Module 2. AC to DC Converters. Version 2 EE IIT, Kharagpur 1

Transcription

1 Module 2 AC to DC Converters erson 2 EE IIT, Kharagpur 1

2 Lesson 1 Sngle Phase Fully Controlled Rectfer erson 2 EE IIT, Kharagpur 2

3 Operaton and Analyss of sngle phase fully controlled converter. Instructonal Objectves On completon the student wll be able to Dfferentate between the constructonal and operaton features of uncontrolled and controlled converters Draw the waveforms and calculate ther average and RMS values of dfferent varables assocated wth a sngle phase fully controlled half wave converter. Explan the operatng prncple of a sngle phase fully controlled brdge converter. Identfy the mode of operaton of the converter (contnuous or dscontnuous) for a gven load parameters and frng angle. Analyze the converter operaton n both contnuous and dscontnuous conducton mode and there by fnd out the average and RMS values of nput/output, voltage/currents. Explan the operaton of the converter n the nverter mode. erson 2 EE IIT, Kharagpur 3

4 1.1 Introducton Sngle phase uncontrolled rectfers are extensvely used n a number of power electronc based converters. In most cases they are used to provde an ntermedate unregulated dc voltage source whch s further processed to obtan a regulated dc or ac output. They have, n general, been proved to be effcent and robust power stages. However, they suffer from a few dsadvantages. The man among them s ther nablty to control the output dc voltage / current magntude when the nput ac voltage and load parameters reman fxed. They are also undrectonal n the sense that they allow electrcal power to flow from the ac sde to the dc sde only. These two dsadvantages are the drect consequences of usng power dodes n these converters whch can block voltage only n one drecton. As wll be shown n ths module, these two dsadvantages are overcome f the dodes are replaced by thyrstors, the resultng converters are called fully controlled converters. Thyrstors are semcontrolled devces whch can be turned ON by applyng a current pulse at ts gate termnal at a desred nstance. However, they cannot be turned off from the gate termnals. Therefore, the fully controlled converter contnues to exhbt load dependent output voltage / current waveforms as n the case of ther uncontrolled counterpart. However, snce the thyrstor can block forward voltage, the output voltage / current magntude can be controlled by controllng the turn on nstants of the thyrstors. Workng prncple of thyrstors based sngle phase fully controlled converters wll be explaned frst n the case of a sngle thyrstor halfwave rectfer crcut supplyng an R or R-L load. However, such converters are rarely used n practce. Full brdge s the most popular confguraton used wth sngle phase fully controlled rectfers. Analyss and performance of ths rectfer supplyng an R-L-E load (whch may represent a dc motor) wll be studed n detal n ths lesson. erson 2 EE IIT, Kharagpur 4

5 1.2 Sngle phase fully controlled halfwave rectfer Resstve load Fg.1. 1(a) shows the crcut dagram of a sngle phase fully controlled halfwave rectfer supplyng a purely resstve load. At ωt = when the nput supply voltage becomes postve the thyrstor T becomes forward based. However, unlke a dode, t does not turn ON tll a gate pulse s appled at ωt = α. Durng the perod < ωt α, the thyrstor blocks the supply voltage and the load voltage remans zero as shown n fg 1.1(b). Consequently, no load current flows durng ths nterval. As soon as a gate pulse s appled to the thyrstor at ωt = α t turns ON. The voltage across the thyrstor collapses to almost zero and the full supply voltage appears across the load. From ths pont onwards the load voltage follows the supply voltage. The load beng purely resstve the load current o s proportonal to the load voltage. At ωt = π as the supply voltage passes through the negatve gong zero crossng the load voltage and hence the load current becomes zero and tres to reverse drecton. In the process the thyrstor undergoes reverse recovery and starts blockng the negatve supply voltage. Therefore, the load voltage and the load current remans clamped at zero tll the thyrstor s fred agan at ωt = 2π + α. The same process repeats there after. From the dscusson above and Fg 1.1 (b) one can wrte For α < ωt π v = v = 2 snωt (1.1) v = = 2 snωt (1.2) R R erson 2 EE IIT, Kharagpur 5

6 Therefore v = = otherwse. 1 2π 1 π OA α (1.3) = v dωt = 2 snωt dωt 2π 2π Or OA = (1+cosα) 2π (1.4) 1 2π 2 ORMS = v dωt 2π (1.5) 1 π 2 2 = 2v α sn ωtdωt 2π 2 π = (1-cos2ωt)dωt 2π α 2 sn2α = π - α + 2π α sn2α = ORMS FF O = = OA π 2π α sn2α π 1- + π 2π (1.6) (1+cosα) Smlar calculaton can be done for. In partculars for pure resstve loads FF o = FF vo Resstve-Inductve load Fg 1.2 (a) and (b) shows the crcut dagram and the waveforms of a sngle phase fully controlled halfwave rectfer supplyng a resstve nductve load. Although ths crcut s hardly used n practce ts analyss does provde useful nsght nto the operaton of fully controlled rectfers whch wll help to apprecate the operaton of sngle phase brdge converters to be dscussed later. 1 2 erson 2 EE IIT, Kharagpur 6

7 As n the case of a resstve load, the thyrstor T becomes forward based when the supply voltage becomes postve at ωt =. However, t does not start conducton untl a gate pulse s appled at ωt = α. As the thyrstor turns ON at ωt = α the nput voltage appears across the load and the load current starts buldng up. However, unlke a resstve load, the load current does not become zero at ωt = π, nstead t contnues to flow through the thyrstor and the negatve supply voltage appears across the load forcng the load current to decrease. Fnally, at ωt = β (β > π) the load current becomes zero and the thyrstor undergoes reverse recovery. From ths pont onwards the thyrstor starts blockng the supply voltage and the load voltage remans zero untl the thyrstor s turned ON agan n the next cycle. It s to be noted that the value of β depends on the load parameters. Therefore, unlke the resstve load the average and RMS output voltage depends on the load parameters. Snce the thyrstors does not conduct over the entre nput supply cycle ths mode of operaton s called the dscontnuous conducton mode. From above dscusson one can wrte. For α ωt β v = v = 2 snωt (1.7) v = otherwse Therefore 1 2π OA = vdωt 2π (1.8) 1 β = 2 snωt dωt 2π α erson 2 EE IIT, Kharagpur 7

8 = (cosα -cosβ) 2π 1 2π 2 ORMS = v dωt 2π (1.9) 1 β 2 2 = 2v α sn ωt dωt 2π β - α sn2α -sn2β = + 2 π 2π OA I OA = = (cosα -cosβ) R 2πR Snce the average voltage drop across the nductor s zero. 1 2 (1.1) However, I ORMS can not be obtaned from ORMS drectly. For that a closed from expresson for wll be requred. The value of β n terms of the crcut parameters can also be found from the expresson of. For α ωt β do R o + L = v = 2snωt dt The general soluton of whch s gven by Where - (ωt-α) tan ϕ 2 =I e + sn(ωt- ϕ ) (1.12) Z ωl 2 2 tanφ = and Z= R +ω L 2 R (1.11) = ωt=α 2 = Z sn(φ - α)e = otherwse. 2 =I + sn(α - φ) Z ( ωt-α) - tanφ + sn(ωt-φ) (1.13) Equaton (1.13) can be used to fnd out I ORMS. To fnd out β t s noted that = ωt=β α-β tanφ sn(φ - α)e = sn(φ - β) (1.14) Exercse 1.1 Equaton (1.14) can be solved to fnd out β Fll n the blank(s) wth approprate word(s) erson 2 EE IIT, Kharagpur 8

9 ) In a sngle phase fully controlled converter the of an uncontrolled converters are replaced by. ) In a fully controlled converter the load voltage s controlled by controllng the of the converter. ) A sngle phase half wave controlled converter always operates n the conducton mode. v) The voltage form factor of a sngle phase fully controlled half wave converter wth a resstve nductve load s compared to the same converter wth a resstve load. v) The load current form factor of a sngle phase fully controlled half wave converter wth a resstve nductve load s compared to the same converter wth a resstve load. Answers: () dodes, thyrstors; () frng angle; () dscontnuous (v) poorer; (v) better. 2) Explan qualtatvely, what wll happen f a free-wheelng dode(cathode of the dode shorted wth the cathode of the thyrstor) s connected across the load n Fg 1.2.(a) Answer: Referrng to Fg 1.2(b), the free wheelng dode wll reman off tll ωt = π snce the postve load voltage across the load wll reverse bas the dode. However, beyond ths pont as the load voltage tends to become negatve the free wheelng dode comes nto conducton. The load voltage s clamped to zero there after. As a result ) Average load voltage ncreases ) RMS load voltage reduces and hence the load voltage form factor reduces. ) Conducton angle of load current ncreases as does ts average value. The load current rpple factor reduces. erson 2 EE IIT, Kharagpur 9

10 1.3 Sngle phase fully controlled brdge converter Fg 1.3 (a) shows the crcut dagram of a sngle phase fully controlled brdge converter. It s one of the most popular converter crcuts and s wdely used n the speed control of separately excted dc machnes. Indeed, the R L E load shown n ths fgure may represent the electrcal equvalent crcut of a separately excted dc motor. The sngle phase fully controlled brdge converter s obtaned by replacng all the dode of the correspondng uncontrolled converter by thyrstors. Thyrstors T 1 and T 2 are fred together whle T 3 and T 4 are fred 18º after T 1 and T 2. From the crcut dagram of Fg 1.3(a) t s clear that for any load current to flow at least one thyrstor from the top group (T 1, T 3 ) and one thyrstor from the bottom group (T 2, T 4 ) must conduct. It can also be argued that nether T 1 T 3 nor T 2 T 4 can conduct smultaneously. For example whenever T 3 and T 4 are n the forward blockng state and a gate pulse s appled to them, they turn ON and at the same tme a negatve voltage s appled across T 1 and T 2 commutatng them mmedately. Smlar argument holds for T 1 and T 2. For the same reason T 1 T 4 or T 2 T 3 can not conduct smultaneously. Therefore, the only possble conducton modes when the current can flow are T 1 T 2 and T 3 T 4. Of coarse t s possble that at a gven moment none of the thyrstors conduct. Ths stuaton wll typcally occur when the load current becomes zero n between the frngs of T 1 T 2 and T 3 T 4. Once the load current becomes zero all thyrstors reman off. In ths mode the load current remans zero. Consequently the converter s sad to be operatng n the dscontnuous conducton mode. Fg 1.3(b) shows the voltage across dfferent devces and the dc output voltage durng each of these conducton modes. It s to be noted that whenever T 1 and T 2 conducts, the voltage across T 3 and T 4 becomes v. Therefore T 3 and T 4 can be fred only when v s negatve.e, over the negatve half cycle of the nput supply voltage. Smlarly T 1 and T 2 can be fred only over the postve half cycle of the nput supply. The voltage across the devces when none of the thyrstors conduct depends on the off state mpedance of each devce. The values lsted n Fg 1.3 (b) assume dentcal devces. Under normal operatng condton of the converter the load current may or may not reman zero over some nterval of the nput voltage cycle. If s always greater than zero then the converter s sad to be operatng n the contnuous conducton mode. In ths mode of operaton of the converter T 1 T 2 and T 3 T 4 conducts for alternate half cycle of the nput supply. erson 2 EE IIT, Kharagpur 1

11 However, n the dscontnuous conducton mode none of the thyrstors conduct over some porton of the nput cycle. The load current remans zero durng that perod Operaton n the contnuous conducton mode As has been explaned earler n the contnuous conducton mode of operaton never becomes zero, therefore, ether T 1 T 2 or T 3 T 4 conducts. Fg 1.4 shows the waveforms of dfferent varables n the steady state. The frng angle of the converter s α. The angle θ s gven by E snθ = 2 (1.15) 1 It s assumed that at t = - T 3 T 4 was conductng. As T 1 T 2 are fred at ωt = α they turn on commutatng T 3 T 4 mmedately. T 3 T 4 are agan fred at ωt = π + α. Tll ths pont T 1 T 2 conducts. The perod of conducton of dfferent thyrstors are pctorally depcted n the second waveform (also called the conducton dagram) of Fg 1.4. erson 2 EE IIT, Kharagpur 11

12 erson 2 EE IIT, Kharagpur 12

13 The dc lnk voltage waveform shown next follows from ths conducton dagram and the conducton table shown n Fg 1.3(b). It s observed that the emf source E s greater than the dc lnk voltage tll ωt = α. Therefore, the load current contnues to fall tll ths pont. However, as T 1 T 2 are fred at ths pont v becomes greater than E and starts ncreasng through R-L and E. At ωt = π θ v agan equals E. Dependng upon the load crcut parameters o reaches ts maxmum at around ths pont and starts fallng afterwards. Contnuous conducton mode wll be possble only f remans greater than zero tll T 3 T 4 are fred at ωt = π + α where upon the same process repeats. The resultng waveform s shown below v. The nput ac current waveform s obtaned from by notng that whenever T 1 T 2 conducts = and = - whenever T 3 T 4 conducts. The last waveform shows the typcal voltage waveform across the thyrstor T 1. It s to be noted that when the thyrstor turns off at ωt = π + α a negatve voltage s appled across t for a duraton of π α. The thyrstor must turn off durng ths nterval for successful operaton of the converter. It s noted that the dc voltage waveform s perodc over half the nput cycle. Therefore, t can be expressed n a Fourer seres as follows. Where α [ ] (1.16) v = + v cos2nωt + v sn2nωt OA an bn n=1 1 π+α 2 2 OA = v α dωt = cosα π (1.17) π 2 π 2 2 cos(2n +1)α cos(2n -1)α v an = v cos2nωt dωt = - π π 2n +1 2n -1 (1.18) 2 π 2 2 sn(2n +1)α sn(2n -1)α v bn = v sn2nωt dωt = - π π 2n +1 2n -1 (1.19) Therefore the RMS value of the nth harmonc OnRMS = v an +vbn 2 (1.2) RMS value of v can of course be completed drectly from. 1 π+α 2 ORMS = v α dωt = π (1.21) Fourer seres expresson of v s mportant because t provdes a smple method of estmatng ndvdual and total RMS harmonc current njected nto the load as follows: The mpedance offered by the load at nth harmonc frequency s gven by 2 Z = R +(2nωL) 2 (1.22) n 1 α 2 onrms 2 onrms OHRMS onrms Z n n=1 I = ; I = I (1.23) From (1.18) (1.23) t can be argued that n an nductve crcut I onrms as fast as 1/n 2. So n practce t wll be suffcent to consder only frst few harmoncs to obtan a reasonably accurate estmate of I OHRMS form equaton Ths method wll be useful, for example, whle calculatng the requred current deratng of a dc motor to be used wth such a converter. erson 2 EE IIT, Kharagpur 13

14 However to obtan the current ratng of the devce to be used t s necessary to fnd out a closed form expresson of. Ths wll also help to establsh the condton under whch the converter wll operate n the contnuous conducton mode. To begn wth we observe that the voltage waveform and hence the current waveform s perodc over an nterval π. Therefore, fndng out an expresson for over any nterval of length π wll be suffcent. We choose the nterval α ωt π + α. In ths nterval d L +R +E= 2 snωt dt The general soluton of whch s gven by (1.24) ( ωt-α) - tanφ 2 snθ = Ie + sn(ωt-φ)- cosφ Z (1.25) ωl Where, Z= R +ω L; tanφ = ; E= 2snθ; R = Zcosφ R Now at steady state = snce ωt=α s perodc over the chosen nterval. Usng ths ωt=π+α boundary condton we obtan ( ωt-α) - 2 2sn(φ - α) tanφ snθ e + sn(ωt-φ) - = π (1.26) Z - tanφ cosφ 1-e The nput current s related to as follows: = for α ωt π + α (1.27) = - otherwse. Fg 1.5 shows the waveform of n relaton to the v waveform. erson 2 EE IIT, Kharagpur 14

15 It wll be of nterest to fnd out a Fourer seres expresson of. However, usng actual expresson for wll lead to exceedngly complex calculaton. Sgnfcant smplfcaton can be made by replacng wth ts average value I. Ths wll be justfed provded the load s hghly nductve and the rpple on s neglgble compared to I. Under ths assumpton the dealzed waveform of becomes a square wave wth transtons at ωt = α and ωt = α + π as shown n Fg s the fundamental component of ths dealzed. Evdently the nput current dsplacement factor defned as the cosne of the angle between nput voltage (v ) and the fundamental component of nput current ( 1 ) waveforms s cosα (laggng). It can be shown that 2 2 I 1RMS = I (1.28) π erson 2 EE IIT, Kharagpur 15

16 and I RMS =I (1.29) I 2 2 Therefore the nput current dstorton factor = 1RMS = (1.3) I π The nput power factor = RMS Actual Power I 1RMScosα = Apparent Power I 2 2 = cosα π RMS (laggng) (1.31) Therefore, the rectfer appears as a laggng power factor load to the nput ac system. Larger the α poorer s the power factor. The nput current also contan sgnfcant amount of harmonc current (3 rd, 5 th, etc) and therefore appears as a harmonc source to the utlty. Exact composton of the harmonc currents can be obtaned by Fourer seres analyss of and s left as an exercse. Exercse 1.2 Fll n the blank(s) wth the approprate word(s). ) A sngle phase fully controlled brdge converter can operate ether n the or conducton mode. ) ) v) In the contnuous conducton mode at least thyrstors conduct at all tmes. In the contnuous conducton mode the output voltage waveform does not depend on the parameters. The mnmum frequency of the output voltage harmonc n a sngle phase fully controlled brdge converter s the nput supply frequency. v) The nput dsplacement factor of a sngle phase fully controlled brdge converter n the contnuous conducton mode s equal to the cosne of the angle. Answer: () contnuous, dscontnuous; () two; () load; (v) twce; (v) frng. 2. A sngle phase fully controlled brdge converter operates n the contnuous conducton mode from a 23, 5HZ sngle phase supply wth a frng angle α = 3. The load resstance and nductances are 1Ω and 5mH respectvely. Fnd out the 6 th harmonc load current as a percentage of the average load current. Answer: The average dc output voltage s 2 2 OA = cosα = olts π OA Average output load current = = Amps R L From equaton (1.18) a3 = 1.25 olts From equaton (1.19) b3 = 35.5 olts RMS = olts, Z 3 = R L + (6 2 π ) = ohms erson 2 EE IIT, Kharagpur 16

17 . 3RMS I 3RMS = =.2756 Amps = 1.54% of IOA Z Operaton n the dscontnuous conducton mode So far we have assumed that the converter operates n contnuous conducton mode wthout payng attenton to the load condton requred for t. In fgure 1.4 the voltage across the R and L component of the load s negatve n the regon π - θ ωt π + α. Therefore contnues to decrease tll a new par of thyrstor s fred at ωt = π + α. Now f the value of R, L and E are such that becomes zero before ωt = π + α the conducton becomes dscontnuous. Obvously then, at the boundary between contnuous and dscontnuous conducton the mnmum value of whch occurs at ωt = α and ωt = π + α wll be zero. Puttng ths condton n (1.26) we obtan the condton for contnuous conducton as. 2sn(φ - α) snθ - sn(φ - α) - (1.32) π - tanφ cosφ 1-e erson 2 EE IIT, Kharagpur 17

18 Fg 1.6 shows waveforms of dfferent varables on the boundary between contnuous and dscontnuous conducton modes and n the dscontnuous conducton mode. It should be stressed that on the boundary between contnuous and dscontnuous conducton modes the load current s stll contnuous. Therefore, all the analyss of contnuous conducton mode apples to ths case as well. However n the dscontnuous conducton mode remans zero for certan nterval. Durng ths nterval none of the thyrstors conduct. These ntervals are shown by hatched lnes n the conducton dagram of Fg 1.6(b). In ths conducton mode starts rsng from zero as T 1 T 2 are fred at ωt = α. The load current contnues to ncrease tll ωt = π θ. After ths, the output voltage v falls below the emf E and decreases tll ωt = β when t becomes zero. Snce the thyrstors cannot conduct current n the reverse drecton remans at zero tll ωt = π + α when T 3 and T 4 are fred. Durng the perod β ωt π + α none of the thyrstors conduct. Durng ths perod v attans the value E. Performance of the rectfer such as OA, ORMS, I OA, I ORMS etc can be found n terms of α, β and θ. For example erson 2 EE IIT, Kharagpur 18

19 1 π+α 1 β π+α OA = v dωt = 2 α α snωt dωt+ 2 β snθ dωt π π (1.33) Or 2 OA = [ cosα -cosβ +snθ(π + α -β)] π (1.34) OA -E OA - 2snθ I OA = = R Zcosφ (1.35) Or 2 I OA = [ cosα -cosβ +snθ(α -β)] π Zcosφ (1.36) It s observed that the performance of the converter s strongly affected by the value of β. The value of β n terms of the load parameters (.e, θ, φ and Z) and α can be found as follows. In the nterval α ωt β do L +R o +E= 2snωt dt = ωt=α From whch the soluton of can be wrtten as ( ωt-α) 2 - tanφ snθ ( ωt-α) - = sn(φ - α)e - { tanφ } + sn(ωt-φ) Z cosφ 1-e Now = ωt=β α-β tanφ snθ α-β tanφ sn(φ - α)e - + sn(β - φ)= cosφ 1-e Gven φ, α and θ, the value of β can be found by solvng equaton (1.37) (1.38) (1.39) Inverter Mode of operaton The expresson for average dc voltage from a sngle phase fully controlled converter n contnuous conducton mode was 2 2 = cosα (1.4) π For α < π/2, d >. Snce the thyrstors conducts current only n one drecton I > always. Therefore power flowng to the dc sde P = I > for α < π/2. However for α > π/2, <. Hence P <. Ths may be nterpreted as the load sde gvng power back to the ac sde and the converter n ths case operate as a lne commutated current source nverter. So t may be temptng to conclude that the same converter crcut may be operated as an nverter by just ncreasng α beyond π/2. Ths mght have been true had t been possble to mantan contnuous conducton for α < π/2 wthout makng any modfcaton to the converter or load connecton. To supply power, the load EMF source can be utlzed. However the connecton of ths source n Fg 1.3 s such that t can only absorb power but can not supply t. In fact, f an attempt s made to supply power to the ac sde (by makng α > π/2) the energy stored n the load nductor wll be exhausted and the current wll become dscontnuous as shown n Fg 1.7 (a). erson 2 EE IIT, Kharagpur 19

20 Therefore for sustaned nverter mode of operaton the connecton of E must be reversed as shown n Fg 1.7(b). Fg 1.8 (a) and (b) below shows the waveforms of the nverter operatng n contnuous conducton mode and dscontnuous conducton mode respectvely. Analyss of the converter remans unaltered from the rectfer mode of operaton provded θ s defned as shown. erson 2 EE IIT, Kharagpur 2

21 Exercse 1.3 Fll n the blank(s) wth the approprate word(s) ) In the dscontnuous conducton mode the load current remans for a part of the nput cycle. ) ) For the same frng angle the load voltage n the dscontnuous conducton mode s compared to the contnuous conducton mode of operaton. The load current rpple factor n the contnuous conducton mode s compared to the dscontnuous conducton mode. erson 2 EE IIT, Kharagpur 21

22 v) In the nverter mode of operaton electrcal power flows from the sde to the sde. v) In the contnuous conducton mode f the frng angle of the converter s ncreased beyond degrees the converter operates n the mode. Answers: () zero; () hgher; () lower; (v) dc, ac; (v) 9, nverter. 2. A 22, 2A, 15 RPM separately excted dc motor has an armature resstance of.75ω and nductance of 5mH. The motor s suppled from a 23, 5Hz, sngle phase supply through a fully controlled brdge converter. Fnd the no load speed of the motor and the speed of the motor at the boundary between contnuous and dscontnuous modes when α = 25. Answer: At no load the average motor torque and hence the average motor armature current s zero. However, snce a converter carres only undrectonal current, zero average armature current mples that the armature current s zero at all tme. From Fg 1.6(b) ths stuaton can occur only when θ = π/2,.e the back emf s equal to the peak of the supply voltage. Therefore, E b = 2 23 = , Under rated condton E no load b = N no load = 15 = 238 RPM 25 At the boundary between contnuous and dscontnuous conducton modes from equaton π/tanφ 1+e snθ =cosφsn( φ - α) 1-e -π/tanφ From the gven data φ = 87.27, α = 25 snθ =.5632 E b = 2 snθ = olts Motor speed N = 15 = 134 RPM. 25 Summary Sngle phase fully controlled converters are obtaned by replacng the dodes of an uncontrolled converter wth thyrstors. In a fully controlled converter the output voltage can be controlled by controllng the frng delay angle (α) of the thyrstors. Sngle phase fully controlled half wave converters always operate n the dscontnuous conducton mode. Half wave controlled converters usually have poorer output voltage form factor compared to uncontrolled converter. Sngle phase fully controlled brdge converters are extensvely used for small dc motor drves. erson 2 EE IIT, Kharagpur 22

23 Dependng on the load condton and the frng angle a fully controlled brdge converter can operate ether n the contnuous conducton mode or n the dscontnuous conducton mode. In the contnuous conducton mode the load voltage depends only on the frng angle and not on load parameters. In the dscontnuous conducton mode the output voltage decreases wth ncreasng load current. However the output voltage s always greater than that n the contnuous conducton mode for the same frng angle. The fully controlled brdge converter can operate as an nverter provded () α > π 2, () a dc power source of sutable polarty exsts on the load sde. References 1) Power Electroncs P.C.Sen; Tata McGraw-Hll ) Power Electroncs; Crcuts, Devces and Applcatons, Second Edton, Muhammad H.Rashd; Prentce-Hall of Inda; ) Power Electroncs; Converters, Applcatons and Desgn Thrd Edton, Mohan, Undeland, Robbns, John Wleys and Sons Inc, 23. Practce Problems and Answers Q1. Is t possble to operate a sngle phase fully controlled half wave converter n the nvertng mode? Explan. Q2. A 22, 2A 15 RPM separately excted dc motor has an armature resstance of.75ω and nductance of 5 mh. The motor s suppled from a sngle phase fully controlled converter operatng from a 23, 5 Hz, sngle phase supply wth a frng angle of α = 3. At what speed the motor wll supply full load torque. Wll the conducton be contnuous under ths condton? Q3. The speed of the dc motor n queston Q2 s controlled by varyng the frng angle of the converter whle the load torque s mantaned constant at the rated value. Fnd the power factor of the converter as a functon of the motor speed. Assume contnuous conducton and rpple free armature current. Q4. Fnd the load torque at whch the dc motor of Q2 wll operate at 2 RPM wth the feld current and α remanng same. Q5. A separately excted dc motor s beng braked by a sngle phase fully controlled brdge converter operatng n the nverter mode as shown n Fg 1.7 (b). Explan what wll happen f a commutaton falure occurs n any one of the thyrstors. erson 2 EE IIT, Kharagpur 23

24 Answers 1. As explaned n secton 1.3.3, the load crcut must contan a voltage source of proper polarty. Such a load crcut and the assocated waveforms are shown n the fgure next. Fgure shows that t s ndeed possble for the half wave converter to operate n the nvertng mode for some values of the frng angle. However, care should be taken such that becomes zero before v exceeds E n the negatve half cycle. Otherwse wll start ncreasng agan and the thyrstor T wll fal to commutate. 2. For the machne to delver full load torque wth rated feld the armature current should be 2 Amps o Assumng contnuous conducton v = cos3 = volts. π For 2 Amps armature current to flow the back emf wll be E b = a I a R a = = volts E snθ = b = erson 2 EE IIT, Kharagpur 24

25 ωla o For the gven machne, tan φ= = 2.944, φ = R a Now from equaton (1.32) 2sn(φ - α) -π/tanφ -sn(φ - α) = e and snθ = cosφ the conducton s contnuous. At 15 RPM the back emf s = 25 volts. The speed at whch the machne delvers rated torques s N r = 15 = 122 RPM To mantan constant load torque equal to the rated value the armature voltage should be N a = ra I a +E rated b rated Nrated N = =.137 N In a fully controlled converter operatng n the contnuous conducton mode 2 2 a = cosα = cosα π - 4 cosα = N Now the power factor from equaton 1.31 s pf = cosα = N π Ths gves the nput power factor as a functon of speed At 2 RPM, E b = 25 = volts 15 Eb o o snθ = =.84, φ = , α =3 2 From equaton 1.32 t can be shown that the conducton wll be dscontnuous. Now from equaton 1.39 α - β tanφ snθ snθ sn ( α - φ) -( α-β) + e +sn( φ - α) = cosφ cosφ.477(α - β) o or [ ] ( α -β) e sn = ( α -β) Solvng whch β (α - β) o e - sn = erson 2 EE IIT, Kharagpur 25

26 from equaton I oav = [ cosα -cosβ +snθ(α -β)] = Amps πzcosφ the load torque should be = 13.38% 2 of the full load torque. 5. Referrng to Fg 1.8 (a) let there be a commutaton falure of T 1 at ωt = α. In that case the conducton mode wll be T 3 T 2 nstead of T 1 T 2 and v wll be zero durng that perod. As a result average value of wll be less negatve and the average armature current wll ncrease. However the converter wll contnue to operate n the nverter mode and the motor wll be braked. erson 2 EE IIT, Kharagpur 26