Lesson 10 SCR Satu Phasa0001

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Lesson

10Single PhaseControlled Rectifier

Version2 EE IIT, Kharagpur



Operation and Analysis of single phase ully controlled converter.

On completion he studentwill be able o

r Differentiate between the constructional and operation features of uncontrolled and

controlled eonvertefs

r Draw the waveforms and calculatetheir averageand RMS values of different variables

associatedwith a singlephase ully controlledhalf wave converter.

o Explain the operatingprinciple of a singlephase ully contolled bridge converter.

r Identifr the modeof operationof the converter(continuousor discontinuous) or a given

loadparameters nd firing angle.

o Analyze he converter operation n both continuousand discontinuousconductionmodeand here by find out the averageand RMS valuesof inpuVoutput,voltage/currents.

r Explain the operationof the converter n the invertermode.

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Single phaseuncontrolled rectifiers af,eextensivelyused n a numberof power electronic basedconverters. In mostcases hey are used o provide an intermediateunregulateddc voltagesource

which is further processedo obtain a regulateddc or ac output. They have, in general,beenprovedto be efficient and robustpower stages.However,they suffer from a few disadvantages.The main among hem is their inability to confrol the output dc voltage current magnitudewhenthe input ac voltage and load parametersemain fixed. They are alsounidirectional in the sensethat they allow electrical power to flow from the ac side to the dc side only. These twodisadvantages re the direct consequences f using power diodes n these converterswhich canblock voltage only in one direction. As will be shown in this module, these wo disadvantagesare overcome f the diodes are replaced by thyristors, the resulting convertersare called fullyconffolled converters.Thyristors are semicontrolleddeviceswhich can be turnedON by applying a currentpulseat itsgate erminal at a desired nstance. However, they oannotbe turnedoff from the gate erminals.

Thereforg the fully controlled converter continues to exhibit load dependentoutput voltage /current waveformsas in the caseof their uncontrolledcounterpart. However, since he thyristorcan block fonvard voltage, the output voltage / current magnitude can be controlled bycontrolling the turn on instantsof the thyristors. Working principle of thyristors basedsinglephase ully controlled converterswill be explained irst in the caseof a singlethyristor halfwaverectifier circuit supplying an R or R-L load. However, such converters me rarely used inpractice.

Full bridge is the most popular configuration used with single phase ully controlled rectifiers.Analysis and performanceof this rectifier supplying an R-L-E load (which may representa dcmotor) will be studied n detail in this lesson.

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t '

. :0 , * i r r , ' , J

t

rr '.{*

iut { i rsuitdi ;rgrrur[} | 1l'rr 'l irrnr"

I h1

[:is. ll l.l: Singlr1rhl.r fullr u*rrttrrllnl r*lf tr*t rr'r;li l irr rlrl]l.ring rrrirlirr'ftrarl

Fig.lO. 1(a) shows the circuit diagram of a single phasefully controlled halfirave rectifiersupplyinga purelyrpsistive oad. At o:t:0 when he input supplyvoltagebecomes ositive he

thyristor T becomes orward biased. However, unlike a diode, it does not turn ON till a gate

pulse is applied at rot - o. During the period 0 < art< q the thyristor blocks the supply voltage

and the load voltageremainszeroas shown n fig 10.1(b). Consequentlyoo load current flowsduring this interval. As soon asa gate pulse s appliedto the thyristor at cot: o it tums ON. The

voltage across he thyristor collapses o almost zero and the full supply voltage appearsacrossthe load. From this point onwardsthe load voltage follows the supply voltage. The load being

purely resistive the load current lo is proportional to the load voltage. At cot= 7r as the supplyvoltage passes hrouglr the negative going zero crossing the load voltage and hencethe loadcurrentbecomes eroand ries to reversedirection. In the process he thyristor undergoesevers€

rccoveryand startsblocking the negative supply voltage.Therefore, he loadvoltageand the load

currentremainsclamped atzero till the thyristor is fired again at rnt= 2n + CI,.The sameprocess

repeats here after.From hediscussion boveand Fig 10.1 b) onecanwrit€

For o<o)t<rr

vo=\ =€Y sinrot

i^:b=.Elr inoot"RR

(10.1)

(10.2)




vo=io=0 other"wise.

Therefore voou*t"

u,orr:jtJir,sinrotdart

0r Voou= v,^!2n(l + cosa)

Vo*rs =

=

V, (- a sin2a\z= - ; l l - - + - IJ2 \ n 2n )

(10.3)

(10.4)

(10.5)

(10.6)Fuo *:(l +cosa)

Similar calculationcanbe done or ro. In particulars or pureresistive oads FF;s= FFys.

- ; - : " l " - X - - - . - - ; , , , " - * : = = = - - . = - : = = . = - 1 * * ;

Fig 10.2(a) and (b) shows he circuit diagramand the waveformsof a singlephase ullycontrolled alfwave ectifiersupplyrng rcsistivenductiveoad. Although hiscircuit is hardlyused n practice ts analysisdoesprovideuseful nsight nto the operation f fully controlledrectifierswhich will help to appreciatehe operationof singlephasebridgeoonverterso bediscussedater.

'l["znlrin2cotdrrrt

2 n h

* fU'"rs2rrrt)dcot

rvtf sin2alnln-o* z l




J. .

{

1

I

. I t , ' r i t t ' r i

f

u r**

|!-*.

I sl {'iruuil rli;rgr:rru

thr! l 'nrrfurntr

Fig. *-!: Sirrglr: h:irrr irll;rcrurlr*lh.tl ;rlItrilr* rrcdlierrupplriug s rt.ri:rtilc ffduslirr t$:N!

As in the case of a rssistive load, the thyristor T becomes orward biased when the supply

voltage becomespositive at rrlt= 0. However, it doesnot start conduction until a gatepulse isapplied at ot: o. As the thyristor turns ON at eot= o the input voltageappearsacross he loadand the load current starts building up. However, unlike a resistive load, the load current doesnot become zera at 0)t = fi, instead t continues o flow ftrough the thyristor and the negativesupply voltage appearsacross he load forcing the load current to decrease.Finally, at rot: p (p> r) the load currentbecomes ero and he thyristor undergoes everse ecovery. From this pointonwards he thyristor starts blocking the supply voltage and the load voltage remains zero untilthe thyristor is turned OI.l again n the next cycle. It is to be noted hat the valueof p dependsonthe load parameters. Thereforeounlike the resistive load the averageand RMS output voltagedependson the load parameters. Since the thyristors does not conduct ovsr the entire inputsupply cycle this modeof operation s called the "discontinuousconductionmodeo'.

From abovediscussionone can unite.For usro t<p

t:vo :v i=J2Ys inco t

vo= 0 otherwise

(10.4

(10.8) f f,"uoa,ot

=*fJ:\ sinrotda:t

Therefore Voou

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* f"{oo'*

=;fi{"oro-cosB)

Since he average oltage drop across he inductor is zero.

Howevero orurasannot be obtained rom Vexl,as irectly. For that a closed rom expression or iowill be required. The valuE of p in terms of the circuit paraneters can also be found from theexpressionof is.

Vo*r, =

sin(g a)e*E = sin(q- F)

Equation 10.14)canbe solved o find out p

Exercise 10.1

Fill in the blank(s)with appropriateword(s)

(10.e)

(10.r0)

(10.1)

(10.12)

(10.13)

(10.14)

-Y(F-o,s in2a-s in2p) i

r - Voou- YIool,t

=ffi

(cosccosP)

For asa l tSp

ni"+r$ =vo .fzv,rinrtdt

Thegeneral olutionofwhich isgivenby

io=1os'# *4-Y ,in(or,-,P)

Where ang=S anOZ=

iol.r=o oli'

Q = 1 ^ a v z Y r s i n ( a - o )"z. JZV,I- t 't-o) Ito:

Z Lsin(g-a)e-*o sin(rot-g)Jio= 0 othenvise.

Equation10.13)anbeusedo findout onus.To indoutp t is notedhatiol*:u 0

a-B

! fz.,?.in2<rrtdrot2n J"

R2+cD2f




i)

ii)

In a singlephase ully controlled converter he _ of an unconfolled convertersarereplacedby _-In a fully controlled converterthe load voltage s confolled by controlling theofthe converter.A single phase half wave controlled converter always operates in theconductionmode.The voltage form factor of a single phasefully conholled half wave converter with aresistive inductive load is compared o the same converter with a resistiveload.

v) The oadcurrent orm factorof a singlephaseully controlled alf waveconverterwith aresistive nductive oad is comparedo the sameconverterwith a resistiveload.

Answers: i) diodes,hyristors; ii) firing angle; iii) discontinuousiv) poorer; v)better.

2) Explainqualitatively,what will happen f a free-wheeling iode(cathodef the diodeshortedwith the cathode f the hyristor) s connectedcrosshe oad n Fig 10.2.(a)

Answer: Refeningo Fig 10.2(b),he reewheeling iodewill remain fftill rot= n since hepositive oadvoltageacrosshe oadwill reverse ias he diode. Howeveroeyond hispointasthe loadvoltage endso become egativehe free wheelingdiodecomesntoconduction.Theloadvoltages clampedo zero hereafter. As aresult

D Averageoadvoltagencreasesii) RMS oadvoltageeduces ndhencehe oadvoltageorm factorreduces.iiD Conduction ngleof load currentncreasess does ts average alue. The oad

current pple actor educes.

iii)

iv)

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I

. , l l r in

r i l . r{h }

l"ig l{f-}: Slngtcpharr I'ullr c*ntrelledhridgc urrrmlcr ruppl.ring:rn t - L - F- uar}-iu t ( irsuit +li*grunr.hi { nnrlustitrn:rhlr

Fig 10.3(a) shows he circuit diagramof a single phase ully conholled bridge converter. It i$one of the mostpopular converter circuits and is widely used n the speedconfol of separatelyexcited dc machines. Indee4 the R-L-E load shown in this figure may represent he electricalequivalentcircuit of a separately xcited dc motor.

The single phase ully controlled bridge converter s obtained by replacing all the diodeof the correspondinguncontrolledconverterby thyristors. ThyristorsT1and T2 aft fired togetherwhile Tr and Ta are ired l80oafterTr andT2. Fromthe cirsuit diagramof Fig 10.3(a) t is clear

that for any load current to flow at least one thyristor from the top group (Tr, T:) and onethyristor from the bottom goup (Tz, T+) must conduct. It can also be argued hat neither TrT:nor TzT+ can conduct simultaneously. For example whenever T3 and Ta are in the forwardblocking stateanda gatepulse is applied to them, they turn ON andat the same ime a negativevoltage is applied acrossT1and T2 commutating hem immediately. Similar argumentholds forTr andTz.

For the same easonTrT+ or TzT: can not conduct simultaneously. Therefore, he onlypossible conduction modes when the current io can flow are T1T2 and T3Ta. Of coarse t ispossible that at a given moment none of the thyristors conduct. This situation will typioaltyoccur when the load current becomeszero in between he firings of T1T2and TrT+. Once theload current becomeszso all thyriston rcmain off. In this modethe load curent remains zero.

Consoquentlyhe converter s said o be operating n the discontinuousconductionmode.Fig 10.3(b) shows he voltage auoss different devicesand the dc output voltage during

eachof theseconductionmodes. It is to be noted hat wheneverT1and T2conducts, he voltageasrossT3andT+ becomesvi. ThereforeT3andT+can be fired only whenv1 s negative .eooverthe negative half cycle of the input supply voltage. Similarly Tr and T2 can be fired only overthe positive half cycle of the input supply. The voltage across he devices when none of thethyristors conduct dependson the off state mpedanceof eachdevice. The values isted in Fig10.3O) assumsdenticaldevices.

Under normal operating condition of the converter the load current may or may notremainzero over some nterval of the input voltage cycle. If io is alwaysgreater han zero thenthe converter is said to

be operatingin

the continuous conduction mode. In this mode ofoperation of the convefts TrTz and TgTaconducts or alternate half cycle of the input supply.

Iflt

I ' l't": 'r,

l' . 1

T'r . it 0 - t , l t

-[ ',] ',1 1 il tt - 1 ,

tu.\ F,t , - ! l

tli:J-

?

r l ] .

!

t , I tI

t:




However, in the discontinuousconduction mode none of the thyristors conduct over someportion of the input cycle. The load currentremainszero during thatperiod.

As hasbeenexplained earlier in the continuousconductionmode of operation o n€vor becomes

zero, therrfore, either TrTz or T3Ta conducts. Fig 10.4 shows the waveforms of different

variables n the steadystate. The firing angle ofthe converter s o. The angle 0 is givenby

sino -L

"l2Vl

(10.15)

It is assumed hat at t = 0 T:T+ was conducting. As TrTz are fired at 0)t = c they tum on

commutating TrT+ immediately. TtT+ are again fired at (0t = n * c,. Till this point TrTz

conducts. The period of conduction of different thyristors arepictorially depicted n the second

waveform (alsocalled the conductiondiagram) of Fig 10.4.

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I. ! r - -

I

I

I

I

il - ll *.*#I

ff e tl!#.

l;ig, I ll.*: tl'a-rufirrm,i in liirrglc phar* l'ultr curr{rrllled hrirlg* trrnt'rrlr"rr in

{rrrt*i$url$F ilsdstqi$il rnudr.

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The dc link voltage waveform shown next follows from this conduction diagram and theconduction able shown n Fig 10.3(b). It is observed hat the emf sourceE is greater han the dclink voltage till ort= u,. Therefore, he load current i6 continues o fall till this point. However,asT1T2are fired at this point v6 becomesgreater hanE and iostarts ncreasing hrough R-L andE. At ot = tt - 0 vo again equalsE. Dependingupon the load circuit parameterso reachestsmaximumat around his point and starts alling afterwards. Continuousconductionmodewill bepossibleonly if i6remainsgreater hanzero till T:T+ are fired at cot= n * a where upon the sameprocess epeats. The resulting iowaveform is shown below vo. The input ac currentwaveform iiis obtained from roby noting that whenever TrTz conducts i = io and ii = - io whenever T3Taconducts. The last waveformshows he typical voltage waveformacn)ss he thyristor T1. It is tobe noted hat whenthe thyristorfums offat cDt r + cra negativevoltage is applied across t for aduration of z - a. The thyristor must turn offduring this interval for successfuloperation of theconverter.

It is notedthat the dc voltagewaveform is periodic over half the input cycle. Thereforeoit can be expressedn a Fourier seriesas follows.

vo = Voav I ["- cos2nort v, sin2not]n= l

where voou= !['."u0 616y1

2tE y. esssT x J t " n

-- -z tr-- cos2ncordart='&]cos(2n+J)g-cos(2n-l)al 60.t9* :;I

vof i

'L2n+r zn- l J

-2 7 - . s in2nar tdo t= , f i , [ s in (2n+ l )c -s in (2n- l )a l , ,0 . ,n ,t= ; -bvorE 'L

zn+r 2n- r I

ThereforeheRMSvalueofthe nth harmonicI --:-

V***:15,Jnl+uil (10.20)

RMS valueof vocanof course e completedirectly rom.

Vo*"s:

(10.16)

(10.17)

(10.21)

Fourierseriesexpressionf vo s importantbecauset providesa simplemethodof estimatingindividualand otal RMS harmonic urrent qiected nto he oadas ollows:

The mpedancefferedby the oadatnthharmonicrequencysgivenby

(r0.22)

(10.23)

z"=rfitr[iff

I-*r,os'r"'*t, Ioo",[t'*-]*

From 10.18) (10.23)t canbearguedhat n an nductive ircuitLnnr,as-+0 as astas /n2. Soin practice t will be sufficient o consideronly first few harmonicso obtaina reasonablyaccurate stimate f lorm'asormequation 0.23.Thismethodwill be useful, or example,whilecalculatinghe equired un€ntderating f a dcmotor o beusedwith sucha converter.

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However to obtain the current rating of the device o be used t is necessaryo find out aclosed form expressionof io. This will also help to establishthe condition under which theconverterwill operate n the continuousconductionmode.

To begin with we observe hat the voltage waveform and hence he current waveform isperiodic over an interval a. Therefore, finding out an expression or iq over any interval of

lengtha will be sufficient. We choose he interval a < rrrt< n * o.

In this interval

r$+ni, +E="'6ysinotdt

Thegeneralolution fwhich sgivenby

io-1r-*f+

(r0,24)

(10.2s)

Where,

Now at steadystate iol.,=o iol*=o*osince o is periodic over the chosen nterval. Using this

boundarycondition we obtain

7= JPlffi; tanq f,t r =.Ev,sino;t =Zcosq

T - (or*)

. JZV, 2sin(9-0)'ffiz | - nP

L l - e * t

+sin(ote) -I4-lcoseJ

The input current i is related o io as follows:

i , = io for u. ( rot (n*c

ii: - io otherwise.

Fig 10.5shows he waveform of ii in relation to the vi wavefonn.

(r0.26)

(r0.27)

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I

I

I

I

I

I

I

f

I

I

I

t

I

I

I

a

I

I

I

l;[*. lt].il Tlru nput rur i'r.rt{arid t' f,il$d;inrr.ulalurtlprrrreril

It will be of interest to find out a Fourier series expressionof i1. However, using actual

expression or ii will leadto exceedinglycomplex salculation. Significant simplification can bemadeby replacing is with its averagevalue Io. This will be ustified providedthe load is highlyinductive andthe ripple on io is negligible compared o Io. Under this assumption he idealizedwaveformofi lbecomesasquarewavewithtransit ionsatort=aandot=o*nasshowninFig10.5. i1 is the fundamental omponent f this idealized t.

Evidently the input current displacement factor defined as the cosine of the anglebetween nput voltage (v1)and the fundamentalcomponentof input current (i11)waveforms iscosa(lagging).

It can be shown hat

(10.28)

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.' fir _ L \ a f.iIRMS

1l



and Lno,s Io (10.29)

Thereforehe nputcurrent istortionactor= rl'ft' =& (10.30)I,*r, 7t

Thenputoweractoroffi :H3242

.=1a16sss (lagging) (10.31)

Thereforeoherectifierappearssa laggingpower actor oad o the nputac systetn.Larger he'o' poorers thepower actor.

The nput current i alsocontainsignificant mountof harmonic unent 3'd,sth,etc) andtherefore ppearssa harmonic ourceo theutility. Exactcomposition f the harmonic urrentscanbe obtained y Fourierseries nalysis f ii and s left asan exercise.

Exercise 0.2

Fill in the blank(s)with the appropriate ord(s).

i) A singlephaseully controlledbridgeconverter an operate ither n the_ orconductionmode.

ii) In the continuousconductionmodeat least --- thyristorsconductat all times.

iii) In the continuous conduction mode the output voltage waveform doesnot dependon theparameters.

iv) Theminimum requency fthe outputvoltageharmonicn asinglephaseully controlledbridgeconverters_ the nputsupply requency.

v) The nputdisplacementactor of a singlephaseully controlledbridgeconvertern thecontinuousonductionmodes equalo thecosine f the angle.

Answer: (i) continuous,iscontinuous;ii) two;(iii) load; iv) twice;(v) firing.

2. A singlephaseully controlledbridge converter peratesn the continuous onductionmode rom a 230V, 50HZ singlephase upplywith a firing angleo = 30o. The loadresisknceand nductancesre10Q and 50mH espectively.Find out the 6fr harmonic

loadcurrentasapercentagefthe averageoadcurrent.

Answer: Theaverage c outputvoltages

v^^.,='&

,,coso= 179.33 oltsoAvf i

Average utputoadcurrentVf* =17.93AmpsRL

Fromequation10.18) ar= 10.25 oltsFromequation10.19) 6 = 35.5Volts

V***=

26.126Volts,Z,==94.78ohms

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r - V o r * t r - n . ,fr** =tr

= Q.2756mps l.5406 f I'AV

' : : : - - . :

So far we have assumedhat the converteroperatesn continuous onductionmodewithoutpayingattentiono the oadcondition equired or it. In figure10.4hevoltageacrosshe R andL componentf the oad s negativenthe regionn- 0 < rotS n * a. Thereforeo continuestodecreaseill a newpairofthyristor s firedat(ot: rtr o. Now f thevalueof & L andE aresuchthat 0 becomeserobeforeot: ,tr+ o theconduction ecomesiscontinuous.Obviouslyhen,at the boundarybetween ontinuous nddiscontinuous onduction he minimum value of iowhichoccurs tcot a andcot fl't-o will bezero.Puttinghis conditionn (10.26)we obtain heconditionor continuous onduction s.

2sin(-E:o)-sin(q r) -

sino= o

l- e *P cosp (t0,32)

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f l f

t l !t l r

I

I

I

I

tI

t l l

t r !t r t

- -t -Il.

t

t

I

t*

l i ig. llt.6: Sirrglrptlx+r-ullr u*utrulNrdrulrrrl*r iu thr: lircuillinu*ut uruluuliunnuilltsiitl {}rl l ir hriuur|:trihrttr#rrru(}u!irr${rl}rnrldirru*lrli lruuurJ$nrltrrtirrlr:

ihl in rlirtuntinu*rs*rlrullrqliurr

Fig 10.6 shows waveforrns of different variables on the boundary between continuous anddiscontinuous onductionmodesand n the discontinuousonductionmode. t shouldbe stressedthat on the boundarybetweencontinuousanddiscontinuousconductionmodes he load curent is

still continuous.Therefore,all the analysisof continuousconductionmode applies o this caseaswell. However in the discontinuousconduction mode 6remains zero for certain nterval. During .

this interval none of the thyristors conduct. These ntervals are shown by hatched ines in theconductiondiagramof Fig 10.60). In this conduction mode o starts ising from zero as TrTzorefired at rot = o. The load cunent continues o increase ill olt = r - 0. After this, the outputvoltage vo falls below the emf E and io decreasesill alt = p when it becomeszero. Sincethethyristors cannotconductcurrent n the reversedirection i6 remainsat zero till ot = n + o whenT: and Ta are fired. During the period F S ot < n * o,none of the thyristorsconduct. During thisperiodvoattains he value E.Performanceof the rectifier such as Voev, Vonr,rs, onv, Ionus etc can be found in termsof o, pand 0. For example




Or

(10.34)

(r0.35)

(10.36)

(10.37)

(10.38)

(10.3e)inlea;e -#[, -"#]. sin(pe) oGiveng, u and 0, the valueof p can be foundby solvingequation10.39.

' ii , - , , i : : . . ' : : i - i ' : l i i - ' i . - ; i 1 , , : ' , . : i ; i : ; t * ; r r ; i :

The expression for average dc voltage from a single phase fully controlled converter incontinuousconductionmodewas

uottEu,"oro

(10.40)

For u < n/2,Vo> 0. Since he thyristorsconducts urrentonly in onedirection o> 0 always.Thereforepowerf lowingtothedcsideP=VoIo>0foracrJZ.Howeverfora>n/2,YoHenceP < 0. Thismay be nterpreted s he oadsidegiving powerback o theacsideand heconvertern this caseoperateas a line commutatedurrentsource nverter. So it.may betempting o concludehat the sameconverter ircuit may be operated s an inverterby justincreasing beyond rl2. Thismight havebeenrue had t beenpossibleo maintain ontinuousconductionor u < c/2withofi makinganymodification o theconverter r loadconnection.Tosupplypower, he loadEMF source anbeutilized. Howeverheconnecfion f this sourcenFig 10.3 s such hat t canonlyabsorb owerbut cannot supplyt. In fact, f anattempts madeto supplypower o theac side bymakinga> nl2) theenergy tored n the oad nductorwill beexhaustednd hecurrentwill become iscontinuoussshownn Fig 10.7 a).

vouu *,f.'n, dcot=*[f f*, sinotdart*{".'.EV

sinodcot]10.33)

Or Voor, [email protected]+silg(a+a-B)ln

| _ VoouE _ Voou.Dv,.ineroAv-R

-za"rq

- ..12V,lonv -::l:lcoso -cosp sine(aB)l

fi LCOSq-

r$+ni" +E=rDysinrotdt

iolrr=o o

From which the solution of io can be written as

t,+l"rn(rpc;e-#

#{, -. ffi }+sinra*e)]

Now iol*=uo

It is observedhat theperformancef the converters stronglyaffectedby the valueof F.Thevalueof B n terms f the oadparametersi.e,0,g mdZ) ando canbe oundas ollows.

ln the nterval c <cot< p

Version EE IIT, Kharagpur19



t' t

rh , t in r * l

l;'ig. ll.T; lilr*r{er nl*rrtruf rrpeatiun nl'ring}r ph:txr ulll s*rlltrullrd *ntrrl*r

lalllutEfirrrrrr lin'n 3 iiijl .t Ii :- l

tlrl {'ir*uil ftrunttli*tt in lhr: ltl rrtt'r ln*tk"

Therefore for sustained nverter mode of operation the connection of E must be reversedas

shown n Fig 10.7(b).

Fig 10.8 (a) and (b) below shows the waveforms of the inverter operating in continuous

conduction mode and discontinuousconduction mode respectively. Analysis of the converter

remainsunaltered rom the rectifier mode of operationprovided0 is definedas shown. .

I




a a !- t - - t f

I

Itrr--.,

f:ig. I{i.l*; tt'rr{:l'ur:tttr in lhr inr erltt ntudt' rl u;rcr;r{ilrn

{ir {-rrrrlirrutttr}

ruri{iil{lius:

{h I di:i*r tr| rtiljri*t$ {ntEluf f *rl.

Exercise 0.3

Fill in the blank(s)with the appropriate ord(s)

i) In the discontinuousonductionmodehe oadcrurent emains for apart of the input cycle.

iD For the same iring angle the load voltage in the discontinuousconduction modecompared o the continuousconductionmodeof operation.

iii) The load curent ripple factor in the continuous conduction mode is

compared o the discontinuousconductionmode.

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iv) In the inverter mode of operationelectrical power flows from the -- sideto the side.

v) In the continuousconductionmode f the firing angleof the converter s increased

beyond degreeshe converteroperatesn the_ mode.

Answers: (i) zero; ii) higher; iii) lower;(iv) dc, ac; (v) 90, nverter.

2. A 220 V, 20A, 1500 RPM separatelyexcited dc motor has an armatureresistanceofA.75Aand nductance f 50mH. The motor s supplied rom a 230V,50Hz, singlephase

supply through a fully controlled bridge converter. Find the no load speedof the motorand the specdof the motor at the boundarybetweencontinuousand discontinuousmodeswhena =25o.

Answer: At no load the averagemotor torque and hence he averagemotor annaturecurrent iszero. However, since a converter carries only unidirectional currenf zrra averugearmafure

current implies that the afinaturecunent is zero at all time. From Fig 10.6(b) his situation canoccur only when 0 = rd2, i.e the back emf is equal o thepeakof the supply voltage. Therefore,

- . r l :Euloor*aJZx230V =325.27, Underratedconditionol,r* 205V

125'?*t5oo=2380 PM.. Nr,oroad t

,0 ,

At theboundarybetweencontinuousand discontinuousconductionmodes rom equation10.32I + ddtNng

sin0 cosrpsin($c)1-p;

From hegivendata = 87 27ou: 25o

sinO 0.5632Eo €Y sino= 183.18 olts

Motorspeed = tl1'lt . t500= 1340RPM.205

--.;._ji*i-€'3rY

r Single phase fully controlled converters me obtained by replacing the diodes of anuncontrolled converterwith thyristors.

o In a fully controlled converter the output voltage can be controlled by controlling the

firing delay angle(a) ofthe thyristors.

. Single phase ully controlled half wave convertersalways operate n the discontinuousconductionmode.

r Half waveconfrolled convertersusually havepooreroutputvoltageform factor comparedto uncontrolledconverter.

. Single phase ully contuolledbridge convertersare extensively used for small dc motordrives.

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Depending on the load condition and the firing angle a fully controlled bridge convertercan operateeither in the continuousconductionmode or in the discontinuousconductionmode.

In the continuousconduction mode he load voltagedependsonly on the firing angleand

not on loadparameters.

In the discontinuousconduction modethe output voltage decreaseswith increasing oadcurrent. However the output voltage is always greater than that in the continuousconductionmodefor the same iring angle.

The fully controlledbridge convertercan operateas an inverterprovided (i) u > /r, (ii) a

dc power sourceof suitablepolarity exists on the load side.

r-;*?*l=*r:**=

l) "PowerElectronics"P.C.Sen;TataMcGraw-Hill 1995

"Power Electronics; Circuits, Devices and Applications", SecondEdition, MuhammadH.Rashid;Prentice-Hall f India; 1994.

"Powsr Electronics; Converters, Applications and Design" Third Edition" Mohan,Undeland,Robbins, ohnWileys and Sons nc,2003.

Is it possible o operatea singlephase ully controlledhalf wave convertern theinvertingmode?Explain.

A220YJ0A 1500RPM separatelyxciteddcmotorhasan armatureesistancef 0.750and nductance f 50 mH. Themotor s supplied rom a singlephaseully contolledconverter peratingrom a 230V, 50FIz,singlephase upplywith a firing angleof o :

30o. At what speed he motor will supplyfull load torque. Will the conductionbecontinuous nderhis condition?

The speed f thedc motor n questionQ2 s confiolledby varying hefiring angleof theconverterwhile the load torque s maintained onstantat the ratedvalue. Find the"powerfactor"of thesonverter s a functionof the motor speed. Assume ontinuousconduction nd ipple reearmatureurrent.

Find he oad orqueatwhich he dc motorof Q2will operate t 2000RPMwith the ieldcurrentandc remaining ame.

A separatelyxciteddc motor s beingbrakedby a singlephaseully controlledbridgeconverter perating n the invertermodeasshown n Fig 10.7 b). Explainwhat will

happenf a commutationailureoccursn anyoneof the hyristors.

2)

3)

Ql .

Q2.

Q3.

44.

Q5.

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5

1', " ' ! r

sitt r'

1. As explained n section 10.3.3, he load circuit must contain a voltage sourceof prop€rpolarity. Sucha loadcircuit and he associatedwaveformsare shown n the figure next.

l ' - is. Fl l1{ ir t

l ; ic. "F:[ I l {h}

Figure showsthat it is indeedpossible for the half wave converter to operate n the invertingmode for somevaluesof the firing angle. However, care should be taken such hat io becomeszero before vi exceedsE in the negativehalf cycle. Otherwise o will start increasingagain andthe thyristor T will fail to

commutate.

2. For the machine o deliver frrll loadtorquewith rated field the armaturecurrentshould be20 Amps.

Assuming ontinuousonductionv=242x230 cos3Oo 179.33 olts.f i

For 20 Amps armaturecurrent o flow the back ernfwill beEt - Vo Iu& = 179.33 20 x 0.75 164.33 olts

sino=Eo/- =0.505.

/ J2V1




For thegivenmachine, an$= gL" = 20,944, h= 87.266"R"

Now from equation 10.32)

2sin(rn"r

, -"*ffi

- in(q a) : ll '2369

*dtho =10.589cosq

.'. heconductionscontinuous.

At 1500RPM hebackemf s 220 2A x 0.75= 205volts..'.Thespeed twhich hemachine eliversatedorques

is N- : 164'33x150 :12A2RPM. 20s

3. To maintain onstantoad orqueequal o theratedvalue hearmature oltage hould eV":r" I , l . *+Eol ,*o+

r\,aed

=0.75x20+205r :0 .137N+15V1500

In a fully controlled onverter peratingn the continuousonductionmode^ 1 .

U,=TV,coso

= 207.073osa

coso6.616 10'nN

0.0724

Nowthepower actor rom equation 0.3 ist^li

p t=T"osa=5.9565

x 10-4N+0.0652

Thisgives he nputpower actorasa functionof speed.

4. AI2000RPM. - = 200Qx 205=273.33 olts1500

Fsin0= =0.84, 6=87.266", =30o

{2vtFromequation 0.32t canbe shownhat he conduction ill bediscontinuous.

Now romequation 0.394[ sino , ,l sinosin[(o-e)-(.-p)]ehe

Lffi*""(e-')lffior ro477(c'p)n.A*.g+t2] - sin[s2.26G"(a-p)] = tz.ot

18.4st5"'0a7(c'p)sin[(a-B)+ 57.266,f=i .6t

Solvingwhichp=140o




) .

.'. fromequation 0.36

r * Jiv,I*" :ffi [cosocosBsin0(aB)l Z.AZAmps

.' . the oad orqueshouldW ?'916

"t00:13.38%ofthe tull load orque.

20

Refening to Fig 10.8(a) let there be a commutationfailure of T1 at cDt= o. In that casethe conduction mode will beT3T2 insteadof Tr Tzandvowill be zeroduring that period.

As a result avsrage value of Vo will be lessnegativeand the averagearmature currentwill increase. However the sonverterwill continueto operate n the inverter mode andthe motor will be braked.


Lesson 10 SCR Satu Phasa0001

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