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Power electronics note


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Power electronics note

  1. 1. Power Electronics Dr.Ali Mohamed Eltamaly Mansoura University Faculty of Engineering
  2. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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: