Lesson 2 Chapter Two Three Phase Uncontrolled Rectifier


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1 Lesson 2 Chapter Two Three Phase Uncontrolled Rectfer. Operatng prncple of three phase half wave uncontrolled rectfer The half wave uncontrolled converter s the smplest of all three phase rectfer topologes. Although not much used n practce t does provde useful nsght nto the operaton of three phase converters. Fg. shows the crcut dagram, conducton table and wave forms of a three phase half wave uncontrolled converter supplyng a resstve nductve load. Fg.: Operaton of the three phase half wave uncontrolled rectfer (a) crcut dagram, (b) conducton table, (c) wave forms For smplcty the load current ( o ) has been assumed to be rpple free. As shown n Fg. (a), n a three phase half wave uncontrolled converter the anode of
2 a dode s connected to each phase voltage source. The cathodes of all three dodes are connected together to form the postve load termnal. The negatve termnal of the load s connected to the supply neutral. Fg. (b) shows the conducton table of the converter. It should be noted that for the type of load chosen the converter always operates n the contnuous conducton mode. The conducton dagram for the dodes (as shown n Fg. (c) second waveform) can be drawn easly from the conducton dagram. Snce the dodes can block only negatve voltage t follows from the conducton table that a phase dode conducts only when that phase voltage s maxmum of the three. (In sgnal electroncs the crcut of Fg. (a) s also known as the maxmum value crcut). Once the conducton dagram s drawn other waveforms of Fg. (c) are easly obtaned from the supply voltage waveforms n conjuncton wth the conducton table. The phase current waveforms of Fg. (c) deserve specal menton. All of them have a dc component whch flows through the ac source. Ths may cause dc saturaton n the ac sde transformer. Ths s one reason for whch the converter confguraton s not preferred very much n practce. From the waveforms of Fg. (c) (60) (6) The output voltage form factor = (62), (63) Input power factor = (64)
3 The harmoncs present n v o and can be found by Fourer seres analyss of the correspondng waveforms of Fg. (c) and s left as an exercse. Exercse Fll n the blank(s) wth the approprate word(s). ) Three phase half wave uncontrolled rectfer uses dodes. ) Three phase half wave uncontrolled rectfer requres phase wre power supply. ) In a three phase half wave uncontrolled rectfer each dode conduct for radans. v) The mnmum frequency of the output voltage rpple n a three phase half wave uncontrolled rectfer s tmes the nput voltage frequency. v) The nput lne current of a three phase half wave uncontrolled rectfer contan component. Answers: () three; () three, four; () 2π/3; (v) three; (v) dc. 2. Assumng rpple free output current, fnd out the, dsplacement factor, dstorton factor and power factor of a three phase half wave rectfer supplyng an R L load. Wth reference to Fg. the expresson for phase current a can be wrtten as a = 0 otherwse. Fundamental component of a can be wrtten as where and
4 φ = 0 splacement factor = cosφ =. R.M.S value of a = I a = storton factor = Power Factor = sp. Factor st. Factor = 2. Three phase full wave uncontrolled converter As has been explaned earler three phase half wave converter suffers from several dsadvantages. Chef among them are dc component n the nput ac current, requrement of neutral connecton and comparatvely lower output voltage. In addton the nput and output waveforms contan lower order harmoncs whch requre heavy flterng. Most of these dsadvantages can be mtgated by usng a three phase full wave brdge rectfer. Ths s probably the most extensvely used rectfer topology from low (>5 KW) to moderately hgh power (> 00 KW) applcatons. In ths secton the operaton of a three phase full wave uncontrolled brdge rectfer wth two dfferent types of loads namely the R L E type load and the capactve load wll be descrbed. 2. Operaton of a 3 phase full wave uncontrolled brdge rectfer supplyng an R L E load Ths type of load may represent a dc motor or a battery. Usually for drvng these loads a varable output voltage s requred. Ths requrement has to be met by usng a varable ac source (e.g a 3 phase varable) snce the average output voltage of an uncontrolled rectfer s constant for a gven ac voltage. It wll also be assumed n the followng analyss that the load sde nductance s large enough to keep the load current contnuous. The relevant condton for contnuous conducton wll be derved but analyss of dscontnuous conducton mode wll not be attempted. Compared to sngle phase converters the cases of dscontnuous conducton n 3 phase brdge converter are neglgble.
5 Fg.2: Operaton of the three phase full wave uncontrolled rectfer (a) crcut dagram, (b) conducton table, (c) wave forms Snce the load current s assumed to be contnuous at least one dode from the top group (, 3 and 5 ) and one dode from the bottom group ( 2, 4 and 6 ) must conduct at all tme. It can be easly verfed that only one dode from each group (ether top or bottom) conducts at a tme and two dodes from the same phase leg never conducts smultaneously. Thus the converter has sx dfferent
6 dode conducton modes. These are 2, 2 3, 3 4, 4 5, 5 6 and 6. Each conducton mode lasts for π/3 rad and each dode conducts for 20º. Fg. 2 (b) shows voltages across dfferent dodes and the output voltage n each of these conducton modes. The tme nterval durng whch a partcular conducton mode wll be effectve can be ascertaned from ths table. For example the conducton mode wll occur when the voltage across all other 2 dodes (.e. v ba, v ca and v cb ) are negatve. Ths mples that 2 conducts n the nterval 0 ωt π/3 as shown n Fg. 2.2 (c). The dodes have been numbered such that the conducton sequence s When a dode stops conducton ts current s commutated to another dode n the same group (top or bottom). Ths way the sequence of conducton modes become, . The conducton dagram n Fg. 2 (c) s constructed accordngly. The output dc voltage can be constructed from ths conducton dagram usng approprate lne voltage segments as specfed n the conducton table. The nput ac lne currents can be constructed from the conducton dagram and the output current. For example a = o for 0 ωt π/3 and 5π/3 ωt 2π a =  o for 2π/3 ωt 4π/3 a = 0 otherwse. (65) The lne current wave forms and ther fundamental components are shown n Fg.2 (c). It s clear from Fg.2 (c) that the dc voltage output s perodc over one sxth of the nput ac cycle. For π/3 ωt 2π/3 (66) (67) (68) (69)
7 I RMS can be found as follows (70) Snce nput dsplacement factor s unty (7) Power factor = dstorton factor = (72) A closed form expresson for o can be found as follows for π/3 ωt 2π/3 The general soluton s gven by (73) (74) Where Now snce the current waveform s perodc over one sxth of the nput ac cycle (75) (76) (77) (78)
8 Exercse 2 Fll n the blank(s) wth the approprate word(s). ) Three phase full wave uncontrolled rectfer uses dodes. ) Three phase full wave uncontrolled rectfer does not requre wre connecton. ) In a three phase full wave uncontrolled rectfer each dode conducts for radans. v) The mnmum frequency of the output voltage rpple n a three phase full wave rectfer s tmes the nput voltage frequency. v) The nput ac lne current of a three phase full wave uncontrolled rectfers supplyng an R L E load contan only harmoncs but no harmonc or component. v) A three phase full wave uncontrolled rectfer supplyng an R L E load normally operates n the conducton mode. Answers: contnuous. () sx; () neutral; () 2π/3; (v) sx; (v) odd, trpler, dc; (v) 2. A 220 V, 500 rpm 20 A separately excted dc motor has armature resstance of Ω and neglgble armature nductance. The motor s suppled from a three phase full wave uncontrolled rectfer connected to a 220 V, 3 phase, 50 Hz supply through a Δ/Y transformer. Fnd out the transformer turns rato so that the converter apples rated voltage to the motor. What s the maxmum torque as a percentage of the rated torque the motor wll be able to supply wthout over heatng. Assume deal transformer and contnuous conducton. Answer: Average output voltage of the converter s V L = 63 Volts. Ths s the lne voltage of the secondary sde of the transformer. The secondary s star connected. So Secondary phase voltage =. Prmary sde s delta connected. So Prmary phase voltage = 220 V. The requred turns rato = Output voltage can be wrtten as Where v hn = nth harmonc voltage magntude.
9 Where E = back emf and r = armature resstance To prevent over heatng I 0RMS = 20 A For the gven converter I 0AV = Amps. In a separately excted dc machne T e I 0AV Maxmum allowable torque = torque. = % of full load 2.2 Operaton of a three phase uncontrolled brdge rectfer supplyng a capactve load A three phase uncontrolled brdge rectfer supplyng a capactve load s a very popular power electronc converter. It s very wdely used as the front end of a varable voltage varable frequency dc ac nverter. Fg.3 (a) shows the power crcut dagram of such a converter. Operaton of the converter can be explaned as follows. The top group dodes (, 3, 5 ) form a Maxmum value crcut and therefore the maxmum of the phase voltages v an, v bn, v cn appears at the postve dc bus. On the other hand, the bottom group dodes ( 2, 4, 6 ) form a Mnmum value crcut. Therefore the mnmum of the phase voltages v an, v bn and v cn appears at the negatve dc bus. Therefore, the output voltage waveform at any nstant s equal to the maxmum of the sx lne voltages v ab, v bc, v ca, v ba, v cb and v ac provded at least one dode from the top group and one from the bottom group conducts at that nstant. None of the dodes wll conducts, however f the
10 output capactor voltage s larger than the maxmum lne voltage. All the sx operatng modes of a 3 phase brdge rectfer namely, 2, 2 3, 3 4, 4 5, 5 6 and 6 appear n that order. In addton an addtonal operatng mode n whch none of the dodes conduct appears n the conducton dagram as shown n Fg.3 (b). urng these perods the output capactor dscharges through the load. As the capactor voltage decreases ts voltage becomes equal to the ncomng lne voltage. At ths nstant the approprate dodes from both the top and the bottom group starts conductng and contnuous to do so tll the sum of the capactor chargng current and the load current becomes zero. Fg.3: Operaton of the three phase full wave uncontrolled brdge rectfer (a) crcut dagram, (b) conducton table, (c) wave forms
11 In the nterval α ωt β (79) (80) (8) (82) Where At ωt = β, = 0 (83) n the nterval β ωt α + π/3 (84) at ωt = α + π/3 (85) (86) Also at ωt = α + π/3
12 (87) From whch the value of α can be found. Equaton 82 gves the expresson of the output current of the rectfer. It s observed that s dscontnuous and contans large rpple. Ths s a major dsadvantage of ths converter. Ths rpple s also reflected n the nput current of the rectfer as shown n Fg 3 (b). However, the dsplacement factor of the converter stll remans unty. The current can be made contnuous by connectng an nductor of approprate value between the rectfer and the capactor. Analyss of such a converter s smlar to a converter supplyng an R L E load where the value of E s. Exercse 3 Fll n the blank(s) wth the approprate word(s) ) A three phase full wave uncontrolled rectfer supplyng a capactve load can operate n the conducton mode. ) The output rpple factor of a three phase full wave uncontrolled rectfer supplyng a capactve load s very low. ) The output rpple factor of a three phase full wave uncontrolled rectfer supplyng a capactve load s very hgh. v) The nput current dsplacement factor of a three phase full wave uncontrolled rectfer supplyng a capactve load s. v) The nput current dstorton factor of a three phase full wave uncontrolled rectfer supplyng a capactve load s very. Answers: () dscontnuous; () voltage; () current; (v) unty; (v) hgh. 2. A three phase full wave rectfer operates from 220 volts, three phase 50 Hz supply and supples a capactve resstve load of 20 Amps. An nductor of neglgble resstance s nserted between the rectfer and the capactor. Assumng the capactor to be large enough so that the output voltage s almost rpple free. Calculate the value of the nductor so that the rectfer output current s contnuous. Answers: The followng fgure shows the crcut arrangement and the correspondng waveforms.
13 Snce the conducton s contnuous and or θ = 72.73º In the nterval Snce v 0 s almost rpple free v 0 = V 0 = Now
14 or For just contnuous conducton L = 0 at ωt = θ or ωl =.0306 Ω or L = 3.28 mh. Practce Problems and Answers Q. A three phase half wave rectfer operates from a three phase 220 V, 50 Hz supply and supples a resstve load rated at 200 Volts KW through an nductance large enough to make the load current rpple free. Fnd out the power consumed by the load? What wll be the load power f the nductor s shorted? Q2. A three phase full wave rectfer operates from a three phase 220 V 50 Hz supply through a three phase Δ/Y transformer and supples a 200 V 500 R.P.M, 50 Amps separately excted dc motor. Fnd out the turns rato of the transformer so that the motor operates at rated speed at full load. If the motor armature resstance s 0.5 Ω fnd out the nductance to be connected n seres wth the motor such that the rectfer operates n the contnuous conducton mode at 50 % of the full load torque. Q3. A three phase full wave rectfer supples a resstve capactve load of 50 Amps from a 220 V. 3 phase 50 Hz supply. Fnd out the value of the load capactance such that the load voltage rpple s less than 5 %. Answers to practce problems. Snce the load current s rpple free the power consumed by the load wll be Now
15 When the nductor s shorted Now from Equ. (2.2) 2. To run at rated speed at full load the motor termnal voltage must be 200 volts. V L = 48. volts Where V L s the secondary lne voltage. Secondary s star connected. So secondary phase voltage Prmary s delta connected. So prmary phase voltage V = 220 V Requred turns rato = At 50% of the full load torque motor current s 25 Amps back Emf = = 87.5 Volts. speed at 50% of full load torque =. At 50% of full load torque the motor operates n the contnuous conducton mode, wth reference to Fg. 2.2 and equaton 2.9.
16 Where θ = 69.64º =.254 rad. At the juncton of contnuous and dscontnuous conducton OR OR Solvng whch φ = 34.65º. ωl = Ω or L =. mh. 3. Assumng lnear rpple. From Fg. 2.3, V 0Mn = Volts V 0AV = V. I 0AV = 50 Amps R = Ω. From Fg V 0Mn occurs at ωt = α V 0Mn =
17 sn α = or α = 72º But from Equaton (2.28) where from whch φ = 3.5º tanφ = 0.066, R = Ω ωc = , C = 8575 μf.
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