdV

o

o

I

I

Ss

u

m diLtdtV sin (2.55)

Then; osm ILuV 2coscosThen; osm ILuV 2cos1

48

Then;m

os

V

ILu

21cos

Then;m

os

V

ILu

21cos 1 (2.56)

Andm

os

V

ILut

21cos

1 1 (2.57)

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

dt

diLv S

srd (2.58)

Then oS

I

I

SS

u

rd ILdiLtdvo

o

2 (2.59)

u

rd tdv is the reduction area in one commutation period t . But we

have two commutation periods t in one period of supply voltage. So the

total reduction per period is: oS

u

rd ILtdv 42 (2.60)

To obtain the average reduction in DC output voltage rdV due tosource inductance we have to divide the above equation by the periodtime 2 . Then;

oSoS

rd ILfIL

V 42

4 (2.61)

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

osm

rdceinducsourcewithoutdcactualdc IfLV

VVV 42

tan (2.62)

To obtain the rms value and Fourier transform of the supply current itis better to move the vertical axis to make the waveform odd or even thiswill greatly simplfy the analysis. So, it is better to move the vertical axisof supply current by 2/u as shown in Fig.2.18. Moveing the vertical axiswill not change the last results. If you did not bleave me keep going in theanalysis without moveing the axis.

49

Fig. 2.18 The old axis and new axis for supply currents.

Fig.2.19 shows a symple drawing for the supply current. This drawinghelp us in getting the rms valuof the supply current. It is clear from thewaveform of supply current shown in Fig.2.19 that we obtain the rmsvalue for only a quarter of the waveform because all for quarter will bethe same when we squaret the waveform as shown in the followingequation:

]2

[2

2/

0

2

2/

22u

uo

os tdItdt

u

II (2.63)

Then;32

2

2283

42 23

2

2 uIuu

u

II oo

s (2.64)

2

u

oI

oI 2

u2u

2

u

22

u

2

usI

2

Fig.2.19 Supply current waveform

50

To obtain the Fourier transform for the supply current waveform youcan go with the classic fourier technique. But there is a nice and easymethod to obtain Fourier transform of such complcated waveform knownas jump technique [ ]. In this technique we have to draw the wave formand its drevatives till the last drivative values all zeros. Then record thejump value and its place for each drivative in a table like the table shownbelow. Then; substitute the table values in (2.65) as following:

2

u

oI

oI2

u

u

Io2

u

Io2

sI

2u

2

u

22

u

2

2

u

2

u

2u

2

u

22

u

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

2u

sI 0 0 0 0

sIu

Io2

u

Io2

u

Io2

u

Io2

51

It is an odd function, then 0no aam

sss

m

sssn tnJ

ntnJ

nb

11

sin1

cos1

(2.65)

2sin

2sin

2sin

2sin

2*

11 un

un

un

un

u

I

nnb o

n

2sin*

82

nu

un

Ib o

n (2.66)

2sin*

81

u

u

Ib o (2.67)

Then;2

sin*2

81

u

u

II o

S (2.68)

32

sin2

32

2cos

2sin4

2cos

32

2

2sin*

2

8

2cos*

21

uu

u

uu

uu

u

uI

u

u

I

u

I

Ipf

o

o

S

S

(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 IfLV

VVV 42

tan

VVactualdc 9703200*005.0*50*4

15556*2

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

52

.285.015556

200*005.0*50**2*21cos

21cos 11 rad

V

ILu

m

os

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

917.0

3285.

2285.0

285.0sin*2

32

sin*2

2cos*1

uu

uu

I

Ipf

S

S

AuI

I oS 85.193

3

285.0

2

200*2

32

2 22

Au

u

II o

S 46.1792285.0

sin*285.0*2

200*82

sin*2

81

%84.40146.17985.193

122

1S

Si I

ITHD

2.5 Three Phase Diode Rectifiers2.5.1 Three-Phase Half Wave Rectifier

Fig.2.21 shows a half wave three-phase diode rectifier circuit withdelta star three-phase transformer. In this circuit, the diode with highestpotential with respect to the neutral of the transformer conducts. As thepotential of another diode becomes the highest, load current is transferredto that diode, and the previously conduct diode is reverse biased “OFFcase”.

Fig.2.21 Half wave three-phase diode rectifier circuit with delta star three-phasetransformer.

53

For the rectifier shown in Fig.2.21 the load voltage, primary diodecurrents and its FFT components are shown in Fig.2.22, Fig.2.23 andFig.2.24 respectively.

Fig.2.22 Secondary and load voltages of half wave three-phase diode rectifier.

Fig.2.23 Primary and diode currents.

6 6

5

54

Fig.2.24 FFT components of primary and diode currents.

By considering the interval from6

to6

5 in the output voltage we can

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

mm

mdc VV

tdtVV 827.02

33sin

2

36/5

6/

(2.70)

R

V

R

VI mm

dc*827.0

**2

33 (2.71)

mmmrms VVtdtVV 8407.08

3*3

2

1sin

2

36/5

6/

2 (2.72)

R

VI m

rms8407.0

(2.73)

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

R

V

R

VII mm

Sr 4854.03

08407 (2.74)

Note that the rms value of diode current has been obtained from therms value of load current divided by 3 because the diode current hasone third pulse of similar three pulses in load current.ThePIV of the diodes is mLL VV 32 (2.75)

Primary current

Diode current

55

Example 7 The rectifier in Fig.2.21 is operated from 460 V 50 Hz supplyat secondary side and the load resistance is R=20 . If the sourceinductance is negligible, determine (a) Rectification efficiency, (b) Formfactor (c) Ripple factor (d) Transformer utilization factor, (e) Peakinverse 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 VV

V 827.02

33,

R

V

R

VI mm

dc0827

2

33

mrms VV 8407.0

R

VI m

rms8407.0

%767.96rmsrms

dcdc

ac

dc

IV

IV

P

P

(b) %657.101dc

rms

V

VFF

(c) %28.1811 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac

(d)R

VII m

rmsS3

8407.0

31

%424.66

3

8407.0*2/*3

/)827.0(

*3

2

R

VV

RV

IV

PTUF

mm

m

SS

dc

(e) The PIV= 3 Vm=650.54V

(f) 06.2

3

8407.0/)(

R

VRV

I

ICF

m

m

S

peakS

56

2.5.2 Three-Phase Half Wave Rectifier With DC Load Current andzero source inductanceIn case of pure DC load current as shown in Fig.2.25, the diode currentand 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 isbetter 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 thenwe can deal with the secondary current as even functions. Then, 0nb ofsecondary current. Values of na can be calculated as following:

32

13/

3/0

oo

ItdIa (2.76)

harmonicstrepleanallfor

nforn

I

nforn

I

tnn

I

dwttnIa

o

o

o

on

0

17,16,11,10,5,43*

,....14,13,8,7,2,13*

sin

cos*1

3/3/

3/

3/

(2.77)

...8sin8

17sin

7

15sin

5

14sin

4

12sin

2

1sin

3

3)( tttttt

IItI OO

s(2.78)

57

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

58

%24.1090924.119

*21

2

3

3/1))((

2

2

2

1 O

o

S

Ss

I

I

I

ItITHD

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

harmonicstrepleanallfor

nforn

I

tnn

ItdtnIb

o

oon

0

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

cos2

sin*2 3/2

0

3/2

0

(2.79)

...8sin8

17sin

7

15sin

5

14sin

4

12sin

2

1sin

3)( tttttt

Iti O

P(2.80)

The rms value of oP II3

2 (2.81)

%983.67133

21

2

33

2

1))((2

2

2

1 O

o

P

PP I

I

I

ItITHD (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

VVV

V mm

dc 613.310827.02

33, AIdc 100

VVV mrms 759.3158407.0 , AIrms 100

%37.98100*759.315

100*613.310

rmsrms

dcdc

ac

dc

IV

IV

P

P

(b) %657.101dc

rms

V

VFF

(c) %28.1811 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac

59

(d) AII rmsS 735.57100*31

31

%52.67735.57*2/*3

100*613.310

*3 mSS

dc

VIV

PTUF

(e) The PIV= 3 Vm=650.54V

(f) 732.1735.57

100)(

S

peakS

I

ICF

2.5.3 Three-Phase Half Wave Rectifier With Source InductanceThe source inductance in three-phase half wave diode rectifier Fig.2.27

will change the shape of the output voltage than the ideal case (withoutsource inductance) as shown in Fig.2.28. The DC component of theoutput voltage is reduced due to the voltage drop on the sourceinductance. To calculate this reduction we have to discuss Fig.2.27 withreference to Fig.2.28. As we see in Fig.2.28 when the voltage bv is goingto be greater than the voltage av at time t (at the arrow in Fig.2.28) thediode D1 will try to turn off, in the same time the diode D2 will try toturn on but the source inductance will slow down this process and makesit done in time t (overlap time or commutation time). The overlap timewill take time t to completely turn OFF D1 and to make D2 carry theentire load current ( oI ). Also in the time t the current in bL will changefrom zero to oI and the current in aL will change from oI to zero. This isvery clear from Fig.2.28. Fig.2.29 shows the equivalent circuit of threephase half wave diode bridge in commutation period t .

Fig.2.27 Three-phase half wave rectifier with load and source inductance.

60

Fig.2.28 Supply current and output voltage for three-phase half waverectifier with pure DC load and source inductance.

Fig.2.29 The equivalent circuit for three-phase half wave diode rectifier incommutation period.

From Fig.2.29 we can get the following equations

01dc

Daa V

dt

diLv (2.83)

02dc

Dbb V

dt

diLv (2.84)

subtract (2.84) from(2.83) we get:

61

012

dt

di

dt

diLvv DD

ba

Multiply the above equation by td the following equation can beobtained: 012 DDba didiLtdvv

substitute the voltage waveforms of av and bv into the above equation

we get: 2132

sinsin DDmm didiLtdtVtV

Then; 216sin3 DDm didiLtdtV

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

o

o

I

D

I

D

u

m didiLtdtV0

2

0

1

6

5

6

56

sin3

Then; om LIuV 266

5cos

66

5cos3

Then; om LIuV 2coscos3

Then; om LIuV 2cos13

Then;m

o

V

LIu

3

2cos1

Then;m

o

V

LIu

3

21cos

Thenm

o

V

LIu

3

21cos 1 (2.85)

m

o

V

LIut

3

21cos

1 1 (2.86)

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

dt

diLv D

rd

62

Then, o

I

D

u

rd LIdiLtdvo

0

6

5

6

5

(2.87)

u

rd tdv6

5

6

5

is the reduction area in one commutation period t . But,

we have three commutation periods, t in one period. So, the totalreduction per period is:

o

u

rd LItdv 3*36

5

6

5

To obtain the average reduction in DC output voltage rdV due to

source inductance we have to divide the total reduction per period by 2as following:

oo

rd ILfLI

V 32

3 (2.88)

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

oceinduc

sourcewithout

dcActualdc ILfVV 3tan

(2.89)

om

Actualdc ILfV

V 32

33 (2.90)

Example 9 Three-phase half-wave diode rectifier connected to 66 kV, 50Hz , 5mH supply to feed a DC load with 500 A DC, fined the average DCoutput voltage.

Solution: Vvm 538892*3

66000

(i) oceinduc

sourcewithout

dcActualdc ILfVV 3tan

VILfV

V om

Actualdc 44190500*005.0*50*32

53889*3*33

2

33

63

2.5 Three-Phase Full Wave Diode RectifierThe three phase bridge rectifier is very common in high power

applications and is shown in Fig.2.30. It can work with or withouttransformer and gives six-pulse ripples on the output voltage. The diodesare numbered in order of conduction sequences and each one conduct for120 degrees. These conduction sequence for diodes is 12, 23, 34, 45, 56,and, 61. The pair of diodes which are connected between that pair ofsupply lines having the highest amount of instantaneous line to linevoltage will conduct. Also, we can say that, the highest positive voltageof any phase the upper diode connected to that phase conduct and thehighest negative voltage of any phase the lower diode connected to thatphase conduct.

2.5.1 Three-Phase Full Wave Rectifier With Resistive LoadIn 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 waveformsfor 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 andprimary currents and PIV of D1. Fig.2.34 shows Fourier Transformcomponents of output DC voltage, diode current secondary current andPrimary current respectively.

For the rectifier shown in Fig.2.30 the waveforms is as shown inFig.2.31. The average output voltage is :-

LLmLLm

mdc VVVV

tdtVV 3505.1654.12333

sin33

3/2

3/

(2.91)

R

V

R

V

R

V

R

VI LLLLmm

dc3505.123654.133 (2.92)

LLmmmrms VVVtdtVV 3516.16554.14

3*9

2

3sin3

33/2

3/

2 (2.93)

R

VI m

rms6554.1 (2.94)

Then the diode rms current is

R

V

R

VI mm

r 9667.03

6554.1 (2.95)

R

VI m

S 29667.0 (2.96)

64

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 voltagesand output DC load voltages.

65

Fig.2.32 Diode currents.

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

66

Fig.2.34 Fourier Transform components of output DC voltage, diode currentsecondary 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 50Hz supply and the load resistance is R=20 . If the source inductance isnegligible, determine (a) The efficiency, (b) Form factor (c) Ripple factor(d) Transformer utilization factor, (e) Peak inverse voltage (PIV) of eachdiode and (f) Crest factor of input current.

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

460

VVV

V mm

dc 226.621654.133

,

AR

V

R

VI mm

dc 0613.31654.133

VVVV mmrms 752.6216554.14

3*923

,

AR

VI m

rms 0876.316554.1

67

%83.99rmsrms

dcdc

ac

dc

IV

IV

P

P

(b) %08.100dc

rms

V

VFF

(c) %411 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac

(d)R

V

R

VII mm

rmsS 352.16554.1

*8165.032

%42.95352.1*2/*3

/)654.1(

*3

2

R

VV

RV

IV

PTUF

mm

m

SS

dc

(e) The PIV= 3 Vm=650.54V

(f) 281.1352.1

/3)(

R

VRV

I

ICF

m

m

S

peakS

2.5.2 Three-Phase Full Wave Rectifier With DC Load CurrentThe supply current in case of pure DC load current is shown in

Fig.2.35. Fast Fourier Transform of Secondary and primary currentsrespectively is shown in Fig2.36.As we see it is odd function, then an=0, and

....,.........15,14,12,10,9,8,6,4,3,2,0

.....),........3(13

2),3(

11

2

)3(7

2),3(

5

2,3

2

cos2

sin*2

1311

751

6/56/

6/5

6/

nforb

Ib

Ib

Ib

Ib

Ib

tnn

I

tdtnIb

n

oo

ooo

o

on

(2.97)

ttttvtI

tI os 13sin

13

111sin

11

17sin

7

15sin

5

1sin

32)( (2.98)

68

%31

25

1

23

1

19

1

17

1

13

1

11

1

7

1

5

1))((

22222222

tITHD s

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

oS II32

, oS II3*2

1

%01.311/3*2

3/21))((

2

2

1S

Ss I

ItITHD

Fig.2.35 The D1 and D2 currents, secondary and primary currents.

69

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, 0na , and

....,.........15,14,12,10,9,8,6,4,3,2,0

......,.........13,11,7,5,13*2

3

2cos

3coscos1

2

sin*sin*2sin*2

1

1

3/21

3/2

3/1

3/

01

nforb

nforn

Ib

nnnn

I

tdtnItdtnItdtnIb

n

n

n

(2.99)

tttttI

tIP 13sin13

111sin

11

17sin

7

15sin

5

1sin

3*2)( 1 (2.100)

%30

25

1

23

1

19

1

17

1

13

1

11

1

7

1

5

1))((

22222222

tITHD P

Power Factor =S

S

S

S

I

I

I

I 11 )0cos(*

70

2.5.4 Three-Phase Full Wave Diode Rectifier With Source InductanceThe source inductance in three-phase diode bridge rectifier Fig.2.37

will change the shape of the output voltage than the ideal case (withoutsource inductance) as shown in Fig.2.31. The DC component of theoutput voltage is reduced. Fig.2.38 shows The output DC voltage ofthree-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 sourceinductance

Let us study the commutation time starts at t=5ms as shown inFig.2.39. At this time cV starts to be more negative than bV so diode D6has to switch OFF and D2 has to switch ON. But due to the sourceinductance will prevent that to happen instantaneously. So it will taketime t to completely turn OFF D6 and to make D2 carry all the load

71

current ( oI ). Also in the time t the current in bL will change from oI

to zero and the current in cL will change from zero to oI . This is veryclear from Fig.2.39. The equivalent circuit of the three phase diode bridgeat commutation time t 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 commutationtime t at mst 5

72

Fig.2.41 Simple circuit of the equivalent circuit of the three phase diode bridgeat commutation time t at mst 5

From Fig.2.41 we can get the following defferntial equations:

061b

Dbdc

Daa V

dt

diLV

dt

diLV (2.101)

021c

Dcdc

Daa V

dt

diLV

dt

diLV (2.102)

Note that, during the time t , 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 thefollowing differential equation:

dt

diLVV

dt

diLVV D

ccaD

bba26 (2.105)

026

dt

diL

dt

diLVV D

cD

bcb (2.106)

The above equation can be written in the following manner:026 DcDbcb diLdiLdtVV (2.107)

026 DcDbcb diLdiLtdVV (2.108)Integrate the above equation during the time t 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

Dc

I

Db

u

cb diLdiLtdVV

73

03

2sin

3

2sin

2/

2/

ocob

u

mm ILILtdtVtV

assume Scb LLL

oS

u

m ILttV 23

2cos

3

2cos

2/

2/

oS

m

IL

uuV

2

3

2

2cos

3

2

2cos

3

2

2cos

3

2

2cos

oSm ILuuV 26

7cos

6cos

6

7cos

6cos

m

oS

V

IL

uuuu

2

2

3

2

3

6

7sinsin

6

7coscos

6sinsin

6coscos

m

oS

V

ILuuuu

23sin5.0cos

2

3sin5.0cos

2

3

m

oS

V

ILu

2cos13

LL

oS

LL

o

m

o

V

IL

V

LI

V

LIu

21

2

21

3

21cos

LL

oS

V

ILu

21cos 1 (2.109)

LL

oS

V

ILut

21cos

1 1 (2.110)

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

dt

diLv D

Srd (2.111)

Multiply (2.111) by td and integrate both sides of the resultantequation we get:

74

oS

I

D

u

rd ILLditdvo

0

2

2

(2.112)

u

rd tdv2

2

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

have six commutation periods t in one period so the total reduction perperiod is:

oS

u

rd ILtdv 662

2

(2.113)

To obtain the average reduction in DC output voltage rdV due tosource inductance we have to divide by the period time 2 . Then,

oo

rd fLILI

V 62

6 (2.114)

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

dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan

(2.115)

Fig.2.42 shows the utility line current with some detailes to help us tocalculate its rms value easly.

oI

oI 3

2

sI

u3

2

26

2 u

Fig.2.42 The utility line current

75

uu

u

do

s tdItdtu

II

0

232

22

uu

uu

Io

233

12 32

2

Then63

2 2 uII o

S (2.116)

Fig.2.43 shows the utility line currents and its first derivative that helpus to obtain the Fourier transform of supply current easily. From Fig.2.43we can fill Table(2.2) as explained before when we study Table (2.1).

26

u

oI

oI26

5 u

267 u 26

11 u

sI

u

Io

u

Io

sI

26u

265 u

26

7 u

2611 u

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

26u

265 u

265 u

267 u

267 u

2611 u

26

11 u

sI 0 0 0 0 0 0 0 0

sIu

Io

u

Io

u

Io

u

Io

u

Io

u

Io

u

Io

u

Io

76

It is an odd function, then 0no aam

sss

m

sssn tnJ

ntnJ

nb

11

sin1

cos1

(2.117)

26

11sin

26

11sin

26

7sin

26

7sin

26

5sin

26

5sin

26sin

26sin*

11

un

un

un

un

un

un

un

un

u

I

nnb o

n

6

11cos

6

7cos

6

5cos

6cos

2sin*

22

nnnnnu

un

Ib o

n (2.118)

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

tu

tu

tu

tu

tu

uti

13sin2

13sin

13

111sin

211

sin11

1

7sin2

7sin

7

15sin

25

sin5

1sin

2sin

34

22

22 (2.119)

Then;2

sin62

1u

u

II o

S 2.120)

The power factor can be calculated from the following equation:

2cos

63

2

2sin

62

2cos

21 u

uI

u

u

I

u

I

Ipf

o

o

S

S

Then;

63

sin*3

uu

upf (2.121)

Example 11 Three phase diode bridge rectifier connected to tree phase33kV, 50 Hz supply has 8 mH source inductance to feed 300A pure DCload current Find;

(i) commutation time and commutation angle.(ii) DC output voltage.(iii) Power factor.(iv) Total harmonic distortion of line current.

77

Solution: (i) By substituting for 50**2 , AId 300 , HL 008.0 ,

VVLL 33000 in (2.109), then oradu 61.14.2549.0

Then, msu

t 811.0 .

(ii) The the actual DC voltage can be obtained from (2.115) as following:

dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan

VVdcactual 43830300*008.*50*633000*35.1(iii) the power factor can be obtained from (2.121) then

9644.0

6

2549.0

3*2549.0

2549.0sin3

63

sin*3

uu

upf Lagging

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

AuI

I ds 929.239

62549.0

3*

300*263

2 22

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

Au

u

II o

S 28.2332

2549.0sin*

2*2549.0*

300*3432*

2sin

2

341

9644.02

2549.0cos*

929.23928.233

2cos*1 u

I

Ipf

s

S Lagging.

%05.24128.233

929.2391

22

1S

Si I

ITHD

2.7 Multi-pulse Diode RectifierTwelve-pulse bridge connection is the most widely used in high

number of pulses operation. Twelve-pulse technique is using in mostHVDC schemes and in very large variable speed drives for DC and ACmotors 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 thiswhich gives a 30o-phase shift will form a twelve-pulse converter. In thiskind of converters, each converter will generate all kind of harmonics

78

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 ofhigher order components. An analysis of the waveform shows that the ACline current can be described by (2.83).

tttttIti dP 25cos25

123cos

23

113cos

13

15cos

11

1cos

32)( (2.83)

%5.13

35

1

35

1

25

1

23

1

13

1

11

1))((

222222

tITHD P

As shown in (10) the THDi is about 13.5%. The waveform of utilityline current is shown in Fig.2.34. Higher pulse number like 18-pulse or24-pulse reduce the THD more and more but its applications very rare. Inall kind of higher pulse number the converter needs special transformer.Sometimes the transformers required are complex, expensive and it willnot be ready available from manufacturer. It is more economic to connectthe small WTG to utility grid without isolation transformer. The mainidea here is to use a six-pulse bridge directly to electric utility withouttransformer. But the THD must be lower than the IEEE-519 1992 limits.

2N :1

32 N :1

Va

c

b

a

bc

a

b

c

Fig.2.33 Twelve-pulse converter arrangement

79

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

80

Problems1- Single phase half-wave diode rectifier is connected to 220 V, 50 Hz

supply to feed 5 pure resistor. Draw load voltage and current anddiode voltage drop waveforms along with supply voltage. Then,calculate (a) The rectfication effeciency. (b) Ripple factor of loadvoltage. (c) Transformer Utilization Factor (TUF) (d) Peak InverseVoltage (PIV) of the diode. (e) Crest factor of supply current.

2- The load of the rectifier shown in problem 1 is become 5 pureresistor and 10 mH inductor. Draw the resistor, inductor voltagedrops, and, load current along with supply voltage. Then, find anexpression for the load current and calculate the conduction angle,

. 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 anthe 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 currentduring 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 positivepolarity terminal are as follows:

tVtVtVVtv d 3cos2sin2cos2 311

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.

81

6- Center tap diode rectifier is connected to 220 V, 50 Hz supply viaunity turns ratio center-tap transformer to feed 5 resistor load.Draw load voltage and currents and diode currents waveforms alongwith 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 factorof supply current.

7- Single phase diode bridge rectifier is connected to 220 V, 50 Hzsupply to feed 5 resistor. Draw the load voltage, diodes currentsand calculate (a) The rectfication effeciency. (b) Ripple factor of loadvoltage. (c) Transformer Utilization Factor (TUF) (d) Peak InverseVoltage (PIV) of the diode. (e) Crest factor of supply current.

8- If the load of rectifier shown in problem 7 is changed to be 5resistor in series with 10mH inductor. Calculate and draw the loadcurrent during the first two periods of supply voltages waveform.

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

10- If the load of problem 7 is changed to be 45 A pure DC. Draw diodediodes currents and supply currents along with supply voltage. Then,calculate (a) The rectfication effeciency. (b) Ripple factor of loadvoltage. (c) Transformer Utilization Factor (TUF) (d) Peak InverseVoltage (PIV) of the diode. (e) Crest factor of supply current. (f)input power factor.

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

12- Three-phase half-wave diode rectifier is connected to 380 V, 50Hzsupply via 380/460 V delta/way transformer to feed the load with 45A DC current. Assuming ideal transformer and zero sourceinductance. Then, draw the output voltage, secondary and primarycurrents along with supply voltage. Then, calculate (a) Rectficationeffeciency. (b) Crest factor of secondary current. (c) TransformerUtilization Factor (TUF). (d) THD of primary current. (e) Inputpower factor.

82

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 10 resistor. Draw the output voltage, diode currentsand supply current of phase a. Then, calculate: (a) The rectficationeffeciency. (b) Ripple factor of load voltage. (c) TransformerUtilization 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 findTHD of supply current and input power factor.

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

17- Single phase diode bridge rectifier is connected to square waveformwith amplitude of 200V, 50 Hz. The supply has 4 mH sourceinductance. The load connected to the rectifier is 45 A pure DCcurrent. Draw, output voltage, diode currents and supply currentalong 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, 1SL

mH and VVd 160 . The input voltage sv has the pulse waveform

shown in the following figure. Plot si and di waveforms and find theaverage value of dI .

+

-SV

dVSi

di

Hzf 50

o60 o60o60o120

o120o120

V200

t

Chapter 3

3.1 IntroductionThe controlled rectifier circuit is divided into three main circuits:-

(1) Power CircuitThis is the circuit contains voltage source, load and switches asdiodes, thyristors or IGBTs.

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

(3) Triggering circuitThis circuit lies between the control circuit and power thyristors.Sometimes this circuit called switch drivers circuit. This circuitcontains buffers, opt coupler or pulse transformers. The mainpurpose of this circuit is to separate between the power circuit andcontrol 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 circuitvoltage, which can be neglected. When the anode voltage is greater thanthe cathode voltage and there is positive pulse applied to its gate, thethyristor will turn on. The thyristor can be naturally turned off if thevoltage of its anode becomes less than its cathode voltage or it can beturned off by using commutation circuit. If the voltage of its anode isbecome positive again with respect to its cathode voltage the thyristorwill not turn on again until gets a triggering pulse to its gate.

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

1- Natural CommutationIf the input voltage is AC, the thyristor current passes through a naturalzero, and a reverse voltage appear across the thyristor, which in turnautomatically turned off the device due to the natural behavior of ACvoltage source. This is known as natural commutation or linecommutation. This type of commutation is applied in AC voltage

81

controller rectifiers and cycloconverters. In case of DC circuits, thistechnique does not work as the DC current is unidirectional and does notchange its direction. Thus the reverse polarity voltage does not appearacross the thyristor. The following technique work with DC circuits:

2- Forced CommutationIn DC thyristor circuits, if the input voltage is DC, the forward current ofthe thyristor is forced to zero by an additional circuit called commutationcircuit to turn off the thyristor. This technique is called forcedcommutation. Normally this method for turning off the thyristor isapplied in choppers.

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

3.2 Half Wave Single Phase Controlled Rectifier3.2.1 Half Wave Single Phase Controlled Rectifier With ResistiveLoadThe circuit with single SCR is similar to the single diode circuit, thedifference being that an SCR is used in place of the diode. Most of thepower electronic applications operate at a relative high voltage and insuch cases; the voltage drop across the SCR tends to be small. It is quiteoften justifiable to assume that the conduction drop across the SCR iszero when the circuit is analyzed. It is also justifiable to assume that thecurrent through the SCR is zero when it is not conducting. It is knownthat the SCR can block conduction in either direction. The explanationand the analysis presented below are based on the ideal SCR model. Allsimulation carried out by using PSIM computer simulation program [ ].

A circuit with a single SCR and resistive load is shown in Fig.3.1. Thesource vs is an alternating sinusoidal source. If tVv ms sin , vs is

positive when 0 < t < , and vs is negative when < t <2 . When vs

starts become positive, the SCR is forward-biased but remains in theblocking state till it is triggered. If the SCR is triggered at t = thenis called the firing angle. When the SCR is triggered in the forward-biasstate, it starts conducting and the positive source keeps the SCR inconduction till t reaches radians. At that instant, the current throughthe circuit is zero. After that the current tends to flow in the reverse

82

direction and the SCR blocks conduction. The entire applied voltage nowappears across the SCR. Various voltages and currents waveforms of thehalf-wave controlled rectifier with resistive load are shown in Fig.3.2 for

=40o. FFT components for load voltage and current of half wave singlephase controlled rectifier with resistive load at =40o are shown inFig.3.3. It is clear from Fig.3.3 that the supply current containes DCcomponent and all other harmonic components which makes the supplycurrent highly distorted. For this reason, this converter does not haveacceptable 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-phasecontrolled rectifier with resistive load at =40o.

83

Fig.3.3 FFT components for load voltage and current of half wave single phasecontrolled rectifier with resistive load at =40o.

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

)cos1(2

))cos(cos(2

)sin(2

1 mmmdc

VVtdtVV (3.1)

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

mdm

VV (3.2)

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

)cos1(5.0dm

dcn V

VV (3.3

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

2

)2sin(1

2)sin(

2

1 2 mmrms

VtdtVV (3.3)

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

R

VI rms

s (3.4)

84

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

(a) Vdm is the maximum output voltage and can be acheaved when0 , The normalized output voltage is shown in equation (3.3) which is

required to be 70%. Then,

7.0)cos1(5.0dm

dcn V

VV . Then, =66.42o =1.15925 rad.

(b) 220mV V

VV

VV mdmdc 02.49*7.0*7.0 , A

R

VI dc

dc 268.315

02.49

2

)2sin(1

2m

rmsV

V ,

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

VV

V mS 56.155

2The rms value of the transformer secondery current is:

AII rmsS 34145.6Then, the rectification efficiency is:

%56.2634145.6*121.95

268.3*02.49

*

*

rmsrms

dcdc

ac

dc

IV

IV

P

P

(b) 94.12202.49

121.95

dc

rms

V

VFF

(c) 6624.1194.11 22FFV

VRF

dc

ac

(d) 1624.034145.6*56.155

268.3*02.49

SS

dc

IV

PTUF

85

(e) The PIV is mV

(f) Creast factor of input current CF is as following:

313.234145.6

6667.14

34145.6)( R

V

I

ICF

m

S

peakS

3.2.2 Half Wave Single Phase Controlled Rectifier With RL LoadA 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 ( t), vs is positive when0 < t < , and vs is negative when < t <2 . When vs starts becomepositive, the SCR is forward-biased but remains in the blocking state tillit is triggered. If the SCR is triggered when t = then it startsconducting and the positive source keeps the SCR in conduction till treaches radians. At that instant, the current through the circuit is notzero and there is some energy stored in the inductor at t = radians.The voltage across the inductor is positive when the current through it isincreasing and it becomes negative when the current through the inductortends to fall. When the voltage across the inductor is negative, it is insuch a direction as to forward bias the SCR. There is current through theload at the instant t = radians and the SCR continues to conduct tillthe energy stored in the inductor becomes zero. After that the currenttends to flow in the reverse direction and the SCR blocks conduction.Fig.3.5 shows the output voltage, resistor, inductor voltages and thyristorvoltage drop waveforms.

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

86

Fig.3.5 Various voltages and currents waveforms for half wave single phasecontrolled rectifier with RL load.

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

ttViRdt

diL m ),(sin* (3.5)

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

ttL

Vi

L

R

dt

di m 0),(sin* (3.6)

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

AdttL

Veeti m

dtL

Rdt

L

R

sin*)(

Then, AdttL

Veeti m

tL

Rt

L

R

sin*)(

87

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

tL

Rm AetLtR

LwR

Vti cossin)(

222 (3.7)

Assume LjRZ . Then, 2222 LRZ ,

cosZR , sinZL andR

Ltan

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

tancossinsincos)(

t

m AettZ

Vti

Then, tansin)(

t

m AetZ

Vti t (3.8)

The value of A can be obtained using the initial condition. Since thediode starts conducting at t = and the current starts building up fromzero, then, i( ) = 0. Then, the value of A is expressed by the followingequation:

)sin(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 t.

RL

tm etZ

Vti /)sin()sin()( t (3.10)

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

RLe /

)(

)sin()sin( (3.11)

)cos(cos*2

sin*2

mmdc

Vtdt

VV (3.12)

R

ZL

88

The rms load voltage is:

)2cos2(cos2

1*

2sin*

2

1 2 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 theaverage voltage across the inductor is zero.

)cos(cos*2 R

VI m

dc (3.14)

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

81.11)10*20*314(10 232Z , .561.0tan 1 RadR

wL

.5236.06180

*30 Rad

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

follows:-

5236.0628.0

/

0374.0)561.0sin(*6283.18

)561.05236.0sin()561.0sin(81.11

220)(

t

RL

t

et

etti

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

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

VVdc 9.59)70766.3cos5236.0(cos*2

220

Then, AR

VI dc

dc 99.510

9.59

Similarly, Vrms can be obtained from equation (3.13)

VVrms 384.111)05236*2cos70766.3*2(cos2

15236.070766.3*

2

220

89

AR

VI rms

rms 1384.1110

384.111

%92.28384.111

9.59

*

*2

2

2

2

rms

dc

rmsrms

dcdc

V

V

IV

IV

3.2.3 Half Wave Single Phase Controlled Rectifier With FreeWheeling DiodeThe 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 inaddition, marked D1. The voltage source vs is an alternating sinusoidalsource. If vs = Vm sin ( t), vs is positive when 0 < t < , and it isnegative when < t <2 . When vs starts become positive, the SCR isforward-biased but remains in the blocking state till it is triggered. Whenthe SCR is triggered in the forward-bias state, it starts conducting and thepositive source keeps the SCR in conduction till t reaches radians. Atthat instant, the current through the circuit is not zero and there is someenergy stored in the inductor at t = radians. In the absence of the freewheeling diode, the inductor would keep the SCR in conduction for partof the negative cycle till the energy stored in it is discharged. But, when afree wheeling diode is present as shown in the circuit shown in Fig.3.7the supply voltage will force D1 to turn on because its anode voltagebecome more positive than its cathode voltage and the current has a paththat offers almost zero resistance. Hence the inductor discharges itsenergy during < t < (2 ) (Assuming continuous conductionmode) through the load. When there is a free wheeling diode, the currentthrough the load tends to be continuous, at least under ideal conditions.When the diode conducts, the SCR remains reverse-biased, because thevoltage vs is negative.

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

90

Fig.3.7 Various voltages and currents waveforms for half wave single phasecontrolled rectifier with RL load and freewheeling diode.

91

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

When the SCR conducts, the driving function for the differentialequation is the sinusoidal function defining the source voltage. During theperiod defined by < t < (2 ), the SCR blocks current and acts asan open switch. On the other hand, the free wheeling diode conductsduring this period, and the driving function can be set to be zero volts.For < t < , equation (3.15) applies whereas equation (3.16) appliesfor the rest of the cycle. As in the previous cases, the solution is obtainedin two parts.

ttViRdt

diL m ),(sin* (3.15)

2,0* tiRdt

diL (3.16)

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

RL

tm eAtZ

Vti /*)sin(*)( t (3.17)

The difference in the solution of (3.17) than (3.8) is how the constantA is evaluated? In the circuit without free-wheeling diode 0i , sincethe current starts build up from zero when the SCR is triggered during thepositive half cycle. Assuming in the steady state the load is continuousand periodical which means that:

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

teBwti

wt

tan

)(

*)( (3.19)

At t then, iB (3.20)To obtain B, we have to obtain i from (3.17) by letting t equal. Then,

tan*)sin(*)( eAZ

Vi m (3.22)

Substitute (3.21) into (3.19) we get the current during2t as following:

92

teeAZ

Vti

t

m tan

)(

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

teAeZ

Vti

tt

m tantan

)(

**)sin(*)( (3.23)

From the above equation we can obtain 2i as following:

teAeZ

Vi m tan

2

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

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

AZ

Vi m )sin(* (3.25)

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

tan2

tan **)sin(*)sin(* eAeZ

VA

Z

V mm

)sin()sin(*1 tantan

2

eZ

VeA m

tan2

tan

1

)sin()sin(

*

e

e

Z

VA m (3.26)

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

Also, the load current in the period of 1212 ntn

.......,3,2,1,0n can be obtained from substituting (3.26) into (3.23).

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

93

Solution: 9539.32)10*100*314(10 232Z

.2625.1tan 1 RadR

wL,

.5236.06180

*30 Rad

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

122 ntn where .......,3,2,1,0n Then,)5236.0(*31845.0*48858.7)2625.1sin(*676.6)( tetti

61457.9)(i , 99246.2)(iDuring the period of 1212 ntn where

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

Then, ,*61457.9)( )(31845.0 teti

VVV

dwtwtVV mmmdc 3372.65)cos1(

2))cos(cos(

2)sin(

2

1

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

VV

tdtVV mmrms 402.108)

2

)2sin((

1

2)sin(

2

1 2/12/1

2

Atde

tdetI

t

tdc

2546.4*61457.9

*48858.7)2625.1sin(*676.62

1

2)(31845.0

)5236.0(*31845.0

Atde

tdetI

t

trms

43288.5*61457.9

*48858.7)2625.1sin(*676.62

1

2/12

2)(31845.0

2)5236.0(*31845.0

94

%2.4743288.5*402.108

2546.4*3372.65

*

*

rmsrms

dcdc

ac

dc

IV

IV

P

P

3.3 Single-Phase Full Wave Controlled Rectifier3.3.1 Single-Phase Center Tap Controlled Rectifier With ResistiveLoadCenter tap controlled rectifier is shown in Fig.3.8. When the upper half ofthe transformer secondary is positive and thyristor T1 is triggered, T1 willconduct and the current flows through the load from point a to point b.When the lower half of the transformer secondary is positive and thyristorT2 is triggered, T2 will conduct and the current flows through the loadfrom point a to point b. So, each half of input wave a unidirectionalvoltage (from a to b ) is applied across the load. Various voltages andcurrents waveforms for center tap controlled rectifier with resistive loadare shown in Fig.3.9 and Fig.3.10.

Fig.3.8 Center tap controlled rectifier with resistive load.

ab

95

Fig.3.9 The output voltgae and thyristor T1 reverse voltage wavforms alongwith the supply voltage wavform.

Fig.3.10 Load current and thyristors currents for Center tap controlled rectifierwith resistive load.

96

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

)cos1())cos(cos()sin(1 mm

mdcVV

tdtVV (3.27)

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

)cos1(5.0dm

dcn V

VV (3.28)

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

2

)2sin(

2)sin(

1 2 mmrms

VtdtVV (3.29)

Example 4 The rectifier shown in Fig.3.8 has load of R=15 and,Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain anaverage output voltage of 70 % of the maximum possible output voltage,calculate:- (a) The delay angle , (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 powerfactor.Solution :

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

7.0)cos1(5.0dm

dcn V

VV , then, =66.42o

(b) 220mV , then, VV

VV mdmdc 04.98

2*7.0*7.0

AR

VI dc

dc 536.615

04.98

2

)2sin(

2m

rmsV

V

at =66.42o Vrms=134.638 VThen, Irms=134.638/15=8.976 A

97

VV

V mS 56.155

2The rms value of the transformer secondery current is:

AI

I rmsS 347.6

2Then, The rectification efficiency is:

%04.53976.8*638.134

536.6*04.98

*

*

rmsrms

dcdc

ac

dc

IV

IV

P

P

(c) 3733.104.98

638.134

dc

rms

V

VFF and,

9413.013733.11 22FFV

VRF

dc

ac

(d) 32479.034145.6*56.155*2

536.6*04.982 SS

dc

IV

PTUF

(e) The PIV is 2 Vm

(f) Creast Factor CF, 313.234145.6

6667.14

34145.6)( R

V

I

ICF

m

S

peakS

3.3.2 Single-Phase Fully Controlled Rectifier Bridge With ResistiveLoadThis section describes the operation of a single-phase fully-controlledbridge rectifier circuit with resistive load. The operation of this circuit canbe understood more easily when the load is pure resistance. The mainpurpose of the fully controlled bridge rectifier circuit is to provide avariable DC voltage from an AC source.

The circuit of a single-phase fully controlled bridge rectifier circuit isshown in Fig.3.11. The circuit has four SCRs. For this circuit, vs is asinusoidal voltage source. When the supply voltage is positive, SCRs T1and T2 triggered then current flows from vs through SCR T1, load resistorR (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

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

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

Fig.3.12 Various voltages and currents waveforms for converter shown inFig.3.11 with resistive load.

99

Fig.3.13 FFT components of the output voltage and supply current for convertershown in Fig.3.11.

The main purpose of this circuit is to provide a controllable DC outputvoltage, which is brought about by varying the firing angle, . Let vs = Vm

sin t, with 0 < t < 360o. If t = 30o when T1 and T2 are triggered,then the firing angle is said to be 30o. In this instance, the other pair istriggered when t = 30+180=210o. When vs changes from positive tonegative value, the current through the load becomes zero at the instant

t = radians, since the load is purely resistive. After that there is nocurrent flow till the other is triggered. The conduction through the load isdiscontinuous.

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

2 radians. Since the output waveform repeats itself every half-cycle, theaverage output voltage is expressed as a function of , as shown inequation (3.27).

)cos1()cos(cos)sin(1 mm

mdcVV

tdtVV (3.27)

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

)cos1(5.0dm

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 mmrms

VtdtVV (3.29)

100

Example 5 The rectifier shown in Fig.3.11 has load of R=15 and,Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain anaverage output voltage of 70% of the maximum possible output voltage,calculate:- (a) The delay angle , (b) The efficiency, (c) Ripple factor ofoutput voltage(d) The transformer utilization factor, (e) The peak inversevoltage (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 =0,The normalized output voltage is shown in equation (3.31) which is

required to be 70%.

Then, 7.0)cos1(5.0dm

dcn V

VV , then, =66.42o

(b) 220mV , then, VV

VV mdmdc 04.98

2*7.0*7.0

AR

VI dc

dc 536.615

04.98

2)2sin(

2m

rmsV

V

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

VV

V 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

dc

IV

IV

P

P

(c) 3733.104.98

638.134

dc

rms

V

VFF

9413.013733.11 22FFV

VRF

dc

ac

101

(d) 4589.0976.8*56.155

536.6*04.98

SS

dc

IV

PTUF

(e) The PIV is Vm

(f) Creast Factor CF, 634.1976.8

6667.14

976.8)( R

V

I

ICF

m

S

peakS

3.3.3 Full Wave Fully Controlled Rectifier With RL Load InContinuous Conduction Mode

The circuit of a single-phase fully controlled bridge rectifier circuit isshown in Fig.3.14. The main purpose of this circuit is to provide avariable DC output voltage, which is brought about by varying the firingangle. The circuit has four SCRs. For this circuit, vs is a sinusoidalvoltage source. When it is positive, the SCRs T1 and T2 triggered thencurrent flows from +ve point of voltage source, vs through SCR T1, loadinductor 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, loadinductor L (from up to down), SCR T4 and back into the +ve point ofvoltage source. Even though the direction of current through the sourcealternates from one half-cycle to the other half-cycle, the current throughthe load remains unidirectional (from up to down). Fig.3.15 showsvarious voltages and currents waveforms for the converter shown inFig.3.14. Fig.3.16 shows the FFT components of load voltage and supplycurrent.

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

102

Fig.3.15 Various voltages and currents waveforms for the converter shown inFig.3.14 in continuous conduction mode.

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

Let vs = Vm sin t, with 0 < t < 360o. If t = 30o when T1 and T2are triggered, then the firing angle is said to be 30o. In this instance theother pair is triggered when t= 210o. When vs changes from a positiveto a negative value, the current through the load does not fall to zerovalue at the instant t = radians, since the load contains an inductor

103

and the SCRs continue to conduct, with the inductor acting as a source.When the current through an inductor is falling, the voltage across itchanges sign compared with the sign that occurs when its current isrising. When the current through the inductor is falling, its voltage is suchthat the inductor delivers power to the load resistor, feeds back somepower to the AC source under certain conditions and keeps the SCRs inconduction forward-biased. If the firing angle is less than the load angle,the energy stored in the inductor is sufficient to maintain conduction tillthe next pair of SCRs is triggered. When the firing angle is greater thanthe load angle, the current through the load becomes zero and theconduction through the load becomes discontinuous. Usually thedescription of this circuit is based on the assumption that the loadinductance is sufficiently large to keep the load current continuous andripple-free.

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

cos2

)sin(1 m

mdcV

tdtVV (3.33)

The normalized output voltage is cosdm

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 (3.35)

Example 6 The rectifier shown in Fig.3.14 has pure DC load current of50 A and, Vs=220 sin 314 t and unity transformer ratio. If it is required toobtain an average output voltage of 70% of the maximum possible outputvoltage, calculate:- (a) The delay angle , (b) The efficiency, (c) Ripplefactor (d) The transformer utilization factor, (e) The peak inverse voltage(PIV) of the thyristor and (f) Crest factor of input current. (g) Inputdisplacement factor.

104

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

required to be 70%. Then, 7.0cosdm

dcn V

VV , then, =45.5731o= 0.7954

(b) 220mV

VV

VV mdmdc 04.98

2*7.0*7.0 ,

2m

rmsV

V

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

VV

V mS 56.155

2The rms value of the transformer secondery current is: AII rmsS 50Then, The rectification efficiency is

%02.6350*563.155

50*04.98

*

*

rmsrms

dcdc

ac

dc

IV

IV

P

P

(c) 587.104.98

563.155

dc

rms

V

VFF

23195.113733.11 22FFV

VRF

dc

ac

(d) 4589.050*56.155

50*04.98

SS

dc

IV

PTUF

(e) The PIV is Vm

(f) Creast Factor CF, 1)(

S

peakS

I

ICF

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

105

Fig.3.17 Load, resistor and inductor voltages waveforms along with supplyvoltage 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 theconverter shown in Fig.3.14 in case of discontinuous conduction mode.

106

Fig.3.18 FFT components of supply current along with supply voltage of theconverter shown in Fig.3.14 in discontinuous conduction mode.

During the period of t the following differential equation can

be obtained:

ttViRdt

diL m ),(sin* (3.35)

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

tan

)(

)sin()sin()(

t

m etZ

Vti (3.36)

Where 2221 )(tan LRZandR

L

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

tan

)(

)sin()sin( e (3.37)

)cos(cos*sin* mmdc

Vtdt

VV (3.38)

The rms load voltage is:

107

)2cos2(cos2

1*

2sin*

1 2 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 theaverage voltage across the inductor is zero.

)cos(cos*R

VI m

dc (3.40)

3.3.5 Single Phase Full Wave Fully Controlled Rectifier With SourceInductance:Full wave fully controlled rectifier with source inductance is shown inFig.3.19. The presence of source inductance changes the way the circuitoperates during commutation time. Let vs = Vm sin wt, with 0 < t <360o. Let the load inductance be large enough to maintain a steadycurrent through the load. Let firing angle be 30o. Let SCRs T3 and T4be in conduction before t < 30o. When T1 and T2 are triggered at t =30o, there is current through the source inductance, flowing in thedirection opposite to that marked in the circuit diagram and hencecommutation of current from T3 and T4 to T1 and T2 would not occurinstantaneously. The source current changes from dcI to dcI due to thewhole of the source voltage being applied across the source inductance.When T1 is triggered with T3 in conduction, the current through T1would 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 thisperiod, the current through T2 would rise from zero to dcI and, thecurrent through T4 would fall from dcI to zero.

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

108

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

Fig.3.20 Output voltage, thyristors current along with supply voltage waveformof a single phase full wave fully controlled rectifier with source inductance.

Fig.3.21 Output voltage, supply current along with supply voltage waveform ofa single phase full wave fully controlled rectifier with source inductance.

109

Let us study the commutation period starts at ut . WhenT1 and T2 triggered, then T1 and T2 switched on and T3 and T4 try toswitch off. If this happens, the current in the source inductance has tochange its direction. But source inductance prevents that to happeninstantaneously. So, it will take time t to completely turn off T3 and T4and to make T1 and T2 carry the entire load current oI which is very

clear from Fig.3.20. Also, in the same time ( t ) the supply currentchanges 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 controlledrectifier during that commutation period. From Fig.3.22 We can get thedifferential equation representing the circuit during the commutation timeas shown in (3.41).

Fig.3.22 The equivalent circuit of single phase fully controlled rectifierduring the commutation period.

0dt

diLv s

ss (3.41)

Multiply the above equation by t then,0sin ssm diLtdtV

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

o

o

I

I

ss

u

m diLtdtV sin

Then, osm ILuV 2coscos . Then,

m

os

V

ILu

2coscos

110

Thenm

os

V

ILu

2coscos 1 (3.42)

Thenm

os

V

ILut

2coscos

1 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,

dt

diLv s

srd (3.44)

Then, os

I

I

ss

u

rd ILdiLtdtvo

o

2 (3.45)

u

rd tdtv 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 42 (3.46)

To obtain the average reduction in DC output voltage rdV due tosource inductance we have to divide the above equation by the period ofsupply voltage waveform, 2 . Then,

osos

rd IfLIL

V 42

4 (3.47)

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

osm

rdceinducsourcewithoutdcactualdc IfLV

VVV 4cos2

tan(3.48)

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

32

2 2 uII o

s (3.49)

111

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

2sin*

2

81

u

u

II o

S (3.50)

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

2cos. 1 u

I

Ifp

s

s (3.51)

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

A half controlled single-phase bridge is shown in Fig.3.23. Thisrectifier uses two SCRs and two diodes in addition to freewheeling diode.When the source voltage is in its positive half cycle and thyristor T2 istriggered and it will conduct with diode D1. When the supply voltage isgoing to negative and thyristor T4 is triggered and it will conduct withdiode D3. This circuit cannot work without freewheeling diode in case ofinductive load. The inductive energy in the load would freewheel throughthe diode D1 and Thyristor T4 or diode D3 and thyristor T2 even if thereis no gate signal. If there is freewheeling diode is connected across theload of a bridge converter, it will remove the negative voltage in the loadvoltage. In this converter the diode D1 and Thyristor T2 work in positivehalf cycle of the supply voltage and D3 and T4 can work in the negativehalf cycle. As an example at 60o firing angle as shown in Fig.3.24, thenthe thyristor T2 triggers at 60o conducts till t = . At t= the voltageacross the freewheeling diode will start to be forward then the storedenergy in the load will take the freewheeling diode as a pass to circulatethe stored reactive power, In this case D1 and T2 will turned off. Thecircuit will still like that till the thyristor T4 gets its triggering pulse atangle 240o. In this case, the voltage across the freewheeling diode willreverse and diode D3 and thyristor T4 will conduct till t=360o. When

t=360o the voltage across the freewheeling diode will be forward againand the load will freewheel the stored reactive power in the load throughthe freewheeling diode and the diode D3 and T4 will turned off and cyclewill repeated again. The average load voltage is shown in shown in thefollowing equation:-

112

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

)cos1()cos(cos)sin(1 mm

mdcVV

tdtVV (3.52)

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

The normalized output voltage is: )cos1(5.0dm

dcn V

VV (3.53)

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

2

)2sin(

2)sin(

1 2 mmrms

VtdtVV (3.54)

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

113

3.3.6 Inverter Mode Of OperationThe 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 theac side. The easiest way to understand the inverter mode of operation is toassume that the DC side of the converter can be replaced by a currentsource of a constant amplitude dI , as shown in Fig.3.25. For a delayangle a greater than 90° but less than 180°, the voltage and currentwaveforms are shown in Fig.3.26. The average value of dv is negative,

given by (3.48), where 90° < < 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 cosSsac IVP is also negative because o90 .

Fig.3.25 Single phase SCR inverter.

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

114

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

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

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

dsdodd ILVVE2

cos (3.55)

The equation is exact if the current is constant at dI ; otherwise, a

value of di at t should be used in (3.55) instead of dI . Fig.3.28shows that for a given value of , for example, 1, the intersection of the

do source voltage 1dd EE , and the converter characteristic at 1,determines the do current 1dI , and hence the power flow 1dP .

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

/t should be greater than the thyristor turn-off time q Otherwise,

the thyristor will prematurely begin to conduct, resulting in the failure ofcurrent to commutate from one thyristor pair to the other, an abnormaloperation that can result in large destructive currents.

u180 (3.56)

115

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

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

made sufficiently large (e.g.,165o) so that di is discontinuous as shown inFig.3.30. Then, 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 Rectifier3.5.1 Three Phase Half Wave Controlled Rectifier With ResistiveLoadFig.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 pulsesfor three thyristor are displaced 120o relative to each other’s. Eachthyristor will conduct for 120o. A thyristor can be fired to conduct whenits anode voltage is positive with respect to its cathode voltage. Themaximum output voltage occurred when =0 which is the same as diodecase. This rectfier has continuous load current and voltage in case of30. However, the load voltage and current will be discontinuous in caseof > 30.

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

117

In case of 30, various voltages and currents of the converter shownin Fig.3.31 are shown in Fig.3.32. Fig.3.33 shows FFT components ofload voltage, secondary current and primary current. As we see the loadvoltage contains high third harmonics and all other triplex harmonics.Also secondary current contains DC component, which saturate thetransformer core. The saturation of the transformer core is the maindrawback of this system. Also the primary current is highly distorted butwithout a DC component. The average output voltage and current areshown in equation (3.57) and (3.58) respectively. The rms output voltageand current are shown in equation (3.59) and (3.60) respectively.

cos675.0cos2

3

cos827.0cos2

33sin

2

36/5

6/

LLLL

mm

mdc

VV

VV

tdtVV(3.57)

cos*827.0

cos**2

33

R

V

R

VI mm

dc (3.58)

2cos8

3

6

13sin

2

36/5

6/

2mmrms VtdtVV (2.57)

R

V

Im

rms

2cos8

3

6

13

(3.60)

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

R

VI

IIm

rmsSr

2/1

2cos8

3

6

1

3 (3.61)

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

118

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

119

Fig.3.33 FFT components of load voltage, secondary current and supply currentfor the converter shown in Fig.3.31 for 30.

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

6cos14775.0

6cos1

2

3sin

2

3

6/m

mmdc V

VtdtVV (3.63)

6cos1

2

3

R

VI m

dc (3.64)

)23/sin(8

1

424

53sin

2

3

6/

2mmrms VtdtVV (2.63)

)23/sin(8

1

424

53

R

VI m

rms (3.66)

Then the diode rms current can be obtaiend as follows:

)23/sin(8

1

424

5

3 R

VIII mrms

Sr (3.67)

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

120

Fig.3.34 Various voltages and currents waveforms for converter shown inFig.3.22 for > 30.

121

Fig.3.35 FFT components of load voltage, secondary current and supply currentfor the converter shown in Fig.3.22 for > 30.

Example 7 Three-phase half-wave controlled rectfier is connected to 380V three phase supply via delta-way 380/460V transformer. The load of

the rectfier is pure resistance of 5 . The delay angle o25 . 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 asfollowing:

VVV LLdc 5.28125cos4602

3cos

2

3

Then; AR

VI dc

dc 3.565

5.281

From (3.59) we can calculate rmsV as following:

2cos8

3

6

1*22cos

8

3

6

13 LLmrms VVV

Then, VVrms 8.29825*2cos8

3

6

1*460*2

122

Then AR

VI rms

rms 76.595

8.298

Then, the rectfication effeciency can be calculated as following:

%75.88100*rmsrms

dcdc

IV

IV

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

AI

I rmsS 5.34

3

76.59

3

%66.57100*5.34*460*3

3.56*5.281

*3 sLL

dcdc

IV

IVTUF

AV

R

VI LLm

peakS 12.755

460*3/2

5

*3/2,

177.25.34

12.75,

S

peakSF I

IC

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

o60Slution: From (3.63) the DC value of the output voltage can be obtainedas following:

VV

V mdc 33.179

36cos1

2

460*3

23

6cos1

2

3

Then; AR

VI dc

dc 87.355

33.179

From (3.65) we can calculate rmsV as following:

V

VV mrms

230)3/23/sin(81

43/

245

*460*2

)23/sin(81

4245

3

Then AR

VI rms

rms 465

230

Then, the rectfication effeciency can be calculated as following

123

%79.60100*rmsrms

dcdc

IV

IV

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

AI

I rmsS 56.26

3

46

3

%4.30100*56.26*460*3

87.35*33.179

*3 sLL

dcdc

IV

IVTUF

AV

R

VI LLm

peakS 12.755

460*3/2

5

*3/2,

83.256.26

12.75,

S

peakSF I

IC

VVPIV LL 54.650460*22

3.5 Three Phase Half Wave Controlled Rectifier With DC LoadCurrent

The Three Phase Half Wave Controlled Rectifier With DC LoadCurrent is shown in Fig.3.36, the load voltage will reverse its directiononly if > 30. However if < 30 the load voltage will be positive all thetime. Then in case of > 30 the load voltage will be negative till the nextthyristor in the sequence gets triggering pulse. Also each thyristor willconduct 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 currentand supply current for the converter shown in Fig.3.36 for > 30 andpure DC current load.

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

124

Fig.3.37 Various voltages and currents waveforms for the converter shown inFig.3.36 for > 30 and pure DC current load.

Fig.3.38 FFT components of load voltage, secondary current and supply currentfor the converter shown in Fig.3.36 for > 30 and pure DC current load.

t=0

125

As explained before the secondary current of transformer contains DCcomponent. Also the source current is highly distorted which make thissystem has less practical significance. The THD of the supply current canbe 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, thenthe waveform can be represented as odd function. So, an=0 and bn can beobtained as the following:-

32

cos12

)sin(2

3/2

0

n

n

ItdtnIb dc

dcn for n=1,2,3,4,…

Then,2

3*

2

n

Ib dc

n for n=1,2,4,5,7,8,10,….. (3.69)

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

equation

......7sin7

15sin

5

14sin

4

12sin

2

1sin

3)( ttttt

Iti dc

p (3.71)

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

21

21

2

p

pp

I

IITHD (3.72)

Where, dcp II *3

2 (3.73)

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

2

31

dcp

II (3.74)

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

%68

2

92

9

3

2

22

22

2

dc

dcdc

I

II

THD (3.75)

126

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

%68....14

1

13

1

11

1

10

1

8

1

7

1

5

1

4

1

2

1 222222222

THD (3.76)

The supply current THD is very high and it is not acceptable by anyelectric utility system. In case of full wave three-phase converter, theTHD 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 380V three phase supply via delta-way 380/460V transformer. The load of

the rectfier draws 100 A pure DC current. The delay angle, o30 .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 A05.121380

460*100 . Then,

AI rmsP 84.983

2*05.121, .

1PI can be obtained from equation (372) where

AI

I dcP 74.81

2

05.121*3

2

31 .

Then, %98.67100*174.81

84.98100*1

22

1

,

P

rmsPI I

ITHD

P

The input power factor can be calculated as following:

LaggingI

IfP

rmsP

P 414.066

cos*84.98

74.81

6cos*.

,

1

3.6 Three Phase Half Wave Controlled Rectifier With Free WheelingDiode

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

127

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

The average output voltage is :-

6cos14775.0

6cos1

2

3sin

2

3

6/

mm

mdc VV

tdtVV (3.77)

6cos1

2

3

R

VI m

dc (3.78)

)23/sin(81

4245

3sin23

6/

2mmrms VtdtVV

(3.79)

)23/sin(8

1

424

53

R

VI m

rms (3.80)

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

128

Fig.3.40 Various voltages and currents waveforms for the converter shown inFig.3.36 for > 30 with RL load and freewheeling diode.

Fig.3.41 FFT components of load voltage, secondary current and supply currentfor the converter shown in Fig.3.36 for > 30 and RL load.

129

3.7 Three Phase Full Wave Fully Controlled Rectifier Bridge3.7.1 Three Phase Full Wave Fully Controlled Rectifier WithResistive Load

Three-phase full wave controlled rectifier shown in Fig.3.42. As wecan see in this figure the thyristors has labels T1, T2,……,T6. The labelof each thyristor is chosen to be identical to triggering sequence wherethyristors are triggered in the sequence of T1, T2,……,T6 which is clearfrom 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

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

)120(sin

)120(sin

)(sin

tVv

tVv

tVv

mc

mb

ma

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

ot 15030 . So, the thyristor T1 is forward bias during this periodand it is ready to conduct at any instant in this period if it gets a pulse onits gate. In Fig.3.44 the firing angle 40 as an example. So, T1 takes a

pulse at ot 70403030 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 .

It can be seen from Fig.3.44 that the voltage bv is the highest positive

voltage of the three phase voltage when t is in the range ofot 270150 . So, the thyristor T3 is forward bias during this period

131

and it is ready to conduct at any instant in this period if it gets a pulse onits gate. In Fig.3.44, the firing angle 40 as an example. So, T3 takes a

pulse at ot 19040150150 .

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

ot 390270 . So, the thyristor T5 is forward bias during this periodand it is ready to conduct at any instant in this period if it gets a pulse onits gate. In Fig.3.44, the firing angle 40 as an example. So, T3 takes a

pulse at ot 310270 .It can be seen from Fig.3.44 that the voltage av is the highest negative

voltage of the three phase voltage when t is in the rangeot 330210 . 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 onits gate. In Fig.338, the firing angle 40 as an example. So, T4 takes a

pulse at ot 25040210210 .

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

ot 450330 or ot 90330 in the next period of supply voltagewaveform. So, the thyristor T6 is forward bias during this period and it isready to conduct at any instant in this period if it gets a pulse on its gate.In Fig.3.44, the firing angle 40 as an example. So, T6 takes a pulse

at ot 370330 .

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

ot 21090 . So, the thyristor T2 is forward bias during this period andit is ready to conduct at any instant in this period if it gets a pulse on itsgate. In Fig.3.44, the firing angle 40 as an example. So, T2 takes a

pulse at ot 13090 .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 voltageconnected to its anode and another one in the lower half (T2, T4, or, T6)

132

which become forward bias at highest negative voltage connected to itscathode. So the load is connected at any time between the highestpositive phase voltage and the highest negative phase voltage. So, theload voltage equal the highest line to line voltage at any time which isclear from Fig.3.45. The following table summarizes the aboveexplanation.

Period, range of wt SCR Pair in conduction

+ 30o to + 90o T1 and T6

+ 90o to + 150o T1 and T2

+ 150o to + 210o T2 and T3

+ 210o to + 270o T3 and T4

+ 270o to + 330o T4 and T5

+ 330o to + 360o and + 0o to + 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 .

The line current waveform is very easy to obtain it by applyingkerchief'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 thisrectifier for 60,40 is shown in Fig.3.46. Fast Fouriertransform (FFT) of output voltage and supply current are shown inFig.3.47.

133

Fig.3.46 The input current of this rectifier of rectifier in Fig.3.42 at60,40 .

Fig.3.46 FFT components of output voltage and supply current of rectifier inFig.3.42 at 60,40 .

134

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

cos33

)6

sin(33

2/

6/

mmdc

VtdtVV (3.81)

The maximum average output voltage for delay angle =0 is

mdm

VV

33 (3.82)

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

cosdm

dcn V

VV (3.83)

The rms value of the output voltage is found from the followingequation:

2cos4

33

2

13)

6sin(3

32/

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 . If o60 the output voltage, andphase 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 .

135

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

3/cos133

)6

sin(33

6/5

6/

mmdc

VtdtVV (3.85)

The maximum average output voltage for delay angle =0 is

mdm

VV

33 (3.86)

The normalized average output voltage is

3/cos1dm

dcn V

VV (3.87)

The rms value of the output voltage is found from the followingequation:

62cos2

4

313

)6

sin(33

6/5

6/

2

m

mrms

V

tdtVV

(3.88)

Example 10 Three-phase full-wave controlled rectifier is connected to380 V, 50 Hz supply to feed a load of 10 pure resistance. If it isrequired to get 400 V DC output voltage, calculate the following: (a) Thefiring angle, (b) The rectfication effeciency (c) The crest factor ofinput current. (d) PIV of the thyristors.Solution: From (3.81) the average voltage is :

VV

V mdc 400cos

380*3

2*33

cos33

.

Then o79.38 , AR

VI dc

dc 4010

400

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

79.38*2cos4

33

2

1*380*

3

2*32cos

4

33

2

13 mrms VV

Then, VVrms 412.412

136

Then, AR

VI rms

rms 24.4110

412.412

Then, %07.94100*24.41*4.412

40*400100*

*

*

rmsrms

dcdc

IV

IV

The crest factor of input current,rmss

peakSF I

IC

,

,

AR

tI peakS 11.53

10

3079.3830sin380*26sin380*2

,

AII rmsrmss 67.3324.41*3

2*

3

2,

Then, 577.167.33

11.53

,

,

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 :

VV

V mdc 150cos

380*3

2*33

cos33

. Then, o73

It is not acceptable result because the above equation valid only for60 . Then we have to use the (3.85) to get dcV as following:

VVdc 1503/cos1380*

3

2*33

. Then, o05.75

Then AR

VI dc

dc 1510

150

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

3005.75*2cos180

*05.75*24

31*380*

3

2*3

62cos2

4

313 mrms VV

137

Then, VVrms 075.198

Then, AR

VI rms

rms 8075.1910

075.198

Then, %35.57100*81.19*075.198

15*150100*

*

*

rmsrms

dcdc

IV

IV

The crest factor of input current,rmss

peakSF I

IC

,

,

AR

tI peakS 97.37

10

3005.7530sin380*26sin380*2

,

AII rmsrmss 1728.168075.19*3

2*

3

2,

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 pureDC Load CurrentThree-phase full wave-fully controlled rectifier with pure DC load currentis shown in Fig.3.48. Fig.3.49 shows various currents and voltage of theconverter shown in Fig.3.48 when the delay angle is less than 60o. As wesee in Fig.3.49, the load voltage is only positive and there is no negativeperiod in the output waveform. Fig.3.50 shows FFT components of

output voltage of rectifier shown in Fig.3.48 for o60 .

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

138

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

Fig.3.50 FFT components of SCR, secondary, primary currents respectively ofrectifier shown in Fig.3.48.

139

In case of the firing angle is greater than o60 , the output voltagecontains 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 .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 ofthree-phase full-wave diode bridge rectifier typically except the phaseshift between the phase voltage and phase current is zero in case of diodebridge but it is in case of three-phase full-wave controlled rectifierwith pure DC load current as shown in Fig.3.53. So, the input powerfactor of three-phase full-wave diode bridge rectifier with pure DC loadcurrent is:

cos1

s

s

I

IrPowerFacto (3.89)

Fig.3.51 Output voltage and supply current waveforms along with three phaseline voltages for the rectifier shown in Fig.3.48 for > 60o with pure DC

current load.

140

Fig.3.52 FFT components of SCR, secondary, primary currents respectively ofrectifier shown in Fig.3.48 for > 60o.

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

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

141

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

Fig.3.54 Output voltage and supply current of rectifier shown n Fig.3.48 withpure DC load and source inductance the waveforms.

Fig.3.55 FFT components of output voltage of rectifier shown in Fig.3.48 for> 60o and there is a source inductance.

142

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 isready to switch as soon as it gets a pulse on its gate. Thyristor T2 gets apulse after that by as shown in Fig.3.56n, so, T6 has to switch OFFand T2 has to switch ON. But due to the source inductance will preventthat to happen instantaneously. So it will take time redut sec to

completely turn OFF T6 and to make T2 carry all the load current ( oI ).

Also in the time t the current in bL will change from oI to zero and thecurrent in cL will change from zero to oI . This is very clear fromFig.3.56n. The equivalent circuit of the three phase full wave controlledrectifier at commutation time t is shown in Fig.3.57n and Fig.3.58n.

Fig.3.56n Waveforms represent the one commutation period.

143

Fig.3.57n The equivalent circuit of the three phase controlled rectifier atcommutation time t shown in Fig.3.56n.

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

061b

Tbdc

Taa V

dt

diLV

dt

diLV (3.90)

021c

Tcdc

Taa V

dt

diLV

dt

diLV (3.91)

Note that, during the time t , 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 thefollowing differential equation:

dt

diLVV

dt

diLVV T

ccaT

bba26 (3.94)

026

dt

diL

dt

diLVV T

cT

bcb (3.95)

The above equation can be written in the following manner:026 TcTbcb diLdiLdtVV (3.96)

026 TcTbcb diLdiLtdVV (3.97)

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

144

00

2

0

6

2/

2/

o

o

I

Tc

I

Tb

u

cb diLdiLtdVV

03

2sin

3

2sin

2/

2/

ocob

u

mm ILILtdtVtV

assume Scb LLL

oS

u

m ILttV 23

2cos

3

2cos

2/

2/

oS

m

IL

uuV

2

3

2

2cos

3

2

2cos

3

2

2cos

3

2

2cos

oSm ILuuV 26

7cos

6cos

67

cos6

cos

m

oS

V

IL

uuuu

2

6

7sinsin

6

7coscos

6sinsin

6coscos

6

7sinsin

6

7coscos

6sinsin

6coscos

m

oS

V

IL

uuuu

2sin5.0cos

2

3sin5.0cos

2

3

sin5.0cos2

3sin5.0cos

2

3

m

oS

V

ILu

2coscos3

LL

oS

LL

o

m

o

V

IL

V

LI

V

LIu

2

2

2

3

2coscos (3.98)

LL

oS

V

ILu

2coscos 1 (3.99)

LL

oS

V

ILut

2coscos

1 1 (3.100)

145

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

the drop across the source inductance.dt

diLv T

Srd (3.101)

Multiply (3.101) by td and integrate both sides of the resultantequation we get:

oS

I

D

u

rd ILLditdvo

0

2

2

(3.102)

u

rd tdv2

2

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

have six commutation periods t in one period so the total reduction perperiod is:

oS

u

rd ILtdv 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 2 . Then,

oo

rd fLILI

V 62

6 (3.104)

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

osLLrdceinducsourcewithoutdcactualdc IfLVVVV 6cos23

tan(3.105)

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

uu

u

do

s tdItdtu

II

0

232

22

uu

uu

Io

233

12 32

2

146

Then63

2 2 uII o

S (3.106)

oI

oI 3

2

sI

u3

2

26

2 u

Fig.3.58n The utility line current

Fig.3.59 shows the utility line currents and its first derivative that helpus to obtain the Fourier transform of supply current easily. From Fig.2.43we can fill Table(3.1) as explained before when we study Table (2.1).

26

u

oI

oI26

5 u

267 u 26

11 u

sI

u

Io

u

Io

sI

26u

265 u

267 u

2611 u

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

147

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

sJ

26u

26u

265 u

265 u

267 u

267 u

2611 u

2611 u

sI 0 0 0 0 0 0 0 0

sIu

Io

u

Io

u

Io

u

Io

u

Io

u

Io

u

Io

u

Io

It is an odd function, then 0no aam

sss

m

sssn tnJ

ntnJ

nb

11

sin1

cos1

(3.107)

26

11sin

26

11sin

26

7sin

26

7sin

26

5sin

26

5sin

26sin

26sin*

11

un

un

un

un

un

un

un

un

u

I

nnb o

n

6

11cos

6

7cos

6

5cos

6cos

2sin*

22

nnnnnu

un

Ib o

n (3.108)

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

tu

tu

tu

tu

tu

uti

13sin2

13sin

13

111sin

2

11sin

11

1

7sin2

7sin

7

15sin

25

sin5

1sin

2sin

34

22

22(3.109)

Then;2

sin62

1u

u

II o

S (3.110)

The power factor can be calculated from the following equation:

2cos

63

2

2sin

62

2cos

21 u

uI

u

u

I

u

I

Ipf

o

o

S

S

Then;2

cos

63

2sin*32

u

uu

u

pf (3.111)

148

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

2cos

u. Another expression for the displacement power factor, by

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

uVVV LLLLdc coscos2

3cos

23

uVV LLdc coscoscos22

23

uVV LLdc coscos2

23 (3.112)

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

uIVP oLLdc coscos*2

23 (3.113)

On the AC side, the AC power is:

11 cos3 SLLac IVP (3.114)Substitute from (3.110) into (3.114) we get the following equation:

11 cos2

sin26

cos2

sin2

343

u

u

IVu

u

IVP oLLo

LLac (3.115)

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

2sin*4

coscoscos 1 u

uu (2.116)

The source inductance reduces the magnitudes of the harmoniccurrents. Fig.3.60a through d show the effects of SL (and hence of u) onvarious harmonics for various values of , where dI is a constant dc.

The harmonic currents are normalized by 1I I, with 0SL , which is

given by (2.98) where os II6

1 in this case. Normally, the DC-side

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

149

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

Table 3.2 Typical and idealized harmonics.

3.7.2 Inverter Mode of OperationOnce again, to understand the inverter mode of operation, we willassume that the do side of the converter can be represented by a currentsource of a constant amplitude dI , as shown in Fig.3.61. For a delayangle a greater than 90° but less than 180°, the voltage and current

150

waveforms are shown in Fig.3.62a. The average value of dV is negativeaccording to (3.81). On the ac side, the negative power implies that thephase angle 1 , between sv and si , is greater than 90°, as shown inFig.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 can be obtained from the characteristics shown in Fig.3.64.

151

Similar to the discussion in connection with single-phase converters,

the extinction angle uo180 must be greater than the thyristorturn-off interval qt 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 StartupAs discussed for start up of a single-phase inverter, the delay anglein the three-phase inverter of Fig.3.63 is initially made sufficientlylarge (e.g., 165°) so that id is discontinuous. Then, is decreased bythe controller such that the desired dI and dP are obtained.

152

Problems1- Single phase half-wave controlled rectifier is connected to 220 V,

50Hz supply to feed 10 resistor. If the firing angle o30draw output voltage and drop voltage across the thyristor alongwith the supply voltage. Then, calculate, (a) The rectficationeffeciency. (b) Ripple factor. (c) Peak Inverse Voltage (PIV) of thethyristor. (d) The crest factor FC of input current.

2- Single phase half-wave controlled rectfier is connected to 220 V,50Hz supply to feed 5 resistor in series with mH10 inductor if

the firing angle o30 .(a) Determine an expression for the current through the load in

the first two periods of supply current, then fiend the DCand rms value of output voltage.

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

3- Solve problem 2 if there is a freewheeling diode is connected inshunt with the load.

4- single phase full-wave fully controlled rectifier is connected to220V, 50 Hz supply to feed 5 resistor, if the firing angle

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

5- In the problem 4, if there is a 5mH inductor is connected in serieswith the 5 resistor. Draw waveforms of output voltage andcurrent, resistor and inductor voltages, diode currents, supplycurrents. Then, find an expression of load current, DC and rmsvalues 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 . Draw the load voltage, thyristor, andload 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.

153

9- Single phase full-wave semi-controlled rectifier is connected to220 V, 50Hz supply to feed 5 resistor in series with 5 mHinductor, the load is connected in shunt with freewheeling diode.Draw the load voltage and current, resistor voltage and inductorvoltage 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 DCload of 1 kW with pure DC current. A 1.5-kVA-isolationtransformer with a source-side voltage rating of 120 V at 50 Hz isused. 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 ifthe 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 = 135°. UsingPSIM, obtain sv , si , dv , and di waveforms in steady state.

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

13-In the three-phase fully controlled rectifier is connected to 460 Vat 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 forwhich dmdc VV 5.0 . Draw dv waveform and identify thedevices that conduct during various intervals. Obtain the DPF, PF,and %THD in the input line current and compare results with afull-bridge converter operating at dmdc VV 5.0 . Assume SL .

154

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 ratingof 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 itsratings. The ac source voltage of nominally 208 V (line to line) isin the range of -10% and +5%. Assume the load current is pureDC, calculate the minimum transformer turns ratio if the DC loadvoltage is to be regulated at a constant value of 300 V. What is thevalue of 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 and if the power flow is 55 kW.

Chapter 4

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

To be precise, the switch-mode inverter in Fig.4.1 is a converterthrough which the power flow is reversible. However, most of the timethe power flow is from the dc side to the motor on the ac side, requiringan inverter mode of operation. Therefore, these switch-mode convertersare often referred to as switch-mode inverters.

To slow down the ac motor in Fig.4.1, the kinetic energy associatedwith the inertia of the motor and its load is recovered and the ac motoracts as a generator. During the so-called braking of the motor, the powerflows from the ac side to the dc side of the switch-mode converter and itoperates in a rectifier mode. The energy recovered during the braking ofthe ac motor can be dissipated in a resistor, which can be switched inparallel with the dc bus capacitor for this purpose in Fig.4.1. However, inapplications where this braking is performed frequently, a betteralternative is regenerative braking where the energy recovered from themotor load inertia is fed back to the utility grid, as shown in the system ofFig.4.2. This requires that the converter connecting the drive to the utilitygrid be a two-quadrant converter with a reversible dc current, which canoperate as a rectifier during the motoring mode of the ac motor and as aninverter 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 asshown in Fig.4.2. There are other reasons for using such a switch-mode

153

rectifier (called a rectifier because, most of the time, the power flowsfrom the ac line input to the dc bus) to interface the drive with the utilitysystem.

Fig.4.1 Switch mode inverter in ac motor drive.

Fig.4.2 Switch-mode converters for motoring and regenerative braking in acmotor 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 tobe a dc voltage source, as was assumed in the block diagrams of Fig.4.1and Fig.4.2. Such inverters are referred to as voltage source inverters(VSIs). The other types of inverters, now used only for very high powerac motor drives, are the current source inverters (CSIs), where the dcinput to the inverter is a dc current source. Because of their limitedapplications, the CSIs are not discussed.

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

154

schemes to pulse-width modulate the inverter switches in order to shapethe output ac voltages to be as close to a sine wave as possible. Out ofthese various PWM schemes, a scheme called the sinusoidal PWM willbe discussed in detail, and some of the other PWM techniques will bedescribed in a separate section at the end of this chapter.2. Square-wave inverters. In these inverters, the input dc voltage iscontrolled in order to control the magnitude of the output ac voltage, andtherefore the inverter has to control only the frequency of the outputvoltage. 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 inverterswith single-phase output, it is possible to control the magnitude and thefrequency of the inverter output voltage, even though the input to theinverter is a constant dc voltage and the inverter switches are not pulse-width modulated (and hence the output voltage wave-shape is like asquare wave). Therefore, these inverters combine the characteristics ofthe previous two inverters. It should be noted that the voltage cancellationtechnique works only with single-phase inverters and not with three-phase inverters.

4.2 BASIC CONCEPTS OF SWITCH-MODE INVERTERSIn this section, we will consider the requirements on the switch-modeinverters. For simplicity, let us consider a single-phase inverter, which isshown in block diagram form in Fig.4.3a, where the output voltage of theinverter is filtered so that ov can be assumed to be sinusoidal. Since theinverter supplies an inductive load such as an ac motor, oi will lag ov , asshown in Fig.4.3b. The output waveforms of Fig.4.3b show that duringinterval 1, ov and oi are both positive, whereas during interval 3, ov and

oi are both negative. Therefore, during intervals 1 and 3, theinstantaneous 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 areof opposite signs during intervals 2 and 4, and therefore op flows fromthe ac side to the dc side of the inverter, corresponding to a rectifier modeof operation. Therefore, the switch-mode inverter of Fig.4.3a must becapable of operating in all four quadrants of the oo vi plane, as shownin Fig.4.3c during each cycle of the ac output. Such a four-quadrant

155

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

Fig.4.3 Single-Phase switch-mode inverter.

Fig.4.4 One-leg switch-mode inverter.

156

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

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

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

Before discussing the PWM behavior, it is necessary to define a fewterms. 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), recognizingthat the inverter output voltage ' will not be a perfect sine wave and willcontain voltage components at harmonic frequencies of 1f . Theamplitude 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:

1f

fm s

f (4.2)

In the inverter of Fig.4.4b, the switches AT and AT are controlledbased on the comparison of controlv and triv and the following outputvoltage results, independent of the direction of oi :

157

tricontrol vv onisTA , dAo Vv2

1 (4.3)

tricontrol vv onisTA , dAo Vv2

1

Fig.4.5 Pulse width modulation.Since the two switches are never off simultaneously, the output voltage

Aov fluctuates between two values ( dV2

1 and dV

2

1). Voltage Aov and

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

158

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

harmonic voltages dhAo VV2

1/ˆ having significant amplitudes are plotted.

This plot (for 0.1am ) 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 benoted that the average output voltage (or more specifically, the outputvoltage 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 Vv

V

V

vV 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 aswitching time period, Eq. (4.4) indicates how the "instantaneousaverage" value of Aov (averaged over one switching time period sT )

varies from one switching time period to the next. This "instantaneousaverage" is the same as the fundamental-frequency component of Aov .

The foregoing argument shows why controlv is chosen to be sinusoidal toprovide a sinusoidal output voltage with fewer harmonics. Let the controlvoltage vary sinusoidally at the frequency 2/11f , which is the

desired (or the fundamental) frequency of the inverter output:tVv controlcontrol 1sinˆ

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 mfor

Vtm

Vt

V

Vv (4.6)

159

Therefore, 12

ˆ1 a

daAo mfor

VmV (4.7)

which shows that in a sinusoidal PWM, the amplitude of thefundamental-frequency component of the output voltage varies linearlywith am (provided 1am .0). Therefore, the range of am from 0 to 1 isreferred to as the linear range.

Fig.4.6 Sinusoidal PWM.

2. The harmonics in the inverter output voltage waveform appear assidebands, centered around the switching frequency and its multiples, thatis, 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 9fm (which is always the case,

except in very high power ratings), the harmonic amplitudes are almostindependent of fm , though mf defines the frequencies at which they

occur. Theoretically, the frequencies at which voltage harmonics occurcan be indicated as: 1fkjmf fh

that is, the harmonic order h corresponds to the kth sideband of j times thefrequency modulation ratio fm :

kmjh f (4.8)

where the fundamental frequency corresponds to 1h . For odd values ofj, the harmonics exist only for even values of k. For even values of j, theharmonics exist only for odd values of k.

160

In Table 8-1, the normalized harmonics dhAo VV21

/ˆ are tabulated as a

function of the amplitude modulation ratio ma, assuming 9fm . 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 ofthe 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 symmetrywith the time origin shown in Fig.4.5b, which is plotted for 15fm .

Therefore, only odd harmonics are present and the even harmonicsdisappear from the waveform of Aov . Moreover, only the coefficients ofthe sine series in the Fourier analysis are finite; those for the cosine seriesare zero. The harmonic spectrum is plotted in Fig.4.5c.Table 1 Generalized Harmonics of Aov , for a Large fm

161

Example 1 In the circuit of Fig.4.4, VVd 300 , 8.0am , 39fm , and

the fundamental frequency is 47 Hz. Calculate the rms values of thefundamental-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 thefrequency modulation ratio fm . Because of the relative ease in filtering

harmonic voltages at high frequencies, it is desirable to use as high aswitching frequency as possible, except for one significant drawback:Switching losses in the inverter switches increase proportionally with theswitching frequency sf . Therefore, in most applications, the switchingfrequency is selected to be either less than 6 kHz or greater than 20 kHzto be above the audible range. If the optimum switching frequency (basedon 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 oftenoutweighed 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 motordrives (where the fundamental frequency of the inverter output may berequired 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. Onthe other hand, fm will be larger than 100 for switching frequencies

higher than 20 kHz. The desirable relationships between the triangularwaveform 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

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 21ff mm

1. Synchronous PWM. For small values of fm , the triangular

waveform signal and the control signal should be synchronized to eachother (synchronous PWM) as shown in Fig.4.5a. This synchronous PWMrequires 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 undesirablein most applications. This implies that the triangular waveform frequencyvaries with the desired inverter frequency (e.g., if the inverter outputfrequency 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 unipolarvoltage switching, to be discussed in the following sections.

4.2.1.2 Large 21ff mm

The amplitudes of subharmonics due to asynchronous PWM are smallat large values of fm . Therefore, at large values of fm , the

asynchronous PWM can be used where the frequency of the triangularwaveform 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, thesubharmonics at zero or close to zero frequency, even though small inamplitude, will result in large currents that will be highly undesirable.Therefore, the asynchronous PWM should be avoided.

4.2.1.3 Overmodulation 0.1am

In the previous discussion, it was assumed that 0.1am ,

corresponding to a sinusoidal PWM in the linear range. Therefore, theamplitude of the fundamental-frequency voltage varies linearly with am ,

as derived in Eq.(4.7). In this range of 0.1am , PWM pushes theharmonics into a high-frequency range around the switching frequency

163

and its multiples. In spite of this desirable feature of a sinusoidal PWM inthe linear range, one of the drawbacks is that the maximum availableamplitude of the fundamental-frequency component is not as high as wewish. This is a natural consequence of the notches in the output voltagewaveform of Fig.4.5b.

To increase further the amplitude of the fundamental-frequencycomponent in the output voltage, ma is increased beyond 1.0, resulting inwhat is called overmodulation. Overmodulation causes the output voltageto contain many more harmonics in the sidebands as compared with thelinear range (with 0.1am ), as shown in Fig.4.7. The harmonics withdominant amplitudes in the linear range may not be dominant duringovermodulation. More significantly, with overmodulation, the amplitudeof the fundamental-frequency component does not vary linearly with theamplitude 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 VV2

1/ˆ depends on fm in the overmodulation

region. This is contrary to the linear range ( 1am .0) where dhAo VV2

1/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 therequirements indicated previously for a small value of fm .

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

164

Fig.4.8 Voltage control by varying am .

The overmodulation region is avoided in uninterruptible powersupplies because of a stringent requirement on minimizing the distortionin the output voltage. In induction motor drives, overmodulation isnormally used.

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

24ˆ

2 1d

Aod V

VV

(4.12)

4.2.2 SQUARE-WAVE SWITCHING SCHEMEIn 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. FromFourier analysis, the peak values of the fundamental-frequency and

165

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

2273.1

2

4ˆ1

ddAo

VVV (4.13)

andh

VV Ao

hAo1

ˆˆ (4.14)

where the harmonic order h takes on only odd values, as shown inFig.4.9b. It should be noted that the square-wave switching is also aspecial case of the sinusoidal PWM switching when am becomes so largethat the control voltage waveform intersects with the triangular waveformin 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 inFig.4.8.

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

Fig.4.9 Square wave switching.

4.3 SINGLE PHASE IVERTERS4.3.1 HALF-BRIDGE INVERTERS (SINGLE PHASE)

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

166

midpotential, with a voltage dV2

1, across each capacitor. Sufficiently

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

Assuming PWM switching, we find that the output voltage waveformwill be exactly as in Fig.4.5b. It should be noted that regardless of theswitch states, the current between the two capacitors C+ and C- (whichhave equal and very large values) divides equally. When T+ is on, eitherT+ or D+ conducts depending on the direction of the output current, and iosplits equally between the two capacitors. Similarly, when the switchT 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 atmidpotential.

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

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

In a half-bridge inverter, the peak voltage and current ratings of theswitches 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 preferredover other arrangements in higher power ratings. With the same dc inputvoltage, the maximum output voltage of the full-bridge inverter is twicethat of the half-bridge inverter. This implies that for the same power, theoutput current and the switch currents are one-half of those for a

167

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 SwitchingThe 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. Withthis type of PWM switching, the output voltage waveform of leg A isidentical to the output of the basic one-leg inverter, which is determinedin the same manner by comparison of controlv and triv in Fig.4.12a. Theoutput 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 inSection 4.2 for the basic one-leg inverter completely applies to this typeof PWM switching. Therefore, the peak of the fundamental-frequency

168

component in the output voltage 1oV can be obtained from Eqs. (4.7),

(4.12), and (4.18) as:0.1ˆ

1 adao mVmV (4.19)

and 0.14ˆ

1 adod mVVV (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. Theamplitudes of harmonics in the output voltage can be obtained by usingTable 1, as illustrated by the following example.

Example 2 In the full-bridge converter circuit of Fig.4.11, VVd 300 ,

am =0.8, 39fm , and the fundamental frequency is 47 Hz. Calculate the

rms values of the fundamental-frequency voltage and some of thedominant 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*

2

1

d

hAo

d

hAod

d

hAodho V

V

V

VV

V

VVV (4.21)

169

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 inthe PWM biopolar voltage-switching scheme.

For simplicity, fictitious L-C high-frequency filters will be used at thedc side as well as at the ac side, as shown in Fig.4.13. The switchingfrequency is assumed to be very high, approaching infinity. Therefore, tofilter out the high-switching-frequency components in ov and di , thefilter components L and C required in both ac and dc-side filters approachzero. This implies that the energy stored in the filters is negligible. Sincethe converter itself has no energy storage elements, the instantaneouspower 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 1 ,

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

1 , then the output current would also be sinusoidal and would lag ov

for an inductive load such as an ac motor:tIi oo 1sin2 (4.23)

where is the angle by which oi lags ov .

170

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,

tItVtitvtiV oooodd 11* sin2sin2 (4.24)

Therefore 21* 2coscos dd

d

oo

d

ood iIt

V

IV

V

IVti (4.25)

tIIti ddd 12* 2cos2 (4.26)

where cosd

ood V

IVI (4.27)

andd

ood V

IVI

2

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 shownin Fig.4.14.

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

In practical systems, the previous assumption of a constant dc voltageas the input to the inverter is not entirely valid. Normally, this dc voltageis obtained by rectifying the ac utility line voltage. A large capacitor isused across the rectifier output terminals to filter the dc voltage. The

171

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

4.3.2.2 PWM with Unipolar Voltage SwitchingIn PWM with unipolar voltage switching, the switches in the two legs ofthe full-bridge inverter of Fig.4.11 are not switched simultaneously, as inthe previous PWM scheme. Here, the legs A and B of the full-bridgeinverter 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 tocontrol the switches in leg A:

dANAtricontrol VVandonTvv (4.29)0ANAtricontrol VandonTvv

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

dBNBtricontrol VVandonTvv (4.30)0BNBtricontrol VandonTvv

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

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

(4.31)

172

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

We notice that when both the upper switches are on, the outputvoltage 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 whenboth bottom switches AT and BT are on.

In this type of PWM scheme, when a switching occurs, the outputvoltage changes between zero and dV or between zero and dV

voltage levels. For this reason, this type of PWM scheme is called PWMwith a unipolar voltage switching, as opposed to the PWM with bipolar(between dV and dV ) voltage-switching scheme described earlier.

173

This scheme has the advantage of "effectively" doubling the switchingfrequency as far as the output harmonics are concerned, compared to thebipolar voltage switching scheme. Also, the voltage jumps in the outputvoltage 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, wherethe lowest harmonics (in the idealized circuit) appear as sidebands oftwice the switching frequency. It is easy to understand this if we choosethe frequency modulation ratio fm to be even ( fm should be odd for

PWM with bipolar voltage switching) in a single-phase inverter. Thevoltage waveforms ANv and BNv are displaced by 180° of thefundamental frequency 1f with respect to each other. Therefore, theharmonic components at the switching frequency in ANv and BNv have

the same phase ( 0.180 fo

BNAN m , 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 outputvoltage BNANo vvv . In addition, the sidebands of theswitching-frequency harmonics disappear. In a similar manner, the otherdominant harmonic at twice the switching frequency cancels out, whileits sidebands dc not. Here also

0.1ˆ1 adao mVmV (4.32)

and 0.14ˆ

1 adod mVVV (4.33)

Example 3 In Example 2, suppose that a PWM with unipolar voltageswitching scheme is used, with 38fm . Calculate the rms values of the

fundamental frequency voltage and some of the dominant harmonics inthe output voltage.Solution: Based on the discussion of unipolar voltage switching, theharmonic order h can be written as

kmjh f2 (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

2/13.212

d

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 VAt 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 voltageswitching of Example 2 shows that, in both cases, thefundamental-frequency voltages are the same for equal am However,with unipolar voltage switching, the dominant harmonic voltagescentered around fm disappear, thus resulting in a significantly lower

harmonic content.

dc-Side Current di . Under conditions similar to those in the circuit ofFig.4.13 for the PWM with bipolar voltage switching, Fig.4.16 shows thedc-side current di for the PWM unipolar voltage-switching scheme,

where 14fm (instead of 15fm for the bipolar voltage switching).

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

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

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

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

175

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

do VV4ˆ

1 (4.36)

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

inverter circuit. It is based on the combination of square-wave switchingand PWM with a unipolar voltage switching. In the circuit of Fig.4.17a,the switches in the two inverter legs are controlled separately (similar toPWM unipolar voltage switching). But all switches have a duty ratio of0.5, 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 becontrolled. During this overlap interval, the output voltage is zero as aconsequence of either both top switches or both bottom switches beingon. With 0, the output waveform is similar to a square-wave inverterwith 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 .

176

It is easier to derive the fundamental and the harmonic frequency

components of the output voltage in terms of21

90o , as is shown

in Fig.4.17b:

(4.37)

where21

90o and h is an odd integer.

Fig.4.17c shows the variation in the fundamental-frequencycomponent as well as the harmonic voltages as a function of . These arenormalized with respect to the fundamental-frequency component for thesquare-wave ( = 0) operation. The total harmonic distortion, which isthe ratio of the rms value of the harmonic distortion to the rms value ofthe fundamental-frequency component, is also plotted as a function of .Because of a large distortion, the curves are shown as dashed for largevalues of .

4.3.2.5 Switch Utilization in Full-Bridge InvertersSimilar to a half-bridge inverter, if a transformer is utilized at the

output of a full-bridge inverter, the transformer leakage inductance doesnot present a problem to the switches.

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

dT VV (4.38)

and peakoT iI , (4.39)

4.3.2.6 Ripple in the Single-Phase Inverter OutputThe ripple in a repetitive waveform refers to the difference between

the instantaneous values of the waveform and its fundamental-frequencycomponent.

177

Fig.4.18a shows a single-phase switch-mode inverter. It is assumed tobe supplying an induction motor load, which is shown by means of asimplified 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 forthe real power transfer to the load.

We can separate the fundamental-frequency and the ripplecomponents in ov and oi by applying the principle of superposition to thelinear 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 theripple frequency, respectively, where the ripple frequency componentconsists of sub-components at various harmonic frequencies.

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

1111 ooLoo LIjEVEV (4.40)

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

rippleripple kdvL

ti0

1 (4.42)

178

where k is a constant and 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 rippleis independent of the power being transferred to the load.

As an example, Fig.4.19a shows the ripple current for a square-waveinverter output. Fig.4.19b shows the ripple current in a PWM bipolarvoltage switching. In drawing Figs.4.19a and 4.19b, thefundamental-frequency components in the inverter output voltages arekept equal in magnitude (this requires a higher value of dV in the PWMinverter). The PWM inverter results in a substantially smaller peak ripplecurrent compared to the square-wave inverter. This shows the advantageof pushing the harmonics in the inverter output voltage to as highfrequencies as feasible, thereby reducing the losses in the load byreducing the output current harmonics. This is achieved by using higherinverter switching frequencies, which would result in more frequentswitching and hence higher switching losses in the inverter. Therefore,from the viewpoint of the overall system energy efficiency, a compromisemust be made in selecting the inverter switching frequency.

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

179

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

with a center tapped primary. We will initially assume that the outputcurrent 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 thedirection of oi , nVv do / , where n is the transformer turns ratiobetween the primary half and the secondary windings, as shown inFig.4.20. Similarly, when 2T is on (and 1T is off), nVv do / . Apush-pull inverter can be operated in a PWM or a square-wave mode andthe waveforms are identical to those in Figs.4.5 and 4.12 for half-bridgeand full-bridge inverters. The output voltage in Fig.4.20 equals:

0.1ˆ1 a

dao m

n

VmV (4.43)

and 0.14ˆ

1 ad

od m

n

VV

n

V (4.44)

In a push-pull inverter, the peak switch voltage and current ratings areniIVV 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 oneswitch in series conducts at any instant of time. This can be important ifthe dc input to the converter is from a low-voltage source, such as abattery, where the voltage drops across more than one switch in serieswould result in a significant reduction in energy efficiency. Also, thecontrol drives for the two switches have a common ground. It is,however, difficult to avoid the dc saturation of the transformer in apush-pull inverter.

180

The output current, which is the secondary current of the transformer,is a slowly varying current at the fundamental output frequency. It can beassumed to be a constant during a switching interval. When a switchingoccurs, the current shifts from one half to the other half of the primarywinding. This requires very good magnetic coupling between these twohalf-windings in order to reduce the energy associated with the leakageinductance of the two primary windings. This energy will be dissipated inthe switches or in snubber circuits used to protect the switches. This is ageneral phenomenon associated with all converters (or inverters) withisolation where the current in one of the windings is forced to go to zerowith every switching. This phenomenon is very important in the design ofsuch converters.

In a pulse-width-modulated push-pull inverter for producingsinusoidal 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 transformerdesigned to operate at the switching frequency in a switch-mode dcpower supply. This will result in a high transformer leakage inductance,which is proportional to the square of the number of turns, provided allother dimensions are kept constant. This makes it difficult to operate asine-wave-modulated PWM push-pull inverter at switching frequencieshigher than approximately 1 kHz.

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

in various single-phase inverters, the circuit conditions are idealized. Wewill assume that max,dV is the highest value of the input voltage, which

establishes the switch voltage ratings. In the PWM mode, the inputremains 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, weassume that there is enough inductance associated with the output load toyield a purely sinusoidal current (an idealized condition indeed for asquare-wave output) with an rms value of max,oI at the maximum load.

If the output current is assumed to be purely sinusoidal, the inverterrms volt-ampere output at the fundamental frequency equals max,1 oo IV at

the maximum rated output, where the subscript 1 designates the

181

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

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

Switch utilization ratio =TT

oo

IqV

IV 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 ofoperation at the maximum rated output. (The maximum switch utilizationoccurs at max,dd VV ).

In practice, the switch utilization ratio would be much smaller than0.16 for the following reasons: (1) switch ratings are chosenconservatively to provide safety margins; (2) in determining the switchcurrent rating in a PWM inverter, one would have to take into account thevariations in the input dc voltage available; and (3) the ripple in theoutput current would influence the switch current rating. Moreover, the

182

inverter may be required to supply a short-term overload. Thus, theswitch utilization ratio, in practice, would be substantially less than the0.16 calculated.

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

Maximum switch utilization ratio = 0.1,81

421

aaa mmm ) (4.54)

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

Example 4 In a single-phase full-bridge PWM inverter, dV varies in arange of 295-325 V. The output voltage is required to be constant at 200V (rms), and the maximum load current (assumed to be sinusoidal) is 10A (rms). Calculate the combined switch utilization ratio (under theseidealized 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) isVAIV oo 200010*200max,1 (4.55)

Therefore, from Eq.(4.46)

Switch utilization ratio = 11.014.14*325*4

2000max,1

TT

oo

IqV

IV

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

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

183

The most frequently used three-phase inverter circuit consists of threelegs, one for each phase, as shown in Fig.4.21. Each inverter leg issimilar to the one used for describing the basic one-leg inverter in Section4.2. Therefore, the output of each leg, for example ANv , (with respect to

the negative dc bus), depends only on dV and the switch status; theoutput voltage is independent of the output load current since one of thetwo switches in a leg is always on at any instant. Here, we again ignorethe blanking time required in practical circuits by assuming the switchesto be ideal. Therefore, the inverter output voltage is independent of thedirection of the load current.

Fig.4.21 Three-phase inverter.

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

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

15fm ).

It should also be noted from Fig.4.22b that an identical amount ofaverage dc component is present in the output voltages ANv and BNv ,which are measured with respect to the negative dc bus. These dccomponents 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 asingle-phase full-bridge inverter utilizing a PWM switching.

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

184

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, centeredaround 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-leginverter can be eliminated from the line-to-line voltage of a three-phaseinverter. 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, mfshould be a multiple of 3 to cancel out the most dominant harmonics inthe 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 ( 0.1am )In the linear region ( 0.1am ), the fundamental-frequency component inthe output voltage varies linearly with the amplitude modulation ratio am .From Figs.4.5b and 4.22b, the peak value of the fundamental-frequencycomponent in one of the inverter legs is

2ˆ

1d

aANV

mV (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

185

(4.57)

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

186

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

4.4.1.2 Overmodulation ( 0.1am )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 magnitudedoes 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, forsufficiently large values of ma, the PWM degenerates into a square-waveinverter waveform. This results in the maximum value of 1LLV equal to0.78 dV as explained in the next section.

In the overmodulation region compared to the region with 0.1am ,more sideband harmonics appear centered around the frequencies ofharmonics mf and its multiples. However, the dominant harmonics maynot have as large an amplitude as with 0.1am . Therefore, the powerloss in the load due to the harmonic frequencies may not be as high in theovermodulation region as the presence of additional sideband harmonicswould suggest. Depending on the nature of the load and on the switchingfrequency, the losses due to these harmonics in overmodulation may beeven 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.

187

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

4.4.2 SQUARE-WAVE OPERATION IN THREE-PHASEINVERTERSIf 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 waveformsare shown in Fig.4.24b. Here, each switch is on for 180° (i.e., its dutyratio is 50%). Therefore, at any instant of time, three switches are on.

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

188

In the square-wave mode of operation, the inverter itself cannotcontrol the magnitude of the output ac voltages. Therefore, the dc inputvoltage must be controlled in order to control the output in magnitude.Here, the fundamental-frequency line-to-line rms voltage component inthe output can be obtained from Eq. (13) for the basic one-leg inverteroperating 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 decreaseinversely proportional to their harmonic order, as shown in Fig.4.24c:

dhLL Vh

V78.0

(4.59)

where ,.......3,2,11 nnh

It should be noted that it is not possible to control the outputmagnitude in a three-phase, square-wave inverter by means of voltagecancellation as described in Section 4.3.2.4.

4.4.3 SWITCH UTILIZATION IN THREE-PHASE INVERTERSWe will assume that max,dV is the maximum input voltage that remains

constant during PWM and is decreased below this level to control theoutput voltage magnitude in a square-wave mode. We will also assumethat there is sufficient inductance associated with the load to yield a puresinusoidal output current with an rms value of max.oI (both in the PWMand the square-wave mode) at maximum loading. Therefore, each switchwould have the following peak ratings:

max,dT VV (4.60)

and max,2 oT II (4.61)

If1LLV is the rms value of the fundamental-frequency line-to-line

voltage component, the three-phase output volt-amperes (rms) at thefundamental 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

189

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

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

maximum of 0.125 for a PWM linear region with 0.1am .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 asin a single-phase inverter. In other words, using the switches withidentical ratings, a three-phase inverter with 50% increase in the numberof switches results in a 50% increase in the output volt-ampere, comparedto a single-phase inverter.

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

mode inverter in a block diagram form. It is assumed to be supplying athree-phase ac motor load. Each phase of the load is shown by means ofits simplified equivalent circuit with respect to the load neutral n. Theinduced 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

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

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

CBAkedt

diLv kn

kkn ,, (4.66)

In a three-phase, three-wire load0CBA iii (4.67a)

and 0CBA iiidt

d (4.67b)

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

0CBA eee (4.68)From the foregoing equations, the following condition for the invertervoltages can be written:

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

CNBNANnN vvvv3

1 (4.70)

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

CNBNANAn vvvv3

1

3

2 (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 componentsof the phase voltage 1AnV and the output current 1Ai are responsible for

the real power transformer since the back-emf teA is assumed to besinusoidal and the load resistance is neglected. Therefore, in a phasorform as shown in Fig.4.25b

111 AAAn ILjEV (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

191

phase-to-load-neutral voltage AnV is shown in Figs.4.26a and 4.26b forsquare-wave and PWM operations, respectively. Both inverters haveidentical 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 isobtained by using Eq.(4.42) and plotted in Fig.4.26. This current ripple isindependent of the power being transferred, that is, the current ripplewould be the same so long as for a given load inductance L, the ripple inthe 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 asquare-wave inverter.

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

4. 5 RECTIFIER MODE OF OPERATION

192

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 forphases B and C are equal in magnitude, but ± 120° displaced with respectto the control voltage of phase A.

193

4.6 PROGRAMMABLE PWM CONVERTERSThis technique combines the square wave switching and PWM to

control the fundamental output voltage as well as to eliminate thedesignated harmonics from the output voltage. This technique providesthe facility to adjust the output voltage and simultaneous optimization ofan objective function. The types of objective functions that technique canoptimize 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 voltage2- 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 dV2

1 is plotted in

Fig.4.34a, where six notches are introduced in the otherwise square-waveoutput, to control the magnitude of the fundamental voltage and toeliminate fifth and seventh harmonics. On a half-cycle basis, each notchprovides 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-wavesymmetry (sometimes it is referred to as odd quarter-wave symmetry).Therefore, only odd harmonics (coefficients of sine series) will bepresent. Since in a three-phase inverter (consisting of three such inverterlegs), the third harmonic and its multiples are canceled out in the output,these harmonics need not be eliminated from the output of the inverterleg by means of waveform notching.

194

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

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

273.14

2/

ˆ1

d

Ao

V

V (4.73)

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

188.12/

ˆmax,1

d

Ao

V

V (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

1

cos1 (4.75)

Where m is the number of jumps in the waveform. Jumps of thewaveform 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 4JTime 0 1 2 3Value 2 -2 2 -2

Then, 321 cos2cos2cos20cos22

nnnnn

bn (4.76)

Where2321

195

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

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

3211 coscoscos14

b (4.77)

05cos5cos5cos14

3215b (4.78)

07cos7cos7cos14

3217b (4.79)

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

1

14

1

7cos7coscos

5cos5cos5cos

coscoscos1

321

321

321b

(4.80)

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

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

196

To allow control over the fundamental output and to eliminate thefifth-, seventh-, eleventh-, and the thirteenth-order harmonics, fivenotches per half-cycle would be needed. In that case, each switch wouldhave 11 times the switching frequency compared with a square-waveoperation.

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

Solution: It is clear that we have only two conditions which are 05b

and 07b . So, we have only two notches per half cycle as shown in thefollowing figure.

1 2 2 1

1

2

22 122

2/d

Ao

V

V

t1

Notch1Notch2

So we need only two variables 1and 2 which can be obtained fromthe following equations:

21 coscos14

nnn

bn (4.81)

05cos5cos14

215b (4.82)

07cos7cos14

217b (4.83)

1

1

7cos7cos

5cos5cos

21

21 (4.84)

By solving the above equation we can get the value of 1 and 2 asfollowing:

o8111.121 and o8458.242

The following table shows the absolute value of each harmonics aftereliminating fifth and seventh harmonics.

197

Harmonicorder

nb for square waven

bn4 nb after eliminating 5th and 7th

21 coscos14

nnn

bn

1 1.27324 1.18713 0.244413 0.205115 0.25468 0.07 0.18189 0.09 0.14147 0.099511 0.11575 0.2122713 0.09794 0.2714115 0.08488 0.2506717 0.07490 0.1688119 0.06701 0.0718321 0.06063 0.00407THD 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 PWMinverter (Four switch inverter) uses four semiconductor switches. Thereduction in number of switches reduces switching losses, system cost andenhances reliability of the system.

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

ab

c

2dV

2dV

S1S3

S2S4

Fig.4.40 Four switch converter.

198

System AnalysisFor 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

(4.85)

The line voltages at the generator terminals can be expressed asfollows:

2

2*43

21

d

d

cb

ab

V

V

SS

SS

V

V (4.86)

Where S1 , S2 , S3 and S4 are the switching functions of switches 1,2,3and 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

andV

SV

(4.88)

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

90sin35.0

90sin35.0

30sin35.0

30sin35.0

14

13

12

11

tV

VS

tV

VS

tV

VS

tV

VS

d

m

d

m

d

m

d

m

(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)

199

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 atleast twice the maximum of input line-to-line voltage to avoid the inputcurrent distortion.

The main disadvantage of four switch converter is it needs for higherdc voltage to give the same line-to-line voltage as the six switch inverterwhich 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

PROBLEMSSINGLE PHASE1- 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 inthe output ov , 0.1am

(a) What is the highest 1oV , that can be obtained and stamped on itsnameplate 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 issupplying 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 Hzand the type of load is as shown in Fig.4.18a with L = 100 mH. If theinverter is operating in a square-wave mode, calculate the peak value ofthe ripple current.

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

4- Repeat Problem 2 but assume that the output voltage is controlledby 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-PHASE6- Consider the problem of ripple in the output current of a three-

phase square-wave inverter. Assume 2201LLV 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 asynchronous PWM mode with 39fm and 8.0am . 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

201

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

5.1 INTRODUCTIONTraditionally, dc motor drives have been used for speed and position

control applications. In the past few years, the use of ac motor servodrives in these applications is increasing. In spite of that, in applicationswhere an extremely low maintenance is not required, dc drives continueto be used because of their low initial cost and excellent driveperformance.

5.2 EQUIVALENT CIRCUIT OF dc MOTORSIn a dc motor, the field flux f is established by the stator, either by

means of permanent magnets as shown in Fig.5-1a, where f stays

constant, or by means of a field winding as shown in Fig.5-1b, where thefield current If controls f . If the magnetic saturation in the flux path can

be neglected, then:

fff Ik (5.1)

where fk is a field constant of proportionality.

The rotor carries in its slots the so-called armature winding, whichhandles the electrical power. This is in contrast to most ac motors, wherethe power-handling winding is on the stator for ease of handling thelarger amount of power. However, the armature winding in a dc machinehas to be on the rotor to provide a "mechanical" rectification of voltagesand currents (which alternate direction as the conductors rotate from theinfluence

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

201

of one stator pole to the next) in the armature-winding conductors, thusproducing a dc voltage and a dc current at the terminals of the armaturewinding. The armature winding, in fact, is a continuous winding, withoutany beginning or end, and it is connected to the commutator segments.These commutator segments, usually made up of copper, are insulatedfrom each other and rotate with the shaft. At least one pair of stationarycarbon brushes is used to make contact between the commutatorsegments (and, hence, the armature conductors), and the stationaryterminals of the armature winding that supply the dc voltage and current.

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

affem ikT (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 speedwm in the presence of a field flux f

mfea ke (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 emmT .The electrical power is calculated as

amfeaae ikieP (using Eq. 5-3) (5.4)

and the mechanical power as:amffemmm IkTP (using Eq. 5-2) (5.5)

In steady state,

me PP (5.6)Therefore, from the foregoing equations

.sec/.. radWb

Vk

WbA

Nmk et (5.7)

In practice, a controllable voltage source tv is applied to the armatureterminals to establish ai . Therefore, the current ai in the armature circuitis determined by tv , the induced back-emf ae , the armature-winding

resistance aR , and the armature-winding inductance aL:

202

dt

diiRev a

aaaat (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 WLm

mem (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 torqueof the load.

Seldom are dc machines used as generators. However, they act asgenerators while braking, where their speed is being reduced. Therefore,it is important to consider dc machines in their generator mode ofoperation. In order to consider braking, we will assume that the flux f

is kept constant and the motor is initially driving a load at a speed of m .

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 energyassociated with the motor load inertia is converted into electrical energyby the dc machine, which now acts as a generator. This energy must besomehow absorbed by the source of v, or dissipated in a resistor.

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

203

magnitude (assuming that f is constant). Ultimately, the generation

stops when the rotor comes to a standstill and all the inertial energy isextracted. If the terminal-voltage polarity is also reversed, the direction ofrotation of the motor will reverse. Therefore, a dc motor can be operatedin either direction and its electromagnetic torque can be reversed forbraking, 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 MOTORSOften in small dc motors, permanent magnets on the stator as shown in

Fig. 5-1a produce a constant field flux f . In steady state, assuming a

constant field flux ( f , Eqs. 5.2, 5.3 and 5.8 result in

aTem IkT (5.10)

mEa kE (5.11)

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

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

emT

at

Em T

k

RV

k

1 (5.13)

The plot of this equation in Fig. 5-4b shows that as the torque isincreased, the torque-speed characteristic at a given tV is essentiallyvertical, except for the droop to the voltage drop IaRa across thearmature-winding resistance. This droop in speed is quite small inintegral horsepower dc motors but may be substantial in small servomotors. More importantly, however, the torque-speed characteristics canbe shifted horizontally

204

Figure 5.4 Permanent-magnet dc motor: (a) equivalent circuit: (b) torque-speedcharacteristics: 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 becontrolled by controlling tv in a permanent-magnet dc motor with aconstant f .

In a continuous steady state, the armature current aI should not exceed

its rated value, and therefore, the torque should not exceed the ratedtorque. Therefore, the characteristics beyond the rated torque are shownas dashed in Fig. 5-4b. Similarly, the characteristic beyond the ratedspeed is shown as dashed, because increasing the speed beyond the ratedspeed would require the terminal voltage tV to exceed its rated value,which is not desirable. This is a limitation of permanent-magnet dcmotors, where the maximum speed is limited to the rated speed of themotor. 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 ispossible to significantly exceed current and torque limits on a short-termbasis. Figure 5.4c also shows the terminal voltage required as a functionof speed and the corresponding aE.

5.4 dc MOTORS WITH A SEPARATELY EXCITED FIELDWINDINGPermanent-magnet dc motors are limited to ratings of a few horsepowerand also have a maximum speed limitation. These limitations can beovercome 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 themost flexibility in controlling the dc motor, the field winding is excitedby a separately controlled dc source fv , as shown in Fig. 5-5a. As

205

indicated by Eq. 5-1, the steady-state value of f 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 and ffkk . Equation (5.14) shows that

in a dc motor with a separately excited field winding, both tV and f can

be controlled to yield the desired torque and speed. As a general practice,to maximize the motor torque capability, f (hence fI ) is kept at its

rated value for speeds less than the rated speed. With f 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-speedcharacteristics are also the same as those for a permanent-magnet dcmotor that were shown in Fig. 5-4b. With f constant and equal to its

rated value, the motor torque-speed capability is as shown in Fig. 5-5b,where this region of constant f is often called the constant-torque

region. The required terminal voltage tV in this region increases linearlyfrom approximately zero to its rated value as the speed increases fromzero 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 f is decreased by decreasing fI . Since aI is not allowed to

exceed its rated value on a continuous basis, the torque capabilitydeclines, since f is reduced in Eq. (5.2). In this so-called field-

weakening region, the maximum power aa IE (equal to mm T ) into themotor is not allowed to exceed its rated value on a continuous basis. Thisregion, also called the constant-power region, is shown in Fig. 5-5b,where emT declines with m and tV , aE , and aI stay constant at their

rated values. It should be emphasized that Fig.5.5b is the plot of themaximum continuous capability of the motor in steady state. Anyoperating point within the regions shown is, of course, permissible. In thefield-weakening region, the speed may be exceeded by 50-100% of itsrated value, depending on the motor specifications.

206

Figure 5-5 Separately excited dc motor: (a) equivalent circuit; (6) continuoustorque-speed capability.

5.5 EFFECT OF ARMATURE CURRENT WAVEFORMIn dc motor drives, the output voltage of the power electronic convertercontains an ac ripple voltage superimposed on the desired dc voltage.Ripple in the terminal voltage can lead to a ripple in the armature currentwith the following consequences that must be recognized: the form factorand torque pulsations.

5.5.1 FORM FACTORThe 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) variesproportionally 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 themotor (i.e., higher heating) and, hence, the lower the motor efficiency.

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

207

Therefore, it is desirable to improve the form factor of the armaturecurrent as much as possible.5.5.2 TORQUE PULSATIONSSince 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 isnot large. This is another reason to minimize the ripple in the armaturecurrent. It should be noted that a high-frequency torque ripple will resultin smaller speed fluctuations, as compared with a low-frequency torqueripple of the same magnitude.

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

In spite of the increasing popularity of ac servo drives, dc servo drives arestill widely used. If it were not for the disadvantages of having acommutator and brushes, the dc motors would be ideally suited for servodrives. 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-SIGNALDYNAMIC PERFORMANCE

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

208

Figure 5-6 Closed-loop position/speed dc servo drive.For analyzing small-signal dynamic performance of the motor-loadcombination around a steady-state operating point, the followingequations can be written in terms of small deviations around their steady-state values:

If we take Laplace transform of these equations, where Laplace variablesrepresent only the small-signal 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 otherinput to zero, the superposition principle yields (note that this is alinearized system)

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

(5.22)

(5.16)

(5.17)

(5.18)

(5.19)

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 inEq. 5.22 is then the motor inertia mJ . Therefore

(5.24)We will define the following constants:

(5.25)

(5.26)Using m and e in the expression for sG1 yields

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

ms by ems 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 timeconstant e , determines how quickly the armature current builds up, as

shown in Fig.5-8, in response to a step change tv in the terminal voltage,where the rotor speed is assumed to be constant.

210

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

builds up in response to a step change tv in the terminal voltage,provided that the electrical time constant e is assumed to be negligibleand, hence, the armature current can change instantaneously. Neglecting

e in Eq. 5-28, the change in speed from the steady-state condition can beobtained as

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

(5.30)where m is the mechanical time constant with which the speed changesin 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 largetransients, the torque produced by the motor is simply max,aT Ik .

5.6.2 POWER ELECTRONIC CONVERTERBased on the previous discussion, a power electronic converter supplyinga 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 fastacceleration and deceleration. The dynamic current limit is generally

211

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

3- For accurate control of position, the average voltage output of theconverter should vary linearly with its control input, independent ofthe 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 formfactor and should minimize the fluctuations in torque and speed of themotor.

5- The converter output should respond as quickly as possible to itscontrol input, thus allowing the converter to be represented essentiallyby a constant gain without a dead time in the overall servo drivetransfer function model.

Figure 5-9Mechanical time constant m ; 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 avery low power range. Therefore, the choice must be made betweenswitch-mode dc-dc converters or the line-frequency-controlledconverters. Here, only the switch-mode dc-dc converters are described.Drives with line-frequency converters can be analyzed in the samemanner.

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

212

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 thetype and filtered by means of a filter capacitor. An energy dissipationcircuit is included to prevent the filter capacitor voltage from becominglarge in case of braking of the dc motor. All four switches in the converter of Fig. 5-10 are switched duringeach cycle of the switching frequency. This results in a true four-quadrantoperation with a continuous-current conduction, where both tV and aI

can smoothly reverse, independent of each other. Ignoring the effect ofblanking time, the average voltage output of the converter varies linearlywith 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 unipolarvoltage-switching scheme can be used. Thus, the converter in Fig.5.6 canbe 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 dcmotor load flows continuously even at small values of aI . However, it isimportant to consider the peak-to-peak ripple in the armature currentbecause of its impact on the torque pulsations and heating of the motor.Moreover, a larger current ripple requires a larger peak current rating ofthe converter switches.

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

213

armature current can be expressed in terms of their dc and the ripplecomponents 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 r

araAart (5.34)

Where aaat IREV (5.35)

Anddt

tdiLtiRtv r

arar (5.36)

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

dt

tdiLtv r

ar (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 theseresults to the dc motor drive, Fig. 5-11a shows the voltage ripple tvr

and the resulting ripple current tir using Eq. 5.37. From thesewaveforms, the maximum peak-to-peak ripple can be calculated as:

sa

dPP fL

VI

2max (5.38)

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

214

Figure 5-11 Ripple ri in the armature current: (a) PWM bipolar voltage

switching, 0tV ; (b) PWM unipolar voltage switching, dt VV 2/1 .

The ripple voltage for a PWM unipolar voltage switching is shown tobe 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 (5.39)

Equations 5-38 and 5-39 show that the maximum peak-to-peak ripplecurrent is inversely proportional to aL and sf . Therefore, carefulconsideration must be given to the selection of sf and aL , where aL canbe increased by adding an external inductor in the series with the motorarmature.5.6.4 CONTROL OF SERVO DRIVES

A servo system where the speed error directly controls the powerelectronic converter is shown in Fig. 5-12a. The current-limiting circuitcomes into operation only when the drive current tries to exceed anacceptable limit max,aI during fast accelerations and decelerations.

During these intervals, the output of the speed regulator is suppressed andthe current is held at its limit until the speed and position approach theirdesired values.

To improve the dynamic response in high-performance servo drives,an internal current loop is used as shown in Fig.5-12b, where thearmature current and, hence, the torque are controlled. The currentcontrol 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 canbe controlled in a similar manner as the current-regulated modulation in adc-to-ac inverter. The only difference is that the reference current insteady state in a dc-dc converter is a dc rather than a sinusoidalwaveform. Either a variable-frequency tolerance band control, or a fixedfrequency control can be used for current control.

5.6.5 NONLINEARITY DUE TO BLANKING TIME

215

In a practical full-bridge dc-dc converter, where the possibility of a shortcircuit across the input dc bus exists, a blanking time is introducedbetween the instant at which a switch turns off and the instant at whichthe other switch in the same leg turns on. The effect of the blanking timeon the output of dc-to-ac full bridge PWM inverters. That analysis is alsovalid for PWM full-bridge dc-dc converters for dc servo drives. Theoutput voltage of the converter is proportional to the motor speed comand the output current ai is proportional to the torque emT , Fig. 5.13. If atan arbitrary speed m , 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 theperformance 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

Fig.5.13 Effect of blanking time.

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

aL , switching frequency sf , blanking time ct , and switching times t, ofthe solid state devices in the dc-dc converter can be summarized asfollows: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 tf 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 DRIVESUnlike servo drives, the response time to speed and torque commands isnot as critical in adjustable-speed drives. Therefore, either switch-modedc-dc converters as discussed for servo drives or the line-frequencycontrolled converters can be used for speed control.

5.7.1 SWITCH-MODE dc-dc CONVERTERIf a four-quadrant operation is needed and a switch-mode converter isutilized, 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 thetwo-quadrant converter shown in Fig. 5-14a can be used. It consists oftwo switches, where one of the switches is on at any time, to keep theoutput voltage independent of the direction of ai . The armature current

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 remainsunipolar. Since ai can flow in both directions, unlike in the single-switchstep-down and step-up dc-dc converters, ai in the circuit of Fig. 5-14awill not become discontinuous. For a single-quadrant operation where the speed remains unidirectionaland braking is not required, the step-down converter shown in Fig. 5-14bcan be used.5.7.2 LINE-FREQUENCY CONTROLLED CONVERTERSIn many adjustable-speed dc drives, especially in large power ratings, itmay be economical to utilize a line-frequency controlled converter of thetype 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 themotor armature may be required to keep the ripple in ai low, to minimizeits 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 deadtime in responding to the changes in the speed control signal, comparedto high-frequency switch-mode dc-dc converters. Once a thyristor or apair of thyristors is triggered on in the circuits of Fig.5.15, the delay angle

that controls the converter output voltage applied to the motorterminals cannot be increased for a portion of the 50-Hz cycle. This may

218

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

The current through these line-frequency controlled converters isunidirectional, but the output voltage can reverse polarity. The two-quadrant operation with the reversible voltage is not suited for dc motorbraking, which requires the voltage to be unidirectional but the current tobe 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, asdepicted in Fig. 5-16b.

An alternative to using two converters is to use one phase-controlledconverter 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 reducedrapidly, since the direction of rotation remains the same, Ea is of the samepolarity as in the motoring mode. Therefore, to let the converter go intoan inverter mode, contactors 1M and 2M are opened and 1R and 2R areclosed. It should be noted that tie contactors switch at zero current whenthe current through them is brought to zero by the converter.

Fig.5.15 Line-frequency-controlled converters for dc motor drives: (a) single-phaseinput, (b) three-phase input.

219

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

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

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

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

controlv or the delay angle , the discontinuous current causes the outputvoltage to go up. This voltage rise causes the motor speed to increase atlow 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 thearmature resistance; additional drop in speed occurs in the phase-controlled converter-driven motors due to commutation voltage dropsacross the ac-side inductance sL , which approximately equal

aS IL /2 in single-phase converters and aS IL /3 in three-phaseconverters. These effects results in poor speed regulation under an open-loop operation.

220

Fig.5.17 Effects of discontinuous ai on m .

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

loop control is shown in Fig.5.18 where the speed command * isgenerated 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, thuspreventing the rotor current from exceeding its rating. The slope of the

dtd / limiter can be adjusted to match the motor-load inertia. The currentlimner in such drives may be just a protective measure, whereby if themeasured current exceeds its rated value, the controller shuts the driveoff. A manual restart may be required. As discussed in Section 5-6, aclosed-loop control can also be implemented.

5.7.5 FIELD WEAKENING IN ADJUSTABLE-SPEED dc MOTORDRIVES

In a dc motor with a separately excited field winding, the drive can beoperated at higher than the rated speed of the motor by reducing the fieldflux f . Since many adjustable speed drives, especially at higher power

ratings, employ a motor with a wound field, this capability can beexploited by controlling the field current and f . The simple line

frequency phase-controlled converter shown in Fig. 5-15 is normally usedto control fI through the field winding, where the current is controlled in

magnitude but always flows in only one direction. If a converter topologyconsisting of only thyristors (such as in Fig. 5-15) is chosen, where theconverter output voltage is reversible, the field current can be decreasedrapidly.

221

Figure 5-18 Open-loop speed control.

5.7.6 POWER FACTOR OF THE LINE CURRENT INADJUSTABLE-SPEED DRIVES

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

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

222

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

223

PROBLEMS1- Consider a permanent-magnet dc servo motor with the followingparameters:

224

Chapter 6

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, novoltage drop when conducting , infinite Z when off.

||

||,

,0

2,

,

,

,0

2,

,

sin

sin

sin

tEe

tEe

tEe

mT

mL

m

224

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

where,2

00 valueAverage)(

2

1

2tdte

aL

For fundamental component :-

point.datumthe

andlfundamentathebetweenanglentdisplacemetan

componentlfundamentaofvaluepeak

sin)(1

,cos)(1

1

111

21

211

2

012

01

b

a

bac

tdttebtdttea LL

For the n-th harmonics,2

0

2

0sin)(

1,cos)(

1tdtntebtdtntea LnLn

For the fundamental of the load voltage,

2sin)(2

12costantan

]2sin)(2[)12(cos2

)2sin)(2(2

)12(cos2

cos)(1

1

1

111

221

1

2

01

b

a

Ec

Eb

Etdttea

m

m

mL

RMS Load Voltage

r.m.s. value of tdteEe LLL2

0

2 )(2

1

]2sin)(2[2

1

2

]2sin)(2[4

)sin(2

1 22,

,22

mL

mmL

EE

EtdtEE

225

with resistive load, the instantaneous load current is :-2,

,|sin

R

tE

R

ei mLL

The r.m.s. load current

]2sin)(2[2

1]2sin)(2[

2

1

2 R

E

R

EI m

L

6.3 Power Dissipation

]2sin)(2[2

valuesr.m.s.theare&where

powerAverage

product.ampvoltousinstantaneofvalueAverage2

1

2

22

2

0

R

E

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 cosIEP

where1

11

11 cos,

1

2currentlfundamentar.m.s.

c

b

R

cI

6.4 Power factor in non-sinusoidal circuits

In general,IE

PPF

sVoltampereApparentPowerAverage

.supplyatcurrentr.m.s.

supply.atvoltager.m.s.

I

E

Definition is true irrespective for any waveform and frequency.Let e and i be periodic in ,2

2

0

22

0

2

2

0

21

;21

21

tdiItdeE

tdieP

The usual aim is to obtain unity PF at the supply.

226

To obtain unity P.F., certain conditions must be observed w.r.t. e( t) andi( t):-(i) e( t) and i( t) must be of the same frequency.(ii) e( t) and i( t) must be of the same wave shape (whatever the

wave shape)(iii) e( t) and i( t) 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 2 but is non-sinusoidal.Average power is obtained by combining in-phase voltage and currentcomponents of the same frequency.

FactorntDisplacemexFactorDistortion

coscos

cos

1111

11

I

I

IE

IE

IE

PPF

IEP

Distortion Factor = 1 for sinusoidal operationDisplacement factor is a measure of displacement between e( t) andi( t).Displacement Factor =1 for sinusoidal resistive operation.Calculation of PF

]2sin)(2[2

1

]2sin)(2[2

1

]2sin)(2[2 2

2

2

R

EE

R

RE

IE

RIPF

R-L load

R

LLRZ 1222 tan,

227

Steady-state at 0 (sinusoidal operation)

Applying gate signal at t = , where > The start of conduction is delayed until .t Subsequent totriggering, let the instantaneous current )( ti consists of hypotheticalsteady-state components )( tiss and transient component )( titrans ,

)()()( tititi transss

Now the t , the instantaneous steady-state component has the value,

)sin()(Z

Ei mss

But at ,t total current 0)( ti . At t = , iss( ) = -itrans( )

)sin()(Z

Eti m

trans

Subsequent to The transient component decay exponentially from it’s instantaneous -

)sin(Z

Em by the constant.R

LZ

tm

trans eZ

Eti )sin()(

For t > ,)(

)sin()(t

L

Rm

trans eZ

Eti

Complete solution for first cycle

228

2)()(

0

)(2,,,,0

)sin()sin(

)sin()[sin()(

tL

Rx

tt

R

xt

t

Rxxm

e

etZ

Eti

Extinction Angle For the current in the interval xt

)()sin()sin()(

tL

Rmm e

Z

Et

Z

Eti

But at 0)(, tit , hence

)()sin()sin(0

xL

Rmm e

Z

Ex

Z

E

229

But cotand0L

R

Z

Em )cot()sin()sin(0 xex

If and are known, x can be calculated. However, this is a

transcendental equation (i.e. cannot be solved explicitly and no way ofobtaining ),(fx ).

Method of solution is by iteration,e.g. If = 600, = 1200 = 2 / 3, 578.0cot

0)

3

2(578.0

00 222)60120sin()60sin( xexx

Rough Approximation :- 0180x

whereLlargeandRsmallfor20~15LRfor15~10

LsmallandRlargefor10~50000

00

For the previous example,

0000

0000

000

2202060180

2251560180

2015,60

xor

x

Load voltage

]2sin2sin)(2[2

]2cos2[cos2

sin)(

1

1

2,,,,0

xxE

b

xE

a

tEte

m

m

xxmL

230

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, PT

N ; r.m.s. load voltage, EL

T

N

Power Factor =T

N

T

N

R

EE

T

N

R

E

EI

P

2

6.6 Advantages and Disadvantages of Integral cycle ControlAdvantages1. Avoids the radio frequency interference created by phase - angle

switching .2. Able to switch the loads with a large thermal time const .Disadvantages1. 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 21- Single phase half-wave diode rectifier is connected to 220 V, 50 Hz

supply to feed 5 pure resistor. Draw load voltage and current anddiode voltage drop waveforms along with supply voltage. Then,calculate (a) The rectfication effeciency. (b) Ripple factor of loadvoltage. (c) Transformer Utilization Factor (TUF) (d) Peak InverseVoltage (PIV) of the diode. (e) Crest factor of supply current.

2- The load of the rectifier shown in problem 1 is become 5 pureresistor and 10 mH inductor. Draw the resistor, inductor voltagedrops, and, load current along with supply voltage. Then, find anexpression for the load current and calculate the conduction angle,

. 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 anthe 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 theload of the rectifier shown in problem 2. Calculate the load currentduring 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 positivepolarity terminal are as follows:

tVtVtVVtv d 3cos2sin2cos2 311

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 viaunity turns ratio center-tap transformer to feed 5 resistor load.Draw load voltage and currents and diode currents waveforms alongwith 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 factorof supply current.

4

7- Single phase diode bridge rectifier is connected to 220 V, 50 Hzsupply to feed 5 resistor. Draw the load voltage, diodes currentsand calculate (a) The rectfication effeciency. (b) Ripple factor of loadvoltage. (c) Transformer Utilization Factor (TUF) (d) Peak InverseVoltage (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 5resistor in series with 10mH inductor. Calculate and draw the loadcurrent during the first two periods of supply voltages waveform.

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

10- If the load of problem 7 is changed to be 45 A pure DC. Draw diodediodes currents and supply currents along with supply voltage. Then,calculate (a) The rectfication effeciency. (b) Ripple factor of loadvoltage. (c) Transformer Utilization Factor (TUF) (d) Peak InverseVoltage (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 ,50Hzsupply. The supply has 4 mH source inductance. The load connectedto the rectifier is 45 A pure DC current. Draw, output voltage, diodecurrents and supply current along with the supply voltage. Then,calculate the DC output voltage, THD of supply current and inputpower factor, and, input power factor and THD of the voltage at thepoint of common coupling.

7

12- Three-phase half-wave diode rectifier is connected to 380 V, 50Hzsupply via 380/460 V delta/way transformer to feed the load with 45 ADC current. Assuming ideal transformer and zero source inductance.Then, draw the output voltage, secondary and primary currents alongwith supply voltage. Then, calculate (a) Rectfication effeciency. (b) Crestfactor 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, 50Hzsupply to feed 10 resistor. Draw the output voltage, diode currentsand supply current of phase a. Then, calculate: (a) The rectficationeffeciency. (b) Ripple factor of load voltage. (c) TransformerUtilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of thediode. (e) Crest factor of supply current.

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

9

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

10

17- Single phase diode bridge rectifier is connected to square waveformwith amplitude of 200V, 50 Hz. The supply has 4 mH sourceinductance. The load connected to the rectifier is 45 A pure DCcurrent. Draw, output voltage, diode currents and supply currentalong 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, 1SL

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

11

Problems Of Chapter 31- Single phase half-wave controlled rectifier is connected to 220 V,

50Hz supply to feed 10 resistor. If the firing angle o30 drawoutput voltage and drop voltage across the thyristor along with thesupply voltage. Then, calculate, (a) The rectfication effeciency. (b)Ripple factor. (c) Peak Inverse Voltage (PIV) of the thyristor. (d) Thecrest factor FC of input current.

2- Single phase half-wave controlled rectifier is connected to 220 V,50Hz supply to feed 5 resistor in series with mH10 inductor if the

firing angle o30 .(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 outputvoltage.

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

12

3- Solve problem 2 if there is a freewheeling diode is connected in shuntwith the load.

4- single phase full-wave fully controlled rectifier is connected to 220V,

50 Hz supply to feed 5 resistor, if the firing angle o40 . Drawthe load voltage and current, thyristor currents and supply current.Then, calculate (a) The rectfication effeciency. (b) Peak InverseVoltage (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 withthe 5 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 outputvoltages.

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 . Draw the load voltage, thyristor, and load currents.Then, calculate (a) the rectfication effeciency. (b) Ripple factor ofoutput voltage. (c) Crest factor of supply current. (d) Use Fourierseries to fiend an expression for supply current. (e) THD of supplycurrent. (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 220V, 50Hz supply to feed 5 resistor in series with 5 mH inductor, theload is connected in shunt with freewheeling diode. Draw the loadvoltage and current, resistor voltage and inductor voltage diodes andthyristor 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 DCload of 1 kW with pure DC current. A 1.5-kVA-isolation transformerwith a source-side voltage rating of 120 V at 50 Hz is used. It has atotal leakage reactance of 8% based on its ratings. The ac sourcevoltage of nominally 120 V is in the range of -10% and +5%. Then,Calculate the minimum transformer turns ratio if the DC load voltageis to be regulated at a constant value of 100 V. What is the value of awhen 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 = 135°. Using PSIM,obtain sv , si , dv , and di waveforms in steady state.

12- In the inverter of Problem 12, vary the delay angle from a value of165° down to 120° and plot di versus . Obtain the delay angle b ,below which di becomes continuous. How does the slope of thecharacteristic in this range depend on SL ?

13-In the three-phase fully controlled rectifier is connected to 460 V at 50Hz 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 for which

dmdc VV 5.0 . Draw dv waveform and identify the devices thatconduct during various intervals. Obtain the DPF, PF, and %THD inthe input line current and compare results with a full-bridge converteroperating 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 12kW. A Y- Y connected isolation transformer has a per-phase rating of5 kVA and an AC source-side voltage rating of 120 V at 50 Hz. It hasa total per-phase leakage reactance of 8% based on its ratings. The acsource voltage of nominally 208 V (line to line) is in the range of-10% and +5%. Assume the load current is pure DC, calculate theminimum transformer turns ratio if the DC load voltage is to beregulated at a constant value of 300 V. What is the value of 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 and if the power flow is 55 kW.

18

Problems Of Chapter 41- In a single-phase full-bridge PWM inverter, the input dc

voltage varies in a range of 295-325 V. Because of the lowdistortion required in the output ov , 0.1am

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

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

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

19

2- Consider the problem of ripple in the output current of asingle-phase full-bridge inverter. Assume 1oV = 220 V at a frequencyof 47 Hz and the type of load is as shown in Fig.18a with L = 100mH. If the inverter is operating in a square-wave mode, calculate thepeak value of the ripple current.

20

3- Repeat Problem 2 with the inverter operating in a sinusoidalPWM 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 asrequired in the PWM inverter of Problem 3.

21

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

22

THREE-PHASE6- Consider the problem of ripple in the output current of a three-

phase square-wave inverter. Assume 2201LLV V at a frequency

of 52 Hz and the type of load is as shown in Fig.25a with L = 100mH. Calculate the peak ripple current defined in Fig.26a.

23

7- Repeat Problem 6 if the inverter of Problem 6 is operating in asynchronous PWM mode with 39fm and 8.0am . Calculate the

peak ripple current defined in Fig.26b.

24

8- In the three-phase, square-wave inverter of Fig.24a, considerthe 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 purelyinductive, where the load resistance, though finite, can be neglected.

26

Problems Of Chapter 51- Consider a permanent-magnet dc servo motor with the followingparameters:

27

28

29

30

31

32

33

Power Electronics (2)Assignment