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Power electronics note Document Transcript

  • 1. Power Electronics Dr.Ali Mohamed Eltamaly Mansoura University Faculty of Engineering
  • 2. Chapter Four 113 Contents 1 Chapter 1 Introduction1.1. Definition Of Power Electronics 11.2 1 Main Task Of Power Electronics1.3 Rectification 21.4 DC-To-AC Conversion 31.5 DC-to-DC Conversion 41.6 AC-TO-AC Conversion 41.7 Additional Insights Into Power Electronics 51.8 Harmonics 71.9 Semiconductors Switch types 12 Chapter 2 17 Diode Circuits or Uncontrolled Rectifier2.1 17 Half Wave Diode Rectifier2.2 29 Center-Tap Diode Rectifier2.3 35 Full Bridge Single-Phase Diode Rectifier2.4 40 Three-Phase Half Wave Rectifier2.5 49 Three-Phase Full Wave Rectifier2.6 56 Multi-pulse Diode Rectifier
  • 3. Fourier Series 114 Chapter 3 59 Scr Rectifier or Controlled Rectifier3.1 59 Introduction3.2 60 Half Wave Single Phase Controlled Rectifier3.3 73 Single-Phase Full Wave Controlled Rectifier3.4 91 Three Phase Half Wave Controlled Rectifier3.5 95 Three Phase Half Wave Controlled Rectifier With DC Load Current3.6 98 Three Phase Half Wave Controlled Rectifier With Free Wheeling Diode3.7 100 Three Phase Full Wave Fully Controlled Rectifier Chapter 4 112 Fourier Series4-1 112 Introduction4-2 113 Determination Of Fourier Coefficients4-3 119 Determination Of Fourier Coefficients Without Integration
  • 4. Chapter 1 Introduction1.1. 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 DC voltage to another DC voltage, and • Cycloconverter and cycloinverter converting an AC voltage to another AC voltage.In addition, SCRs and other power semiconductor devices are used 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,
  • 5. 2 Chapter One• Double three-phase, three-pulse star-connected rectifiers with inter-phase transformer (IPT),• Three-phase semi-controlled bridge rectifier,• Three-phase fully-controlled bridge rectifier, and ,• Double three-phase fully controlled bridge rectifiers with IPT. Apart from the configurations listed above, there are series-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 specific application like electroplating1.3 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.
  • 6. Introduction 3Either 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
  • 7. 4 Chapter OnekHz, 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 the control circuitry.• Microcontrollers and digital signal processors for use in sophisticated applications.• Many power electronic circuits have an electrical machine as their load. In AC variable speed drive, it may be a reluctance
  • 8. Introduction 5 motor, an induction motor or a synchronous motor. In a DC variable speed drive, it is usually a DC shunt motor.• In a circuit such as an inverter, a transformer may be connected at its output and the transformer may have to operate with a nonsinusoidal waveform at its input.• A pulse transformer with a ferrite core is used commonly to transfer the gate signal to the power semiconductor device. A ferrite-cored transformer with a relatively higher power output is also used in an application such as a high frequency inverter.• Many power electronic systems are operated with negative feedback. A linear controller such as a PI controller is used in relatively simple applications, whereas a controller based on digital or state-variable feedback techniques is used in more sophisticated applications.• Computer simulation is often necessary to optimize the design of a power electronic system. In order to simulate, knowledge of software package such as MATLAB, Pspice, Orcad,…..etc. and the know-how to model nonlinear systems may be necessary. The study of power electronics is an exciting and a 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 Harmonics The invention of the semiconductor controlled rectifier (SCR orthyristor) 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
  • 9. 6 Chapter One“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 Components There are many bad effects of harmonics on the power systemcomponents. 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 the load current and square harmonics frequency,• The hysterics losses will increase,• The loading capability is derated by harmonic currents , and,• Possible resonance may occur between transformer inductance and line capacitor.In Capacitors• The life expectancy decreases due to increased dielectric losses that cause additional heating, reactive power increases due to harmonic voltages, and,• Over voltage can occur and resonance may occur resulting in harmonic magnification.In Cables• Additional heating occurs in cables due to harmonic currents because of skin and proximity effects which are function of frequency, and,• The I2R losses increase.In Switchgear• Changing the rate of rise of transient recovery voltage, and,• Affects the operation of the blowout.In Relays• Affects the time delay characteristics, and,
  • 10. Introduction 7• False tripping may occurs.In Motors• Stator and rotor I2R losses increase due to the flow of harmonic currents,• In the case of induction motors with skewed rotors the flux changes in both the stator and rotor and high frequency can produce substantial iron losses, and,• Positive sequence harmonics develop shaft torque that aid shaft rotation; negative sequence harmonics have opposite effect.In Generators• Rotor and stator heating ,• Production of pulsating or oscillating torques, and,• Acoustic noise.In Electronic Equipment• Unstable operation of firing circuits based on zero voltage crossing,• Erroneous operation in measuring equipment, and,• Malfunction of computers allied equipment due to the presence of ac supply harmonics.1.7.2 Harmonic Standards It should be clear from the above that there are serious effects on thepower 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 ω canbe expressed as in the following equation. a0 ∞ f (t ) = + ∑ (a n cos(nωt ) + bn sin( nωt ) ) (1.1) 2 n=1
  • 11. 8 Chapter One 2π 1 where a n = π ∫ f (t ) cos (nωt ) dωt (1.2) 0 2π 1 and bn = π ∫ f (t ) sin (nωt ) dωt (1.3) 0 Each frequency component n has the following value f n (t ) = a n cos ( nωt ) + bn sin (nωt ) (1.4) fn(t) can be represented as a phasor in terms of its rms value as shown in the following equation a n + bn 2 2 Fn = e jϕ n (1.5) 2 − bn Where ϕ n = tan −1 (1.6) an The amount of distortion in the voltage or current waveform is qualified by means of an Total Harmonic Distortion (THD). The THD in current and voltage are given as shown in (1.7) and (1.8) respectively. 2 Is − I s1 2 ∑ I sn 2 n≠n THDi = 100 * = 100 * (1.7) I s1 I s1 Vs2 − Vs2 ∑Vsn 2 1 n≠n THDv = 100 * = 100 * (1.8) Vs1 Vs1 Where THDi & THDv The Total Harmonic Distortion in the current and voltage waveforms Current and voltage limitations included in the update IEE 519 1992 are shown in Table(1.1) and Table(1.2) respectively [2].Table (1.1) IEEE 519-1992 current distortion limits for general distributionsystems (120 to 69kV) the maximum harmonic current distortion in percent of I L Individual Harmonic order (Odd Harmonics) I SC / I L n<11 11≤ n<17 17≤ n<23 23≤ n<35 35≤ n< TDD <20 4.0 2.0 1.5 0.6 0.3 5.0 20<50 7.0 3.5 2.5 1.0 0.5 8.0 50<100 10.0 4.5 4.0 1.5 0.7 12.0 100<1000 12.0 5.5 5.0 2.0 1.0 15.0 >1000 15.0 7.0 6.0 2.5 1.4 20.0
  • 12. Introduction 9 ∞ 100 Where; TDD (Total Demand Distortion) = I ML ∑ I n2 , n=2 Where I ML is the maximum fundamental demand load current (15 or30min demand). I SC is the maximum short-circuit current at the point of commoncoupling (PCC). I L is the maximum demand load current at the point of commoncoupling (PCC). Table (1.2) Voltage distortion limitsBus voltage at PCC Individual voltage distortion (%) THDv (%)69 kV and blow 3.0 5.069.001 kV through 161kV 1.5 2.5161.001kV and above 1 1.51.8 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.
  • 13. 10 Chapter One 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 diode can withstand only so much voltage before it breaks down. So if the PIV is exceeded than the PIV rated for the diode, then the diode will conduct in both forward and reverse bias and the diode will be immediately destroyed. • Maximum Average Current: Is the average current that the diode can carry.It is convenient for simplicity in discussion and quite useful in 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
  • 14. Introduction 11current 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. Incase 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
  • 15. 12 Chapter Onecentral 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.
  • 16. Introduction 13In 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 gatecurrent 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.
  • 17. 14 Chapter OneThe 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.
  • 18. Introduction 15 Fig.1.6 Operating characteristics of TRIAC.ele146DIAC DIAC is like a TRIAC without a gate terminal. DIAC conducts currentin 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.
  • 19. 16 Chapter One Fig.1.8 DIAC v-i characteristics1.9 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 pnp Fig.1.9 The electric symbol of npn and pnp transistors.
  • 20. Introduction 17Fig.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 V BB representinga forward bias voltage and Vcc 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 β as shown in ( ) Iβ= C IBCurrent and voltage analysis of NPN transistors is shown if Fig.1.11. It isclear from Fig.1.11 that:V Rb = V BB − V BE = I B * R BThen, the base current can be obtained as shown in the followingequation: V − V BEI B = BB RB
  • 21. 18 Chapter OneThe voltage on RC resistor are:V RC = I C * RCVCE = VCC − I C * RCFig.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 VCC increases the voltage VCE gradually as shown in thesaturation region. When VCE become more than 0.7 V, the base to collector junctionbecome reverse bias and the transistor moves to linear region. In linearregion I C approximately constant for the same amount of base currentwhen VCE increases.When VCE 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 MOSFET The power MOSFET has two important advantages over than BJT,First of them, is its need to very low operating gate current, the second of
  • 22. Introduction 19them, 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 theadvantage 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:
  • 23. 20 Chapter One1.12 Power Junction Field Effect Transistors This device is also sometimes known as the static induction transistor(SIT). It is effectively a JFET transistor with geometry changes to 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 Thyristor This device is essentially a modification of the SIT. The drain of theSIT 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 Thyristors The MOS-controlled thyristor (MCT) is a relatively new device whichis 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.
  • 24. Introduction 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.
  • 25. Chapter 2Diode Circuits or Uncontrolled Rectifier2.1 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 Parameters In most rectifier applications, the power input is sine-wave voltageprovided 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
  • 26. Diode Circuits or Uncontrolled Rectifier 23between 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, Vdc ,The average value of the output current, I dc ,The rms value of the output voltage, Vrms ,The rms value of the output current, I rmsThe output DC power, Pdc = Vdc * I dc (2.1)The output AC power, Pac = Vrms * I rms (2.2) PThe effeciency or rectification ratio is defiend as η = dc (2.3) Pac 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:-Vac = Vrms − Vdc 2 2 (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. V FF = rms (2.5) Vdc 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. Vac Vrms − Vdc 2 2 2 VrmsRF = = = 2 − 1 = FF 2 − 1 (2.6) Vdc Vdc Vdc The Transformer Utilization Factor (TUF) is defiend as:- PTUF = dc (2.7) VS I S
  • 27. 24 Chapter Two Where VS and I S 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. I S − I S1 2 2 2 ISTHDi = 2 = 2 −1 (2.8.a) I S1 I S1 VS2 − VS21 VS2THDv = = −1 (2.8.b) VS21 VS21 where I S1 and VS1 are the fundamental component of the input currentand voltage, I S and VS 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:- I S ( peak ) CF = (2.9) IS In general, power factor in non-sinusoidal circuits can be obtained asfollowing: Real Power P PF = = = cos φ (2.10) Apparent Voltamperes VS I S Where, φ 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. P V I1 cos φ1 I S1PF = = = cos φ = Distortion Factor * Displaceme nt Faactor (2.11) 1 VS I S VS I S IS Where φ1 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 v(ωt ) and i (ωt ) .
  • 28. Diode Circuits or Uncontrolled Rectifier 252.3 Single-Phase Half-Wave Diode Rectifier Most of the power electronic applications operate at a relative highvoltage 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: π 1 VVdc = 2π∫Vm sin ωt dωt = m π (2.12) 0Where, Vm is the maximum value of supply voltage. Because the load is resistor, the average or DC component of loadcurrent is: V VI dc = dc = m (2.13) R π R The root mean square (rms) value of a load voltage is defined as: π 1 VVrms = ∫ Vm sin 2 ωt dωt = m 2 (2.14) 2π 2 0 Similarly, the root mean square (rms) value of a load current is definedas: V VI rms = rms = m (2.15) R 2R
  • 29. 26 Chapter Two It is clear that the rms value of the transformer secondary current, I Sis the same as that of the load and diode currents VThen I S = I D = m (2.15) 2R Where, I D 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.
  • 30. Diode Circuits or Uncontrolled Rectifier 27Example 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 Vdc is defiend as: π 1 V VVdc = 2π ∫ Vm sin(ωt ) dωt = m (− cos π − cos(0)) = m 2π π 0 Vdc VmThen, I dc = = R πR π 1 V Vm VVrms = 2π ∫ (Vm sin ωt ) 2 = m , 2 I rms = 2R and, VS = m 2 0The rms value of the transformer secondery current is the same as that of Vthe load: I S = m Then, the efficiency or rectification ratio is: 2R Vm Vm * Pdc Vdc * I dc π πRη= = = = 40.53% Pac Vrms * I rms Vm Vm * 2 2R Vm V π(b) FF = rms = 2 = = 1.57 Vdc Vm 2 π Vac(c) RF = = FF 2 − 1 = 1.57 2 − 1 = 1.211 Vdc Vm Vm P π π R(d) TUF = dc = = 0.286 = 28.6% VS I S Vm Vm 2 2R(e) It is clear from Fig2.2 that the PIV is Vm . I S ( peak ) Vm / R(f) Creast Factor CF, CF = = =2 IS Vm / 2 R
  • 31. 28 Chapter Two2.3.2 Half Wave Diode Rectifier With R-L Load In case of RL load as shown in Fig.2.3, The voltage source, VS is analternating sinusoidal voltage source. If vs = Vm sin (ωt ) , v s is positivewhen 0 < ω t < π, and vs is negative when π < ω t <2π. When v s startsbecoming positive, the diode starts conducting and the source keeps thediode in conduction till ω t reaches π radians. At that instant defined byω t =π 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 vs changes from a positive to a negative value, the voltageacross the diode changes its direction and there is current through the loadat the instant ω t = π 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 < ω t < β, where β > π ( β 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 β < ω t < 2π, the diode blocks current and acts as anopen switch. For this period, there is no equation defining the behavior ofthe circuit.For 0 < ω t < β, the following differential equation defines the circuit: di L + R * i = Vm sin (ωt ), 0 ≤ ωt ≤ β (2.17) dtDivide the above equation by L we get: di R V + * i = m sin (ωt ), 0 ≤ ωt ≤ β (2.18) dt L L The instantaneous value of the current through the load can beobtained from the solution of the above equation as following:
  • 32. Diode Circuits or Uncontrolled Rectifier 29 R ⎡ R ⎤ −∫ dt ∫ dt Vmi (t ) = e ⎢ L ⎢e ∫ L * L sin ωt dt + A⎥ ⎥ (2.19) ⎣ ⎦Where A is a constant. − t⎡ ⎤ R R t VThen; i (t ) = e ⎢ ∫ L ⎢ e L * m sin ωt dt + A⎥ L ⎥ (2.20) ⎣ ⎦By integrating (2.20) (see appendix) we get: R Vm − ti (t ) = (R sin ωt − ωL cosωt ) + Ae L (2.21) R 2 + w 2 L2 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.
  • 33. 30 Chapter TwoAssume Z∠φ = R + j wLThen Z 2 = R 2 + w2 L2 , Z ωL wLR = Z cos φ , ωL = Z sin φ and tan φ = RSubstitute these values into (2.21) we get the following equation: Φ R R V − ti (t ) = m (cos φ sin ωt − sin φ cosωt ) + Ae L Z R V − tThen, i (t ) = m sin (ωt − φ ) + Ae L (2.22) ZThe above equation can be written in the following form: R ωt − ωt − V Vi (t ) = m sin (ωt − φ ) + Ae ω L = m sin (ωt − φ ) + Ae tan φ (2.23) Z Z The value of A can be obtained using the initial condition. Since thediode starts conducting at ω t = 0 and the current starts building up fromzero, i (0 ) = 0 (discontinuous conduction). The value of A is expressed bythe following equation: V A = m sin (φ ) ZOnce the value of A is known, the expression for current is known. Afterevaluating A, current can be evaluated at different values of ωt . ⎛ ωt ⎞ − Vm ⎜ tan φ ⎟i (ωt ) = ⎜ sin (ωt − φ ) + sin (φ )e Z ⎜ ⎟ (2.24) ⎟ ⎝ ⎠ Starting from ω t = π, as ωt increases, the current would keepdecreasing. For some value of ωt , say β, the current would be zero. If ω t> β, the current would evaluate to a negative value. Since the diodeblocks current in the reverse direction, the diode stops conducting whenωt reaches β. The value of β can be obtained by substituting thati (ωt ) = 0 wt = β into (2.24) we get: ⎛ β ⎞ − Vm ⎜ ⎟ ⎜ sin (β − φ ) + sin (φ )e tan φi(β ) = ⎟=0 (2.25) Z ⎜ ⎟ ⎝ ⎠
  • 34. Diode Circuits or Uncontrolled Rectifier 31 The value of β 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). β V VVdc 2π ∫ = m * sin ωt dωt = m * (1 − cos β ) 2π (2.26) 0 β 1 VmVrms = * ∫ (Vm sin ωt ) 2 dwt = * β + 0.5(1 − sin( 2 β ) (2.27) 2π 2 π 02.3.3 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 vs be defined as Vm sin (ωt ) which is positivewhen 0 < ωt < π radians and it is negative when π < ω t < 2π radians.When vs is positive, diode D1 conducts and the output voltage, vobecome positive. This in turn leads to diode D2 being reverse-biasedduring this period. During π < wt < 2π, the voltage vo 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 vowould 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 vs which is negative. Fig.2.5 Half wave diode rectifier with free wheeling diode.
  • 35. 32 Chapter Two 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 vs 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 π < ω t < 2π,diode D1 blocks current and acts as an open switch. On the other hand,diode D2 conducts during this period, the driving function can be set tobe zero volts. For 0 < ω t < π, the differential equation (2.18) applies. Thesolution of this equation will be as obtained before in (2.20) or (2.23). ⎛ ωt ⎞ − Vm ⎜ tan φ ⎟i (ωt ) = sin (ωt − φ ) + sin (φ ) e 0 < ωt < π (2.28) Z ⎜ ⎜ ⎟ ⎟ ⎝ ⎠ For the negative half-cycle ( π < ωt < 2π ) 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. diL + R* i = 0 for π < ωt < 2π (2.29) dt The solution of (2.29) is given by the following equation: ωt − π − tan φi (ωt ) = B e (2.30) The constant B can be obtained from the boundary condition wherei (π ) = B is the starting value of the current in π < ωt < 2π and can beobtained from equation (2.23) by substituting ωt = π π V −Then, i(π ) = m (sin(π − φ ) + sin (φ ) e tan φ ) = B Z
  • 36. Diode Circuits or Uncontrolled Rectifier 33The above value of i (π ) can be used as initial condition of equation(2.30). Then the load current during π < ωt < 2π is shown in thefollowing equation. ⎛ π ⎞ ωt −π − − Vm ⎜ tan φ ⎟i (ωt ) = sin (π − φ ) + sin (φ ) e e tan φ for π < ωt < 2π Z ⎜ ⎟ (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.
  • 37. 34 Chapter Two For the period 2π < ωt < 3π the value of i (2π ) 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 ω t byωt − 2π and the solution is given by equation (2.32). This period( 2π < ωt < 3π ) differ than the period 0 < wt < π in the way to get theconstant A where in the 0 < ωt < π the initial value was i (0) = 0 but inthe case of 2π < ωt < 3π the initial condition will be i (2π ) that givenfrom (2.31) and is shown in (2.33). ωt − 2π − Vi (ωt ) = m sin (ωt − 2π − φ ) + Ae tan φ for 2π < ωt < 3π (2.32) Z The value of i (2π ) can be obtained from (2.31) and (2.32) as shownin (2.33) and (2.34) respectively. ⎛ π ⎞ π − − Vm ⎜ tan φ ⎟i (2π ) = sin (π − φ ) + sin (φ ) e ⎟e tan φ (2.33) Z ⎜⎜ ⎟ ⎝ ⎠ Vi (2π ) = m sin (− φ ) + A (2.34) Z By equating (2.33) and (2.34) the constant A in 2π < ωt < 3π can beobtained from the following equation: V A = i (2π ) + m sin (φ ) (2.35) Z Then, the general solution for the period 2π < ωt < 3π is given byequation (2.36): ωt − 2π Vm ⎛ V ⎞ − 2π < ωt < 3π (2.36)i (ωt ) = sin (ωt − 2π − φ ) + ⎜ i(2π ) + m sin (φ )⎟e tan φ Z ⎝ Z ⎠ Where i (2π ) can be obtained from equation (2.33).Example 2 A diode circuit shown in Fig.2.3 with R=10 Ω, L=20mH, andVS=220 2 sin314t. (a) Determine the expression for the current though the load in the period 0 < ωt < 2π and determine the conduction angle β . (b) If we connect free wheeling diode through the load as shown in Fig.2.5 Determine the expression for the current though the load in the period of 0 < ωt < 3π .
  • 38. Diode Circuits or Uncontrolled Rectifier 35Solution: (a) For the period of 0 < ωt < π , the expression of the loadcurrent can be obtained from (2.24) as following: −3 −1 ωL −1 314 * 20 *10φ = tan = tan = 0.561 rad . and tan φ = 0.628343 R 10 Z = R 2 + (ωL) 2 = 10 2 + (314 * 20 *10 − 3 ) 2 = 11.8084Ω ⎛ ωt ⎞ − Vm ⎜ ⎟i (ωt ) = sin (ωt − φ ) + sin (φ ) e tan φ Z ⎜ ⎜ ⎟ ⎟ ⎝ ⎠ = 220 2 11.8084 [ ] sin (ωt − 0.561) + 0.532 * e −1.5915 ωti (ωt ) = 26.3479 sin (ωt − 0.561) + 14.0171* e −1.5915 ωtThe value of β can be obtained from the above equation by substitutingfor i ( β ) = 0 . Then, 0 = 26.3479 sin (β − 0.561) + 14.0171 * e −1.5915 β By using the numerical analysis we can get the value of β. Thesimplest method is by using the simple iteration technique by assumingΔ = 26.3479 sin (β − 0.561) + 14.0171 * e −1.5915 β and substitute differentvalues for β in the region π < β < 2π till we get the minimum value of Δthen the corresponding value of β is the required value. The narrowintervals mean an accurate values of β . The following table shows therelation between β and Δ: β Δ 1.1 π 6.49518 1.12 π 4.87278 1.14 π 3.23186 1.16 π 1.57885 1.18 π -0.079808 1.2 π -1.73761 It is clear from the above table that β ≅ 1.18 π rad. The current in β < wt < 2π 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
  • 39. 36 Chapter Two0 < ωt < π (D1 “ON”, D2 “OFF”), the second case when π < ωt < 2π(D1 “OFF” and D2 “ON”) and the last one when 2π < ωt < 3π (D1“ON”, D2 “OFF”). In the first part ( 0 < ωt < π ) the expression for the load current can be obtained as In case (a). Then:i ( wt ) = 26.3479 sin (ωt − 0.561) + 14.0171 * e −1.5915 wt for 0 < ωt < πthe current at ωt = π is starting value for the current in the next part.Theni (π ) = 26.3479 sin (π − 0.561) + 14.0171 * e −1.5915 π = 14.1124 A In the second part π < ωt < 2π , the expression for the load current can be obtained from (2.30) as following: ωt −π − tan φi (ωt ) = B ewhere B = i (π ) = 14.1124 AThen i (ωt ) = 14.1124 e −1.5915(ωt −π ) for ( π < ωt < 2π ) The current at ωt = 2π is starting value for the current in the next part.Theni (2π ) = 0.095103 A In the last part ( 2π < ωt < 3π ) the expression for the load current can be obtained from (2.36): ωt − 2π − ⎛ ⎞ i (ωt ) = m sin (ωt − 2π − φ ) + ⎜ i (2π ) + m sin (φ )⎟e V V tan φ Z ⎝ Z ⎠∴ i (ωt ) = 26.3479 sin (ωt − 6.8442) + (0.095103 + 26.3479 * 0.532)e −1.5915(ωt − 2π )∴ i (ωt ) = 26.3479 sin (ωt − 6.8442) + 14.1131e −1.5915(ωt − 2π ) for( 2π < ωt < 3π )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:
  • 40. Diode Circuits or Uncontrolled Rectifier 37Advantages • The need for center-tapped transformer is eliminated, • The output is twice that of the center tapped circuit for the same secondary voltage, and, • The peak inverse voltage is one half of the center-tap circuit.Disadvantages • It requires four diodes instead of two, in full wave circuit, and, • There are always two diodes in series are conducting. Therefore, total voltage drop in the internal resistance of the diodes and losses are increased. The following sections explain and analyze these rectifiers.2.4.1 Center-Tap Diode Rectifier With Resistive Load In the center tap full wave rectifier, current flows through the load inthe 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:
  • 41. 38 Chapter Two π 1 2 Vm π∫ mVdc = V sin ωt dωt = (2.36) π 0 2 VmI dc = (2.37) π R π 1 (V sin ωt ) Vm π∫ mVrms = 2 dω t = (2.38) 2 0 VmI rms = (2.39) 2 RPIV of each diode = 2Vm (2.40) VVS = m (2.41) 2The rms value of the transformer secondery current is the same as that ofthe diode: V IS = ID = m (2.41) 2R Fig.2.9 Various current and voltage waveforms for center-tap diode rectifier with resistive load.
  • 42. Diode Circuits or Uncontrolled Rectifier 39Example 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 2 Vm 2 Vm * Pdc Vdc * I dc π πRη= = = = 81.05% Pac Vrms * I rms Vm Vm * 2 2R Vm V(b) FF = rms = 2 = π = 1.11 Vdc 2 Vm 2 2 π Vac(c) RF = = FF 2 − 1 = 1.112 − 1 = 0.483 Vdc 2 Vm 2 Vm Pdc π π R(d) TUF = = = 0.5732 2 VS I S V V 2 m m 2 2R(e) The PIV is 2Vm Vm I S ( peak )(f) Creast Factor of secondary current, CF = = R =2 IS Vm 2R2.4.2 Center-Tap Diode Rectifier With R-L Load Center-tap full wave rectifier circuit with RL load is shown in Fig.2.10.Various voltage and current waveforms for Fig.2.10 is shown in Fig.2.11.An expression for load current can be obtained as shown below: It is assumed that D1 conducts in positive half cycle of VS and D2conducts in negative half cycle. So, the deferential equation defines thecircuit is shown in (2.43). di L + R * i = Vm sin(ωt ) (2.43) dt The solution of the above equation can be obtained as obtained beforein (2.24)
  • 43. 40 Chapter Two Fig.2.10 Center-tap diode rectifier with R-L load Fig.2.11 Various current and voltage waveform for Center-tap diode rectifier with R-L load ⎛ ωt ⎞ − Vm ⎜ tan φ ⎟i (ωt ) = ⎜ sin (ωt − φ ) + sin (φ )e Z ⎜ ⎟ for 0 < ωt < π (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)
  • 44. Diode Circuits or Uncontrolled Rectifier 41 ωt − π − Vi (ωt ) = m sin (ωt − π − φ ) + Ae tan φ (2.45) ZThe value of constant A can be obtained from initial condition. If weassume that i(π)=i(2π)=i(3π)=……..=Io (2.46)Then the value of I o can be obtained from (2.44) by letting ωt = π ⎛ π ⎞ − Vm ⎜ tan φ ⎟I o = i (π ) = ⎜ sin (π − φ ) + sin (φ )e Z ⎜ ⎟ (2.47) ⎟ ⎝ ⎠ Then use the value of I o as initial condition for equation (2.45). So wecan obtain the value of constant A as following: π −π − Vi (π ) = I o = m sin (π − π − φ ) + Ae tan φ Z VThen; A = I o + m sin (φ ) (2.48) ZSubstitute (2.48) into (2.45) we get: ωt − π − ⎛ ⎞i (ωt ) = m sin (ωt − π − φ ) + ⎜ I o + m sin (φ )⎟e tan φ , then, V V Z ⎝ Z ⎠ ⎡ ωt −π ⎤ ωt −π − −i (ωt ) = Vm ⎢ sin (ωt − π − φ ) + sin (φ )e tan φ ⎥ + I e tan φ (for π < ωt < 2π ) (2.49) Z ⎢ ⎥ o ⎢ ⎣ ⎥ ⎦ In the next half cycle 2π < ωt < 3π 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: ⎡ ωt − 2π ⎤ ωt − 2π − −i (ωt ) = Vm ⎢ sin (wt − 2π − φ ) + sin (φ )e tan φ ⎥ + I e tan φ (for 2π < ωt < 3π )(2.50) Z ⎢ ⎥ o ⎢ ⎣ ⎥ ⎦2.4.3 Single-Phase Full Bridge Diode Rectifier With Resistive Load Another alternative in single-phase full wave rectifier is by using fourdiodes 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”).
  • 45. 42 Chapter Two In positive half cycle the supply voltage forces diodes D1 and D2 to be"ON". In same time it forces diodes D3 and D4 to be "OFF". So, 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-phase diode rectifier.
  • 46. Diode Circuits or Uncontrolled Rectifier 43Example 4 The rectifier shown in Fig.2.12 has a purely resistive load ofR=15 Ω and, VS=300 sin 314 t and unity transformer ratio. Determine (a)The efficiency, (b) Form factor, (c) Ripple factor, (d) TUF, (e) The peakinverse voltage, (PIV) of each diode, (f) Crest factor of input current, and,(g) Input power factor.Solution: Vm = 300 V π 1 2 Vm 2 Vm π∫ mVdc = V sin ωt dωt = = 190.956 V , I dc = = 12.7324 A π π R 0 1/ 2 ⎡1 π ⎤ VVrms =⎢ (Vm sin ωt )2 dωt ⎥ ∫ = Vm = 212.132 V , I rms = m = 14.142 A ⎢π 0 ⎣ ⎥ ⎦ 2 2R Pdc V I(a) η = = dc dc = 81.06 % Pac Vrms I rms V(b) FF = rms = 1.11 Vdc Vac Vrms − Vdc 2 2 2 Vrms(c) RF = = = 2 − 1 = FF 2 − 1 = 0.482 Vdc Vdc Vdc Pdc 190.986 *12.7324(d) TUF = = = 81 % VS I S 212.132 * 14.142(e) The PIV= Vm =300V I S ( peak ) 300 / 15(f) CF = = = 1.414 IS 14.142 Re al Power I2 *R(g) Input power factor = = rms =1 Apperant Power VS I S2.4.4 Full Bridge Single-phase Diode Rectifier with DC Load Current The full bridge single-phase diode rectifier with DC load current isshown 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 I S = I o
  • 47. 44 Chapter Two Fig.2.14 Full bridge single-phase diode rectifier with DC load current. Fig.2.15 Various current and voltage waveforms for full bridge single-phase diode rectifier with DC load current. The supply current in case of pure DC load current is shown inFig.2.15, as we see it is odd function, then an coefficients of Fourierseries equal zero, an = 0 , and π 2 2 Io [− cos nωt ]π π∫bn = I o * sin nωt dωt = nπ 0 (2.51) 0 = 2 Io [cos 0 − cos nπ ] = 4 I o for n = 1, 3, 5, ............. nπ nπThen from Fourier series concepts we can say: 4 Io 1 1 1 1i (t ) = * (sin ωt + sin 3ωt + sin 5ωt + sin 7ωt + sin 9ωt + ..........) (2.52) π 3 5 7 9
  • 48. Diode Circuits or Uncontrolled Rectifier 45 2 2 2 2 2 2 2 ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞∴ THD( I s (t )) = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 46% ⎝ 3 ⎠ ⎝ 5 ⎠ ⎝ 7 ⎠ ⎝ 9 ⎠ ⎝ 11 ⎠ ⎝ 13 ⎠ ⎝ 15 ⎠or we can obtain THD ( I s (t )) as the following: 4 IoFrom (2.52) we can obtain the value of is I S1 = 2π 2 ⎛ ⎞ 2 ⎜ ⎟ 2 ⎛ IS ⎞ ⎜ Io ⎟ −1 = ⎛ 2π ⎞∴ THD ( I s (t )) = ⎜ ⎟ ⎜ ⎟ − 1 = 48.34% ⎜ I ⎟ −1 = ⎜ 4 I ⎟ ⎜ 4 ⎟ ⎝ S1 ⎠ ⎜ o ⎟ ⎝ ⎠ ⎝ 2π ⎠Example 5 solve Example 4 if the load is 30 A pure DCSolution: From example 4 Vdc= 190.986 V, Vrms=212.132 V I dc = 30 A and I rms = 30 A P V I(a) η = dc = dc dc = 90 % Pac Vrms I rms V(b) FF = rms = 1.11 Vdc Vac Vrms − Vdc 2 2 2 Vrms(c) RF = = = 2 − 1 = FF 2 − 1 = 0.482 Vdc Vdc Vdc Pdc 190.986 *30(d) TUF = = = 90 % VS I S 212.132 * 30(e) The PIV=Vm=300V I 30(f) CF = S ( peak ) = =1 IS 30 4 Io 4 * 30(g) I S1 = = = 27.01A 2π 2π Re al PowerInput Power factor= = Apperant Power VS I S1 * cos φ I * cos φ 27.01 = = S1 = *1 = 0.9 Lag VS I S IS 30
  • 49. 46 Chapter Two2.4.5 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 currentiS from a value of I o to − I o (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 bridge rectifier with source inductance.
  • 50. Diode Circuits or Uncontrolled Rectifier 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 Δt to completely turn OFF D1 and D2 and to make D3 andD4 carry the entire load current ( I o ). Also in the time Δt the supplycurrent will change from I o to − I o which is very clear in Fig.2.16.Fig.2.17 shows the equivalent circuit of the diode bridge at time Δt . Fig.2.17 The equivalent circuit of the diode bridge at commutation time Δt .From Fig.2.17 we can get the following equations diVS − Ls S = 0 (2.53) dtMultiply the above equation by dωt then,VS dωt = ωLs diS (2.54) Integrate both sides of the above equation during the commutationperiod ( Δt sec or u rad.) we get the following:VS dωt = ωLs diSπ +u −Io ∫ Vm sin ωt dωt = ωLs ∫ diS (2.55) π IoThen; Vm [cos π − cos(π + u )] = −2ωLs I oThen; Vm [− 1 + cos(u )] = −2ωLs I o
  • 51. 48 Chapter Two 2ωLs I oThen; cos(u ) = 1 − Vm ⎛ 2ωLs I o ⎞Then; u = cos −1 ⎜1 − ⎜ ⎟ (2.56) ⎝ Vm ⎟ ⎠ u 1 ⎛ 2ωLs I o ⎞And Δt = = cos −1 ⎜1 − ⎜ ⎟ (2.57) ω ω ⎝ Vm ⎟ ⎠ It is clear that the DC voltage reduction due to the source inductance isthe drop across the source inductance. divrd = Ls S (2.58) dt π +u −IoThen ∫ vrd dω t = ∫ ω LS diS = −2ω LS I o (2.59) π Ioπ +u ∫ vrd dω t is the reduction area in one commutation period Δt . But we πhave two commutation periods Δt in one period of supply voltage. So the π +utotal reduction per period is: 2 ∫ vrd dω t = −4 ω LS I o (2.60) π To obtain the average reduction in DC output voltage Vrd due tosource inductance we have to divide the above equation by the periodtime 2π . Then; − 4ω LS I oVrd = = −4 f LS I o (2.61) 2π The DC voltage with source inductance tacking into account can becalculated as following: 2VVdc actual = Vdc without sourceinduc tan ce − Vrd = m − 4 fLs I o (2.62) π 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 u / 2 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.
  • 52. Diode Circuits or Uncontrolled Rectifier 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: π u/2 2 2 2 ⎛ 2I o ⎞Is = ∫ ωt ⎟ dωt + ∫ I o dωt ] 2 [ ⎜ (2.63) π 0 ⎝ u ⎠ u/2 2I o ⎡ 4 u 3 π u ⎤ 2 2I o ⎡π u ⎤ 2Then; I s = ⎢ + − ⎥= − (2.64) π ⎢ 3u 2 8 2 2 ⎥ ⎣ ⎦ π ⎢ 2 3⎥ ⎣ ⎦ Is u Io π 2π u − u π+ 2 2 π u 2 u u 2π − 2 − Io π− 2 2 Fig.2.19 Supply current waveform
  • 53. 50 Chapter Two 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: Is u Io π 2π u u − u π+ 2 2 2 u u π− 2π − 2 2 − Io ′ Is 2Io u π u u − u π+ 2 2 2 u u π− 2π − 2I o 2 2 − u Fig.2.20 Supply current and its first derivative.Table(2.1) Jumb value of supply current and its first derivative. Js u u u u − π− π+ 2 2 2 2 Is 0 0 0 0 ′ Is 2Io − 2I o − 2Io 2I o u u u u
  • 54. Diode Circuits or Uncontrolled Rectifier 51It is an odd function, then ao = an = 0 ⎡m 1 m ⎤ ∑ ∑ 1bn = ⎢ J s cos nωt s − ′ J s sin nωt s ⎥ (2.65) nπ⎢ s =1 ⎣ n s =1 ⎥ ⎦ 1 ⎡ − 1 2I o ⎛ ⎛ u⎞ ⎛u⎞ ⎛ u⎞ ⎛ u ⎞ ⎞⎤bn = ⎢ * ⎜ sin n⎜ − ⎟ − sin n⎜ ⎟ − sin n⎜ π − ⎟ + sin n⎜ π + ⎟ ⎟⎥ nπ ⎣ n u ⎝ ⎝ 2⎠ ⎝2⎠ ⎝ 2⎠ ⎝ 2 ⎠ ⎠⎦ 8I nubn = 2 o * sin (2.66) n πu 2 8I ub1 = o * sin (2.67) πu 2 8I o uThen; I S1 = * sin (2.68) 2 πu 2 8I o u * sin I ⎛u⎞ 2 πu 2 ⎛u⎞pf = S1 * cos⎜ ⎟ = cos⎜ ⎟ IS ⎝2⎠ 2I o ⎡π u ⎤ 2 ⎝2⎠ − ⎥ π ⎢ 2 3⎦ ⎣ (2.69) ⎛ u⎞ ⎛u⎞ 4 sin ⎜ ⎟ cos⎜ ⎟ = ⎝ 2 ⎠ ⎝ 2 ⎠ = 2 sin (u ) ⎡π u ⎤ ⎡π u ⎤ u π⎢ − ⎥ u π⎢ − ⎥ ⎣ 2 3⎦ ⎣ 2 3⎦Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz,source inductance X s = 5 mH 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), Vm = 11000 * 2 = 15556V 2VVdc actual = Vdc without sourceinduc tan ce − Vrd = m − 4 fLs I o π 2 *15556Vdc actual = − 4 * 50 * 0.005 * 200 = 9703V π(ii) From (2.56) the commutation angle u can be obtained as following:
  • 55. 52 Chapter Two ⎛ 2ωLs I o ⎞ 2 * 2 * π * 50 * 0.005 * 200 ⎞u = cos −1 ⎜1 − ⎜ ⎟ = cos −1 ⎛1 − ⎟ ⎜ ⎟ = 0.285 rad . ⎝ Vm ⎠ ⎝ 15556 ⎠The input power factor can be obtained from (2.69) as following I ⎛u⎞ 2 * sin (u ) 2 * sin (0.285)pf = S1 * cos⎜ ⎟ = = = 0.917 IS ⎝2⎠ ⎡ π u⎤ ⎡ π .285 ⎤ u π ⎢ − ⎥ 0.285 π ⎢ − ⎣ 2 3⎦ ⎣2 3 ⎥ ⎦ 2I o ⎡π u ⎤ 2 2 * 200 2 ⎡ π 0.285 ⎤IS = − ⎥= ⎢ − 3 ⎥ = 193.85 A π ⎢ 2 3⎦ ⎣ π ⎣2 ⎦ 8I o u 8 * 200 ⎛ 0.285 ⎞I S1 = * sin = * sin ⎜ ⎟ = 179.46 A 2 πu 2 2 π * 0.285 ⎝ 2 ⎠ 2 2 ⎛ IS ⎞ ⎟ −1 = ⎛ 193.85 ⎞ ⎜THDi = ⎜ ⎟ ⎜ ⎟ − 1 = 40.84% ⎝ I S1 ⎠ ⎝ 179.46 ⎠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-phase transformer.
  • 56. Diode Circuits or Uncontrolled Rectifier 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. π 5π 6 6Fig.2.22 Secondary and load voltages of half wave three-phase diode rectifier. Fig.2.23 Primary and diode currents.
  • 57. 54 Chapter Two Primary current Diode current Fig.2.24 FFT components of primary and diode currents. π 5π By considering the interval from to in the output voltage we can 6 6calculate the average and rms output voltage and current as following: 5π / 6 3 3 3 VmVdc = 2π ∫ Vm sin ωt dωt = 2π = 0.827Vm (2.70) π /6 3 3 Vm 0.827 * VmI dc = = (2.71) 2 *π * R R 5π / 6 ∫ (Vm sin ωt ) 3 1 3* 3Vrms = 2 dωt = + Vm = 0.8407 Vm (2.72) 2π 2 8π π /6 0.8407 VmI rms = (2.73) R Then the diode rms current is equal to secondery current and can beobtaiend as following: 08407 Vm Vm Ir = IS = = 0.4854 (2.74) R 3 R 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 2 VLL = 3 Vm (2.75)
  • 58. Diode Circuits or Uncontrolled Rectifier 55Example 7 The rectifier in Fig.2.21 is operated from 460 V 50 Hz supplyat secondary side and the load resistance is R=20 Ω. 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: 460(a) VS = = 265.58 V , Vm = 265.58 * 2 = 375.59 V 3 3 3 VmVdc = = 0.827 Vm , 2π 3 3 Vm 0827 Vm I dc = = 2π R RVrms = 0.8407 Vm 0.8407 Vm I rms = R P V Iη = dc = dc dc = 96.767 % Pac Vrms I rms V(b) FF = rms = 101.657 % Vdc Vac Vrms − Vdc 2 2 2 Vrms(c) RF = = = 2 − 1 = FF 2 − 1 = 18.28 % Vdc Vdc Vdc 1 0.8407 Vm(d) I S = I rms = 3 3 R Pdc (0.827Vm ) 2 / RTUF = = = 66.424 % 3 * VS I S 0.8407 Vm 3 * Vm / 2 * 3R(e) The PIV= 3 Vm=650.54V I S ( peak ) Vm / R(f) CF = = = 2.06 IS 0.8407 Vm 3R
  • 59. 56 Chapter Two2.5.2 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 onecoefficient an or bn respectively. If we put Y-axis at point ωt = 30o thenwe can deal with the secondary current as even functions. Then, bn = 0 ofsecondary current. Values of an can be calculated as following: π /3 1 Ia0 = 2π ∫ I o dωt = 3o (2.76) −π / 3 π /3 1an = π ∫ I o * cos nωt dwt −π / 3 = o [sin nωt ]−π //3 I π 3 nπ I = o * 3 for n = 1,2,7,8,13,14,.... (2.77) nπ I = − o * 3 for n = 4,5,10,11,16,17 nπ = 0 for all treplean harmonics IO 3I O ⎛ 1 1 1 1 1 ⎞I s (t ) = + ⎜ sin ωt + sin 2ωt − sin 4ωt − sin 5ωt + sin 7ωt + sin 8ωt − −... ⎟ (2.78) 3 π ⎝ 2 4 5 7 8 ⎠
  • 60. Diode Circuits or Uncontrolled Rectifier 57 New axis Fig.2.26 Primary and secondary current waveforms and FFT components of three-phase half wave rectifier with dc load current
  • 61. 58 Chapter Two 2 ⎛ ⎞ 2 ⎜ ⎟ ⎛I ⎞ ⎜ Io / 3 ⎟ 2 *π 2THD( I s (t )) = ⎜ S ⎟ ⎜I ⎟ −1 = ⎜ −1 = − 1 = 1.0924 = 109.24% ⎝ S1 ⎠ ⎜ 3I O ⎟ ⎟ 9 ⎜ π 2 ⎟ ⎝ ⎠It is clear that the primary current shown in Fig.2.26 is odd, then, an=0, 2π / 3bn = 2 ∫ I o * sin nωt dωt = 2I o [− cos nωt ] 20π / 3 π nπ 0 3I o = for n = 1,2,4,5,7,8,10,11,13,14,.... (2.79) nπ = 0 for all treplean harmonics 3I O ⎛ 1 1 1 1 1 ⎞iP (t ) = ⎜ sin ωt + sin 2ωt + sin 4ωt + sin 5ωt + sin 7ωt + sin 8ωt − −... ⎟ (2.80) π ⎝ 2 4 5 7 8 ⎠ 2The rms value of I P = Io (2.81) 3 2 ⎛ 2 ⎞ 2 ⎜ Io ⎟ 2 ⎛I ⎞ 3 ⎟ − 1 = ⎛ 2π ⎞ − 1 = 67.983% (2.82)THD ( I P (t )) = ⎜ P ⎟ ⎜I ⎟ −1 = ⎜ ⎜ ⎟ ⎜ 3I O ⎟ ⎝3 3⎠ ⎝ P1 ⎠ ⎜ ⎟ ⎝ π 2 ⎠Example 8 Solve example 7 if the load current is 100 A pure DC 460Solution: (a) VS = = 265.58 V , Vm = 265.58 * 2 = 375.59 V 3 3 3 VmVdc = = 0.827 Vm = 310.613V , I dc = 100 A 2πVrms = 0.8407 Vm = 315.759 V , I rms = 100 A P V I 310.613 * 100η = dc = dc dc = = 98.37 % Pac Vrms I rms 315.759 *100 V(b) FF = rms = 101.657 % Vdc Vac Vrms − Vdc 2 2 2 Vrms(c) RF = = = 2 − 1 = FF 2 − 1 = 18.28 % Vdc Vdc Vdc
  • 62. Diode Circuits or Uncontrolled Rectifier 59 1 1(d) I S = I rms = *100 = 57.735 A 3 3 Pdc 310.613 * 100TUF = = = 67.52 % 3 * VS I S 3 * Vm / 2 * 57.735(e) The PIV= 3 Vm=650.54V I S ( peak ) 100(f) CF = = = 1.732 IS 57.7352.5.3 Three-Phase Half Wave Rectifier With Source Inductance The source inductance in three-phase half wave diode rectifier Fig.2.27will 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 vb is goingto be greater than the voltage va 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 Δt (overlap time or commutation time). The overlap timewill take time Δt to completely turn OFF D1 and to make D2 carry theentire load current ( I o ). Also in the time Δt the current in Lb will changefrom zero to I o and the current in La will change from I o 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 Δt . Fig.2.27 Three-phase half wave rectifier with load and source inductance.
  • 63. 60 Chapter Two Fig.2.28 Supply current and output voltage for three-phase half wave rectifier with pure DC load and source inductance. Fig.2.29 The equivalent circuit for three-phase half wave diode rectifier in commutation period.From Fig.2.29 we can get the following equations diva − La D1 − Vdc = 0 (2.83) dt divb − Lb D 2 − Vdc = 0 (2.84) dt subtract (2.84) from(2.83) we get:
  • 64. Diode Circuits or Uncontrolled Rectifier 61 ⎛ di di ⎞ va − vb + L⎜ D 2 − D1 ⎟ = 0 ⎝ dt dt ⎠ Multiply the above equation by dωt the following equation can beobtained: (va − vb )dωt + ωL(diD 2 − diD1 ) = 0 substitute the voltage waveforms of va and vb into the above equation ⎛ ⎛ 2π ⎞ ⎞we get: ⎜Vm sin (ωt ) − Vm sin ⎜ ωt − ⎟ ⎟dωt = ωL(diD1 − diD 2 ) ⎝ ⎝ 3 ⎠⎠ ⎛ ⎛ π ⎞⎞ Then; ⎜ 3 Vm sin ⎜ ωt + ⎟ ⎟dωt = ωL(diD1 − diD 2 ) ⎝ ⎝ 6 ⎠⎠ Integrating both parts of the above equation we get the following: 5π +u 6 ⎛0 Io ⎞ ⎛ π⎞ ⎜ ⎟ ∫ ⎝ 6⎠ ⎜I ∫ ∫ 3 Vm sin ⎜ ωt + ⎟dωt = ωL⎜ diD1 − diD 2 ⎟ ⎟ 5π ⎝ o 0 ⎠ 6 ⎛ ⎛ 5π π ⎞ ⎛ 5π π ⎞⎞ Then; 3 Vm ⎜ cos⎜ + ⎟ − cos⎜ + u + ⎟ ⎟ = −2ωLI o ⎝ ⎝ 6 6⎠ ⎝ 6 6 ⎠⎠ Then; 3 Vm (cos(π ) − cos(π + u )) = −2ωLI o Then; 3 Vm (− 1 + cos(u )) = −2ωLI o 2ωLI o Then; 1 − cos(u ) = 3 Vm 2ωLI o Then; cos(u ) = 1 − 3 Vm ⎛ 2ωLI o ⎞ Then u = cos −1 ⎜1 − ⎜ ⎟ (2.85) ⎝ 3 Vm ⎟⎠ u 1 ⎛ 2ωLI o ⎞ Δt = = cos −1 ⎜1 − ⎜ ⎟ (2.86) ω ω ⎝ 3 Vm ⎟ ⎠ It is clear that the DC voltage reduction due to the source inductance isequal to the drop across the source inductance. Then; di vrd = L D dt
  • 65. 62 Chapter Two 5π +u 6 Io Then, ∫ vrd dωt = ∫ ωL diD = ωLI o (2.87) 5π 0 6 5π +u 6 ∫ vrd dωt is the reduction area in one commutation period Δt . But, 5π 6we have three commutation periods, Δt in one period. So, the totalreduction per period is: 5π +u 6 3* ∫ vrd dωt = 3ωLI o 5π 6 To obtain the average reduction in DC output voltage Vrd due tosource inductance we have to divide the total reduction per period by 2πas following: 3ωLI o Vrd = = 3 f L Io (2.88) 2πThen, the DC component of output voltage due to source inductance is:Vdc Actual = Vdc without source − 3 f L Io (2.89) induc tan ce 3 3 VmVdc = − 3 f L Io (2.90) Actual 2πExample 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. ⎛ 66000 ⎞Solution: vm = ⎜ ⎟ * 2 = 53889V ⎝ 3 ⎠(i) Vdc Actual = Vdc without source − 3 f L Io induc tan ce 3 3 Vm 3 * 3 * 53889Vdc = − 3 f L Io = − 3 * 50 * 0.005 * 500 = 44190V Actual 2π 2π
  • 66. Diode Circuits or Uncontrolled Rectifier 632.5 Three-Phase Full Wave Diode Rectifier The three phase bridge rectifier is very common in high powerapplications 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 Load In the circuit of Fig.2.30, the AC side inductance LS is neglected andthe 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 :- 2π / 3 3 3 3 Vm 3 2 VLLVdc = π ∫ 3 Vm sin ωt dωt = π = π = 1.654Vm = 1.3505VLL (2.91) π /3 3 3 Vm 1.654Vm 3 2 VLL 1.3505VLLI dc = = = = (2.92) π R R πR R 2π / 3 ∫ ( )2 dωt = 3 3 9* 3Vrms = 3 Vm sin ωt + Vm = 1.6554 Vm = 1.3516VLL (2.93) π 2 4π π /3 1.6554 VmI rms = (2.94) RThen the diode rms current is 1.6554 Vm VmIr = = 0.9667 (2.95) R 3 R VmI S = 0.9667 2 (2.96) R
  • 67. 64 Chapter Two IL Ip 3 Is 1 5 VL a b c 4 6 2 Fig.2.30 Three-phase full wave diode bridge rectifier. Fig.2.31 shows complete waveforms for phase and line to line input voltages and output DC load voltages.
  • 68. Diode Circuits or Uncontrolled Rectifier 65 Fig.2.32 Diode currents. Fig.2.33 Secondary and primary currents and PIV of D1.
  • 69. 66 Chapter Two Fig.2.34 Fourier Transform components of output DC voltage, diode current secondary current and Primary current respectively of three-phase full wave diode bridge rectifier.Example 10 The rectifier shown in Fig.2.30 is operated from 460 V 50Hz supply and the load resistance is R=20 Ω. 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. 460Solution: (a) VS = = 265.58 V , Vm = 265.58 * 2 = 375.59 V 3 3 3 VmVdc = = 1.654Vm = 621.226 V , π 3 3 Vm 1.654VmI dc = = = 31.0613 A π R R 3 9* 3Vrms = + Vm = 1.6554 Vm = 621.752 V , 2 4π 1.6554 VmI rms = = 31.0876 A R
  • 70. Diode Circuits or Uncontrolled Rectifier 67 Pdc V Iη= = dc dc = 99.83 % Pac Vrms I rms V(b) FF = rms = 100.08 % Vdc Vac Vrms − Vdc 2 2 2 Vrms(c) RF = = = 2 − 1 = FF 2 − 1 = 4 % Vdc Vdc Vdc 2 1.6554 Vm V(d) I S = I rms = 0.8165 * = 1.352 m 3 R R Pdc (1.654Vm ) 2 / RTUF = = = 95.42 % 3 * VS I S V 3 * Vm / 2 *1.352 m R(e) The PIV= 3 Vm=650.54V I 3 Vm / R(f) CF = S ( peak ) = = 1.281 IS Vm 1.352 R2.5.2 Three-Phase Full Wave Rectifier With DC Load Current The supply current in case of pure DC load current is shown inFig.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 5π / 6 2bn = π ∫ I o * sin nωt dωt π /6 = 2 Io [− cos nωt ]ππ/ 66 5 / nπ 2 Io 2 Io 2 Iob1 = 3, b5 = (− 3 ), b7 = (− 3 ) (2.97) π 5π 7π 2 Io 2 Iob11 = ( 3 ), b13 = ( 3 ),............. 11π 13πbn = 0, for n = 2,3,4,6,8,9,10,12,14,15,............. 2 3I o ⎛ 1 1 1 1 ⎞I s (t ) = ⎜ sin ωt − sin 5v ωt − sin 7ωt + sin 11ωt + sin 13ωt ⎟ (2.98) π ⎝ 5 7 11 13 ⎠
  • 71. 68 Chapter Two 2 2 2 2 2 2 2 2 ⎛1⎞ ⎛1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞THD ( I s (t )) = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ 5 ⎠ ⎝ 7 ⎠ ⎝ 11 ⎠ ⎝ 13 ⎠ ⎝ 17 ⎠ ⎝ 19 ⎠ ⎝ 23 ⎠ ⎝ 25 ⎠ = 31%Also THD ( I s (t )) can be obtained as following: 2 2* 3IS = I o , I S1 = Io 3 π 2 ⎛I ⎞ 2/3THD ( I s (t )) = ⎜ S ⎟ − 1 = ⎜ ⎟ − 1 = 31.01% ⎝ I S1 ⎠ 2*3/π 2 Fig.2.35 The D1 and D2 currents, secondary and primary currents.
  • 72. Diode Circuits or Uncontrolled Rectifier 69Fig2.36 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, an = 0 , and 2π / 3 2⎛ ⎞ π /3 π ⎜ I * sin nωt dωt + 2 I * sin nωt dωt + I1 * sin nωt dωt ⎟bn = π⎜ ∫ 1 ∫ 1 ∫ ⎟ ⎝ 0 π /3 2π / 3 ⎠ 2I ⎛ π 2π ⎞ (2.99) = 1 ⎜1 − cos nπ + cos n − cos n ⎟ nπ ⎝ 3 3 ⎠ 2 * 3I1bn = for n = 1,5,7,11,13,............... nπbn = 0, for n = 2,3,4,6,8,9,10,12,14,15,............. 2 * 3I1 ⎛ 1 1 1 1 ⎞I P (t ) = ⎜ sin ωt + sin 5ωt + sin 7ωt + sin 11ωt + sin 13ωt ⎟ (2.100) π ⎝ 5 7 11 13 ⎠ 2 2 2 2 2 2 2 2 ⎛1⎞ ⎛1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞THD ( I P (t )) = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ 5 ⎠ ⎝ 7 ⎠ ⎝ 11 ⎠ ⎝ 13 ⎠ ⎝ 17 ⎠ ⎝ 19 ⎠ ⎝ 23 ⎠ ⎝ 25 ⎠ = 30% I S1 IPower Factor = * cos(0) = S1 IS IS
  • 73. 70 Chapter Two2.5.4 Three-Phase Full Wave Diode Rectifier With Source Inductance The source inductance in three-phase diode bridge rectifier Fig.2.37will 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 source inductance Let us study the commutation time starts at t=5ms as shown inFig.2.39. At this time Vc starts to be more negative than Vb 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 Δt to completely turn OFF D6 and to make D2 carry all the load
  • 74. Diode Circuits or Uncontrolled Rectifier 71current ( I o ). Also in the time Δt the current in Lb will change from I oto zero and the current in Lc will change from zero to I o . This is veryclear from Fig.2.39. The equivalent circuit of the three phase diode bridgeat commutation time Δt at t = 5ms is shown in Fig.2.40 and Fig.2.41. Fig.2.39 Waveforms represent the commutation period at time t=5ms.Fig.2.40 The equivalent circuit of the three phase diode bridge at commutation time Δt at t = 5ms
  • 75. 72 Chapter TwoFig.2.41 Simple circuit of the equivalent circuit of the three phase diode bridge at commutation time Δt at t = 5msFrom Fig.2.41 we can get the following defferntial equations: di diVa − La D1 − Vdc − Lb D 6 − Vb = 0 (2.101) dt dt diD1 diVa − La − Vdc − Lc D 2 − Vc = 0 (2.102) dt dt diD1Note that, during the time Δt , iD1 is constant so = 0 , substitute this dtvalue in (2.101) and (2.102) we get the following differential equations: diVa − Vb − Lb D 6 = Vdc (2.103) dt diVa − Vc − Lc D 2 = Vdc (2.104) dtBy equating the left hand side of equation (2.103) and (2.104) we get thefollowing differential equation: diD 6 diVa − Vb − Lb = Va − Vc − Lc D 2 (2.105) dt dt diD 6 diVb − Vc + Lb − Lc D 2 = 0 (2.106) dt dtThe above equation can be written in the following manner:(Vb − Vc )dt + Lb diD6 − Lc diD 2 = 0 (2.107)(Vb − Vc )dω t + ω Lb diD 6 − ω Lc diD 2 = 0 (2.108) Integrate the above equation during the time Δt with the help ofFig.2.39 we can get the limits of integration as shown in the following:π / 2+u 0 Io ∫ (Vb − Vc )dω t + ∫ ω Lb diD6 − ∫ ω Lc diD 2 = 0 π /2 Io 0
  • 76. Diode Circuits or Uncontrolled Rectifier 73π / 2+u ⎛ ⎛ 2π ⎞ ⎛ 2π ⎞ ⎞ ∫ ⎜Vm sin ⎜ ω t − ⎟ − Vm sin ⎜ ω t + ⎟ ⎟dω t + ωLb (− I o ) − ωLc I o = 0 π /2 ⎝ ⎝ 3 ⎠ ⎝ 3 ⎠⎠assume Lb = Lc = LS π / 2+u ⎡ ⎛ 2π ⎞ ⎛ 2π ⎞⎤Vm ⎢− cos⎜ ω t − ⎟ + cos⎜ ω t + ⎟ = 2ω LS I o ⎣ ⎝ 3 ⎠ ⎝ 3 ⎠⎥π / 2 ⎦ ⎡ ⎛π 2π ⎞ ⎛π 2π ⎞ ⎛ π 2π ⎞ ⎛ π 2π ⎞ ⎤Vm ⎢− cos⎜ + u − ⎟ + cos⎜ + u + ⎟ + cos⎜ − ⎟ − cos⎜ + ⎟ ⎣ ⎝2 3 ⎠ ⎝2 3 ⎠ ⎝2 3 ⎠ ⎝2 3 ⎠⎥⎦ = 2ω LS I o ⎡ ⎛ π⎞ ⎛ 7π ⎞ ⎛ −π ⎞ ⎛ 7π ⎞⎤Vm ⎢− cos⎜ u − ⎟ + cos⎜ u + ⎟ + cos⎜ ⎟ − cos⎜ ⎟⎥ = 2ω LS I o ⎣ ⎝ 6⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠⎦⎡ ⎛π ⎞ ⎛π ⎞ ⎛ 7π ⎞ ⎛ 7π ⎞ 3⎤⎢− cos(u ) cos⎜ ⎟ − sin (u ) sin ⎜ ⎟ + cos(u ) cos⎜ ⎟ − sin (u ) sin ⎜ 3 ⎟+ + ⎥⎣ ⎝6⎠ ⎝6⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ 2 2 ⎦ 2ω LS I o = Vm⎡ ⎤ 2ω LS I o cos(u ) − 0.5 sin (u ) − cos(u ) + 0.5 sin (u ) + 3 ⎥ = 3 3⎢−⎣ 2 2 ⎦ Vm 2ω LS I o 3[1 − cos(u )] = Vm 2ω LI o 2ω LI o 2 ω LS I ocos(u ) = 1 − =1− =1− 3 Vm 2 VLL VLL ⎡ 2ω LS I o ⎤u = cos −1 ⎢1 − ⎥ (2.109) ⎣ VLL ⎦ u 1 ⎡ 2ω LS I o ⎤Δt = = cos −1 ⎢1 − ⎥ (2.110) ω ω ⎣ VLL ⎦It is clear that the DC voltage reduction due to the source inductance isthe drop across the source inductance. divrd = LS D (2.111) dtMultiply (2.111) by dωt and integrate both sides of the resultantequation we get:
  • 77. 74 Chapter Twoπ +u2 Io ∫ vrd dω t = ∫ ω LdiD = ω LS I o (2.112) π 0 2π +u2 ∫ vrd dω t is the reduction area in one commutation period Δt . But we π 2have six commutation periods Δt in one period so the total reduction perperiod is: π +u 26 ∫ vrd dω t = 6ω LS I o (2.113) π 2 To obtain the average reduction in DC output voltage Vrd due tosource inductance we have to divide by the period time 2π . Then, 6ω LI oVrd = = 6 fLI o (2.114) 2π The DC voltage without source inductance tacking into account can becalculated as following:Vdc actual = Vdc without sourceinduc tan ce − Vrd = 1.35VLL − 6 fLI d (2.115)Fig.2.42 shows the utility line current with some detailes to help us tocalculate its rms value easly. Is u Io 2π +u 3 2π u 2π + 6 2 − Io 3 Fig.2.42 The utility line current
  • 78. Diode Circuits or Uncontrolled Rectifier 75 ⎡ π u ⎤ + ⎢ ⎛I u 2 3 2 ⎥ 2I o ⎡ 1 3 π u 2 ⎤ ⎢ ⎜ o ωt ⎞ dωt + I d dωt ⎥ = 2Is = ∫ ∫ ⎢ 2 u + 3 + 2 − u⎥ 2 ⎟ π⎢ ⎝u ⎠ ⎥ π ⎣ 3u ⎦ 0 u ⎢ ⎥ ⎣ ⎦ 2I o ⎡π u ⎤ 2Then I S = − (2.116) π ⎢ 3 6⎥ ⎣ ⎦ 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). u Is Io 11π u 7π u − − 6 2 6 2 π u 5π u − − 6 2 6 2 − Io Is′ u Io u 5π u 7π u − − 6 2 6 2 π u 11π u − − I 6 2 − o 6 2 u Fig.2.43 The utility line currents and its first derivative.Table(2.2) Jumb value of supply current and its first derivative.Js π u π u 5π u 5π u 7π u 7π u 11π u 11π u − + − + − + − + 6 2 6 2 6 2 6 2 6 2 6 2 6 2 6 2Is 0 0 0 0 0 0 0 0 ′Is Io − Io − Io Io − Io Io Io − Io u u u u u u u u
  • 79. 76 Chapter TwoIt is an odd function, then ao = an = 0 ⎡m 1 m ⎤ ∑ ∑ 1bn = ⎢ J s cos nωt s − J s sin nωt s ⎥ ′ (2.117) nπ ⎢ s =1 ⎣ n s =1 ⎥ ⎦ 1 ⎡ − 1 Io ⎛ ⎛π u ⎞ ⎛π u ⎞ ⎛ 5π u ⎞ ⎛ 5π u ⎞bn = ⎢ * ⎜ sin n⎜ − ⎟ − sin n⎜ + ⎟ − sin n⎜ − ⎟ + sin n⎜ + ⎟ nπ ⎣ n u ⎝ ⎝ 6 2⎠ ⎝ 6 2⎠ ⎝ 6 2⎠ ⎝ 6 2⎠ ⎛ 7π u ⎞ ⎛ 7π u ⎞ ⎛ 11π u ⎞ ⎛ 11π u ⎞ ⎞ ⎤ − sin n⎜ − ⎟ + sin n⎜ + ⎟ + sin n⎜ − ⎟ − sin n⎜ + ⎟ ⎟⎥ ⎝ 6 2⎠ ⎝ 6 2⎠ ⎝ 6 2⎠ ⎝ 6 2 ⎠ ⎠⎦ 2I o nu ⎡ nπ 5nπ 7 nπ 11nπ ⎤bn = ⎢cos 6 − cos 6 − cos 6 + cos 6 ⎥ (2.118) * sin n πu 2 2 ⎣ ⎦Then, the utility line current can be obtained as in (2.119). 4 3 ⎡ ⎛u⎞ ⎛ 5u ⎞ ⎛ 7u ⎞i (ω t ) = ⎢sin ⎜ ⎟ sin (ωt ) − 2 sin ⎜ 2 ⎟ sin (5ωt ) − 2 sin ⎜ 2 ⎟ sin (7ωt ) + 1 1 π u ⎣ ⎝2⎠ 5 ⎝ ⎠ 7 ⎝ ⎠ (2.119) ⎛ 11u ⎞ ⎛ 13u ⎞ ⎤ ⎟ sin (11ωt ) + 2 sin ⎜ ⎟ sin (13ωt ) − − + + ⎥ 1 1 + sin ⎜ 112 ⎝ 2 ⎠ 13 ⎝ 2 ⎠ ⎦ 2 6 Io ⎛ u ⎞Then; I S1 = sin ⎜ ⎟ 2.120) πu ⎝2⎠The power factor can be calculated from the following equation: 2 6 Io ⎛ u ⎞ sin ⎜ ⎟ I S1 ⎛u⎞ πu ⎝2⎠ ⎛u⎞pf = cos ⎜ ⎟ = cos ⎜ ⎟ IS ⎝2⎠ 2I o ⎡π u ⎤ 2 ⎝2⎠ − ⎥ π ⎢ 3 6⎦ ⎣ 3 * sin (u )Then; pf = (2.121) ⎡π u ⎤ u π⎢ − ⎥ ⎣ 3 6⎦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.
  • 80. Diode Circuits or Uncontrolled Rectifier 77Solution: (i) By substituting for ω = 2 * π * 50 , I d = 300 A , L = 0.008 H ,VLL = 33000V in (2.109), then u = 0.2549 rad . = 14.61 o uThen, Δt = = 0.811 ms . ω(ii) The the actual DC voltage can be obtained from (2.115) as following:Vdc actual = Vdc without sourceinduc tan ce − Vrd = 1.35VLL − 6 fLI dVdcactual =1.35 * 33000 − 6 * 50 * .008 * 300 = 43830V(iii) the power factor can be obtained from (2.121) then 3 * sin (u ) 3 sin (0.2549 ) pf = = = 0.9644 Lagging ⎡π u ⎤ ⎡ π 0.2549 ⎤ u π ⎢ − ⎥ 0.2549 * π ⎢ − ⎣ 3 6⎦ ⎣3 6 ⎥ ⎦(iv) The rms value of supply current can be obtained from (2.116)asfollowing 2I d ⎡π u ⎤ 2 2 * 300 2 ⎛ π 0.2549 ⎞Is = − ⎥= *⎜ − ⎟ = 239.929 A π ⎢ 3 6⎦ ⎣ π ⎝3 6 ⎠The rms value of fundamental component of supply current can beobtained from (2.120) as following: 4 3 Io ⎛ u ⎞ 4 3 * 300 ⎛ 0.2549 ⎞I S1 = sin ⎜ ⎟ * 2 3 = * sin ⎜ ⎟ = 233.28 A πu 2 ⎝2⎠ π * 0.2549 * 2 ⎝ 2 ⎠ I ⎛ u ⎞ 233.28 ⎛ 0.2549 ⎞ pf = S1 * cos⎜ ⎟ = * cos⎜ ⎟ = 0.9644 Lagging. Is ⎝ 2 ⎠ 239.929 ⎝ 2 ⎠ 2 2 ⎛I ⎞ ⎛ 239.929 ⎞THDi = ⎜ S ⎟ − 1 = ⎜ ⎜I ⎟ ⎟ − 1 = 24.05% ⎝ S1 ⎠ ⎝ 233.28 ⎠2.7 Multi-pulse Diode Rectifier Twelve-pulse bridge connection is the most widely used in highnumber 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
  • 81. 78 Chapter Twodescribed above but some will cancel, being equal in amplitude but 180oout 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). ⎛ ⎞ cos (25ωt )⎟ (2.83) 2 3 I ⎜ cos (ωt ) − 1 cos (5ωt ) + cos (13ωt ) − 1 1 cos (23ωt ) + 1i (t ) = πP d ⎝ 11 13 23 25 ⎠ 2 2 2 2 2 2 ⎛1⎞ ⎛1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞THD ( I P (t )) = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ 11 ⎠ ⎝ 13 ⎠ ⎝ 23 ⎠ ⎝ 25 ⎠ ⎝ 35 ⎠ ⎝ 35 ⎠ = 13.5% 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 a a1 Vd b1 c c1 b 2 3 N :1 a2 b2 c2 Fig.2.33 Twelve-pulse converter arrangement
  • 82. Diode Circuits or Uncontrolled Rectifier 79 (a) Utility input current. (b) FFT components of utility current. Fig.2.34 Simulation results of 12.pulse system.
  • 83. 80 Chapter Two Problems1- Single phase half-wave diode rectifier is connected to 220 V, 50 Hz supply to feed 5Ω pure resistor. Draw load voltage and current and diode voltage drop waveforms along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.2- The load of the rectifier shown in problem 1 is become 5Ω pure resistor and 10 mH inductor. Draw the resistor, inductor voltage drops, and, load current along with supply voltage. Then, find an expression for the load current and calculate the conduction angle, β . Then, calculate the DC and rms value of load voltage.3- In the rectifier shown in the following figure assume VS = 220V , 50Hz, L = 10mH and Ed = 170V . Calculate and plot the current an the diode voltage drop along with supply voltage, vs . vdiode + vL - i + - + vs Ed -4- Assume there is a freewheeling diode is connected in shunt with the load of the rectifier shown in problem 2. Calculate the load current during two periods of supply voltage. Then, draw the inductor, resistor, load voltages and diode currents along with supply voltage.5- The voltage v across a load and the current i into the positive polarity terminal are as follows: v(ωt ) = Vd + 2 V1 cos(ωt ) + 2 V1 sin (ωt ) + 2 V3 cos(3ωt ) i (ωt ) = I d + 2 I1 cos(ωt ) + 2 I 3 cos(3ωt − φ ) Calculate the following: (a) The average power supplied to the load. (b) The rms value of v(t ) and i (t ) . (c) The power factor at which the load is operating.
  • 84. Diode Circuits or Uncontrolled Rectifier 816- Center tap diode rectifier is connected to 220 V, 50 Hz supply via unity turns ratio center-tap transformer to feed 5Ω resistor load. Draw load voltage and currents and diode currents waveforms along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.7- Single phase diode bridge rectifier is connected to 220 V, 50 Hz supply to feed 5Ω resistor. Draw the load voltage, diodes currents and calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.8- If the load of rectifier shown in problem 7 is changed to be 5Ω resistor in series with 10mH inductor. Calculate and draw the load current during the first two periods of supply voltages waveform.9- Solve problem 8 if there is a freewheeling diode is connected in shunt with the load.10- If the load of problem 7 is changed to be 45 A pure DC. Draw diode diodes currents and supply currents along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current. (f) input power factor.11- Single phase diode bridge rectifier is connected to 220V ,50Hz supply. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor, and, input power factor and THD of the voltage at the point of common coupling.12- Three-phase half-wave diode rectifier is connected to 380 V, 50Hz supply via 380/460 V delta/way transformer to feed the load with 45 A DC current. Assuming ideal transformer and zero source inductance. Then, draw the output voltage, secondary and primary currents along with supply voltage. Then, calculate (a) Rectfication effeciency. (b) Crest factor of secondary current. (c) Transformer Utilization Factor (TUF). (d) THD of primary current. (e) Input power factor.
  • 85. 82 Chapter Two13- 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 10Ω resistor. Draw the output voltage, diode currents and supply current of phase a. Then, calculate: (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.15- Solve problem 14 if the load is 45A pure DC current. Then find THD of supply current and input power factor.16- If the supply connected to the rectifier shown in problem 14 has a 5 mH source inductance and the load is 45 A DC. Find, average DC voltage, and THD of input current.17- Single phase diode bridge rectifier is connected to square waveform with amplitude of 200V, 50 Hz. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor.18- In the single-phase rectifier circuit of the following figure, LS = 1 mH and Vd = 160V . The input voltage vs has the pulse waveform shown in the following figure. Plot is and id waveforms and find the average value of I d . id + iS Vd VS - f = 50 Hz 200V 120o ωt o o 60o 120 60o 60o 120
  • 86. Chapter 3 Thyristor Converters or Controlled Converters3.1 IntroductionThe controlled rectifier circuit is divided into three main circuits:- (1) Power Circuit This is the circuit contains voltage source, load and switches as diodes, thyristors or IGBTs. (2) Control Circuit This circuit is the circuit, which contains the logic of the firing of switches that may, contains amplifiers, logic gates and sensors. (3) Triggering circuit This circuit lies between the control circuit and power thyristors. Sometimes this circuit called switch drivers circuit. This circuit contains buffers, opt coupler or pulse transformers. The main purpose of this circuit is to separate between the power circuit and control circuit. The thyristor is normally switched on by applying a pulse to its gate.The forward drop voltage is so small with respect to the power 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
  • 87. SCR Rectifier or Controlled Rectifier 81controller 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 vs = Vm sin (ωt ) , vs ispositive when 0 < ω t < π, and vs is negative when π < ω t <2π. When vsstarts become positive, the SCR is forward-biased but remains in theblocking state till it is triggered. If the SCR is triggered at ω t = α, then αis 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 ω t reaches π radians. At that instant, the current throughthe circuit is zero. After that the current tends to flow in the reverse
  • 88. 82 Chapter Threedirection 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 forα=40o. FFT components for load voltage and current of half wave singlephase controlled rectifier with resistive load at α=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-phase controlled rectifier with resistive load at α=40o.
  • 89. SCR Rectifier or Controlled Rectifier 83Fig.3.3 FFT components for load voltage and current of half wave single phase controlled rectifier with resistive load at α =40o. The average voltage, Vdc , across the resistive load can be obtained byconsidering the waveform shown in Fig.3.2. π 1 V VVdc = 2π ∫ Vm sin(ωt ) dωt = m (− cos π + cos(α )) = m (1 + cos α ) (3.1) 2π 2π α The maximum output voltage and can be acheaved when α = 0 whichis the same as diode case which obtained before in (2.12). VVdm = m (3.2) πThe normalized output voltage is the DC voltage devideded by maximumDC voltage, Vdm which can be obtained as shown in equation (3.3). VVn = dc = 0.5 (1 + cos α ) (3.3 Vdm The rms value of the output voltage is shown in the followingequation:- π Vm 1⎛ sin(2 α ) ⎞ ∫ (Vm sin(ω t )) 1Vrms = 2 dω t = ⎜π − α + ⎟ (3.3) 2π 2 π⎝ 2 ⎠ α The rms value of the transformer secondery current is the same as thatof the load: V I s = rms (3.4) R
  • 90. 84 Chapter ThreeExample 1 In the rectifier shown in Fig.3.1 it has a load of R=15 Ω and,Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain anaverage output voltage of 70% of the maximum possible output voltage,calculate:- (a) The firing angle, α, (b) The efficiency, (c) Ripple factor (d)Transformer utilization factor, (e) Peak inverse voltage (PIV) of thethyristor and (f) The crest factor of input current.Solution: (a) Vdm is the maximum output voltage and can be acheaved whenα = 0 , The normalized output voltage is shown in equation (3.3) which isrequired to be 70%. Then, VVn = dc = 0.5 (1 + cos α ) = 0.7 . Then, α=66.42o =1.15925 rad. Vdm(b) Vm = 220 V V V 49.02Vdc = 0.7 * Vdm = 0.7 * m = 49.02 V , I dc = dc = = 3.268 A π R 15 Vm 1 ⎛ sin(2 α ) ⎞Vrms = ⎜π − α + ⎟, 2 π⎝ 2 ⎠at α=66.42 , Vrms=95.1217 V. Then, Irms=95.1217/15=6.34145 A o VVS = m = 155.56 V 2The rms value of the transformer secondery current is:I S = I rms = 6.34145 AThen, the rectification efficiency is: P V *Iη = dc = dc dc Pac Vrms * I rms 49.02 * 3.268 = = 26.56% 95.121 * 6.34145 V 95.121 π(b) FF = rms = = = 1.94 Vdc 49.02 2 2 V(c) RF = ac = FF 2 − 1 = 1.94 2 − 1 = 1.6624 Vdc P 49.02 * 3.268(d) TUF = dc = = 0.1624 VS I S 155.56 * 6.34145
  • 91. SCR Rectifier or Controlled Rectifier 85(e) The PIV is Vm(f) Creast factor of input current CF is as following: Vm I S ( peak ) R 14.6667 CF = = = = 2.313 IS 6.34145 6.341453.2.2 Half Wave Single Phase Controlled Rectifier With RL LoadA circuit with single SCR and RL load is shown in Fig.3.4. The source vsis an alternating sinusoidal source. If vs = Vm sin ( ω t), vs is positive when0 < ω t < π, and vs is negative when π < ω t <2π. When vs starts becomepositive, the SCR is forward-biased but remains in the blocking state tillit is triggered. If the SCR is triggered when ω t = α, then it startsconducting and the positive source keeps the SCR in conduction till ω treaches π radians. At that instant, the current through the circuit is notzero and there is some energy stored in the inductor at ω t = π radians.The voltage across the inductor is positive when the current through it 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 ω t = π 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.
  • 92. 86 Chapter Three Fig.3.5 Various voltages and currents waveforms for half wave single phase controlled rectifier with RL load. It is assumed that the current flows for α < ω t < β, where 2π > β > π .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 α < ω t < β , equation (3.5) applies. di L + R * i = Vm sin (ω t ), α ≤ ωt ≤ β (3.5) dt Divide the above equation by L we get the following equation: di R V + * i = m sin (ωt ), 0 ≤ ωt ≤ β (3.6) dt L L The instantaneous value of the current through the load can beobtained from the solution of the above equation as following: R ⎡ R ⎤ −∫ dt ∫ dt Vmi (ωt ) = e L ⎢ ⎢ ∫ e L * L sin ωt dt + A⎥ ⎥ ⎣ ⎦ − t⎡ ⎤ R R t VmThen, i (ωt ) = e L ⎢ ⎢ ∫ eL * L sin ωt dt + A⎥ ⎥ ⎣ ⎦
  • 93. SCR Rectifier or Controlled Rectifier 87Where, A is constant. By integrating the above equation we get: R Vm − ti (ωt ) = (R sin ωt − ωL cos ωt ) + Ae L (3.7) R 2 + w 2 L2Assume Z∠φ = R + j ωL . Then, Z 2 = R 2 + ω 2 L2 , Z ωL ωLR = Z cos φ , ωL = Z sin φ and tan φ = R ΦSubstitute that in (3.7) we get the following equation: ωt R − Vmi (ωt ) = (cos φ sin ωt − sin φ cos ωt ) + Ae tan φ Z ωt − VThen, i (ωt ) = m sin (ωt − φ ) + Ae tan φ α < ωt < β (3.8) Z The value of A can be obtained using the initial condition. Since thediode starts conducting at ω t = α and the current starts building up fromzero, then, i(α) = 0. Then, the value of A is expressed by the followingequation: A = − sin(α − φ ) (3.9) Once the value of A is known, the expression for current is known.After evaluating A, current can be evaluated at different values of ω t. ωt −α ⎞ Vm ⎛ ⎜ sin(ωt − φ ) − sin(α − φ )e − ωL / R ⎟i (ωt ) = α < ωt < β (3.10) Z ⎜ ⎜ ⎟ ⎟ ⎝ ⎠ When the firing angle α and the extinction angle β are known, theaverage and rms output voltage at the cathode of the SCR can beevaluated. We know that, i=0 when ω t=β substitute this condition inequation (3.10) gives equation (3.11) which used to determine β. Oncethe value of A, α and the extinction angle β 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. ( β −α ) −sin( β − φ ) = sin(α − φ )e ωL / R (3.11) β V VVdc 2π ∫ = m * sin ωt dωt = m * (cos α − cos β ) 2π (3.12) α
  • 94. 88 Chapter Three The rms load voltage is: β V * (Vm sin ωt )2 dωt = m * (β − α ) − 1 (cos 2 β − cos 2α ) (3.13) 1Vrms = 2π ∫ 2 π 2 α 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. VI dc = m * (cos α − cos β ) (3.14) 2π RExample 2 A thyristor circuit shown in Fig.3.4 with R=10 Ω, L=20mH,and VS=220 sin314t and the firing angle α is 30o. Determine theexpression for the current though the load. Also determine therectification efficiency of this rectifier.Solution: wLZ = 10 2 + (314 * 20 *10 − 3 ) 2 = 11.81 Ω , φ = tan −1 = 0.561 Rad . R π πα = 30 * = = 0.5236 Rad . 180 6 From equation (3.10) the current α < ω t < β can be obtained as ωt − α ⎞ 220 ⎛⎜ sin(ωt − 0.561) − sin(0.5236 − 0.561)e − ωL / R ⎟ i (ωt ) =follows:- 11.81 ⎜ ⎜ ⎟ ⎟ ⎝ ⎠ ( =18.6283* sin(ωt − 0.561) + 0.0374e − 0.628 (ωt − 0.5236 ) )The excitation angle β can be obtained from equation (3.11) as:sin( β − 0.561) = sin(05236 − 0561)e −0.628( β − 0.5236) Then, by using try and error technique which explained in Chapter 2we can get the value of β , β = 3.70766 Rad.The DC voltage can be obtained from equation (3.12) 220Vdc = * (cos 0.5236 − cos 3.70766) = 59.9V 2π Vdc 59.9Then, I dc = = = 5.99 A R 10 Similarly, Vrms can be obtained from equation (3.13) 220Vrms = * (3.70766 − 0.5236) − 1 (cos 2 * 3.70766 − cos 2 * 05236) = 111.384 V 2 π 2
  • 95. SCR Rectifier or Controlled Rectifier 89 Vrms 111.384I rms = = = 11.1384 A R 10 Vdc * I dc 2 Vdc 59.9 2η= = 2 = = 28.92% Vrms * I rms Vrms 111.384 23.2.3 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 ( ω t), vs is positive when 0 < ω t < π, and it isnegative when π < ω t <2π. 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 ω t reaches π radians. Atthat instant, the current through the circuit is not zero and there is someenergy stored in the inductor at ω t = π 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 π < ω t < (2π + α) (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.
  • 96. 90 Chapter Three Fig.3.7 Various voltages and currents waveforms for half wave single phase controlled rectifier with RL load and freewheeling diode.
  • 97. SCR Rectifier or Controlled Rectifier 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 π < ω t < (2π + α), 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 α < ω t < π , 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. di L + R * i = Vm sin (ωt ), α ≤ ωt ≤ π (3.15) dt di L + R * i = 0, π ≤ ω t ≤ 2π + α (3.16) dt The solution of (3.15) is the same as obtained before in (3.8) which isshown in (3.17). ω t −α V −i (t ) = m * sin(ω t − φ ) + A * e ωL / R α <ωt <π (3.17) Z The difference in the solution of (3.17) than (3.8) is how the constantA is evaluated? In the circuit without free-wheeling diode i (α ) = 0 , 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: i (α ) = i (2π + α ) = i (2nπ + α ) (3.18) The solution of (3.16) can be obtained as following: ( wt − π ) − tan φ i ( wt ) = B * e π < ωt < 2π + α (3.19) At ωt = π then, B = i (π ) (3.20) To obtain B, we have to obtain i (π ) from (3.17) by letting ωt equalπ . Then, π −α − V i (π ) = m * sin(φ ) + A * e tan φ (3.22) Z Substitute (3.21) into (3.19) we get the current duringπ < ωt < (2π + α ) as following:
  • 98. 92 Chapter Three ⎛ − π −α ⎞ − (ωt −π ) ⎜ Vm ⎟ tan φi (ω t ) = ⎜ * sin(φ ) + A * e tan φ ⎟*e π < ωt < 2π + α (3.22) ⎜ Z ⎟ ⎝ ⎠ (ωt −π ) ωt − α − − V∴ i (ω t ) = m * sin(φ ) * e tan φ + A * e tan φ π < ωt < 2π + α (3.23) Z From the above equation we can obtain i (2π + α ) as following: α +π 2π − − Vi (2π + α ) = m * sin(φ ) * e tan φ + A * e tan φ π < ωt < 2π + α (3.24) Z From (3.17) we can obtain ι (α ) by letting ωt = α as following:i (α ) = m * sin(α − φ ) + A V (3.25) Z Substitute (3.24) and (3.25) into (3.18) we get the following: α +π 2π − −Vm V * sin(α − φ ) + A = m * sin(φ ) * e tan φ + A * e tan φ Z Z ⎛ 2π ⎞ ⎛ α +π ⎞ − − ⎜ tan φ ⎟ Vm ⎜ ⎟A⎜1 − e ⎟ Z= * ⎜ sin(φ )e tan φ − sin(α − φ ) ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ α +π ⎞ − ⎜ tan φ ⎟ ⎜ sin(φ )e − sin(α − φ ) ⎟ V ⎜ ⎟∴A= m *⎝ ⎠ (3.26) Z ⎛ 2π ⎞ − ⎜ ⎟ ⎜ 1 − e tan φ ⎟ ⎜ ⎟ ⎝ ⎠ Then, the load current in the period of 2nπ + α < ωt < (2n + 1)π wheren = 0,1, 2, 3, ....... can be obtained from substituting (3.26) into (3.17).Also, the load current in the period of (2n + 1)π < ωt < (2n + 1)π + αn = 0,1, 2, 3, ....... can be obtained from substituting (3.26) into (3.23).Example 3 A thyristor circuit shown in Fig.3.6 with R=10 Ω, L=100 mH,and VS=220 sin314t V, and the firing angle α is 30o. Determine theexpression for the current though the load. Also determine therectification efficiency of this converter.
  • 99. SCR Rectifier or Controlled Rectifier 93Solution: Z = 10 2 + (314 *100 *10 − 3 ) 2 = 32.9539 Ω wLφ = tan −1 = 1.2625 Rad . , R π πα = 30 * = = 0.5236 Rad . 180 6From (3.26) we can obtain A , A=7.48858From (3.17) we can obtain i (ωt ) in the period of2nπ + α < ωt < (2n + 1)π where n = 0,1, 2, 3, ....... Then,i (ωt ) = 6.676 * sin(ωt − 1.2625) + 7.48858 * e −0.31845*(ωt − 0.5236)i (π ) = 9.61457 , i (α ) = 2.99246During the period of (2n + 1)π < ωt < (2n + 1)π + α wheren = 0,1, 2, 3, ....... we can obtain the solution from (3.18) where B = i (π ) .Then, i (ωt ) = 9.61457 * e −0.31845(ωt −π ) , π 1 V VVdc = 2π ∫ Vm sin( wt ) dwt = m ( − cos π + cos(α )) = m (1 + cos α ) = 65.3372 V 2π 2π αThe rms value of the output voltage is shown in the following equation:- 1/ 2 ⎡ 1 π ⎤ V ⎡1 sin( 2 α ) ⎤ 1/ 2Vrms = ⎢ ∫ (Vm sin(ωt ) )2 dω t ⎥ = m ⎢ (π − α + )⎥ =108.402 V ⎢ 2π α ⎣ ⎥ ⎦ 2 ⎣π 2 ⎦ π 1 ⎡I dc = 2π ⎢ ∫( ⎢ 6.676 * sin(ω t − 1.2625) + 7.48858 * e − 0.31845*(ω t − 0.5236) dω t ) ⎣α 2π + α ⎤ + ∫( ) 9.61457 * e − 0.31845(ω t −π ) dω t ⎥ = 4.2546 A ⎥ π ⎦ 1 ⎡⎡ πI rms = ( ⎢ ⎢ ∫ 6.676 * sin(ω t − 1.2625) + 7.48858 * e − 0.31845*(ω t − 0.5236) 2π ⎢ ⎢ ) 2 dω t + ⎣ ⎣α 1/ 2 2π +α ⎤⎤ ∫ (9.61457 * e ) − 0.31845(ω t −π ) 2 dω t ⎥ ⎥ = 5.43288 A π ⎥⎥ ⎦⎦
  • 100. 94 Chapter Three Pdc V *Iη= = dc dc Pac Vrms * I rms 65.3372 * 4.2546 = = 47.2 % 108.402 * 5.432883.3 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. b a Fig.3.8 Center tap controlled rectifier with resistive load.
  • 101. SCR Rectifier or Controlled Rectifier 95 Fig.3.9 The output voltgae and thyristor T1 reverse voltage wavforms along with the supply voltage wavform.Fig.3.10 Load current and thyristors currents for Center tap controlled rectifier with resistive load.
  • 102. 96 Chapter Three The average voltage, Vdc, across the resistive load is given by: π 1 Vm Vm π∫ mVdc = V sin(ω t ) dω t = (− cos π − cos(α )) = (1 + cosα ) (3.27) π π α Vdm is the maximum output voltage and can be acheaved when α=0 inthe above equation. The normalized output voltage is: VVn = dc = 0.5 (1 + cos α ) (3.28) VdmFrom the wavfrom of the output voltage shown in Fug.3.9 the rms outputvoltage can be obtained as following: π Vm sin( 2 α ) ∫ (Vm sin(ω t )) dω t = 1Vrms = 2 π −α + (3.29) π 2π 2 αExample 4 The rectifier shown in Fig.3.8 has load of R=15 Ω and,Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain anaverage output voltage of 70 % of the maximum possible output voltage,calculate:- (a) The delay angle α, (b) The efficiency, (c) The ripple factor(d) The transformer utilization factor, (e) The peak inverse voltage (PIV)of the thyristor and (f) The crest factor of input current. (g) Input powerfactor.Solution : (a) Vdm is the maximum output voltage and can be acheaved when α=0,the normalized output voltage is shown in equation (3.28) which isrequired to be 70%. Then: VVn = dc = 0.5 (1 + cos α ) = 0.7 , then, α=66.42o Vdm 2 Vm(b) Vm = 220 , then, Vdc = 0.7 *Vdm = 0.7 * = 98.04 V π Vdc 98.04I dc = = = 6.536 A R 15 Vm sin( 2 α )Vrms = π −α + 2π 2at α=66.42o Vrms=134.638 VThen, Irms=134.638/15=8.976 A
  • 103. SCR Rectifier or Controlled Rectifier 97 VmVS = = 155.56 V 2The rms value of the transformer secondery current is: II S = rms = 6.347 A 2Then, The rectification efficiency is: P V *Iη = dc = dc dc Pac Vrms * I rms 98.04 * 6.536 = = 53.04% 134.638 * 8.976 V 134.638(c) FF = rms = = 1.3733 and, Vdc 98.04 V RF = ac = FF 2 − 1 = 1.37332 − 1 = 0.9413 Vdc Pdc 98.04 * 6.536(d) TUF = = = 0.32479 2VS I S 2 *155.56 * 6.34145(e) The PIV is 2 Vm Vm I S ( peak ) R 14.6667(f) Creast Factor CF, CF = = = = 2.313 IS 6.34145 6.341453.3.2 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
  • 104. 98 Chapter Threegates. 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 in Fig.3.11 with resistive load.
  • 105. SCR Rectifier or Controlled Rectifier 99Fig.3.13 FFT components of the output voltage and supply current for converter shown in Fig.3.11. The main purpose of this circuit is to provide a controllable DC outputvoltage, which is brought about by varying the firing angle, α. Let vs = Vmsin ω t, with 0 < ω t < 360o. If ω t = 30o when T1 and T2 are triggered,then the firing angle α is said to be 30o. In this instance, the other pair istriggered when ω t = 30+180=210o. When vs changes from positive tonegative value, the current through the load becomes zero at the instantω t = π 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 ( ω t), where ω t varies from 0 to2π radians. Since the output waveform repeats itself every half-cycle, theaverage output voltage is expressed as a function of α, as shown inequation (3.27). π 1 Vm [− cos π − (− cos(α ) )] = Vm (1 + cosα ) (3.27) π∫ mVdc = V sin(ω t ) dω t = π π α Vdm is the maximum output voltage and can be acheaved when α=0,The normalized output voltage is: VdcVn = = 0.5 (1 + cos α ) (3.28) VdmThe rms value of output voltage is obtained as shown in equation (3.29). π Vm sin(2 α ) 1 (V sin(ω t ) ) π∫ mVrms = 2 dω t = π −α + (3.29) α 2π 2
  • 106. 100 Chapter ThreeExample 5 The rectifier shown in Fig.3.11 has load of R=15 Ω and,Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain anaverage output voltage of 70% of the maximum possible output voltage,calculate:- (a) The delay angle α, (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 α=0, The normalized output voltage is shown in equation (3.31) which isrequired to be 70%. V Then, Vn = dc = 0.5 (1 + cos α ) = 0.7 , then, α=66.42o Vdm 2 Vm(b) Vm = 220 , then, Vdc = 0.7 *Vdm = 0.7 * = 98.04 V π Vdc 98.04I dc = = = 6.536 A R 15 Vm sin( 2 α )Vrms = π −α + 2π 2At α=66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A VVS = m = 155.56 V 2The rms value of the transformer secondery current is:I S = I rms = 8.976 AThen, The rectification efficiency is P V *Iη = dc = dc dc Pac Vrms * I rms 98.04 * 6.536 = = 53.04% 134.638 * 8.976 V 134.638(c) FF = rms = = 1.3733 Vdc 98.04 V RF = ac = FF 2 − 1 = 1.37332 − 1 = 0.9413 Vdc
  • 107. SCR Rectifier or Controlled Rectifier 101 Pdc 98.04 * 6.536(d) TUF = = = 0.4589 VS I S 155.56 * 8.976(e) The PIV is Vm Vm I S ( peak ) 14.6667(f) Creast Factor CF, CF = = R = = 1.634 IS 8.976 8.9763.3.3 Full Wave Fully Controlled Rectifier With RL Load In Continuous Conduction Mode The circuit of a single-phase fully controlled bridge rectifier circuit 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.
  • 108. 102 Chapter Three Fig.3.15 Various voltages and currents waveforms for the converter shown in Fig.3.14 in continuous conduction mode. Fig.3.16 FFT components of load voltage and supply current in continuous conduction mode. Let vs = Vm sin ω t, with 0 < ω t < 360o. If ω t = 30o when T1 and T2are triggered, then the firing angle is said to be 30o. In this instance theother pair is triggered when ω t= 210o. When vs changes from a positiveto a negative value, the current through the load does not fall to zerovalue at the instant ω t = π radians, since the load contains an inductor
  • 109. SCR Rectifier or Controlled Rectifier 103and 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 α, 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). π +α 1 2VmVdc = π ∫ Vm sin(ωt ) dωt = π cos α (3.33) α Vdc The normalized output voltage is Vn = = cos α (3.34) Vdm The rms value of output voltage is obtained as shown in equation(3.35). π +α π +α Vm Vm ∫ (Vm sin(ω t )) dω t = 1Vrms = ∫ (1 − cos(2ω t ) dω t = 2 (3.35) π 2π 2 α αExample 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 α, (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.
  • 110. 104 Chapter ThreeSolution: (a) Vdm is the maximum output voltage and can be acheaved whenα=0. The normalized output voltage is shown in equation (3.30) which is Vrequired to be 70%. Then, Vn = dc = cosα = 0.7 , then, α=45.5731o= 0.7954 Vdm(b) Vm = 220 2 Vm VmVdc = 0.7 *Vdm = 0.7 * = 98.04 V , Vrms = π 2At α=45.5731 Vrms=155.563 V. Then, Irms=50 A o VVS = m = 155.56 V 2The rms value of the transformer secondery current is: I S = I rms = 50 AThen, The rectification efficiency is P V *I 98.04 * 50η = dc = dc dc = = 63.02% Pac Vrms * I rms 155.563 * 50 V 155.563(c) FF = rms = = 1.587 Vdc 98.04 V RF = ac = FF 2 − 1 = 1.37332 − 1 = 1.23195 Vdc P 98.04 * 50(d) TUF = dc = = 0.4589 VS I S 155.56 * 50(e) The PIV is Vm I S ( peak )(f) Creast Factor CF, CF = =1 IS3.3.4 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).
  • 111. SCR Rectifier or Controlled Rectifier 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 the converter shown in Fig.3.14 in case of discontinuous conduction mode.
  • 112. 106 Chapter Three Fig.3.18 FFT components of supply current along with supply voltage of the converter shown in Fig.3.14 in discontinuous conduction mode.During the period of α ≤ ω t ≤ β the following differential equation canbe obtained: di L + R * i = Vm sin (ω t ), α ≤ ω t ≤ β (3.35) dtThe solution of the above equation is given as in the following: ⎛ (ω t −α ) ⎞ Vm ⎜ − tan φ ⎟i (ω t ) = ⎜ sin(ω t − φ ) − sin(α − φ )e ⎟ (3.36) Z ⎜ ⎟ ⎝ ⎠ ωLWhere φ = tan −1 and Z 2 = R 2 + (ω L) 2 R But i=0 when ω t=β substitute this condition in equation (3.36) givesequation (3.37) which used to determine β. Once the value of A, α andthe extinction angle β 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. ( β −α ) − tan φsin( β − φ ) = sin(α − φ )e (3.37) β Vm VmVdc = π ∫ * sin ω t dω t = π * (cos α − cos β ) (3.38) α The rms load voltage is:
  • 113. SCR Rectifier or Controlled Rectifier 107 β Vm * (Vm sin ω t )2 dω t = (β − α ) − 1 (cos 2 β − cos 2α ) (3.39) 1Vrms = π ∫ 2π * 2 α 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. VmI dc = * (cos α − cos β ) (3.40) πR3.3.5 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 < ω t <360o. Let the load inductance be large enough to maintain a steadycurrent through the load. Let firing angle α be 30o. Let SCRs T3 and T4be in conduction before ω t < 30o. When T1 and T2 are triggered at ω t =30o, there is current through the source inductance, flowing in 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 − I dc to I dc 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 I dc and the current through T3 would fall from I dc to zero. Similar process occurs with the SCRs T2 and T4. During thisperiod, the current through T2 would rise from zero to I dc and, thecurrent through T4 would fall from I dc to zero.Fig.3.19 single phase full wave fully controlled rectifier with source inductance
  • 114. 108 Chapter Three 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 waveform of a single phase full wave fully controlled rectifier with source inductance.Fig.3.21 Output voltage, supply current along with supply voltage waveform of a single phase full wave fully controlled rectifier with source inductance.
  • 115. SCR Rectifier or Controlled Rectifier 109 Let us study the commutation period starts at α < ω t < α + u . 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 Δt to completely turn off T3 and T4and to make T1 and T2 carry the entire load current I o which is veryclear from Fig.3.20. Also, in the same time ( Δt ) the supply currentchanges from − I o to I o which is very clear from Fig.3.21. Fig.3.22shows 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 rectifier during the commutation period. dis v s − Ls =0 (3.41) dt Multiply the above equation by ωt then, Vm sin ω t dωt − ω Ls dis = 0 Integrate the above equation during the commutation period we get thefollowing: α +u Io ∫Vm sin ω t dω t = ω Ls ∫ dis α −Io Then, Vm [cos α − cos(α + u )] = 2ωLs I o . Then, 2ωLs I ocos(α + u ) = cos α − Vm
  • 116. 110 Chapter Three ⎡ 2ωLs I o ⎤ Then u = cos −1 ⎢cos(α ) − −α Vm ⎥ (3.42) ⎣ ⎦ u 1⎧ ⎡ 2ω Ls I o ⎤ ⎫ Then Δt = = ⎨cos −1 ⎢cos(α ) − − ⎥ −α⎬ (3.43) ω ω⎩ ⎣ Vm ⎦ ⎭ It is clear that the DC voltage reduction due to the source inductance,vrd is the drop across the source inductor. Then, di vrd = Ls s (3.44) dt α +u Io Then, ∫ vrd t dω t = ω Ls ∫ dis = 2ω Ls I o (3.45) α −Io α +u ∫ vrd t dω t is the reduction area in one commutation period. But we αhave two commutation periods in one period of supply voltage waveform.So, the total reduction per period is shown in (3.46): α +u 2 ∫Vrd t dω t = 4ω Ls I o (3.46) α To obtain the average reduction in DC output voltage Vrd due tosource inductance we have to divide the above equation by the period ofsupply voltage waveform, 2π . Then, 4ω Ls I o Vrd = = 4 fLs I o (3.47) 2π The DC voltage with source inductance taking into account can becalculated as following: 2V Vdc actual = Vdc without sourceinduc tan ce − Vrd = m cosα − 4 fLs I o (3.48) π The rms value of supply current is the same as obtained before insingle phase full bridge diode rectifier in (2.64) 2I o ⎡π u ⎤ 2 Is = − (3.49) π ⎢ 2 3⎥ ⎣ ⎦
  • 117. SCR Rectifier or Controlled Rectifier 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 I s1 is shown in (2.68) as following: 8I o u I S1 = * sin (3.50) 2 πu 2 The power factor of this rectifier is shown in the following: I ⎛ u⎞ p. f = s1 cos⎜ α + ⎟ (3.51) Is ⎝ 2⎠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 ω t =π. At ω t=π 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 ω t=360o. Whenω 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:-
  • 118. 112 Chapter Three Fig.3.23 Single-phase half controlled bridge rectifier (semi bridge converter). π 1 Vm [− cos π − (− cos(α ))] = Vm (1 + cosα ) (3.52) π∫ mVdc = V sin(ω t ) dω t = π π α Vdm is the maximum output voltage and can be acheaved when α=0, VThe normalized output voltage is: Vn = dc = 0.5 (1 + cos α ) (3.53) Vdm The rms value of output voltage is obtained as shown in the followingequation:- π Vm sin(2 α ) 1 (Vm sin(ω t ) ) π∫Vrms = 2 dω t = π −α + (3.54) α 2π 2Fig.3.24 Various voltages and currents waveforms for the converter shown in Fig.3.23.
  • 119. SCR Rectifier or Controlled Rectifier 1133.3.6 Inverter Mode Of OperationThe thyristor converters can also operate in an inverter mode, where Vdhas 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 I d , 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 vd is negative,given by (3.48), where 90° < α < 180°. Therefore, the average powerPd = Vd * I d is negative, that is, it flows from the DC to the AC side. Onthe AC side, Pac = Vs I S1 cosφ1 is also negative because φ > 90 o . Fig.3.25 Single phase SCR inverter. Fig.3.26 Waveform output from single phase inverter assuming DC load current.
  • 120. 114 Chapter Three There are several points worth noting here. This inverter mode of operation is possible since there is a source of energy on the DC side. On the ac side, the ac voltage source facilitates the commutation of current from one pair of thyristors to another. The power flows into this AC source. Generally, the DC current source is not a realistic DC siderepresentation of systems where such a mode of operation may beencountered. Fig.3.27 shows a voltage source Ed 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 Ld allows us to assume id 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 Ld is zero, 2 E d = Vd = Vdo cos α − ω Ls I d (3.55) π The equation is exact if the current is constant at I d ; otherwise, avalue of id at ω t = α should be used in (3.55) instead of I d . Fig.3.28shows that for a given value of α , for example, α1 , the intersection of thedo source voltage Ed = Ed 1 , and the converter characteristic at α1 ,determines the do current I d 1 , and hence the power flow Pd1 . During the inverter mode, the voltage waveform across one of thethyristors is shown in Fig.3.29. An extinction angle γ 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 intervaltγ = γ / ω should be greater than the thyristor turn-off time τ 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. γ = 180 − (α + u ) (3.56)
  • 121. SCR Rectifier or Controlled Rectifier 115 Fig.3.27 SCR inverter with a DC voltage source. Fig.3.28 Vd versus I d in SCR inverter with a DC voltage source. Fig.3.29 Voltage across a thyristor in the inverter mode.
  • 122. 116 Chapter ThreeInverter startup For startup of the inverter in Fig.3.25, the delay angle α is initiallymade sufficiently large (e.g.,165o) so that id is discontinuous as shown inFig.3.30. Then, α is decreased by the controller such that the desired I dand Pd 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 α=0 which is the same as diodecase. This rectfier has continuous load current and voltage in case of α ≤30. However, the load voltage and current will be discontinuous in caseof α > 30. Fig.3.31 Three phase half wave controlled rectifier with resistive load.
  • 123. SCR Rectifier or Controlled Rectifier 117 In case of α ≤ 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. 5π / 6 +α 3 3 3 VmVdc = 2π ∫Vm sin ω t dω t = 2π cosα = 0.827Vm cosα π / 6 +α (3.57) 3 = VLL cosα = 0.675VLL cosα 2π 3 3 Vm 0.827 * VmI dc = cos α = cos α (3.58) 2 *π * R R 5π / 6 +α 3 1 3 ∫ (Vm sin ω t ) dω t = 3 Vm cos 2α (2.57) 2Vrms = + 2π 6 8π π / 6 +α 1 3 3 Vm + cos 2α 6 8πI rms = (3.60) R Then the thyristor rms current is equal to secondery current and can beobtaiend as follows: 1/ 2 ⎛1 3 ⎞ Vm ⎜ + ⎜ 6 8 π cos 2α ⎟ ⎟ II r = I S = rms = ⎝ ⎠ (3.61) 3 RThe PIV of the diodes is 2 V LL = 3 Vm (3.62)
  • 124. 118 Chapter ThreeFig.3.32 Voltages and currents waveforms for rectifier shown in Fig.3.31 at α ≤ 30.
  • 125. SCR Rectifier or Controlled Rectifier 119Fig.3.33 FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.31 for α ≤ 30. In case of α > 30, various voltages and currents of the rectifier 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 :- π 3 3 Vm ⎡ ⎛π ⎞⎤ ⎡ ⎛π ⎞⎤ (3.63)Vdc = 2π ∫ Vm sin ω t dωt = 2π ⎢1 + cos ⎜ 6 + α ⎟⎥ = 0.4775Vm ⎢1 + cos ⎜ 6 + α ⎟⎥ ⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦ π / 6 +α 3 Vm ⎡ ⎛π ⎞⎤I dc = ⎢1 + cos ⎜ 6 + α ⎟⎥ 2π R ⎣ (3.64) ⎝ ⎠⎦ π 5 α ∫ (Vm sin ω t ) 3 1Vrms = 2 dω t = 3 Vm − + sin(π / 3 + 2α ) (2.63) 2π 24 4π 8 π π / 6 +α 3 Vm 5 α 1I rms = − + sin(π / 3 + 2α ) (3.66) R 24 4π 8 π Then the diode rms current can be obtaiend as follows: I V 5 α 1I r = I S = rms = m − + sin(π / 3 + 2α ) (3.67) 3 R 24 4π 8 πThe PIV of the diodes is 2 VLL = 3 Vm (3.68)
  • 126. 120 Chapter Three Fig.3.34 Various voltages and currents waveforms for converter shown in Fig.3.22 for α > 30.
  • 127. SCR Rectifier or Controlled Rectifier 121Fig.3.35 FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.22 for α > 30.Example 7 Three-phase half-wave controlled rectfier is connected to 380V three phase supply via delta-way 380/460V transformer. The load ofthe rectfier is pure resistance of 5 Ω . The delay angle α = 25o . Calculate:The rectfication effeciency (b) Transformer Utilization Factor (TUF) (c)Crest Factor C F of the input current (d) PIV of thyristors Solution: From (3.57) the DC value of the output voltage can be obtained as following: 3 3Vdc = VLL cos α = 460 cos 25 = 281.5V 2π 2π V 281.5 Then; I dc = dc = = 56.3 A R 5 From (3.59) we can calculate Vrms as following: 1 3 1 3Vrms = 3 Vm + cos 2α = 2 VLL * + cos 2α 6 8π 6 8π cos (2 * 25) = 298.8 V 1 3 Then, Vrms = 2 * 460 * + 6 8π
  • 128. 122 Chapter Three Vrms 298.8 Then I rms = = = 59.76 A R 5 Then, the rectfication effeciency can be calculated as following: V Iη = dc dc *100 = 88.75% Vrms I rmsThe rms value of the secondary current can be calculated as following: I 59.76I S = rms = = 34.5 A 3 3 Vdc I dc 281.5 * 56.3TUF = = *100 = 57.66% 3 VLL * I s 3 * 460 * 34.5I S , peak = Vm = ( ) 2 / 3 *VLL = ( ) 2 / 3 * 460 = 75.12 A R 5 5 I S , peak 75.12CF = = = 2.177 IS 34.5PIV = 2 VLL = 2 * 460 = 650.54 VExample 8 Solve the previous example (evample 7) if the firing angleα = 60 oSlution: From (3.63) the DC value of the output voltage can be obtainedas following: ⎛ 2⎞ 3⎜ ⎟ ⎜ 3 ⎟ * 460 ⎡ 3 Vm ⎡ ⎛ π ⎞ ⎤ ⎝ ⎠ ⎛ π π ⎞⎤Vdc = 2π ⎣⎢1 + cos ⎜ 6 + α ⎟⎥ = 2π ⎢1 + cos ⎜ 6 + 3 ⎟⎥ = 179.33 V ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦ V 179.33 Then; I dc = dc = = 35.87 A R 5 From (3.65) we can calculate Vrms as following: 5 α 1 Vrms = 3 Vm − + sin(π / 3 + 2α ) 24 4π 8 π 5 π /3 1 = 2 * 460 * − + sin(π / 3 + 2π / 3 ) = 230V 24 4π 8 π V 230 Then I rms = rms = = 46 A R 5Then, the rectfication effeciency can be calculated as following
  • 129. SCR Rectifier or Controlled Rectifier 123 Vdc I dcη= *100 = 60.79 % Vrms I rmsThe rms value of the secondary current can be calculated as following: I 46I S = rms = = 26.56 A 3 3 Vdc I dc 179.33 * 35.87TUF = = *100 = 30.4 % 3 VLL * I s 3 * 460 * 26.56I S , peak = Vm = ( ) 2 / 3 *VLL = ( ) 2 / 3 * 460 = 75.12 A R 5 5 I S , peak 75.12CF = = = 2.83 IS 26.56PIV = 2 VLL = 2 * 460 = 650.54 V3.5 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 α > 30. However if α < 30 the load voltage will be positive all thetime. Then in case of α > 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 α > 30 andpure DC current load. Fig.3.36 Three Phase Half Wave Controlled Rectifier With DC Load Current
  • 130. 124 Chapter Three t=0 Fig.3.37 Various voltages and currents waveforms for the converter shown in Fig.3.36 for α > 30 and pure DC current load.Fig.3.38 FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.36 for α > 30 and pure DC current load.
  • 131. SCR Rectifier or Controlled Rectifier 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:- 2π / 3 2 2 I dc ⎛ 2nπ ⎞bn = π ∫ I dc sin(nω t ) dω t = πn ⎝ ⎜1 − cos 3 ⎠ ⎟ for n=1,2,3,4,… 0 2 I dc 3Then, bn = * for n=1,2,4,5,7,8,10,….. (3.69) πn 2And b n = 0 For n=3,6,9,12,….. (3.70) Then the source current waveform can be expressed as the followingequation 3I dc ⎡ 1 1 1 1 ⎤i p (ωt ) = ⎢sin ωt + 2 sin 2ωt + 4 sin 4ωt + 5 sin 5ωt + 7 sin 7ωt + ......⎥ (3.71) π ⎣ ⎦ 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:- I 2 − I 21 p pTHD = (3.72) I 21 p 2Where, I p = * I dc (3.73) 3 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) 3II p1 = dc (3.74) 2π Substitute equations (3.73) and (3.74) into equation (3.72), then, 2 2 9 2 I dc − I dc 3 2π 2THD = = 68 % (3.75) 9 2 I dc 2π 2
  • 132. 126 Chapter Three 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:- 2 2 2 2 2 2 2 2 2 ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞THD = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + .... ≅ 68 % (3.76) ⎝ 2⎠ ⎝4⎠ ⎝5⎠ ⎝7⎠ ⎝8⎠ ⎝ 10 ⎠ ⎝ 11 ⎠ ⎝ 13 ⎠ ⎝ 14 ⎠ 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 ofthe rectfier draws 100 A pure DC current. The delay angle, α = 30 o .Calculate: (a) THD of primary current. (b) Input power factor.Solution: The voltage ratio of delta-way transformer is 380/460V. Then, 460the peak value of primary current is 100 * = 121.05 A . Then, 380 2 I P, rms = 121.05 * = 98.84 A . 3 I P1 can be obtained from equation (372) where 3I 3 *121.05 I P1 = dc = = 81.74 A . 2π 2π 2 2 ⎛ I P, rms ⎞ ⎟ − 1 *100 = ⎛ 98.84 ⎞Then, (THD )I P = ⎜⎜ ⎜ ⎟ − 1 *100 = 67.98 % ⎝ I P1 ⎟ ⎠ ⎝ 81.74 ⎠The input power factor can be calculated as following: I ⎛ π ⎞ 81.74 ⎛π π ⎞P. f = P1 * cos⎜ α + ⎟ = * cos⎜ + ⎟ = 0.414 Lagging I P , rms ⎝ 6 ⎠ 98.84 ⎝6 6⎠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
  • 133. SCR Rectifier or Controlled Rectifier 127load voltage, secondary current and supply current for the convertershown in Fig.3.39 for α > 30 and RL load is shown in Fig.3.41. In case offiring angle α 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 α 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 :- π 3 3 Vm ⎡ ⎛π ⎞⎤ ⎡ ⎛π ⎞⎤V = ∫ V sin ωt dωt = 1 + cos ⎜ + α ⎟ = 0.4775V 1 + cos ⎜ + α ⎟ (3.77) 2π ⎢ ⎥ m⎢ ⎥ 2π dc m π / 6+α ⎣ ⎝6 ⎠⎦ ⎣ ⎝6 ⎠⎦ 3 Vm ⎡ ⎛π ⎞⎤I dc = ⎢1 + cos ⎜ 6 + α ⎟⎥ 2π R ⎣ (3.78) ⎝ ⎠⎦ π 5 α ∫ (Vm sin ωt ) 3 1Vrms = 2 dωt = 3 Vm − + sin(π / 3 + 2α ) 2π 24 4π 8 π π / 6 +α (3.79) 3 Vm 5 α 1I rms = − + sin(π / 3 + 2α ) (3.80) R 24 4π 8 π Fig.3.39 Three-phase half wave controlled rectifier with free wheeling diode.
  • 134. 128 Chapter Three Fig.3.40 Various voltages and currents waveforms for the converter shown in Fig.3.36 for α > 30 with RL load and freewheeling diode.Fig.3.41 FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.36 for α > 30 and RL load.
  • 135. SCR Rectifier or Controlled Rectifier 1293.7 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.
  • 136. 130 Chapter Three 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:va = Vm sin (ω t )vb = Vm sin (ω t − 120)vc = Vm sin (ω t + 120) It can be seen from Fig.3.44 that the voltage va is the highest positivevoltage of the three phase voltage when ωt is in the range30 < ω t < 150 o . 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 α = 40 as an example. So, T1 takes apulse at ω t = 30 + α = 30 + 40 = 70 o as shown in Fig.3.44. Also, it is clearfrom Fig3.38 that thyristor T1 or any other thyristor remains on for 120o . Fig.3.44 Phase voltages and thyristors currents of three-phase full wave controlled rectifier at α = 40 o . It can be seen from Fig.3.44 that the voltage vb is the highest positivevoltage of the three phase voltage when ωt is in the range of150 < ωt < 270 o . So, the thyristor T3 is forward bias during this period
  • 137. SCR Rectifier or Controlled Rectifier 131and 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 α = 40 as an example. So, T3 takes apulse at ω t = 150 + α = 150 + 40 = 190 o . It can be seen from Fig.3.44 that the voltage vc is the highest positivevoltage of the three phase voltage when ωt is in the range270 < ω t < 390 o . 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 α = 40 as an example. So, T3 takes apulse at ωt = 270 + α = 310 o . It can be seen from Fig.3.44 that the voltage va is the highest negativevoltage of the three phase voltage when ω t is in the range210 < ω t < 330 o . So, the thyristor T4 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.338, the firing angle α = 40 as an example. So, T4 takes apulse at ω t = 210 + α = 210 + 40 = 250o . It can be seen from Fig.3.44 that the voltage vb is the highest negativevoltage of the three phase voltage when ω t is in the range330 < ω t < 450 o or 330 < ω t < 90 o 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 α = 40 as an example. So, T6 takes a pulseat ω t = 330 + α = 370 o . It can be seen from Fig.3.44 that the voltage vc is the highest negativevoltage of the three phase voltage when ω t is in the range90 < ωt < 210 o . 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 α = 40 as an example. So, T2 takes apulse at ωt = 90 + α = 130 o . From the above explanation we can conclude that there is twothyristor 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)
  • 138. 132 Chapter Threewhich 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 α + 30o to α + 90o T1 and T6 α + 90 to α + 150 o o T1 and T2 α + 150 to α + 210 o o T2 and T3 α + 210o to α + 270o T3 and T4 α + 270o to α + 330o T4 and T5 α + 330o to α + 360o and α + 0o to α + 30o T5 and T6Fig.3.45 Output voltage along with three phase line to line voltages of rectifier in Fig.3.42 at α = 40o . The line current waveform is very easy to obtain it by applyingkerchiefs current law at the terminals of any phase. As an example I a = I T 1 − I T 4 which is clear from Fig.3.42. The input current of thisrectifier for α = 40, (α ≤ 60) is shown in Fig.3.46. Fast Fouriertransform (FFT) of output voltage and supply current are shown inFig.3.47.
  • 139. SCR Rectifier or Controlled Rectifier 133 Fig.3.46 The input current of this rectifier of rectifier in Fig.3.42 at α = 40, (α ≤ 60 ) . Fig.3.46 FFT components of output voltage and supply current of rectifier in Fig.3.42 at α = 40, (α ≤ 60 ) .
  • 140. 134 Chapter Three 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. π / 2 +α 3 π 3 3 VmVdc = ∫ π π / 6 +α 3 Vm sin(ω t + ) dω t = 6 π cos α (3.81) The maximum average output voltage for delay angle α=0 is 3 3 VmVdm = (3.82) π The normalized average output voltage is as shown in (3.83) VdcVn = = cos α (3.83) Vdm The rms value of the output voltage is found from the followingequation: π / 2 +α 3 ⎛ 2 π ⎞ ⎛1 3 3 ⎞Vrms = π ∫ 3 ⎜Vm sin(ω t + ) ⎟ dω t = 3 Vm ⎝ 6 ⎠ ⎜ + ⎜2 ⎝ 4π cos 2α ⎟ (3.84) ⎟ ⎠ π / 6 +α In the converter shown in Fig.3.42 the output voltage will becontinuous only and only if α ≤ 60 o . If α > 60 o 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 α = 75o .
  • 141. SCR Rectifier or Controlled Rectifier 135 The average and rms values of output voltage is shown in thefollowing equation: 5π / 6 πVdc = 3 ∫ 3 Vm sin(ω t + ) dω t = 3 3 Vm [1 + cos ( π / 3 + α )](3.85) π 6 π π / 6 +α The maximum average output voltage for delay angle α=0 is 3 3 VmVdm = (3.86) π The normalized average output voltage isVn = dc = [1 + cos ( π / 3 + α )] V (3.87) Vdm The rms value of the output voltage is found from the followingequation: 5π / 6 2 3 ⎛ π ⎞Vrms = π ⎝∫ 3 ⎜Vm sin(ω t + ) ⎟ dω t 6 ⎠ π / 6 +α (3.88) 3 ⎛ ⎛ π ⎞⎞ = 3Vm 1 − ⎜ 2α − cos⎜ 2α + ⎟ ⎟ 4π ⎝ ⎝ 6 ⎠⎠Example 10 Three-phase full-wave controlled rectifier is connected to380 V, 50 Hz supply to feed a load of 10 Ω pure resistance. If it isrequired to get 400 V DC output voltage, calculate the following: (a) Thefiring angle, α (b) The rectfication effeciency (c) The crest factor ofinput current. (d) PIV of the thyristors.Solution: From (3.81) the average voltage is : 2 3 3* * 380 3 3 Vm 3Vdc = cos α = cos α = 400V . π π Vdc 400Then α = 38.79 o , = I dc = = 40 A R 10 From (3.84) the rms value of the output voltage is: ⎛1 3 3 ⎞ 2 ⎛1 3 3 ⎞Vrms = 3 Vm ⎜ ⎟ ⎜ 2 + 4 π cos 2 α ⎟ = 3 * 3 * 380 * ⎜ + ⎜2 cos (2 * 38.79 )⎟ ⎟ ⎝ ⎠ ⎝ 4π ⎠Then, Vrms = 412.412 V
  • 142. 136 Chapter Three Vrms 412.412Then, I rms = = = 41.24 A R 10 V *I 400 * 40Then, η = dc dc *100 = *100 = 94.07% Vrms * I rms 412.4 * 41.24 I S , peakThe crest factor of input current, C F = I s, rms ⎛ π⎞ 2 * 380 sin ⎜ ωt + ⎟ ⎝ 6⎠ 2 * 380 sin (30 + 38.79 + 30)I S , peak = = = 53.11 A R 10 2 2I s , rms = *I rms = * 41.24 = 33.67 A 3 3 I S , peak 53.11Then, C F = = = 1.577 I s, rms 33.67The PIV= 3 Vm=537.4VExample 11 Solve the previous example if the required dc voltage is 150V.Solution: From (3.81) the average voltage is : 2 3 3* * 380 3 3 Vm 3Vdc = cos α = cos α = 150V . Then, α = 73o π πIt is not acceptable result because the above equation valid only forα ≤ 60 . Then we have to use the (3.85) to get Vdc as following: 2 3 3* * 380Vdc == 3 [1 + cos ( π / 3 + α )] = 150V . Then, α = 75.05o π V 150Then I dc = dc = = 15 A R 10From (3.88) the rms value of the output voltage is: 3 ⎛ ⎛ π ⎞⎞Vrms = 3Vm 1 − ⎜ 2α − cos⎜ 2α + ⎟ ⎟ ⎜ 4π ⎝ ⎝ 6 ⎠⎟ ⎠ 2 ⎛ 3 ⎛ π ⎞⎞ = 3* * 380 * ⎜ 4 π ⎜ 2 * 75.05 * 180 − cos (2 * 75.05 + 30 )⎟ ⎟ ⎜1 − ⎟ 3 ⎝ ⎝ ⎠⎠
  • 143. SCR Rectifier or Controlled Rectifier 137Then, Vrms = 198.075 V V 198.075Then, I rms = rms = = 19.8075 A R 10 V *I 150 *15Then, η = dc dc *100 = *100 = 57.35 % Vrms * I rms 198.075 *19.81 I S , peakThe crest factor of input current, C F = I s , rms ⎛ π⎞ 2 * 380 sin ⎜ ωt + ⎟ ⎝ 6⎠ 2 * 380 sin (30 + 75.05 + 30)I S , peak = = = 37.97 A R 10 2 2I s , rms = *I rms = *19.8075 = 16.1728 A 3 3 I S , peak 37.97Then, C F = = = 2.348 I s, rms 16.1728The PIV= 3 Vm=537.4V3.7.1 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 ofoutput voltage of rectifier shown in Fig.3.48 for α < 60o . Fig.3.48 Three phase full wave fully controlled rectifier with pure dc load current
  • 144. 138 Chapter ThreeFig.3.49 Output voltage and supply current waveforms along with three phase linevoltages for the rectifier shown in Fig.3.48 for α < 60o with pure DC current load.Fig.3.50 FFT components of SCR, secondary, primary currents respectively of rectifier shown in Fig.3.48.
  • 145. SCR Rectifier or Controlled Rectifier 139 In case of the firing angle is greater than 60 o , the output voltagecontains negaive portion as shown in Fig.3.51. Fig.3.52 shows FFTcomponents of output voltage of rectifier shown in Fig.3.48 for α > 60 o .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 α 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: I PowerFactor = s1 cos α (3.89) Is Fig.3.51 Output voltage and supply current waveforms along with three phase line voltages for the rectifier shown in Fig.3.48 for α > 60o with pure DC current load.
  • 146. 140 Chapter ThreeFig.3.52 FFT components of SCR, secondary, primary currents respectively of rectifier shown in Fig.3.48 for α > 60o.Fig.3.53 Phase a voltage, current and fundamental components of phase a of threephase full bridge fully controlled rectifier with pure DC current load and α > 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
  • 147. SCR Rectifier or Controlled Rectifier 141diode 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 with pure DC load and source inductance the waveforms.Fig.3.55 FFT components of output voltage of rectifier shown in Fig.3.48 for α > 60o and there is a source inductance.
  • 148. 142 Chapter Three Let us study the commutation time shown in Fig.3.56n. At this time Vcstarts to be more negative than Vb 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 α 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 Δt sec = u red tocompletely turn OFF T6 and to make T2 carry all the load current ( I o ).Also in the time Δt the current in Lb will change from I o to zero and thecurrent in Lc will change from zero to I o . This is very clear fromFig.3.56n. The equivalent circuit of the three phase full wave controlledrectifier at commutation time Δt is shown in Fig.3.57n and Fig.3.58n. α u Va Vc Vb π /2 Io Io Io Io Fig.3.56n Waveforms represent the one commutation period.
  • 149. SCR Rectifier or Controlled Rectifier 143 Fig.3.57n The equivalent circuit of the three phase controlled rectifier at commutation time Δt shown in Fig.3.56n.From Fig.3.57n we can get the following defferntial equations: di diVa − La T 1 − Vdc − Lb T 6 − Vb = 0 (3.90) dt dt di diVa − La T 1 − Vdc − Lc T 2 − Vc = 0 (3.91) dt dt diNote that, during the time Δt , iT 1 is constant so T 1 = 0 , substitute this dtvalue in (3.90) and (3.91) we get the following differential equations: diVa − Vb − Lb T 6 = Vdc (3.92) dt diVa − Vc − Lc T 2 = Vdc (3.93) dtBy equating the left hand side of equation (3.92) and (3.93) we get thefollowing differential equation: di diVa − Vb − Lb T 6 = Va − Vc − Lc T 2 (3.94) dt dt di diVb − Vc + Lb T 6 − Lc T 2 = 0 (3.95) dt dtThe above equation can be written in the following manner:(Vb − Vc )dt + Lb diT 6 − Lc diT 2 = 0 (3.96)(Vb − Vc )dω t + ω Lb diT 6 − ω Lc diT 2 = 0 (3.97) Integrate the above equation during the time Δt with the help ofFig.3.56n we can get the limits of integration as shown in the following:
  • 150. 144 Chapter Threeπ / 2 +α + u 0 Io ∫ (Vb − Vc )dω t + ∫ ω Lb diT 6 − ∫ ω Lc diT 2 = 0 π / 2 +α Io 0π / 2 +α + u ⎛ ⎛ 2π ⎞ ⎛ 2π ⎞ ⎞ ∫ ⎜Vm sin ⎜ ω t − ⎟ − Vm sin ⎜ ω t + ⎟ ⎟dω t + ωLb (− I o ) − ωLc I o = 0 π / 2 +α ⎝ ⎝ 3 ⎠ ⎝ 3 ⎠⎠assume Lb = Lc = LS π / 2 +α + u ⎡ ⎛ 2π ⎞ ⎛ 2π ⎞⎤Vm ⎢− cos⎜ ω t − ⎟ + cos⎜ ω t + ⎟ = 2ω LS I o ⎣ ⎝ 3 ⎠ ⎝ 3 ⎠⎥ π / 2 +α ⎦ ⎡ ⎛π 2π ⎞ ⎛π 2π ⎞ ⎛π 2π ⎞ ⎛π 2π ⎞⎤Vm ⎢− cos⎜ + α + u − ⎟ + cos⎜ + α + u + ⎟ + cos⎜ + α − ⎟ − cos⎜ + α + ⎟⎥ ⎣ ⎝2 3 ⎠ ⎝2 3 ⎠ ⎝2 3 ⎠ ⎝2 3 ⎠⎦ = 2ω LS I o ⎡ ⎛ π⎞ ⎛ 7π ⎞ ⎛ π⎞ ⎛ 7π ⎞⎤Vm ⎢− cos⎜ α + u − ⎟ + cos⎜ α + u + ⎟ + cos⎜ α − ⎟ − cos⎜ α + ⎟⎥ = 2ω LS I o ⎣ ⎝ 6⎠ ⎝ 6 ⎠ ⎝ 6⎠ ⎝ 6 ⎠⎦⎡ ⎛π ⎞ ⎛π ⎞ ⎛ 7π ⎞ ⎛ 7π ⎞⎢− cos(α + u )cos⎜ 6 ⎟ − sin (α + u )sin ⎜ 6 ⎟ + cos(α + u )cos⎜ 6 ⎟ − sin (α + u )sin ⎜ 6 ⎟⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ π π 7π 7π ⎤ 2ω LS I o+ cos α cos + sin α sin − cos α cos + sin α sin ⎥ = 6 6 6 6 ⎦ Vm⎡ 3 cos(α + u ) − 0.5 sin (α + u ) − cos(α + u ) + 0.5 sin (α + u ) 3⎢−⎣ 2 2 3 3 ⎤ 2ω LS I o cos α + 0.5 sin α cos α − 0.5 sin α ⎥ = 2 2 ⎦ Vm 2ω LS I o 3[cos α − cos(α + u )] = Vm 2ω LI o 2ω LI o 2 ω LS I ocos(α ) − cos(α + u ) = = = (3.98) 3 Vm 2 VLL VLL ⎡ 2ω LS I o ⎤u = cos −1 ⎢cos(α ) − ⎥ −α (3.99) ⎣ VLL ⎦ u 1⎧ ⎪ ⎡ 2ω LS I o ⎤ ⎫ ⎪Δt = = ⎨cos −1 ⎢cos(α ) − ⎥ −α ⎬ (3.100) ω ω⎪ ⎩ ⎣ VLL ⎦ ⎪ ⎭
  • 151. SCR Rectifier or Controlled Rectifier 145 It is clear that the DC voltage reduction due to the source inductance is dithe drop across the source inductance. vrd = LS T (3.101) dtMultiply (3.101) by dω t and integrate both sides of the resultantequation we get:π +α + u2 Io ∫ vrd dω t = ∫ ω LdiD = ω LS I o (3.102) π 0 +α 2π +α + u2 ∫ vrd dω t is the reduction area in one commutation period Δt . But we π +α 2have six commutation periods Δt in one period so the total reduction perperiod is: π +α + u 26 ∫ vrd dω t = 6ω LS I o (3.103) π +α 2 To obtain the average reduction in DC output voltage Vrd due tosource inductance we have to divide by the period time 2π . Then, 6ω LI oVrd = = 6 fLI o (3.104) 2π The DC voltage without source inductance tacking into account can becalculated as following: 3 2Vdc = Vdc VLL cosα − 6 fLs I o (3.105) − Vrd = actual π without sourceinduc tan ce Fig.3.58n shows the utility line current with some detailes to help us tocalculate its rms value easly. ⎡ π u ⎤ + ⎢u ⎛ I 2 3 2 ⎥ 2I o ⎡ 1 3 π u 2 ⎤ ⎢ ⎜ o ωt ⎞ dωt + I d dωt ⎥ = 2Is = ∫ ∫ ⎢ 2 u + 3 + 2 − u⎥ 2 ⎟ π⎢ ⎝u ⎠ ⎥ π ⎣ 3u ⎦ 0 u ⎢ ⎥ ⎣ ⎦
  • 152. 146 Chapter Three 2I o ⎡π u ⎤ 2Then I S = − (3.106) π ⎢ 3 6⎥ ⎣ ⎦ Is u Io 2π +u 3 2π u 2π + 6 2 − Io 3 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). u Is Io 11π u 7π u − − 6 2 6 2 π u 5π u − − 6 2 6 2 − Io ′ Is u Io u 5π u 7π u − − 6 2 6 2 π u 11π u − − I 6 2 − o 6 2 u Fig.3.59 The utility line currents and its first derivative.
  • 153. SCR Rectifier or Controlled Rectifier 147Table(3.1) Jumb value of supply current and its first derivative.Js π u π u 5π u 5π u 7π u 7π u 11π u 11π u − + − + − + − + 6 2 6 2 6 2 6 2 6 2 6 2 6 2 6 2Is 0 0 0 0 0 0 0 0 ′Is Io − Io − Io Io − Io Io Io − Io u u u u u u u uIt is an odd function, then ao = an = 0 ⎡m 1 m ⎤ ∑ ∑ 1bn = ⎢ J s cos nωt s − ′ J s sin nωt s ⎥ (3.107) nπ ⎢ s =1 ⎣ n s =1 ⎥ ⎦ 1 ⎡ −1 Io ⎛ ⎛π u ⎞ ⎛π u ⎞ ⎛ 5π u ⎞ ⎛ 5π u ⎞bn = ⎢ * ⎜ sin n⎜ − ⎟ − sin n⎜ + ⎟ − sin n⎜ − ⎟ + sin n⎜ + ⎟ nπ ⎣ n u ⎝ ⎝ 6 2⎠ ⎝ 6 2⎠ ⎝ 6 2⎠ ⎝ 6 2⎠ ⎛ 7π u ⎞ ⎛ 7π u ⎞ ⎛ 11π u ⎞ ⎛ 11π u ⎞ ⎞⎤ − sin n⎜ − ⎟ + sin n⎜ + ⎟ + sin n⎜ − ⎟ − sin n⎜ + ⎟ ⎟⎥ ⎝ 6 2⎠ ⎝ 6 2⎠ ⎝ 6 2⎠ ⎝ 6 2 ⎠ ⎠⎦ 2I o nu ⎡ nπ 5nπ 7nπ 11nπ ⎤bn = ⎢cos 6 − cos 6 − cos 6 + cos 6 ⎥ (3.108) * sin n πu 2 2 ⎣ ⎦Then, the utility line current can be obtained as in (3.109). 4 3 ⎡ ⎛u⎞ ⎛ 5u ⎞ ⎛ 7u ⎞i (ω t ) = ⎢sin ⎜ 2 ⎟ sin (ωt ) − 2 sin ⎜ 2 ⎟ sin (5ωt ) − 2 sin ⎜ 2 ⎟ sin (7ωt ) + 1 1 πu ⎣ ⎝ ⎠ 5 ⎝ ⎠ 7 ⎝ ⎠ (3.109) ⎛ 11u ⎞ ⎛ 13u ⎞ ⎤ ⎟ sin (11ωt ) + 2 sin ⎜ ⎟ sin (13ωt ) − − + + ⎥ 1 1 + sin ⎜ 11 2 ⎝ 2 ⎠ 13 ⎝ 2 ⎠ ⎦ 2 6 Io ⎛ u ⎞Then; I S1 = sin ⎜ ⎟ (3.110) πu ⎝2⎠ The power factor can be calculated from the following equation: 2 6 Io ⎛ u ⎞ sin ⎜ ⎟ I S1 ⎛u⎞ πu ⎝2⎠ ⎛ u⎞pf = cos ⎜ ⎟ = cos ⎜α + ⎟ IS ⎝2⎠ 2 I o ⎡π u ⎤ 2 ⎝ 2⎠ − ⎥ π ⎢ 3 6⎦ ⎣ ⎛u⎞ 2 3 * sin ⎜ ⎟Then; pf = ⎝ 2 ⎠ cos⎛ α + u ⎞ ⎜ ⎟ (3.111) ⎡π u ⎤ ⎝ 2⎠ u π⎢ − ⎥ ⎣ 3 6⎦
  • 154. 148 Chapter ThreeNote, 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 ⎛ u⎞cos⎜ α + ⎟ . Another expression for the displacement power factor, by ⎝ 2⎠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: VLL (cosα − cos(α + u )) 3 2 3Vdc = VLL cos α − π 2π VLL [2 cos α − (cos α − cos(α + u ))] 3 2∴Vdc = 2π VLL [cos α + cos(α + u )] 3 2∴Vdc = (3.112) 2πThen the DC power output from the rectifier is Pdc = Vdc I o . Then, VLL * I o [cos α + cos(α + u )] 3 2Pdc = (3.113) 2πOn the AC side, the AC power is: Pac = 3 VLL I S1 cos φ1 (3.114)Substitute from (3.110) into (3.114) we get the following equation: 4 3 Io ⎛ u ⎞ 6 2 VLL I o ⎛ u ⎞Pac = 3 VLL sin ⎜ ⎟ cos φ1 = sin ⎜ ⎟ cos φ1 (3.115) πu 2 ⎝2⎠ πu ⎝2⎠By equating (3.113) and (3.115) we get the following: u [cos α + cos (α + u )]cos φ1 = (2.116) ⎛u⎞ 4 * sin ⎜ ⎟ ⎝2⎠ The source inductance reduces the magnitudes of the harmoniccurrents. Fig.3.60a through d show the effects of LS (and hence of u) onvarious harmonics for various values of α , where I d is a constant dc.The harmonic currents are normalized by I1 I, with LS = 0 , which is 6given by (2.98) where I s1 = I o in this case. Normally, the DC-side πcurrent is not a constant DC. Typical and idealized harmonics are shownin Table 3.2.
  • 155. SCR Rectifier or Controlled Rectifier 149 Fig.3.60 Normalized harmonic current in the presence of LS [ ].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 I d , as shown in Fig.3.61. For a delayangle a greater than 90° but less than 180°, the voltage and current
  • 156. 150 Chapter Threewaveforms are shown in Fig.3.62a. The average value of Vd is negativeaccording to (3.81). On the ac side, the negative power implies that thephase angle φ1 , between vs and is , 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 givenE d and α can be obtained from the characteristics shown in Fig.3.64.
  • 157. SCR Rectifier or Controlled Rectifier 151 Similar to the discussion in connection with single-phase converters, ( )the extinction angle γ = 180o − α − u must be greater than the thyristorturn-off interval ω t q in the waveforms of Fig.3.54, where v5 is thevoltage across thyristor 5. Fig.3.63 Three phase SCR inverter with a DC voltage source.Fig.3.64 Vd versus I d of Three phase SCR inverter with a DC voltage source.Inverter StartupAs discussed for start up of a single-phase inverter, the delay angle αin the three-phase inverter of Fig.3.63 is initially made sufficientlylarge (e.g., 165°) so that id is discontinuous. Then, α is decreased bythe controller such that the desired I d and Pd are obtained.
  • 158. 152 Chapter ThreeProblems 1- Single phase half-wave controlled rectifier is connected to 220 V, 50Hz supply to feed 10 Ω resistor. If the firing angle α = 30 o draw output voltage and drop voltage across the thyristor along with the supply voltage. Then, calculate, (a) The rectfication effeciency. (b) Ripple factor. (c) Peak Inverse Voltage (PIV) of the thyristor. (d) The crest factor C F of input current. 2- Single phase half-wave controlled rectfier is connected to 220 V, 50Hz supply to feed 5Ω resistor in series with 10mH inductor if the firing angle α = 30 o . (a) Determine an expression for the current through the load in the first two periods of supply current, then fiend the DC and rms value of output voltage. (b) Draw the waveforms of load, resistor, inductor voltages and load current. 3- Solve problem 2 if there is a freewheeling diode is connected in shunt with the load. 4- single phase full-wave fully controlled rectifier is connected to 220V, 50 Hz supply to feed 5Ω resistor, if the firing angle α = 40 o . Draw the load voltage and current, diode currents and supply current. Then, calculate (a) The rectfication effeciency. (b) Peak Inverse Voltage (PIV) of the thyristor. (c) Crest factor of supply current. 5- In the problem 4, if there is a 5mH inductor is connected in series with the 5Ω resistor. Draw waveforms of output voltage and current, resistor and inductor voltages, diode currents, supply currents. Then, find an expression of load current, DC and rms values of output voltages. 6- Solve problem 5 if the load is connected with freewheeling diode. 7- Single phase full wave fully controlled rectifier is connected to 220V, 50 Hz supply to feed the load with 47 A pure dc current. The firing angle α = 40 o . Draw the load voltage, thyristor, and load currents. Then, calculate (a) the rectfication effeciency. (b) Ripple factor of output voltage. (c) Crest factor of supply current. (d) Use Fourier series to fiend an expression for supply current. (e) THD of supply current. (f) Input power factor. 8- Solve problem 7 if the supply has a 3 mH source inductance.
  • 159. SCR Rectifier or Controlled Rectifier 153 9- Single phase full-wave semi-controlled rectifier is connected to 220 V, 50Hz supply to feed 5Ω resistor in series with 5 mH inductor, the load is connected in shunt with freewheeling diode. Draw the load voltage and current, resistor voltage and inductor voltage diodes and thyristor currents. Then, calculate Vdc and Vrms of the load voltages. If the freewheeling diode is removed, explain what will happen? 10- The single-phase full wave controlled converter is supplying a DC load of 1 kW with pure DC current. A 1.5-kVA-isolation transformer with a source-side voltage rating of 120 V at 50 Hz is used. It has a total leakage reactance of 8% based on its ratings. The ac source voltage of nominally 120 V is in the range of -10% and +5%. Then, Calculate the minimum transformer turns ratio if the DC load voltage is to be regulated at a constant value of 100 V. What is the value of a when VS = 120 V + 5%. 11- In the single-phase inverter of, VS = 120 V at 50 Hz, LS = 1.2 mH, Ld = 20 mH, Ed = 88 V, and the delay angle α = 135°. Using PSIM, obtain vs , is , vd , and id waveforms in steady state. 12- In the inverter of Problem 12, vary the delay angle α from a value of 165° down to 120° and plot id versus α . Obtain the delay angle α b , below which id becomes continuous. How does the slope of the characteristic in this range depend on LS ? 13- In the three-phase fully controlled rectifier is connected to 460 V at 50 Hz and Ls = 1mH . Calculate the commutation angle u if the load draws pure DC current at Vdc = 515V and Pdc = 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 VLL and I o . 15- Consider the three-phase, half-controlled converter shown in the following figure. Calculate the value of the delay angle α for which Vdc = 0.5Vdm . Draw vd waveform and identify the devices that conduct during various intervals. Obtain the DPF, PF, and %THD in the input line current and compare results with a full-bridge converter operating at Vdc = 0.5Vdm . Assume LS .
  • 160. 154 Chapter Three 16- Repeat Problem 15 by assuming that diode D f is not present in the converter. 17- The three-phase converter of Fig.3.48 is supplying a DC load of 12 kW. A Y- Y connected isolation transformer has a per-phase rating of 5 kVA and an AC source-side voltage rating of 120 V at 50 Hz. It has a total per-phase leakage reactance of 8% based on its ratings. The ac source voltage of nominally 208 V (line to line) is in the range of -10% and +5%. Assume the load current is pure DC, calculate the minimum transformer turns ratio if the DC load voltage is to be regulated at a constant value of 300 V. What is the value of α when VLL = 208 V +5%. 18- In the three-phase inverter of Fig.3.63, VLL = 460 V at 60 Hz, E = 550 V, and LS = 0.5 mH. Assume the DC-side current is pure DC, Calculate α and γ if the power flow is 55 kW.
  • 161. Chapter 4 SWITCH-MODE dc-ac INVERTERS: dc SINUSOIDAL ac4.1 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
  • 162. SWITCH-MODE dc-ac 153rectifier (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 ac motor drive. In this chapter, we will discuss inverters with single-phase and three-phase ac outputs. The input to switch-mode inverters will be assumed 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
  • 163. 154 Chapter Threeschemes 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 vo can be assumed to be sinusoidal. Since theinverter supplies an inductive load such as an ac motor, io will lag vo , asshown in Fig.4.3b. The output waveforms of Fig.4.3b show that duringinterval 1, vo and io are both positive, whereas during interval 3, vo andio are both negative. Therefore, during intervals 1 and 3, theinstantaneous power flow po = vo * I o is from the dc side to the ac side,corresponding to an inverter mode of operation. In contrast, vo and io areof opposite signs during intervals 2 and 4, and therefore po 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 io − vo plane, as shownin Fig.4.3c during each cycle of the ac output. Such a four-quadrant
  • 164. SWITCH-MODE dc-ac 155inverter is reversible and vo can be of either polarity independent of thedirection of io . 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.
  • 165. 156 Chapter Three 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 Vd 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 SCHEME In inverter circuits, we would like the inverter output to be sinusoidalwith 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 Vtri . Before discussing the PWM behavior, it is necessary to define a fewterms. The triangular waveform Vtri in Fig.4.5a is at a switchingfrequency f s which establishes the frequency with which the inverterswitches are switched ( f s is also called the carrier frequency). Thecontrol signal vcontrol is used to modulate the switch duty ratio and has afrequency f1 , which is the desired fundamental frequency of the invertervoltage output ( f 1 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 f 1 . Theamplitude modulation ratio ma is defined as Vˆma = control (4.1) ˆ Vtriwhere V ˆ is the peak amplitude of the control signal. The amplitude control ˆVtri of the triangular signal is generally kept constant.The frequency modulation ratio m f is defined as: fsmf = (4.2) f1 In the inverter of Fig.4.4b, the switches TA + and TA− are controlledbased on the comparison of vcontrol and vtri and the following outputvoltage results, independent of the direction of io :
  • 166. SWITCH-MODE dc-ac 157 1 vcontrol > vtri TA + is on , v Ao = Vd (4.3) 2 1 vcontrol < vtri TA − is on , v Ao = − Vd 2 Fig.4.5 Pulse width modulation. Since the two switches are never off simultaneously, the output voltagev Ao 1 1 fluctuates between two values ( Vd and − Vd ). Voltage v Ao and 2 2its fundamental frequency component (dashed curve) are shown inFig.4.5b, which are drawn for m f = 15 and ma = 0.8.
  • 167. 158 Chapter ThreeThe harmonic spectrum of v Ao under the conditions indicated inFigs.4.5a and Fig.4.5b is shown in Fig.4.5c, where the normalized ˆ( ) 1harmonic voltages V Ao h / Vd having significant amplitudes are plotted. 2This plot (for ma ≤ 1.0 ) shows three items of importance: 1. The peak amplitude of the fundamental-frequency component( ) ˆ V Ao 1 is ma times Vd / 2 . This can be explained by first considering aconstant vcontrol , as shown in Fig.4.6a. This results in an outputwaveform v Ao . 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 Ts = 1 / f s ) v Ao dependson the ratio of v ˆ to V for a given V : control tri d vcontrol VdV Ao = vcontrol ≤Vtri ˆ (4.4) ˆ Vtri 2 Let us assume (though this assumption is not necessary) that vcontrolvaries very little during a switching time period, that is, m f is large, asshown in Fig.4.6b. Therefore, assuming vcontrol to be constant over aswitching time period, Eq. (4.4) indicates how the "instantaneousaverage" value of v Ao (averaged over one switching time period Ts )varies from one switching time period to the next. This "instantaneousaverage" is the same as the fundamental-frequency component of v Ao .The foregoing argument shows why vcontrol is chosen to be sinusoidal toprovide a sinusoidal output voltage with fewer harmonics. Let the controlvoltage vary sinusoidally at the frequency f1 = ω1 / 2π , which is thedesired (or the fundamental) frequency of the inverter output:vcontrol = Vcontrol sin ω1t ˆwhere Vˆ control ≤V ˆ tri (4.5) Using Eqs. (4.4) and (4.5) and the foregoing arguments, which showthat the fundamental-frequency component (v Ao )1 varies sinusoidally andin phase with vcontrol as a function of time, results in ˆ(v Ao )1 = Vcontrol sin ω1t Vd = ma sin ω1t Vd ˆ for ma ≤ 1 (4.6) Vtri 2 2
  • 168. SWITCH-MODE dc-ac 159Therefore, (VˆAo )1 = ma V2d for ma ≤ 1 (4.7) which shows that in a sinusoidal PWM, the amplitude of thefundamental-frequency component of the output voltage varies linearlywith ma (provided ma ≤ 1 .0). Therefore, the range of ma 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 m f , 2m f , 3m f , and so on. This general patternholds true for all values of ma in the range 0 to 1.For a frequency modulation ratio m f ≤ 9 (which is always the case,except in very high power ratings), the harmonic amplitudes are almostindependent of m f , though mf defines the frequencies at which theyoccur. Theoretically, the frequencies at which voltage harmonics occur ( )can be indicated as: f h = jm f ± k f1that is, the harmonic order h corresponds to the kth sideband of j times thefrequency modulation ratio m f : ( )h = j mf ± k (4.8)where the fundamental frequency corresponds to h = 1 . 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.
  • 169. 160 Chapter Three ˆIn Table 8-1, the normalized harmonics V Ao ( )h / 1 Vd 2 are tabulated as afunction of the amplitude modulation ratio ma, assuming m f ≥ 9 . Onlythose 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 ofFig.4.4 1v AN = v Ao + Vd (4.9) 2 Therefore, the harmonic voltage components in v AN and v Ao are thesame:( ) ( ) V AN h = V Ao h ˆ ˆ (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 ma . 3. The harmonic m f should be an odd integer. Choosing m f as anodd 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 m f = 15 .Therefore, only odd harmonics are present and the even harmonicsdisappear from the waveform of v Ao . 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 v Ao , for a Large m f
  • 170. SWITCH-MODE dc-ac 161Example 1 In the circuit of Fig.4.4, Vd = 300V , ma = 0.8 , m f = 39 , andthe fundamental frequency is 47 Hz. Calculate the rms values of thefundamental-frequency voltage and some of the dominant harmonics inv Ao 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 m f . Because of the relative ease in filteringharmonic 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 f s . 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 f s of 20 kHz orgreater. 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 m fmay be 9 or even less for switching frequencies of less than 2 kHz. Onthe other hand, m f will be larger than 100 for switching frequencieshigher than 20 kHz. The desirable relationships between the triangularwaveform signal and the control voltage signal are dictated by how largem f is. In the discussion here, m f = 21 is treated as the borderline
  • 171. 162 Chapter Threebetween large and small, though its selection is somewhat arbitrary. Here,it is assumed that the amplitude modulation ratio ma is less than 1.4.2.1.1 Small m f (m f ≤ 21) 1. Synchronous PWM. For small values of m f , the triangularwaveform signal and the control signal should be synchronized to eachother (synchronous PWM) as shown in Fig.4.5a. This synchronous PWMrequires that m f be an integer. The reason for using the synchronousPWM is that the asynchronous PWM (where m f is not an integer) resultsin 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 vcontrol , is 65.42 Hz and m f = 15,the triangular wave frequency should be exactly 15 x 65.42 = 981.3 Hz). m 2. m f should be an odd integer. As discussed previously, f shouldbe an odd integer except in single-phase inverters with PWM unipolarvoltage switching, to be discussed in the following sections. (4.2.1.2 Large m f m f > 21) The amplitudes of subharmonics due to asynchronous PWM are smallat large values of m f . Therefore, at large values of m f , theasynchronous PWM can be used where the frequency of the triangularwaveform is kept constant, whereas the frequency of vcontrol varies,resulting in noninteger values of m f (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 ma > 1.0 In the previous discussion, it was assumed that ma ≤ 1.0 ,corresponding to a sinusoidal PWM in the linear range. Therefore, theamplitude of the fundamental-frequency voltage varies linearly with ma ,as derived in Eq.(4.7). In this range of ma ≤ 1.0 , PWM pushes theharmonics into a high-frequency range around the switching frequency
  • 172. SWITCH-MODE dc-ac 163and 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 ma ≤ 1.0 ), 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 ma . Figure 4.8 shows the normalized peak ( ) ˆamplitude of the fundamental-frequency component V Ao h / Vd as a 1 2function of the amplitude modulation ratio ma . Even at reasonably large ( )ˆ 1values of m f , V Ao h / Vd depends on m f in the overmodulation 2 ( ) 1region. This is contrary to the linear range ( ma ≤ 1 .0) where V Ao h / Vd ˆ 2varies linearly with ma , almost independent of mf (provided m f > 9).With overmodulation regardless of the value of m f , it is recommendedthat a synchronous PWM operation be used, thus meeting therequirements indicated previously for a small value of m f .Fig.4.7 Harmonics due to overmodulation, drawn for ma = 2.5 and m f = 15 .
  • 173. 164 Chapter Three Fig.4.8 Voltage control by varying ma . 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 ma , 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 ma > 1Vd 2 ( ) < V Ao 1 < ˆ 4 Vd π 2 (4.12)4.2.2 SQUARE-WAVE SWITCHING SCHEME In the square-wave switching scheme, each switch of the inverter legof 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
  • 174. SWITCH-MODE dc-ac 165harmonic components in the inverter output waveform can be obtainedfor a given input Vd as:(VˆAo )1 = π V2d = 1.273⎛ V2d ⎞ 4 ⎜ ⎟ (4.13) ⎝ ⎠and (V ) = Ao 1 ˆ (Vˆ ) (4.14) Ao h h 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 ma becomes so largethat the control voltage waveform intersects with the triangular waveformin Fig.4.5a only at the zero crossing of vcontrol . Therefore, the outputvoltage is independent of ma 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 Vd 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
  • 175. 166 Chapter Three 1midpotential, with a voltage Vd , across each capacitor. Sufficiently 2large 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 vo = v Ao . 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 directionof io , and io splits equally between the two capacitors. Therefore, thecapacitors C+ and C_ are "effectively" connected in parallel in the path ofio . This also explains why the junction o in Fig.4.10 stays atmidpotential. Since io must flow through the parallel combination of C+ and C_, ioin 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:VT = Vd (4.15)and IT = io, peak (4.16)4.3.2 FULL-BRIDGE INVERTERS (SINGLE PHASE) A full-bridge inverter is shown in Fig.4.11. This inverter consists oftwo 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
  • 176. SWITCH-MODE dc-ac 167half-bridge inverter. At high power levels, this is a distinct advantage,since it requires less paralleling of devices. Fig.4.10 Half-bridge inverter. Fig.4.11 Single-phase full-bridge inverter.4.3.2.1 PWM with Bipolar Voltage Switching The diagonally opposite switches (TA+, TB-) and (TA-, TB+) from the twolegs 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 vcontrol and vtri in Fig.4.12a. Theoutput of inverter leg B is negative of the leg A output; for example, 1 1when TA+ is on and v Ao is equal to + Vd is also on and vBo = − Vd . 2 2Therefore:v Bo (t ) = −v Ao (t ) (4.17)and vo (t ) = v Ao (t ) − vBo (t ) = 2v Ao (t ) (4.18)The vo 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
  • 177. 168 Chapter Three ( ) ˆcomponent in the output voltage Vo1 can be obtained from Eqs. (4.7),(4.12), and (4.18) as:Vo1 = maVd ˆ (ma ≤ 1.0) (4.19) 4and Vd < Vo1 < Vd ˆ (ma > 1.0) (4.20). π Fig4.12 PWM with bipolar voltage switching. In Fig.4.12b, we observe that the output voltage vo switches between− Vd and + Vd voltage levels. That is the reason why this type ofswitching 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, Vd = 300V ,ma =0.8, m f = 39 , and the fundamental frequency is 47 Hz. Calculate therms values of the fundamental-frequency voltage and some of thedominant harmonics in the output voltage vo if a PWM bipolar voltage-switching scheme is used.Solution: From Eq.(4.18), the harmonics in vo can be obtained bymultiplying 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 ( ) Vˆ ( ) Vˆ(Vo )h = 1 * 2 * Vd * Ao h = Vd * Ao h = 212.13 Ao h ( ) Vˆ (4.21) 2 2 Vd / 2 2 Vd / 2 Vd / 2
  • 178. SWITCH-MODE dc-ac 169Therefore, the rms voltages are as follows:Fundamental: Vo1 = 212.13 x 0. 8 = 169.7 V at 47 Hz (Vo )37 = 212.13 x 0.22 = 46.67 V at 1739 Hz (Vo )39 = 212.13 x 0.818 = 173.52 V at 1833 Hz - (Vo )41 = 212.13 x 0.22 = 46.67 V at 1927 Hz (Vo )77 = 212.13 x 0.314 = 66.60 V at 3619 Hz (Vo )79 = 212.13 x 0.314 = 66.60 V at 3713 Hz etc.dc-Side Current id It is informative to look at the dc-side current id 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 vo and id , 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, vo in Fig.4.13 is a pure sine wave atthe fundamental output frequency ω1 ,vo1 = vo = 2 Vo sin ω1t (4.22)If the load is as shown in Fig.4.13, where eo is a sine wave at frequencyω1 , then the output current would also be sinusoidal and would lag vofor an inductive load such as an ac motor: (io = 2 I o sin ω1t − φ ) (4.23)where φ is the angle by which io lags vo .
  • 179. 170 Chapter ThreeOn the dc side, the L-C filter will filter the high-switching-frequency *components in id and id would only consist of the low-frequency and dccomponents. Assuming that no energy is stored in the filters,Vd id (t ) = vo (t )io (t ) = 2 Vo sin ω1t 2 I o sin (ω1t − φ ) * (4.24) V I V ITherefore id (t ) = o o cos φ − o o cos(2ω1t − φ ) = I d + id 2 * (4.25) Vd Vdid (t ) = I d − 2 I d 2 cos(2ω1t − φ ) * (4.26) V Iwhere I d = o o cos φ (4.27) Vd 1 Vo I oand I d 2 = (4.28) 2 Vd * Equation (4.26) for id shows that it consists of a dc component I d ,which is responsible for the power transfer from Vd on the dc side of the *inverter to the ac side. Also, id contains a sinusoidal component at twice *the fundamental frequency. The inverter input current id consists of idand 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 bipolar voltage 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
  • 180. SWITCH-MODE dc-ac 171ripple 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 vtri with vcontrol and− vcontrol , respectively. As shown in Fig.4.15a, the comparison of vcontrolwith the triangular waveform results in the following logic signals tocontrol the switches in leg A:vcontrol > vtri TA+ on and VAN = Vd (4.29)vcontrol < vtri TA− on and VAN = 0The 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, − vcontrol iscompared with the same triangular waveform, which yields thefollowing:− vcontrol > vtri TB + on and VBN = Vd (4.30)− vcontrol < vtri TB − on and VBN = 0 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 io . The waveforms of Fig.4.15 show that there are four combinations ofswitch on-states and the corresponding voltage levels: (4.31)
  • 181. 172 Chapter Three 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 T A + and DB + or D A + and TB + depending on the direction of io . During thisinterval, the input current id is zero. A similar condition occurs whenboth bottom switches TA − and TB − are on. In this type of PWM scheme, when a switching occurs, the outputvoltage changes between zero and + Vd or between zero and − Vdvoltage levels. For this reason, this type of PWM scheme is called PWMwith a unipolar voltage switching, as opposed to the PWM with bipolar(between + Vd and − Vd ) voltage-switching scheme described earlier.
  • 182. SWITCH-MODE dc-ac 173This 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 Vd as compared to 2Vd in theprevious scheme. The advantage of "effectively" doubling the switching frequencyappears 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 m f to be even ( m f should be odd forPWM with bipolar voltage switching) in a single-phase inverter. Thevoltage waveforms v AN and vBN are displaced by 180° of thefundamental frequency f1 with respect to each other. Therefore, theharmonic components at the switching frequency in v AN and v BN havethe same phase ( φ AN − φ BN = 180o. m f = 0 , since the waveforms are 180°displaced and m f is assumed to be even). This results in the cancellationof the harmonic component at the switching frequency in the outputvoltage vo = v AN − vBN . 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 alsoVo1 = ma Vd ˆ (ma ≤ 1.0) (4.32) 4and Vd < Vo1 < Vd ˆ (ma > 1.0) (4.33) πExample 3 In Example 2, suppose that a PWM with unipolar voltageswitching scheme is used, with m f = 38 . Calculate the rms values of thefundamental 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 ( )h = j 2m f ± k (4.34)where the harmonics exist as sidebands around 2m f and the multiples of2m f . Since h is odd, k in Eq.(34) attains only odd values. From Example2,
  • 183. 174 Chapter Three (V Ao )h(Vo )h = 212.13 (4.35) Vd / 2Using Eq.(35) and Table 1, we find that the rms voltages are as follows:At fundamental or 47 Hz: Vo1 = 0.8 x 212.13 = 169.7 VAt h = 2m f - 1 = 75 or 3525 Hz: (Vo )75 = 0.314 x 212.13 = 66.60 VAt h = 2m f + 1 = 77 or 3619 Hz: (Vo )77 = 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 ma However,with unipolar voltage switching, the dominant harmonic voltagescentered around m f disappear, thus resulting in a significantly lowerharmonic content.dc-Side Current id . 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 id for the PWM unipolar voltage-switching scheme,where m f = 14 (instead of m f = 15 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 unipolar voltage switching.4.3.2.3 Square-Wave Operation The full-bridge inverter can also be operated in a square-wave mode.Both types of PWM discussed earlier degenerate into the samesquare-wave mode of operation, where the switches ( T A + , TB − ) and( TB + , T A − ) are operated as two pairs with a duty ratio of 0.5.
  • 184. SWITCH-MODE dc-ac 175 As is the case in the square-wave mode of operation, the outputvoltage magnitude given below is regulated by controlling the input dcvoltage: 4Vo1 = Vd ˆ (4.36) π4.3.2.4 Output Control by Voltage Cancellation This type of control is feasible only in a single-phase, full-bridgeinverter 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 v ANand v BN 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 α = 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 α .
  • 185. 176 Chapter Three It is easier to derive the fundamental and the harmonic frequency 1components of the output voltage in terms of β = 90o − α , as is shown 2in Fig.4.17b: (4.37) 1where β = 90o − α and h is an odd integer. 2 Fig.4.17c shows the variation in the fundamental-frequencycomponent as well as the harmonic voltages as a function of α . These arenormalized with respect to the fundamental-frequency component for thesquare-wave ( α = 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 α .Because of a large distortion, the curves are shown as dashed for largevalues of α .4.3.2.5 Switch Utilization in Full-Bridge Inverters Similar to a half-bridge inverter, if a transformer is utilized at theoutput 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:VT = Vd (4.38)and IT = io, peak (4.39)4.3.2.6 Ripple in the Single-Phase Inverter Output The ripple in a repetitive waveform refers to the difference betweenthe instantaneous values of the waveform and its fundamental-frequencycomponent.
  • 186. SWITCH-MODE dc-ac 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) eo .Since eo (t ) 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 vo and io by applying the principle of superposition to thelinear circuit of Fig.4.18a. Let vo = vo1 + vripple and io = io1 + iripple .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,Vo1 = Eo + VL1 = Eo + jω1LI o1 (4.40) Fig.4.18 Single-phase inverter: (a) circuit; (6) fundamental- frequency components; (c) ripple frequency components: (d) fundamental-frequency phasor diagram. Since the superposition principle is valid here, all the ripple in v,appears across L, where vripple (t ) = vo − vo1 (4.41)The output current ripple can be calculated asiripple (t ) = ∫ vripple (ζ )dζ + k 1 t (4.42) L 0
  • 187. 178 Chapter Threewhere k is a constant and ζ is a variable of integration. With a properly selected time origin t = 0, the constant k in Eq.(4.42)will be zero. Therefore, Eqs.(4.41) and (4.42) show that the current 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 Vd 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) PWM bipolar voltage switching.
  • 188. SWITCH-MODE dc-ac 1794.3.3 PUSH-PULL INVERTERS Fig.4.20 shows a push-pull inverter circuit. It requires a transformerwith a center tapped primary. We will initially assume that the outputcurrent io flows continuously. With this assumption, when the switch T1is on (and T2 is off), T1 would conduct for a positive value of io , and D1would conduct for a negative value of io . Therefore, regardless of thedirection of io , vo = Vd / n , where n is the transformer turns ratiobetween the primary half and the secondary windings, as shown inFig.4.20. Similarly, when T2 is on (and T1 is off), vo = −Vd / n . 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: VVo1 = ma d ˆ (ma ≤ 1.0) (4.43) n Vand d < Vo1 < ˆ 4 Vd (ma > 1.0) (4.44) n π nIn a push-pull inverter, the peak switch voltage and current ratings areVT = 2Vd IT = io, peak / n (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.
  • 189. 180 Chapter Three The output current, which is the secondary current of the transformer,is a slowly varying current at the fundamental output frequency. It can 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 INVERTERS Since the intent in this section is to compare the utilization of switchesin various single-phase inverters, the circuit conditions are idealized. Wewill assume that Vd , max is the highest value of the input voltage, whichestablishes the switch voltage ratings. In the PWM mode, the inputremains constant at Vd , max In the square-wave mode, the input voltage isdecreased below Vd , max to decrease the output voltage from its maximumvalue. 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 I o, max at the maximum load. If the output current is assumed to be purely sinusoidal, the inverterrms volt-ampere output at the fundamental frequency equals Vo1I o, max atthe maximum rated output, where the subscript 1 designates the
  • 190. SWITCH-MODE dc-ac 181fundamental-frequency component of the inverter output. With VT and I T as the peak voltage and current ratings of a switch, the combinedutilization of all the switches in an inverter can be defined as V ISwitch utilization ratio = o1 o, max (4.46) qVT ITwhere q is the number of switches in an inverter. To compare the utilization of switches in various single-phaseinverters, we will initially compare them for a square-wave mode ofoperation at the maximum rated output. (The maximum switch utilizationoccurs at Vd = Vd , max ). 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
  • 191. 182 Chapter Threeinverter 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 (π / 4 )ma as compared to the square-wave switching: 1 π 1Maximum switch utilization ratio = ma = ma , ma ≤ 1.0 ) (4.54) 2π 4 8 Therefore, the theoretical maximum switch utilization ratio in a PWMswitching is only 0.125 at ma = 1 , as compared with 0.16 in square –wave inverter. Example 4 In a single-phase full-bridge PWM inverter, Vd 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 VT = Vd , ma = 325V I T = 2 I o = 2 *10 = 14.14 q = no. of switches = 4The maximum output volt-ampere (fundamental frequency) isVo1I o, max = 200 *10 = 2000 VA (4.55)Therefore, from Eq.(4.46) Vo1 I o, max 2000Switch utilization ratio = = = 0.11 qVT I T 4 * 325 *14.144.4 THREE-PHASE INVERTERS In applications such as uninterruptible ac power supplies and ac motordrives, 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.
  • 192. SWITCH-MODE dc-ac 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 v AN , (with respect tothe negative dc bus), depends only on Vd 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 INVERTERS Similar to the single-phase inverters, the objective inpulse-width-modulated three-phase inverters is to shape and control thethree-phase output voltages in magnitude and frequency with anessentially constant input voltage Vd . 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 form f = 15 ). It should also be noted from Fig.4.22b that an identical amount ofaverage dc component is present in the output voltages v AN and v BN ,which are measured with respect to the negative dc bus. These dccomponents are canceled out in the line-to-line voltages, for example inv AB 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
  • 193. 184 Chapter Threelegs, for example v AB in Fig.4.22b, are identical to the harmonics in v Aoin Fig.4.5, where only the odd harmonics exist as sidebands, centeredaround m f and its multiples, provided mf is odd. Only considering theharmonic at m f (the same applies to its odd multiples), the phasedifference between the mf harmonic in v AN and v BN is (120 m f )°. Thisphase difference will be equivalent to zero (a multiple of 360°) if m f isodd and a multiple of 3. As a consequence, the harmonic at m f issuppressed in the line-to-line voltage v AB . The same argument applies inthe suppression of harmonics at the odd multiples of m f if m f is chosento be an odd multiple of 3 (where the reason for choosing m f to be anodd multiple of 3 is to keep m f odd and, hence, eliminate evenharmonics). 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 m f , to eliminate the even harmonics, a synchronizedPWM 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 m f , the comments in Section 4.2.1.2 for asingle-phase PWM apply.3. During overmodulation ( ma > 1.0), regardless of the value of m f , theconditions pertinent to a small m f should be observed.4.4.1.1 Linear Modulation ( ma ≤ 1.0 )In the linear region ( ma ≤ 1.0 ), the fundamental-frequency component inthe output voltage varies linearly with the amplitude modulation ratio ma .From Figs.4.5b and 4.22b, the peak value of the fundamental-frequencycomponent in one of the inverter legs is( ) ˆ V V AN 1 = ma d 2 (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
  • 194. SWITCH-MODE dc-ac 185 (4.57) Fig.4.22 Three-phase PWM waveforms and harmonic spectrum.
  • 195. 186 Chapter Three 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 ( ma > 1.0 ) In PWM overmodulation, the peak of the control voltages are allowedto 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 voltageVLL1 is plotted as a function of ma . 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 VLL1 equal to0.78 Vd as explained in the next section. In the overmodulation region compared to the region with ma ≤ 1.0 ,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 ma ≤ 1.0 . 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 v LL for a large and odd m f that is a multiple of 3.
  • 196. SWITCH-MODE dc-ac 187 Fig.4.23 Three-phase inverter VLL1(rms ) / Vd as a function of ma .4.4.2 SQUARE-WAVE OPERATION IN THREE-PHASEINVERTERSIf the input dc voltage Vd is controllable, the inverter in Fig.4.24a can beoperated in a square-wave mode. Also for sufficiently large values of ma ,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).
  • 197. 188 Chapter Three 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 loadand contains harmonics (6n ± 1; n = 1, 2, . . .), whose amplitudes decreaseinversely proportional to their harmonic order, as shown in Fig.4.24c: (VLL )h = 0.78 Vd (4.59) h where h = n ± 1 (n = 1, 2, 3,.......) 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 Vd , max is the maximum input voltage that remainsconstant 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 I o. max (both in the PWMand the square-wave mode) at maximum loading. Therefore, each switchwould have the following peak ratings:VT = Vd , max (4.60)and I T = 2 I o, max (4.61) If VLL1 is the rms value of the fundamental-frequency line-to-linevoltage component, the three-phase output volt-amperes (rms) at thefundamental frequency at the rated output is (VA)3 − phase = 3 VLL1 I O, max (4.62)Therefore, the total switch utilization ratio of all six switches combined is
  • 198. SWITCH-MODE dc-ac 189 (4.63)In the PWM linear region (ma ≤ 1.0 ) using Eq.(4.57) and noting that themaximum switch utilization occurs at Vd = Vd , max (4.64) In the square-wave mode, this ratio is 1 / 2π ≅ 0.16 compared to amaximum of 0.125 for a PWM linear region with ma = 1.0 . In practice, the same derating in the switch utilization ratio applies asdiscussed in Section 4.3.4 for single-phase inverters. Comparing Eqs.(4.54) and (4.64), we observe that the maximumswitch 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 OUTPUT Figure 4.25a shows a three-phase, three-leg, voltage source, switch-mode inverter in a block diagram form. It is assumed to be supplying 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 e A (t ), eB (t ) , and ec (t ) are assumed to be sinusoidal. Fig.4.25 Three-phase inverter: (a) circuit diagram; (b) phasor diagram (fundamental frequency).
  • 199. 190 Chapter Three Under balanced operating conditions, it is possible to express theinverter phase output voltages v AN , and so on (with respect to the loadneutral n), in terms of the inverter output voltages with respect to thenegative dc bus N: vkn = vkN − vnN (k = A, B, C ) (4.65)Each phase voltage can be written as divkn = L k + ekn (k = A, B, C ) (4.66) dtIn a three-phase, three-wire loadi A + iB + iC = 0 (4.67a)and (i A + iB + iC ) = 0 d (4.67b) dt Similarly, under balanced operating conditions, the three back-emfsare a balanced three-phase set of voltages, and thereforee A + eB + eC = 0 (4.68)From the foregoing equations, the following condition for the invertervoltages can be written:v An + v Bn + vCn = 0 (4.69)Using Eqs. (4.65) through (4.69),vnN = (v AN + vBN + vCN ) 1 (4.70) 3Substituting vnN from Eq.(4.70) into Eq.(4.65), we can write thephase-to-neutral voltage for phase A asv An = v AN − (vBN + vCN ) 2 1 (4.71) 3 3Similar equations can be written for phase B and C voltages. Similar to the discussion in Section 4.3.2.6 for the ripple in thesingle-phase inverter output, only the fundamental-frequency componentsof the phase voltage V An1 and the output current i A1 are responsible forthe real power transformer since the back-emf e A (t ) is assumed to besinusoidal and the load resistance is neglected. Therefore, in a phasorform as shown in Fig.4.25b V An1 = E A + jω1 L I A1 (4.72) By using the principle of superposition, all the ripple in v An appearsacross the load inductance L. Using Eq.(4.71), the waveform for the
  • 200. SWITCH-MODE dc-ac 191phase-to-load-neutral voltage V An is shown in Figs.4.26a and 4.26b forsquare-wave and PWM operations, respectively. Both inverters haveidentical magnitudes of the fundamental-frequency voltage componentV An1 , which requires a higher Vd in the PWM operation. The voltageripple vripple (= v An − v An1 ) 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 m f , the current ripplein 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) square wave: (b) PWM. 4. 5 RECTIFIER MODE OF OPERATION
  • 201. 192 Chapter Three Fig.4.37 Operation modes: (a) circuit; (b) inverter mode; (c) rectifier mode: (d) constant I A . V An 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.
  • 202. SWITCH-MODE dc-ac 1934.6 PROGRAMMABLE PWM CONVERTERS This technique combines the square wave switching and PWM tocontrol 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 voltage 2- About 50% reduction in switching frequency compared to conventional sine PWM. 3- Suitable for higher voltage and high power inverter systems, where switching frequency is a limitation. 4- Higher voltage gain due to over modulation. 5- Reduced size of dc link filter components. 6- Selective elimination of lower order harmonics guarantee avoidance of resonance with external line filtering networks. 1 The voltage v Ao , of an inverter leg, normalized by Vd is plotted in 2Fig.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.
  • 203. 194 Chapter Three 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: ( ) ˆ V Ao 1 4 = = 1.273 (4.73) Vd / 2 πBecause of the notches to eliminate 5th and 7th harmonics, the maximumavailable fundamental amplitude is reduced. It can be shown that: ( ) ˆ V Ao 1, max = 1.188 (4.74) Vd / 2 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, 1 ⎡m ⎤ bn = ⎢ ∑ J s cos nωt ⎥ (4.75) nπ ⎢ s =1 ⎣ ⎥ ⎦ 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. Js J1 J2 J3 J4Time 0 α1 α2 α3Value 2 -2 2 -2Then, bn = 2 [2 cos n0 − 2 cos nα1 + 2 cos nα 2 − 2 cos nα 3 ] (4.76) nπ π Where α1 < α 2 < α 3 < 2
  • 204. SWITCH-MODE dc-ac 195 The above equation has three variables α1, α 2 , and α 3 so we needthree equation to obtain them. The First equation can be obtained byassigning a specific value to the amplitude of the fundamental componentb1 . Another two equations can be obtained by equating b5 , and b7 byzero to eliminate fifth and seventh harmonics. So, the following equationscan be obtained. b1 = [1 − cos α1 + cos α 2 − cos α 3 ] 4 (4.77) π b5 = [1 − cos 5α1 + cos 5α 2 − cos 5α 3 ] = 0 4 (4.78) π b7 = [1 − cos 7α1 + cos 7α 2 − cos 7α 3 ] = 0 4 (4.79) π The above equation can be rearranged to be in the following form: ⎡ π b1 ⎤ ⎡ cos α1 − cos α 2 cos α 3 ⎤ ⎢1 − ⎢cos 5α − cos 5α 4 ⎥ ⎢ cos 5α 3 ⎥ = ⎢ 1 ⎥ ⎥ (4.80) 1 2 ⎢ ⎥ ⎣ cos α1 ⎢ − cos 7α 2 cos 7α 3 ⎥ ⎦ ⎢ 1 ⎥ ⎢ ⎣ ⎥ ⎦ These equations are nonlinear having multiple solution depending thevalue of b1 . Computer programs help us in solving the above equations.The required values of α1 , α 2 , and α 3 are plotted in Fig.4.39 as afunction of the normalized fundamental in the output voltage. Fig.4.39 The required values of α1 , α 2 , and α 3 .
  • 205. 196 Chapter Three 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 b5 = 0and b7 = 0 . So, we have only two notches per half cycle as shown in thefollowing figure. V Ao Vd / 2 Notch1 Notch2 π + α1 π + α2 ω1t α1 α2 π − α 2 π − α1 π 2π 2π − α 2 2π − α1 So we need only two variables α1 and α 2 which can be obtained fromthe following equations: bn = 4 [1 − cos nα1 + cos nα 2 ] (4.81) nπ b5 = [1 − cos 5α1 + cos 5α 2 ] = 0 4 (4.82) π b7 = [1 − cos 7α1 + cos 7α 2 ] = 0 4 (4.83) π ⎡ cos 5α1 − cos 5α 2 ⎤ ⎡1⎤ ⎢ ⎥=⎢⎥ (4.84) ⎣cos 7α1 − cos 7α 2 ⎦ ⎣1⎦ By solving the above equation we can get the value of α1 and α 2 asfollowing: α1 = 12.8111o and α 2 = 24.8458o The following table shows the absolute value of each harmonics aftereliminating fifth and seventh harmonics.
  • 206. SWITCH-MODE dc-ac 197Harmonic bn for square wave bn = 4 bn after eliminating 5th and 7thorder nπ bn = 4 [1 − cos nα1 + cos nα 2 ] nπ1 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 existingconfiguration 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. S3 S1 Vd a 2 b Vd c 2 S4 S2 Fig.4.40 Four switch converter.
  • 207. 198 Chapter Three• System Analysis For the proposed converter Fig.4.40, the switching requirements canbe stated as follows. Let the input three-phase generated voltages are:Vab = 3 *Vm Sin(ω i t + 30)Vbc = 3 *Vm Sin(ω i t + 270) (4.85)Vca = 3 *Vm Sin(ω i t + 150) The line voltages at the generator terminals can be expressed asfollows: ⎡ Vd ⎤⎡Vab ⎤ ⎡ S1 S 2 ⎤ ⎢ 2 ⎥⎢V ⎥ = ⎢ S S ⎥ * ⎢ V ⎥ (4.86)⎣ cb ⎦ ⎣ 3 4 ⎦ ⎢− d ⎥ ⎣ 2⎦ Where S1 , S2 , S3 and S4 are the switching functions of switches 1,2,3and 4 respectively. Vd is the DC-link voltage.But, S 2 = 1 − S1 and S 4 = 1 − S3 (4.87) ⎛V ⎞ Vab = (2 S1 − 1) ⎜ d ⎟ and ⎝ 2 ⎠Then, (4.88) ⎛ Vd ⎞ Vcb = (2 S3 − 1) ⎜ ⎟ ⎝ 2 ⎠ Then from (4.85), (4.86), (4.87) and (4.88) we get the followingequation: VS1 = 0.5 + 3 m sin (ω1t + 30 ) Vd sin (ω1t + 30 ) VmS 2 = 0 .5 − 3 Vd sin (ω1t + 90 ) VmS 3 = 0 .5 + 3 Vd sin (ω1t + 90 ) VmS 4 = 0 .5 − 3 Vd (4.89) oThen, the shift angle for switching signal of leg ‘a’ is 30 and for leg ‘c’ Vis 90o. Then, Vab = ma d ∠30 o (4.90) 2 2
  • 208. SWITCH-MODE dc-ac 199 VdVbc = ma ∠270 o (4.91) 2 2 VdVca = ma1 ∠150 o (4.92) 2 2 VdThen, VLL = ma (4.93) 2 2 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: V VLL = ma d (four switch converter) (4.94) 2 2 V VLL = 3 ma d (six switch converter) (4.95) 2 2
  • 209. 200 Chapter Three PROBLEMS SINGLE PHASE 1- In a single-phase full-bridge PWM inverter, the input dc voltagevaries in a range of 295-325 V. Because of the low distortion required inthe output vo , ma ≤ 1.0 (a) What is the highest Vo1 , 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,Vo1, max I o1, max = 2000VA , where io 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 Vo1 = 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 m f = 21 and ma = 0.8. Assume a bipolar voltage switching. 4- Repeat Problem 2 but assume that the output voltage is controlledby voltage cancellation and Vd has the same value as required in thePWM inverter of Problem 3. 5- Calculate and compare the peak values of the ripple currents inProblems 2 through 4. THREE-PHASE 6- Consider the problem of ripple in the output current of a three-phase square-wave inverter. Assume (VLL )1 = 220 V at a frequency of 52Hz 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 m f = 39 and ma = 0.8 . Calculate thepeak ripple current defined in Fig.4.26b. 8- In the three-phase, square-wave inverter of Fig.4.24a, consider theload to be balanced and purely resistive with a load-neutral n. Draw the
  • 210. SWITCH-MODE dc-ac 201steady-state v An , u A, i DA + , and id waveforms, where iDA+ is the currentthrough DA + . 9- Repeat Problem 8 by assuming that the toad is purely inductive,where the load resistance, though finite, can be neglected.
  • 211. Chapter 5 dc MOTOR DRIVES5.1 INTRODUCTION Traditionally, dc motor drives have been used for speed and positioncontrol 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 MOTORS In a dc motor, the field flux φ f is established by the stator, either bymeans of permanent magnets as shown in Fig.5-1a, where φ f staysconstant, or by means of a field winding as shown in Fig.5-1b, where thefield current If controls φ f . If the magnetic saturation in the flux path canbe neglected, then:φf =kf If (5.1)where k f 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 field winding.
  • 212. dc MOTOR DRIVES 201of 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 φ f and the armature current ia :Tem = k f φ f ia (5.2)where k f is the torque constant of the motor. In the armature circuit, aback-emf is produced by the rotation of armature conductors at a speedwm in the presence of a field flux φ fea = k eφ f ω m (5.3)where ke is the voltage constant of the motor. In SI units, k f and ke are numerically equal, which can be shown byequating the electrical power ea ia and the mechanical power ω mTem .The electrical power is calculated as Pe = ea ia = k eφ f ω m ia (using Eq. 5-3) (5.4)and the mechanical power as:Pm = ω mTem = k f φ f ω m I a (using Eq. 5-2) (5.5)In steady state, Pe = Pm (5.6)Therefore, from the foregoing equations ⎡ Nm ⎤ ⎡ V ⎤kt ⎢ ⎥ = k e ⎢Wb . rad / sec . ⎥ (5.7) ⎣ A.Wb ⎦ ⎣ ⎦ In practice, a controllable voltage source vt is applied to the armatureterminals to establish ia . Therefore, the current ia in the armature circuitis determined by vt , the induced back-emf ea , the armature-windingresistance Ra , and the armature-winding inductance: La
  • 213. 202 Chapter Five diavt = e a + Ra i a + a (5.8) dtEquation 5-8 is illustrated by an equivalent circuit in Fig.5-2. Fig.5.2 A DC motor equivalent circuit. The interaction of Tem with the load torque, determines how the motorspeed builds up: dω mTem = J + Bω m + TWL (t ) (5.9) dtwhere J and B are the total equivalent inertia and damping, respectively,of the motor load combination and TWL 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 φ fis kept constant and the motor is initially driving a load at a speed of ω m .To reduce the motor speed, if vt is reduced below ea in Fig.5.2, then thecurrent ia will reverse in direction. The electromagnetic torque Temgiven 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, ea decreases in
  • 214. dc MOTOR DRIVES 203magnitude (assuming that φ f is constant). Ultimately, the generationstops 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 MOTORS Often in small dc motors, permanent magnets on the stator as shown inFig. 5-1a produce a constant field flux φ f . In steady state, assuming aconstant field flux ( φ f , Eqs. 5.2, 5.3 and 5.8 result inTem = kT I a (5.10)Ea = k Eω m (5.11)Vt = E a + Ra I a (5.12) where kT = k f φ f and K E = K eφ f . Equations 5-10 through 5-12correspond to the equivalent circuit of Fig.5-4a. From the aboveequations, it is possible to obtain the steady-state speed wm as a functionof Tem for a given Vt : 1 ⎛ R ⎞ωm = ⎜Vt − a Tem ⎟ ⎜ ⎟ (5.13) kE ⎝ kT ⎠ The plot of this equation in Fig. 5-4b shows that as the torque isincreased, the torque-speed characteristic at a given Vt 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
  • 215. 204 Chapter FiveFigure 5.4 Permanent-magnet dc motor: (a) equivalent circuit: (b) torque-speed characteristics: Vt 5 > Vt 4 > Vt 3 > Vt 2 > Vt1 where Vt 4 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 vt in a permanent-magnet dc motor with aconstant φ f . In a continuous steady state, the armature current I a should not exceedits 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 Vt 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. E a5.4 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 v f , as shown in Fig. 5-5a. As
  • 216. dc MOTOR DRIVES 205indicated by Eq. 5-1, the steady-state value of φ f is controlled byI f = V f / R f , where R f is the resistance of the field winding.Since (~f is controllable, Eq. 5-13 can be written as follows: (5.14)recognizing that k E = k eφ f and k = k f φ f . Equation (5.14) shows thatin a dc motor with a separately excited field winding, both Vt and φ f canbe controlled to yield the desired torque and speed. As a general practice,to maximize the motor torque capability, φ f (hence I f ) is kept at itsrated value for speeds less than the rated speed. With φ f at its ratedvalue, 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 φ f constant and equal to itsrated value, the motor torque-speed capability is as shown in Fig. 5-5b,where this region of constant φ f is often called the constant-torqueregion. The required terminal voltage Vt in this region increases linearlyfrom approximately zero to its rated value as the speed increases fromzero to its rated value. The voltage Vt and the corresponding E a areshown in Fig. 5-5b. To obtain speeds beyond its rated value, Vt is kept constant at its ratedvalue and φ f is decreased by decreasing I f . Since I a is not allowed toexceed its rated value on a continuous basis, the torque capabilitydeclines, since φ f is reduced in Eq. (5.2). In this so-called field-weakening region, the maximum power E a I a (equal to ω m Tm ) 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 Tem declines with ω m and Vt , E a , and I a stay constant at theirrated 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.
  • 217. 206 Chapter Five Figure 5-5 Separately excited dc motor: (a) equivalent circuit; (6) continuous torque-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 ia is a pure dc. The more iadeviates 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 ia , whereas the losses in the 2resistance of the armature winding depend on I a (rms). Therefore, thehigher 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.
  • 218. dc MOTOR DRIVES 207Therefore, it is desirable to improve the form factor of the armaturecurrent as much as possible.5.5.2 TORQUE PULSATIONSSince the instantaneous electromagnetic torque Tem (t ) developed by themotor is proportional to the instantaneous armature current ia (t ) , a ripplein ia 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 DRIVES In servo applications, the speed and accuracy of response is important.In spite of the increasing popularity of ac servo drives, dc servo drives 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 Tem in Eq. 5-2 can becontrolled linearly by controlling the armature current ia 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.
  • 219. 208 Chapter Five 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: (5.16) (5.17) (5.18) (5.19)If we take Laplace transform of these equations, where Laplace variablesrepresent only the small-signal Δ values in Eqs.5.16 through 5.19, (5.20) These equations for the motor-load combination can be represented bytransfer function blocks, as shown in Fig.5.7. The inputs to the motor-load combination in Fig. 5.7 are the armature terminal voltage Vt (s ) andthe load torque TWL (s ) . 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)
  • 220. dc MOTOR DRIVES 209 (5.23) Fig.5.7 Block diagram representation of the motor and load (without ay feedback). As a simplification to gain better insight into the dc motor behavior,the friction term, which is usually small, will be neglected by setting B =0 in Eq. 5.22. Moreover, considering just the motor without the load, J inEq. 5.22 is then the motor inertia J m . Therefore (5.24)We will define the following constants: (5.25) (5.26)Using τ m and τ e in the expression for G1 (s ) yields (5.27)Since in general τ m >> τ e , it is a reasonable approximation to replacesτ m by s (τ m + τ e ) in the foregoing expression. Therefore (5.28) The physical significance of the electrical and the mechanical timeconstants of the motor should also be understood. The electrical timeconstant τ e , determines how quickly the armature current builds up, asshown in Fig.5-8, in response to a step change Δvt in the terminal voltage,where the rotor speed is assumed to be constant.
  • 221. 210 Chapter Five Figure 5-8 Electrical time constant Te; speed cam is assumed to be constant. The mechanical time constant τ m determines how quickly the speedbuilds up in response to a step change Δvt in the terminal voltage,provided that the electrical time constant τ e is assumed to be negligibleand, hence, the armature current can change instantaneously. Neglectingτ e in Eq. 5-28, the change in speed from the steady-state condition can beobtained as (5.29)recognizing that Vt (s ) = Δvt / s . From Eq. (5.29) (5.30)where τ 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 kT I a, max .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 fast acceleration and deceleration. The dynamic current limit is generally
  • 222. dc MOTOR DRIVES 211 several times higher than the continuous steady-state current rating of the motor.3- For accurate control of position, the average voltage output of the converter should vary linearly with its control input, independent of the load on the motor. This item is further discussed in Section 5-6-5.4- The converter should produce an armature current with a good form factor and should minimize the fluctuations in torque and speed of the motor.5- The converter output should respond as quickly as possible to its control input, thus allowing the converter to be represented essentially by a constant gain without a dead time in the overall servo drive transfer function model.Figure 5-9Mechanical time constant τ 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-
  • 223. 212 Chapter Fivebridge). 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 Vt and I acan smoothly reverse, independent of each other. Ignoring the effect ofblanking time, the average voltage output of the converter varies linearlywith the input control voltage vcontrol , independent of the load:Vt = k c vcontrol (5.31)where k c 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 k c given by Eq. 5.31. Figure 5-10 A dc motor servo drive; four-quadrant operation.5.6.3 RIPPLE IN THE ARMATURE CURRENT iaTe current through a PWM full-bridge dc-dc converter supplying a dcmotor load flows continuously even at small values of I a . 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 ea (t ) = E a . The terminal voltage and the
  • 224. dc MOTOR DRIVES 213armature current can be expressed in terms of their dc and the ripplecomponents asvt (t ) = Vt + vr (t ) (5.32)ia (t ) = I a + ir (t ) (5.33)where vr (t ) and ir (t ) are the ripple components in vt and ia ,respectively. Therefore, in the armature circuit, from Eq. 5.8, di (t )Vt + vr (t ) = Ea + R A [I a + ir (t )] + La r (5.34) dtWhere Vt = E a + Ra I a (5.35) di (t )And vr (t ) = Ra ir (t ) + La r (5.36) dtAssuming that the ripple current is primarily determined by the armatureinductance La and Ra has a negligible effect, from Eq. 5-36 di (t )vr (t ) ≅ La r (5.37) dt 2The additional heating in the motor is approximately Ra I r where I r isthe rms value of the ripple current ir . The ripple voltage is maximum when the average output voltage in adc is zero and all switches operate at equal duty ratios. Applying theseresults to the dc motor drive, Fig. 5-11a shows the voltage ripple vr (t )and the resulting ripple current ir (t ) using Eq. 5.37. From thesewaveforms, the maximum peak-to-peak ripple can be calculated as: V(ΔI P − P )max = d (5.38) 2 La f swhere Vd is the input dc voltage to the full-bridge converter.
  • 225. 214 Chapter Five Figure 5-11 Ripple ir in the armature current: (a) PWM bipolar voltage switching, Vt = 0 ; (b) PWM unipolar voltage switching, Vt = 1 / 2Vd . The ripple voltage for a PWM unipolar voltage switching is shown tobe maximum when the average output voltage is 1 / 2Vd . Applying thisresult to a dc motor drive, Fig.5.11b shows ir (t ) waveform, where V(ΔI P − P )max = d (5.39) 8La f s Equations 5-38 and 5-39 show that the maximum peak-to-peak ripplecurrent is inversely proportional to La and f s . Therefore, carefulconsideration must be given to the selection of f s and La , where La 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 I a , max 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 ia with its reference value ia produced by the speed regulator.The current ia is inherently controlled from exceeding the current rating *of the drive by limiting the reference current ia to I a , max . 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
  • 226. dc MOTOR DRIVES 215In 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 ia is proportional to the torque Tem , Fig. 5.13. If atan arbitrary speed ω m , the torque and, hence, ia are to be reversed, thereis a dead zone in vcontrol as shown in Fig. 5-13, during which ia and Temremain 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 ia . Fig.5.12 Control of servo drives: (a) no internal current-control loop; (b) internal current-control loop.
  • 227. 216 Chapter Five Fig.5.13 Effect of blanking time.5.6.6 SELECTION OF SERVO DRIVE PARAMETERS Based on the foregoing discussion, the effects of armature inductance La , switching frequency f s , blanking time t c , 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 La / f s .2- The dead zone in the transfer function of the converter, which degrades the servo performance, is proportional to f s t Δ .3- Switching losses in the converter are proportional to f s t c . All these factors need to be considered simultaneously in the selectionof 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 ia . The armature current
  • 228. dc MOTOR DRIVES 217can reverse, and a negative value of I a corresponds to the braking modeof operation, where the power flows from the dc motor to Vd . The outputvoltage Vt can be controlled in magnitude, but it always remainsunipolar. Since ia can flow in both directions, unlike in the single-switchstep-down and step-up dc-dc converters, ia 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 ia 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 angleα that controls the converter output voltage applied to the motorterminals cannot be increased for a portion of the 50-Hz cycle. This may
  • 229. 218 Chapter Fivenot 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 M 1 and M 2 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 M 1 and M 2 are opened and R1 and R2 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-phase input, (b) three-phase input.
  • 230. dc MOTOR DRIVES 219 Fig.5.16 Line-frequency-controlled converters for four-quadrant operation: (a) back-to-back converters for four-quadrant operation (without circulating current); (b) converter operation modes; (c) contactors for four-quadrant operation.5.7.3 EFFECT OF DISCONTINUOUS ARMATURE CURRENT In line-frequency phase-controlled converters and single-quadrantstep-down switch mode dc-dc converters, the output current can becomediscontinuous at light loads on the motor. For a fixed control voltagevcontrol or the delay angle α , the discontinuous current causes the outputvoltage to go up. This voltage rise causes the motor speed to increase atlow values of I a (which correspond to low torque load), as showngenerically by Fig.5.17. With a continuously flowing ia , the drop inspeed at higher torques is due to the voltage drop Ra I a across thearmature resistance; additional drop in speed occurs in the phase-controlled converter-driven motors due to commutation voltage dropsacross the ac-side inductance Ls , which approximately equal(2ωLS / π )I a in single-phase converters and (3ωLS / π )I a in three-phaseconverters. These effects results in poor speed regulation under an open-loop operation.
  • 231. 220 Chapter Five Fig.5.17 Effects of discontinuous ia on ω m .5.7.4 CONTROL OF ADJUSTABLE-SPEED DRIVES The type of control used depends on the drive requirements. An open-loop control is shown in Fig.5.18 where the speed command ω * isgenerated by comparing the drive output with its desired value (which,e.g., may be temperature in case of a capacity modulated heat pump). Ad / dt limiter allows the speed command to change slowly, thuspreventing the rotor current from exceeding its rating. The slope of thed / dt 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 φ f . Since many adjustable speed drives, especially at higher powerratings, employ a motor with a wound field, this capability can beexploited by controlling the field current and φ f . The simple linefrequency phase-controlled converter shown in Fig. 5-15 is normally usedto control I f through the field winding, where the current is controlled inmagnitude 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.
  • 232. dc MOTOR DRIVES 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 I s1 of the line current as a function of speed isshown in Fig. 5-196. Figure 5-19c shows I s1 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, I s1 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.
  • 233. 222 Chapter Five Fig.5.19 Line current in adjustable-speed dc drives: (a) drive capability; (b) switch-mode converter drive; (c) line-frequency thyristor converter drive.
  • 234. dc MOTOR DRIVES 223PROBLEMS1- Consider a permanent-magnet dc servo motor with the followingparameters:
  • 235. 224 Chapter Five
  • 236. Chapter 6 A.C. Voltage Regulators6.1 Single-phase control of load voltage using thyristor switching A. Resistive load Assumptions :a. Triggering circuit provides pulse train to gate thyristor any point on wave 0-180o.b. Ideal supply, i.e. zero Z, voltage remains sinusoidal in spite of non- sinusoidal pulses of current drawn from supply . c. Thyristors are ideal, i.e. no dissipation when conducting, no voltage drop when conducting , infinite Z when off. e = E m sin ωt π , 2π α , π +α e L = E m sin ωt | +| α , π +α 0, π π , 2π α , π +α eT = E m sin ωt | +| α , π +α 0, π
  • 237. 224 Chapter Six6.2 Harmonics analysis of the load voltage (and current) The load voltage eL can be expressed as :- a0 ∞ a ∞ e L (ωt ) = + ∑ a n cos nωt + bn sin nωt = 0 + ∑ C n sin(n ωt + ψ ) 2 n =1 2 n =1 a0 1 2πwhere, = 2 2π 0 ∫ eL (ωt ) dωt = Average valueFor fundamental component :- 1 2π 1 2π a1 = ∫ e L (ωt ) cos ωt dωt , b1 = ∫ e L (ωt ) sinωt dωt π 0 π 0c1 = a12 + b12 = peak value of fundamental component a1ψ 1 = tan −1 = displacement angle between the fundamental and b1 the datum point.For the n-th harmonics, 1 2π 1 2π a n = ∫ e L (ωt ) cos nωt dωt , π ∫0 bn = e L (ωt ) sin nωt dωt π 0For the fundamental of the load voltage, 1 2π Em π ∫0 a1 = e L (ωt ) cos ωt dωt = (cos 2α − 1) 2π Em b1 = (2 (π − α ) + sin 2α ) 2π E c1 = m (cos 2α − 1) 2 + [2 (π − α ) + sin 2α ]2 2π a cos 2α − 1 ψ 1 = tan −1 1 = tan −1 b1 2(π − α ) + sin 2αRMS Load Voltage 1 2πr.m.s. value of eL = E L = 2π ∫0 eL 2 (ωt ) dωt 1 π ,2π E 2 2π ∫α ,α +π EL 2 = ( E m sin ωt ) 2 dωt = m [ 2 (π − α ) + sin 2α ] 4π E 1 EL = m [2(π − α ) + sin 2α ] 2 2π
  • 238. A.C. Voltage Regulators 225with resistive load, the instantaneous load current is :- e E sin ωt π ,2π iL = L = m |α ,π +α R RThe r.m.s. load current E 1 E 1 IL = m [ 2(π − α ) + sin 2α ] = [2(π − α ) + sin 2α ] 2 R 2π R 2π6.3 Power Dissipation 1 2π P= 2π 0 ∫eL i dt = Average value of instantaneous volt − amp product. = Average power EL 2 = I L2 R = where I L & E L are the r.m.s. values R E2 = [2(π − α ) + sin 2α ] 2πRIn terms of harmonic components:- 1 2 P = R ( I12 + I 32 + I 5 2 + .......) = ( E L1 + E L3 + E L5 + .......) 2 2 RIn terms of fundamental components: P = E I1 cos ψ 1where I1 = r.m.s. fundamental current = c1 1 , cosψ 1 = b1 2 R c16.4 Power factor in non-sinusoidal circuits Average Power PIn general, PF = = Apparent Voltamperes E I E = r.m.s. voltage at supply. I = r.m.s. current at supply.Definition is true irrespective for any waveform and frequency.Let e and i be periodic in 2π , 1 2π P= 2π 0 ∫ e i dωt 1 2π 2 1 2π 2 E= 2π 0 ∫ e dωt ; I = 2π 0 i dω t ∫The usual aim is to obtain unity PF at the supply.
  • 239. 226 Chapter SixTo obtain unity P.F., certain conditions must be observed w.r.t. e(ω t) andi(ωt):-(i) e(ω t) and i(ωt) must be of the same frequency.(ii) e(ω t) and i(ωt) must be of the same wave shape (whatever the wave shape)(iii) e(ω t) and i(ωt) must be in time-phase at every instant of the cycle.Power Factor in systems with sinusoidal voltage (at supply) but non-sinusoidal currenti (wt ) is periodic in 2π but is non-sinusoidal.Average power is obtained by combining in-phase voltage and currentcomponents of the same frequency. P = E I1 cos ψ 1 P E I1 cos ψ 1 I1 PF = = = cos ψ 1 EI EI I = Distortion Factor x Displacement FactorDistortion Factor = 1 for sinusoidal operationDisplacement factor is a measure of displacement between e(ω t) andi(ωt).Displacement Factor =1 for sinusoidal resistive operation.Calculation of PF E2 R [2(π − α ) + sin 2α ] I 2R 2π R 2 1 PF = = = [2(π − α ) + sin 2α ] EI E 1 2π E [2(π − α ) + sin 2α ] R 2πR-L load ωL Z = R 2 + ω 2 L2 , φ = tan −1 R
  • 240. A.C. Voltage Regulators 227Steady-state at α = 0 (sinusoidal operation)Applying gate signal at ωt = α, where α > φ The start of conduction is delayed until ωt = α . Subsequent totriggering, let the instantaneous current i (ωt ) consists of hypotheticalsteady-state components iss (ωt ) and transient component itrans (ωt ) ,∴ i (ωt ) = iss (ωt ) + itrans (ωt )Now the ωt = α , the instantaneous steady-state component has the value, Eiss (α ) = m sin(α − φ ) ZBut at ωt = α , total current i (ωt ) = 0 . At ωt = α, iss(α) = -itrans(α) E ∴ itrans (ωt ) = − m sin(α − φ ) ZSubsequent to α The transient component decay exponentially from it’s instantaneous -Em L sin(α − φ ) by the constant Z = Z R. t − Em − itrans (ωt ) = sin(α − φ )e τ Z R E − (ωt −α )For ωt > α, itrans (ωt ) = − m sin(α − φ )e ωL ZComplete solution for first cycle
  • 241. 228 Chapter Six R E − (ωt +π −α ) i (ωt ) = m [sin(ωt − φ ) 0,−α ,, π ,+2π + sin(α + φ )e ωt x π x α x −π 0 Z R R − (ωt −α ) − (ωt −π −α ) ωt ωL 2π − sin(α − φ )e α x + sin(α − φ )l π +αExtinction Angle For the current in the interval α ≤ ωt ≤ x R E E − (ωt −α ) i (ωt ) = m sin(ωt − φ ) − m sin(α − φ )e ωL Z ZBut at ωt = α , i (ωt ) = 0 , hence R E E − ( x −α ) 0 = m sin( x − φ ) − m sin(α − φ )e ωL Z Z
  • 242. A.C. Voltage Regulators 229 Em RBut ≠ 0 and = cot φ ∴ 0 = sin( x − φ ) − sin(α − φ )e − cot( x −α ) Z ωL If α and φ are known, x can be calculated. However, this is atranscendental equation (i.e. cannot be solved explicitly and no way ofobtaining x = f (α , φ ) ).Method of solution is by iteration,e.g. If φ = 600, α = 1200 = 2π / 3, cot φ = 0.578 2π − 0.578( x − ) sin( x − 60 ) = sin(120 − 60)e 0 0 3 ⇒ x = 2220Rough Approximation :- x = 180 0 + φ − Δ Δ = 5 0 ~ 10 0 for large R and small ωLwhere = 10 0 ~ 15 0 for R = ωL = 150 ~ 20 0 for small R and large ωLFor the previous example, φ = 60 0 , Δ = 150 − 20 0 ∴ x = 180 0 + 60 0 − 15 0 = 225 0 or x = 180 0 + 60 0 − 20 0 = 220 0Load voltage eL (ωt ) = Em sin ωt 0,−π,π x α π x , ,2 α + Em a1 = [cos 2α − cos 2 x] 2π E b1 = m [2( x − α ) − sin 2 x + sin 2α ] 2π
  • 243. 230 Chapter SixA.C. voltage control using Integral cycle switchingN = no. of conducting cycles.T = Total ‘ON’+’OFF’ cycles, representing a control period.Analytical Properties. N NPower, P ∝ ; r.m.s. load voltage, EL ∝ T T E2 N P NPower Factor = = R T = EI E N T E R T6.6 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.
  • 244. Problems Of Chapter 21- Single phase half-wave diode rectifier is connected to 220 V, 50 Hz supply to feed 5Ω pure resistor. Draw load voltage and current and diode voltage drop waveforms along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.2- The load of the rectifier shown in problem 1 is become 5Ω pure resistor and 10 mH inductor. Draw the resistor, inductor voltage drops, and, load current along with supply voltage. Then, find an expression for the load current and calculate the conduction angle, β . Then, calculate the DC and rms value of load voltage. 1
  • 245. 3- In the rectifier shown in the following figure assume VS = 220V , 50Hz, L = 10mH and Ed = 170V . Calculate and plot the current an the diode voltage drop along with supply voltage, vs . vdiode + vL - i + - + vs Ed -4- Assume there is a freewheeling diode is connected in shunt with the load of the rectifier shown in problem 2. Calculate the load current during two periods of supply voltage. Then, draw the inductor, resistor, load voltages and diode currents along with supply voltage. 2
  • 246. 5- The voltage v across a load and the current i into the positive polarity terminal are as follows: v(ωt ) = Vd + 2 V1 cos(ωt ) + 2 V1 sin (ωt ) + 2 V3 cos(3ωt ) i (ωt ) = I d + 2 I1 cos(ωt ) + 2 I 3 cos(3ωt − φ ) Calculate the following: (a) The average power supplied to the load. (b) The rms value of v(t ) and i (t ) . (c) The power factor at which the load is operating.6- Center tap diode rectifier is connected to 220 V, 50 Hz supply via unity turns ratio center-tap transformer to feed 5Ω resistor load. Draw load voltage and currents and diode currents waveforms along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current. 3
  • 247. 7- Single phase diode bridge rectifier is connected to 220 V, 50 Hz supply to feed 5Ω resistor. Draw the load voltage, diodes currents and calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current. 4
  • 248. 8- If the load of rectifier shown in problem 7 is changed to be 5Ω resistor in series with 10mH inductor. Calculate and draw the load current during the first two periods of supply voltages waveform.9- Solve problem 8 if there is a freewheeling diode is connected in shunt with the load.10- If the load of problem 7 is changed to be 45 A pure DC. Draw diode diodes currents and supply currents along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current. (f) input power factor. 5
  • 249. 11- Single phase diode bridge rectifier is connected to 220V ,50Hz supply. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor, and, input power factor and THD of the voltage at the point of common coupling. 6
  • 250. 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. 7
  • 251. 14- Three-phase full bridge diode rectifier is connected to 380V, 50Hz supply to feed 10Ω resistor. Draw the output voltage, diode currents and supply current of phase a. Then, calculate: (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.15- Solve problem 14 if the load is 45A pure DC current. Then find THD of supply current and input power factor. 8
  • 252. 16- If the supply connected to the rectifier shown in problem 14 has a 5 mH source inductance and the load is 45 A DC. Find, average DC voltage, and THD of input current. 9
  • 253. 17- Single phase diode bridge rectifier is connected to square waveform with amplitude of 200V, 50 Hz. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor.18- In the single-phase rectifier circuit of the following figure, LS = 1 mH and Vd = 160V . The input voltage vs has the pulse waveform shown in the following figure. Plot is and id waveforms and find the average value of I d . id + iS Vd VS - f = 50 Hz 200V 120o ωt o o o o 60 120 60o 60 120 10
  • 254. Problems Of Chapter 31- Single phase half-wave controlled rectifier is connected to 220 V, 50Hz supply to feed 10 Ω resistor. If the firing angle α = 30 o draw output voltage and drop voltage across the thyristor along with the supply voltage. Then, calculate, (a) The rectfication effeciency. (b) Ripple factor. (c) Peak Inverse Voltage (PIV) of the thyristor. (d) The crest factor C F of input current.2- Single phase half-wave controlled rectifier is connected to 220 V, 50Hz supply to feed 5Ω resistor in series with 10mH inductor if the firing angle α = 30 o .(a) Determine an expression for the current through the load in the first two periods of supply current, then fiend the DC and rms value of output voltage.(b) Draw the waveforms of load, resistor, inductor voltages and load current. 11
  • 255. 3- Solve problem 2 if there is a freewheeling diode is connected in shunt with the load.4- single phase full-wave fully controlled rectifier is connected to 220V, 50 Hz supply to feed 5Ω resistor, if the firing angle α = 40 o . Draw the load voltage and current, thyristor currents and supply current. Then, calculate (a) The rectfication effeciency. (b) Peak Inverse Voltage (PIV) of the thyristor. (c) Crest factor of supply current. 12
  • 256. 5- In the problem 4, if there is a 5mH inductor is connected in series with the 5Ω resistor. Draw waveforms of output voltage and current, resistor and inductor voltages, thyristor currents, supply currents. Then, find an expression of load current, DC and rms values of output voltages.6- Solve problem 5 if the load is connected with freewheeling diode. 13
  • 257. 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 α = 40 o . Draw the load voltage, thyristor, and load currents. Then, calculate (a) the rectfication effeciency. (b) Ripple factor of output voltage. (c) Crest factor of supply current. (d) Use Fourier series to fiend an expression for supply current. (e) THD of supply current. (f) Input power factor.8- Solve problem 7 if the supply has a 3 mH source inductance. 14
  • 258. 9- Single phase full-wave semi-controlled rectifier is connected to 220 V, 50Hz supply to feed 5Ω resistor in series with 5 mH inductor, the load is connected in shunt with freewheeling diode. Draw the load voltage and current, resistor voltage and inductor voltage diodes and thyristor currents. Then, calculate Vdc and Vrms of the load voltages. If the freewheeling diode is removed, explain what will happen?10- The single-phase full wave controlled converter is supplying a DC load of 1 kW with pure DC current. A 1.5-kVA-isolation transformer with a source-side voltage rating of 120 V at 50 Hz is used. It has a total leakage reactance of 8% based on its ratings. The ac source voltage of nominally 120 V is in the range of -10% and +5%. Then, Calculate the minimum transformer turns ratio if the DC load voltage is to be regulated at a constant value of 100 V. What is the value of a when VS = 120 V + 5%. 15
  • 259. 11- In the single-phase inverter of, VS = 120 V at 50 Hz, LS = 1.2 mH, Ld = 20 mH, Ed = 88 V, and the delay angle α = 135°. Using PSIM, obtain vs , is , vd , and id waveforms in steady state.12- In the inverter of Problem 12, vary the delay angle α from a value of 165° down to 120° and plot id versus α . Obtain the delay angle α b , below which id becomes continuous. How does the slope of the characteristic in this range depend on LS ?13- In the three-phase fully controlled rectifier is connected to 460 V at 50 Hz and Ls = 1mH . Calculate the commutation angle u if the load draws pure DC current at Vdc = 515V and Pdc = 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 VLL and I o .15- Consider the three-phase, half-controlled converter shown in the following figure. Calculate the value of the delay angle α for which Vdc = 0.5Vdm . Draw vd waveform and identify the devices that conduct during various intervals. Obtain the DPF, PF, and %THD in the input line current and compare results with a full-bridge converter operating at Vdc = 0.5Vdm . Assume LS . 16
  • 260. 16- Repeat Problem 15 by assuming that diode D f is not present in the converter.17- The three-phase converter of Fig.3.48 is supplying a DC load of 12 kW. A Y- Y connected isolation transformer has a per-phase rating of 5 kVA and an AC source-side voltage rating of 120 V at 50 Hz. It has a total per-phase leakage reactance of 8% based on its ratings. The ac source voltage of nominally 208 V (line to line) is in the range of -10% and +5%. Assume the load current is pure DC, calculate the minimum transformer turns ratio if the DC load voltage is to be regulated at a constant value of 300 V. What is the value of α when VLL = 208 V +5%.18- In the three-phase inverter of Fig.3.63, VLL = 460 V at 60 Hz, E = 550 V, and LS = 0.5 mH. Assume the DC-side current is pure DC, Calculate α and γ if the power flow is 55 kW. 17
  • 261. Problems Of Chapter 4 1- In a single-phase full-bridge PWM inverter, the input dcvoltage varies in a range of 295-325 V. Because of the lowdistortion required in the output vo , ma ≤ 1.0 (a) What is the highest Vo1 , that can be obtained and stamped onits nameplate as its voltage rating? (b) Its nameplate volt-ampere rating is specified as 2000 VA, thatis, Vo1, max I o1, max = 2000VA , where io is assumed to be sinusoidal.Calculate the combined switch utilization ratio when the inverter issupplying its rated volt-amperes. 18
  • 262. 2- Consider the problem of ripple in the output current of asingle-phase full-bridge inverter. Assume Vo1 = 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. 19
  • 263. 3- Repeat Problem 2 with the inverter operating in a sinusoidalPWM mode, with m f = 21 and ma = 0.8. Assume a bipolar voltageswitching. 4- Repeat Problem 2 but assume that the output voltage iscontrolled by voltage cancellation and Vd has the same value asrequired in the PWM inverter of Problem 3. 20
  • 264. 5- Calculate and compare the peak values of the ripple currentsin Problems 2 through 4. 21
  • 265. THREE-PHASE 6- Consider the problem of ripple in the output current of a three-phase square-wave inverter. Assume (VLL )1 = 220 V at a frequencyof 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. 22
  • 266. 7- Repeat Problem 6 if the inverter of Problem 6 is operating in asynchronous PWM mode with m f = 39 and ma = 0.8 . Calculate thepeak ripple current defined in Fig.26b. 23
  • 267. 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 v An , u A, i DA + , and id waveforms, where iDA+is the current through DA + . 24
  • 268. 9- Repeat Problem 8 by assuming that the toad is purelyinductive, where the load resistance, though finite, can be neglected. 25
  • 269. Problems Of Chapter 51- Consider a permanent-magnet dc servo motor with the followingparameters: 26
  • 270. 27
  • 271. 28
  • 272. 29
  • 273. 30
  • 274. 31
  • 275. 32
  • 276. Power Electronics (2) Assignment 33