Mdule AC t DC Cnverters Versn EE IIT, Kharagpur 1
Lessn 9 Sngle Phase Uncntrlled Rectfer Versn EE IIT, Kharagpur
Operatn and Analyss f sngle phase uncntrlled rectfers Instructnal Objectves On cmpletn the student wll be able t Classfy the rectfers based n ther number f phases and the type f devces used. Defne and calculate the characterstc parameters f the vltage and current wavefrms. Analyze the peratn f sngle phase uncntrlled half wave and full wave rectfers supplyng resstve, nductve, capactve and back emf type lads. Calculate the characterstc parameters f the nput/utput vltage/current wavefrms asscated wth sngle phase uncntrlled rectfers. Versn EE IIT, Kharagpur 3
9.1 Intrductn One f the frst and mst wdely used applcatn f pwer electrnc devces have been n rectfcatn. Rectfcatn refers t the prcess f cnvertng an ac vltage r current surce t dc vltage and current. Rectfers specally refer t pwer electrnc cnverters where the electrcal pwer flws frm the ac sde t the dc sde. In many stuatns the same cnverter crcut may carry electrcal pwer frm the dc sde t the ac sde where upn they are referred t as nverters. In ths lessn and subsequent nes the wrkng prncple and analyss f several cmmnly used rectfer crcuts supplyng dfferent types f lads (resstve, nductve, capactve, back emf type) wll be presented. Pnts f nterest n the analyss wll be. Wavefrms and characterstc values (average, RMS etc) f the rectfed vltage and current. Influence f the lad type n the rectfed vltage and current. Harmnc cntent n the utput. Vltage and current ratngs f the pwer electrnc devces used n the rectfer crcut. Reactn f the rectfer crcut upn the ac netwrk, reactve pwer requrement, pwer factr, harmncs etc. Rectfer cntrl aspects (fr cntrlled rectfers nly) In the analyss, fllwng smplfyng assumptns wll be made. The nternal mpedance f the ac surce s zer. Pwer electrnc devces used n the rectfer are deal swtches. The frst assumptn wll be relaxed n a latter mdule. Hwever, unless specfed therwse, the secnd assumptn wll reman n frce. Rectfers are used n a large varety f cnfguratns and a methd f classfyng them nt certan categres (based n cmmn characterstcs) wll certanly help ne t gan sgnfcant nsght nt ther peratn. Unfrtunately, n cnsensus exsts amng experts regardng the crtera t be used fr such classfcatn. Fr the purpse f ths lessn (and subsequent lessns) the classfcatn shwn n Fg 9.1 wll be fllwed. Versn EE IIT, Kharagpur 4
Ths Lessn wll be cncerned wth sngle phase uncntrlled rectfers. 9. Termnlges Certan terms wll be frequently used n ths lessn and subsequent lessns whle characterzng dfferent types f rectfers. Such cmmnly used terms are defned n ths sectn. Let f be the nstantaneus value f any vltage r current asscated wth a rectfer crcut, then the fllwng terms, characterzng the prpertes f f, can be defned. Peak value f f ( ˆf ): As the name suggests ˆf =f ver all tme. Average (DC) value f f(f av ) : Assumng f t be perdc ver the tme perd T 1 T F av = f(t)dt T.(9.1) RMS (effectve) value f f(f RMS ) : Fr f, perdc ver the tme perd T, 1 T F RMS = f (t)d T t..(9.) Frm factr f f(f FF ) : Frm factr f f s defned as FRMS f =. (9.3) F FF Rpple factr f f(f RF ) : Rpple factr f f s defned as av FRMS -Fav f RF = = fff -1 Fav max.(9.4) Versn EE IIT, Kharagpur 5
Rpple factr can be used as a measure f the devatn f the utput vltage and current f a rectfer frm deal dc. Peak t peak rpple f f( ˆf pp ): By defntn ˆf =f -f n Over perd T (9.5) pp max m Fundamental cmpnent f f(f 1 ): It s the RMS value f the snusdal cmpnent n the Furer seres expressn f f wth frequency 1/T. 1 F 1 = ( f A1+ fb1)...(9.6) T where f () t A1 = f t cs π dt T (9.7) T T f () t B1 = f t sn π dt T.(9.8) T K th harmnc cmpnent f f(f K ): It s the RMS value f the snusdal cmpnent n the Furer seres expressn f f wth frequency K/T. 1 F K = ( f AK +fbk) (9.9) T where f AK = f(t) csπkt T dt T...(9.1) T f BK = f(t) snπkt T dt T (9.11) Crest factr f f(c f ) : By defntn ˆf C f = (9.1) F RMS Dstrtn factr f f(df f ) : By defntn F1 DF f =..(9.13) F RMS Ttal Harmnc Dstrtn f f(thd f ): The amunt f dstrtn n the wavefrm f f s quantfed by means f the ndex Ttal Harmnc Dstrtn (THD). By defntn THD = f α K= K 1 Fk F1..(9.14) Frm whch t can be shwn that THD = f 1-DF DF f f (9.15) Versn EE IIT, Kharagpur 6
Dsplacement Factr f a Rectfer (DPF): If v and are the per phase nput vltage and nput current f a rectfer respectvely, then the Dsplacement Factr f a rectfer s defned as. DPF = csφ (9.16) Where φ s the phase angle between the fundamental cmpnents f v and. Pwer factr f a rectfer (PF): As fr any ther equpment, the defntn f the pwer factr f a rectfer s Actual pwer nput t the Rectfer PF =.(9.17) Apparent pwer nput t the Rectfer f the per phase nput vltage and current f a rectfer are v and respectvely then VIcsφ 1 1 PF = (9.18) V I RMS RMS If the rectfer s suppled frm an deal snusdal vltage surce then V =V s, In terms f THD 1 I PF = I PF = RMS 1 RMS csφ =DF DPF DPF 1+THD 1..(9.19)...(9.) Majrty f the rectfers use ether ddes r thyrstrs (r cmbnatn f bth) n ther crcuts. Whle desgnng these cmpnents standard manufacturer s specfcatns wll be referred t. Hwever, certan terms are used n relatn t the rectfer as a system. They are defned next. Pulse number f a rectfer (p): Refers t the number f utput vltage/current pulses n a sngle tme perd f the nput ac supply vltage. Mathematcally, pulse number f a rectfer s gven by Tme perd f the nput supply vltage p = Tme perd f the mnum rder harmnc n the utput vltage/current.. Classfcatn f rectfers can als be dne n terms f ther pulse numbers. Pulse number f a rectfer s always an ntegral multple f the number f nput supply phases. Cmmutatn n a rectfer: Refers t the prcess f transfer f current frm ne devce (dde r thyrstr) t the ther n a rectfer. The devce frm whch the current s transferred s called the ut gng devce and the devce t whch the current s transferred s called the ncmng devce. The ncmng devce turns n at the begnnng f cmmutatn whle the ut gng devce turns ff at the end f cmmutatn. Cmmutatn falure: Refers t the stuatn where the ut gng devce fals t turn ff at the end f cmmutatn and cntnues t cnduct current. Frng angle f a rectfer (α): Used n cnnectn wth a cntrlled rectfer usng thyrstrs. It refers t the tme nterval frm the nstant a thyrstr s frward based t the nstant when a gate pulse s actually appled t t. Ths tme nterval s expressed n radans by multplyng t wth Versn EE IIT, Kharagpur 7
the nput supply frequency n rad/sec. It shuld be nted that dfferent thyrstrs n a rectfer crcut may have dfferent frng angles. Hwever, n the steady state peratn, they are usually the same. Extnctn angle f a rectfer (γ): Als used n cnnectn wth a cntrlled rectfer. It refers t the tme nterval frm the nstant when the current thrugh an utgng thyrstr becmes zer (and a negatve vltage appled acrss t) t the nstant when a pstve vltage s reappled. It s expressed n radans by multplyng the tme nterval wth the nput supply frequency (ω) n rad/sec. The extnctn tme (γ/ω) shuld be larger than the turn ff tme f the thyrstr t avd cmmutatn falure. Overlap angle f a rectfer (μ): The cmmutatn prcess n a practcal rectfer s nt nstantaneus. Durng the perd f cmmutatn, bth the ncmng and the utgng devces cnduct current smultaneusly. Ths perd, expressed n radans, s called the verlap angle μ f a rectfer. It s easly verfed that α + μ + γ = π radan. Exercse 9.1 Fll n the blank(s) wth the apprprate wrd(s). ) In a rectfer, electrcal pwer flws frm the sde t the sde. ) Uncntrlled rectfers emply where as cntrlled rectfers emply n ther crcuts. ) Fr any wavefrm Frm factr s always than r equal t unty. v) The mnmum frequency f the harmnc cntent n the Furer seres expressn f the utput vltage f a rectfer s equal t ts. v) THD s the specfcatn used t descrbe the qualty f wavefrms where as Rpple factr serves the same purpse fr fr wavefrms. v) Input pwer factr f a rectfer s gven by the prduct f the factr and the factr. v) The sum f frng angle, Extnctn angle and verlap angle f a cntrlled rectfer s always equal t. Answers: () ac, dc; () ddes, thyrstrs; () greater; (v) pulse number; (v) ac, dc; (v) dsplacement, dstrtn; (v) π 9.3 Sngle phase uncntrlled half wave rectfer Ths s the smplest and prbably the mst wdely used rectfer crcut albet at relatvely small pwer levels. The utput vltage and current f ths rectfer are strngly nfluenced by the type f the lad. In ths sectn, peratn f ths rectfer wth resstve, nductve and capactve lads wll be dscussed. Versn EE IIT, Kharagpur 8
Fg 9. shws the crcut dagram and the wavefrms f a sngle phase uncntrlled half wave rectfer. If the swtch S s clsed at at t =, the dde D becmes frward based n the the nterval < ωt π. If the dde s assumed t be deal then Fr < ωt π v = v = V sn ωt v D = v v = (9.1) Snce the lad s resstve V = v R = snωt..(9.) R = Fr ωt > π, v becmes negatve and D becmes reverse based. S n the nterval π < ωt π = = v = R =...(9.3) v D = v v = v = V snωt Frm these relatnshps 1 π 1 π V V AV = v dωt = snωtdωt = π V π.(9.4) π 1 π V V DRMS = V sn ωtdωt = π...(9.5) Versn EE IIT, Kharagpur 9
It s evdent frm the wavefrms f v and n Fg 9. (b) that they cntan sgnfcant amunt f harmncs n addtn t the dc cmpnent. Rpple factr f v s gven by VDRM -VDAV 1 v RF = = π -4 VDAV Wth a resstve lad rpple factr f wll als be same. (9.6) Because f such hgh rpple cntent n the utput vltage and current ths rectfer s seldm used wth a pure resstve lad. The rpple factr f utput current can be reduced t same extent by cnnectng an nductr n seres wth the lad resstance as shwn n Fg 9.3 (a). As n the prevus case, the dde D s frward based when the swtch S s turned n. at ωt =. Hwever, due t the lad nductance ncreases mre slwly. Eventually at ωt = π, v becmes zer agan. Hwever, s stll pstve at ths pnt. Therefre, D cntnues t cnduct beynd ωt = π whle the negatve supply vltage s supprted by the nductr tll ts current becmes zer at ωt = β. Beynd ths pnt, D becmes reverse based. Bth v and remans zer tll the begnnng f the next cycle where upn the same prcess repeats. Frm the precedng dscussn Fr ωt β v D = v = v = (9.7) fr β ωt π Versn EE IIT, Kharagpur 1
v = = = v D = v v = v 1 π 1 β V AV = v dωt = Vsnωtdωt π π (9.8) V 1-csβ r V AV = π ( )... (9.9) 1 β V RMS = V sn ωtdωt π V = ( ) 1 V β -snβ β - snβ =..(9.3) π π Frm factr f the vltage wavefrm s VRMS β -snβ v OFF = = π.(9.31) VAV π(1- csβ) The rpple factr. π(β -snβ) v RF = voff -1= -1 (9.3) (1- csβ) All these quanttes are functns f β whch can be fund as fllws. Fr ωt β d v = Vsnωt =L +R dt.(9.33) (ωt = ) = (ωt = β) = The slutn s gven by - ωt tanφ V =I e + sn(ωt-φ) (9.34) Z where tanφ = ωl R and Z = R +ω L..(9.35) Puttng the ntal cndtns f (9.33) ωt - V tanφ = snφe +sn( ωt-φ Z ) (9.36) β - V tanφ (ωt =β)= snφe +sn( β - φ ) = Z β - tanφ r snφe =sn( φ -β).(9.37) β as a functn f φ can be btaned by slvng equatn 9.37. Versn EE IIT, Kharagpur 11
It can be shwn that β ncreases wth φ. Frm Equatn (9.9), V AV decreases wth ncreasng β whle V RMS ncreases wth β. Therefre, wth ncreasng φ (and hence ncreasng L) the frm factr and the rpple factr f v wrsens. Hwever, the rpple factr f decreases wth ncreasng L. Therefre, n certan applcatns, where a smth dc current s f prme mprtance (e.g. the feld supply f a dc mtr) ths cnfguratn f the rectfer s preferred. The prblem f pr frm factr (rpple factr) f the utput vltage can be slved t sme extent by cnnectng a capactr acrss the lad resstance f Fg 9. (a). Ths sngle phase half wave rectfer supplyng a capactve lad s shwn n Fg 9.5 (a). Crrespndng wavefrms are shwn n Fg 9.5 (b). If the capactr was ntally dscharged the dde D s frward based when the swtch S s turned n at ωt =. The utput vltage fllws the nput vltage. The dde D carres bth the capactr chargng current and the lad current. At ωt = β the sum f these tw currents becmes zer and tends t grw n the negatve drectn. At ths pnt the dde becmes Versn EE IIT, Kharagpur 1
reverse based and dscnnects the lad (alng wth the capactr) frm the supply. The capactr then dscharges wth the lad current. Dde D des nt becme frward based tll the nput supply vltage becmes equal t the capactr vltage n the next cycle at ωt = (π + φ). The same prcess repeats thereafter. Frm the precedng dscussn Fr π + φ ωt π + β v = v = Vsnωt 1..(9.38) dv v = c + =c + dt R V r [ ] = ωrccsωt+snωt R V = 1+ω RC cs(ωt-θ)..(9.39) R -1 1 where θ =tan ωrc ( ) 1 Versn EE IIT, Kharagpur 13
At ωt = β+ π, = s β θ = π/ r β = θ + π/ π -1 1 r β = +tan ωrc.(9.4) Agan fr β ω t π + φ dv v =, C + =, v( ωt =β ) = Vcsθ. dt R -(ωt-β) tanθ v = Vcsθ e..(9.41) at ωt =π + φ, v = Vsnφ r r Vsn ϕ = Vcsθ e 3π -( +φ-θ) tanθ snφ =csθ e π -(π+φ- -θ) tanθ 3π -( -θ) tanθ -φ tanθ snφ = csθ e e...(9.4) Frm whch φ can be slved. Peak t peak rpple n v s ˆv = V(1-sn ϕ ).(9.43) pp As c α, θ and β and φ π/ and ˆv pp Therefre, a very large capactr helps t mprve the rpple factr f the utput vltage f ths rectfer. Hwever, as ndcated by Equatn (9.39) the peak current thrugh the dde ncreases prprtnately. It s als nterestng t bserve that unlke the prevus cases the peak reverse vltage appearng acrss D s gven by. vd max = V +vm V (9.44) Ths s smetmes referred t as the peak nverse vltage ratng (PIV) f the dde. Exercse 9. 1. Fll n the blank(s) wth the apprprate wrd(s). ) The rpple factr f the utput vltage and current wavefrms f a sngle phase uncntrlled half wave rectfer s than unty. ) Wth an nductve lad, the rpple factr f the utput f the half wave rectfer mprves but that f the utput becmes prer. ) In bth sngle phase half wave and full wave rectfers the frm factr f the utput vltage appraches wth capactve lads prvded the capactance s enugh. v) The PIV ratng f the rectfer dde used n a sngle phase half wave rectfer supplyng a capactve lad s apprxmately the nput supply vltage. v) The % THD f the nput current f the rectfers supplyng capactve lads s. Answers: () greater; () current, vltage; () unty, large; (v) duble, peak; (v) hgh. Versn EE IIT, Kharagpur 14
. An unregulated dc. pwer supply f average value 1 V and peak t peak rpple f % s t be desgned usng a sngle phase half wave rectfer. Fnd ut the requred nput vltage, the utput capactance and the dde RMS current and PIV ratngs. The equvalent lad resstance s 5 hms. Answer: Frm equatn 9.43. ˆv pp = V (1-sn ϕ) =. 1 =.4 V. V max = V = 1 +.4 = 13.V V = 9.33V sn ϕ =.818 r ϕ = 54.9 =.96 rad. Then frm equatn 9.4 -(3 π + ϕ - θ) tanθ.818 = csθ e (5.67 - θ) tanθ r.818 e = csθ Frm whch θ.35 1 tan θ = =.3553, R = 5Ω, C = 179 μf ωrc PIV f the dde = V = 6.4V. 1 RMS. Dde current = β V ( ) 1 9.35 dωt = 1+ω RC cs(ωt-θ)dωt π ϕ R π 54.9 1 β -ϕ 1 1 = 7.43 + sn(β - θ)- sn( -θ).8564 π ϕ = 4 4 Amps. 9.4 Sngle phase uncntrlled full wave rectfer Sngle phase uncntrlled half wave rectfers suffer frm pr utput vltage and/r nput current rpple factr. In addtn, the nput current cntans a dc cmpnent whch may cause prblem (e.g. Transfrmer saturatn etc) n the pwer supply system. The utput dc vltage s als relatvely less. Sme f these prblems can be addressed usng a full wave rectfer. They use mre number f ddes but prvde hgher average and rms utput vltage. There are tw types f full wave uncntrlled rectfers cmmnly n use. If a splt pwer supply s avalable (e.g. utput frm a splt secndary transfrmer) nly tw dde wll be requred t prduce a full wave rectfer. These are called splt secndary rectfers and are cmmnly used as the nput stage f a lnear dc vltage regulatr. Hwever, f n splt supply s avalable the brdge cnfguratn f the full wave rectfer s used. Ths s the mre cmmnly used full wave uncntrlled rectfer cnfguratn. Bth these cnfguratns are analyzed next. Versn EE IIT, Kharagpur 15
9.4.1 Splt supply sngle phase uncntrlled full wave rectfer. Versn EE IIT, Kharagpur 16
Fg 9.6 shws the crcut dagram and wavefrms f a sngle phase splt supply, uncntrlled full wave rectfer supplyng an R L lad. The splt pwer supply can be thught f t have been btaned frm the secndary f a center tapped deal transfrmer (.e. n nternal mpedance). When the swtch s clsed at the pstve gng zer crssng f v 1 the dde D 1 s frward based and the lad s cnnected t v 1. The currents and 1 start rsng thrugh D 1. When v 1 reaches ts negatve gng zer crssng bth and 1 are pstve whch keeps D 1 n cnductn. Therefre, the vltage acrss D s v CB =v -v1. Beynd the negatve gng zer crssng f v, D becmes frward based and the current cmmutates t D frm D 1. The lad vltage v becmes equal t v and D 1 starts blckng the vltage v AB =v1 -v. The current hwever cntnues t ncrease thrugh D tll t reaches the steady state level after several cycles. Steady state wavefrms f the varables are shwn n Fg 9.6 (b) frm ωt = nwards. It shuld be nted that the current, nce started, always remans pstve. Ths mde f peratn f the rectfer s called the Cntnuus cnductn mde f peratn. Ths shuld be cmpared wth the wavefrm f Fg 9.3 (b) fr the half wave rectfer where remans zer fr sme duratn f the nput supply wavefrm. Ths mde s called the dscntnuus cnductn mde f peratn. Frm the abve dscussn Fr ω t<π v = v 1 = 1...(9.45) fr π ωt<π v = v =...(9.46) Snce v s perdc ver an nterval π 1 π V π V V AV = v dωt = snωtdωt = π π..(9.47) π 1 π V RMS = V sn ωt dωt =V π...(9.48) VRMS π v FF = =.(9.49) VAV RF FF π -8 v = v -1=..(9.5) Bth the frm factr and the rpple factr shws cnsderable mprvement ver ther half wave cunter parts. Versn EE IIT, Kharagpur 17
Versn EE IIT, Kharagpur 18
The sngle phase full wave rectfer stll des nt ffer a smth dc vltage. Wth resstve lad, cnsderable rpple current wll flw nt the lad. Ths prblem can be slved by cnnectng a capactr acrss the lad resstance just as n the case f a half wave rectfer. If the capactr was ntally dscharged, the dde D 1 s frward based when the swtch S s turned n at ωt =. The dde D remans reverse based. The utput vltage fllws the nput vltage. D 1 carres bth the capactr chargng current and the lad current. At ωt = β the sum f these tw currents becmes zer and tends t grw n the negatve drectn. At ths pnt the dde D 1 becmes reverse based and dscnnects the lad alng wth the capactr frm the supply. The capactr then dscharges thrugh the lad untl at ωt = π + φ, v becmes greater than v and frward bases D. D 1 nw remans reverse based. D cnducts up t ωt = π + β. The same prcess repeats thereafter. Frm the dscussn abve Fr π + φ ω t π + β v = v =- Vsnωt dv v = c + =C + (9.51) dt R V r [ ] =- ωrccsωt+snωt R 1 ( ) ( ) V = -1 1 1+ω RC csπ + θ - ωt where θ =tan.(9.5) R ωrc π π π -1 1 at ωt = π + β, 1 = s β -θ = r β = θ + rβ = +tan ωrc (9.53) Agan fr β ω t π + φ dv v 1 = C + = v ( ωt =β ) = Vsnβ = Vcsθ.(9.54) dt R v = Vcsθ e at ωt = π + φ, v = r r -( ωt-β) tanθ snφ Vsnφ = Vcsθ e snφ =csθ e π ( ).(9.55) π ( +θ-π-φ ) tanθ - +φ-θ tanθ π ( ) - -θ tanθ -φtanθ snφ = csθ e e (9.56) Frm whch φ can be slved. Peak t peak rpple n v s ˆv = V(1-snφ).(9.57) pp It can be shwn that fr the same R and C, ˆv pp gven by Equatn (9.57) s smaller than that gven by Equatn (9.43) fr the half wave rectfer. The dde PIV ratngs reman equal t V hwever. Versn EE IIT, Kharagpur 19
Exercse 9.3 1. Fll n the blank(s) wth the apprprate wrd(s). ) The utput vltage frm factr f a sngle phase full wave rectfer s. ) The utput vltage f a sngle phase full wave rectfer supplyng an nductve lad s f the lad parameters. ) The peak t peak utput vltage rpple f a sngle phase splt supply full wave rectfer supplyng a capactve lad s cmpared t an equvalent half wave rectfer. Answers: () π ; () ndependent ; () smaller.. An unregulated dc pwer supply s bult arund a sngle phase splt supply full wave rectfer usng the same nput vltage and utput capactr fund n the prblem f Exercse 9.. The lad resstance s 5 Ω. Fnd ut the average utput vltage, the peak t peak rpple n the utput vltage and the RMS current ratngs f the ddes. Answer: Frm the gven data C = 179 μf, R = 5 Ω, θ =.35 Frm equatn 9.56 Sn θ = cs θ e -(π/ + φ θ) tan θ -.3553(1.5353+ ϕ) -.3553ϕ Or sn θ =.99937 e =.946316 e Frm whch φ = 65.33 V = 9.33 vlts. ˆv pp = V (1-sn ϕ) = 1. vlts. ˆv pp V AV = VMax - = V -.6V = 13. -.6V = 1.6V. % rpple = 9.5% 1 π +θ V β RMS dde current = ( ) 1 dωt = 1+ω RC cs(ωt-θ)dωt π ϕ R π ϕ 1 β -ϕ 1 = 7.43 - sn( ϕ -θ).53 π = 4 3 Amps. Versn EE IIT, Kharagpur
9.4. Sngle phase uncntrlled full brdge rectfer Versn EE IIT, Kharagpur 1
The splt supply full wave sngle phase rectfer ffers as gd perfrmance as pssble frm a sngle phase rectfer n terms f the utput vltage frm factr and rpple factr. They have a few dsadvantages hwever. These are They requre a splt pwer supply whch s nt always avalable. Each half f the splt pwer supply carres current fr nly ne half cycle. Hence they are underutlzed. Versn EE IIT, Kharagpur
The rat f the requred dde PIV t the average ut put vltage s rather hgh. These prblems can be mtgated by usng a sngle phase full brdge rectfer as shwn n Fg 9.8 (a). Ths s ne f the mst ppular rectfer cnfguratn and are used wdely fr applcatns requrng dc. pwer utput frm a few hundred watts t several kl watts. Fg 9.8 (a) shws the rectfer supplyng an R-L-E type lad whch may represent a dc. mtr r a strage battery. These rectfers are als very wdely used wth capactve lads partcularly as the frnt end f a varable frequency vltage surce nverter. Hwever, n ths sectn analyss f ths rectfer supplyng an R-L-E lad wll be presented. Its peratn wth a capactve lad s very smlar t that f a splt supply rectfer and s left as an exercse. When the swtch S s turned n at the pstve gng zer crssng f v n current flws n the crcut tll v crsses E at pnt A. Beynd ths pnt, D 1 & D are frward based by v and current starts ncreasng thrugh them tll the pnt B. After pnt B, v falls belw E and starts decreasng. Nw dependng n the values f R, L & E ne f the fllwng stuatns may arse. may becme zer befre the negatve gng zer crssng f v at pnt C. may cntnue t flw beynd C and becme zer befre the pnt D. may stll be nn zer at pnt D. It shuld be nted that f > ether D 1 D r D 3 D 4 must cnduct. Fg 9.4 (b) shws the wavefrms fr the thrd stuatn. If > at pnt C the negatve gng nput vltage reverse bases D1 & D. Current cmmutates t D3 and D4 as shwn n the asscated cnductn Dagram n Fg 9.8 (b). It shws pctrally the cnductn nterval f dfferent devces. The current cntnues t decrease up t the pnt D beynd whch t agan ncreases. It shuld be nted that n ths mde f cnductn always reman greater than zer. Cnsequently, ths s called the cntnuus cnductn mde f peratn f the rectfer. In the ther tw stuatns the mde f peratn wll be dscntnuus. The steady state wavefrms f the rectfer under cntnuus cnductn mde s shwn t the rght f the pnt ωt = n Fg 9.4 (b). Frm ths fgure and precedng dscussn Fr < ωt π v = v = V sn ωt (9.58) = fr π < ωt π v =- v =- V sn ωt (9.59) = - 1 π V AV = V sn ωt d ωt = V π π (9.6) 1 π V RMS = V sn ωt d ωt = V π (9.61) Versn EE IIT, Kharagpur 3
V π V RMS v OFF = = AV v = v -1 = RF OFF π -8 (9.6) Fndng ut the characterzng quanttes fr wll be dffcult wng t ts cmplcated wavefrm. Cnsderable, smplfcatn s acheved (wthut sgnfcant lss f accuracy) by replacng the actual wavefrm by ts average value I AV = V AV / R. Fg 9.9 shws the apprxmate nput current wave frm and ts fundamental cmpnent. Frm Fg 9.9 Dsplacement angle φ = Input dsplacement factr (DPF) = cs ϕ = 1 (9.63) Il Dstrtn factr (DF l ) = = (9.64) I π Pwer Factr (PF) = AV DPF DF = l π 1 - DF π - 8 % TH D = 1 = 1 DF (9.65) (9.66) Versn EE IIT, Kharagpur 4
The exact analytcal expressn fr (and hence ) can be btaned as fllws. fr < ωt π Ld v = V sn ωt = R + + E (9.67) dt = (steady state perdc bundary cnd.) ωt= ωt=π The general slutn can be wrtten as ωt - V snθ tanϕ = I e + sn( ωt- ϕ ) - Z csϕ ωl E where tan ϕ = ; Z = R + ω L ; sn θ = R V (9.68) Frm the bundary cndtn π V snθ - V snθ tan I - ϕ sn ϕ + = Ie + sn ϕ- Z csϕ Z csϕ I = V snϕ Z ϕ 1 - e - π tan ωt V snϕ - snθ tanϕ = π e + sn tan ( ωt- ) - - Z ϕ ϕ 1 - e csϕ (9.69) (9.7) Frm whch the cndtn fr cntnuus cnductn can be btaned. fr cntnuus cnductn fr all <ω t π hence Mn r ωt=θ Cndtn fr cntnuus cnductn s snϕ - θ tan sn θ ϕ e = sn ( -θ) + -π ϕ tanϕ 1 - e cs ϕ (9.71) If the parameters f the lad (.e, R, L &E) are such that the left hand sde f equatn 9.71 s less than the rght hand sde cnductn f the rectfer becmes dscntnuus.e, the lad current becmes zer fr a part f the nput cycle. Dscntnuus cnductn mde f peratn f ths rectfer s dscussed next. Versn EE IIT, Kharagpur 5
Versn EE IIT, Kharagpur 6
Fg. 9.1(b) shws the wavefrms f dfferent varables under dscntnuus cnductn mde f peratn. In ths mde f peratn D 1 D are nt frward based tll v exceed E at ωt = θ. Cnsequently, n current flws nt the lad tll ths tme. After ωt = θ, the lad s cnnected t the nput surce thrugh D 1 D and starts buldng up. Beynd ωt = π - θ, starts decreasng and becmes zer at ωt = β < π. D 1 D are reverse based at ths pnt. D 3 D 4 are frward based at ωt = π + θ when starts ncreasng agan. Thus nne f the ddes cnduct durng the nterval β < ω + π + θ and remans zer durng ths perd. Frm the precedng dscussn Versn EE IIT, Kharagpur 7
fr θ < ωt β < π fr π v = v = = V sn ωt < π + θ < ω + π + β < π v = - v = - V sn ωt = - v = E = = ther wse (9.7) (9.73) (9.74) 1 1 V = v d ωt = V sn ωt+ E d ωt AV π+θ β π+θ π θ π θ β V OR V AV = [ cs θ - cs β + ( π + θ -β) sn θ] (9.75) π β can be fund n the fllwng manner fr θ < ω t β Ld v = sn ωt =R + + E dt = = ωt=θ ωt=β (9.76) The general slutn s ωt-tanϕ - V sn θ = I e + sn( ωt- φ )- Z cs ϕ where tan ϕ = ωl, Z = R + ω L, snθ = E R Frm the ntal cndtn ωt=θ = V I = sn ( ϕ -θ) + snθ Z csϕ V (9.77) (9.78) ωt-θ -ωt-θ V - snθ tan tan = sn ( ϕ -θ) e - ( 1- e ) ϕ ϕ + sn ( ωt - ϕ Z csϕ ) (9.79) Puttng ωt = β = n Equatn 9.79. θ-β θ-β snθ tanϕ tanϕ sn ( β - ϕ) = 1 - e - sn ( ϕ - θ) e (9.8) csϕ Frm whch β can be slved. Exercse 9.4 1. Fll n the blank(s) wth the apprprate wrd(s). ) The average utput vltage f a full wave brdge rectfer and a splt supply full wave rectfer are prvded the nput vltages are. Versn EE IIT, Kharagpur 8
) ) v) Fr the same nput vltage the brdge rectfer uses the number f ddes used n a splt supply rectfer wth the PIV ratng. Fr cntnuus cnductn, the lad mpedance f a brdge rectfer shuld be. In the cnductn mde the utput vltage f a brdge rectfer s f lad parameters. Answers: () equal, equal; () duble, half; () nductve; (v) cntnuus, ndependent.. A battery s t be charged usng a full brdge sngle phase uncntrlled rectfer. On full dscharge the battery vltage s 1. V. and n full charge t s 1.7 vlts. The battery nternal resstance s.1ω. Fnd ut the nput vltage t the rectfer s that the battery chargng current under full charge cndtn s 1% f the chargng current under fully dscharged cndtn. Assume cntnuus cnductn under all chargng cndtn and fnd ut the nductance t be cnnected n seres wth the battery fr ths cndtn. Answer: Let the rectfer nput vltage be V and the chargng current under fully dscharged cndtn be I. Then assumng cntnuus cnductn V -.1I = 1. and V -.1I = 1.7 π π.9i =.5 V I = 7.78 Amps and V = 14.415 vlts. If cnductn s cntnuus at full charge cndtn t wll be cntnuus fr all ther chargng cndtns. Fr cntnuus cnductn snϕ -θ tanϕ snθ e =sn( ϕ -θ)+ -π tanϕ 1-e cs ϕ E Frm gven data snθ = =.63, θ = 38.535 V Frm whch φ = 86.5 tan ϕ = ωl = 16.35 r ωl = 1.635 hms R L = 5. mh. References [1] P.C. Sen, Pwer Electrncs, Tata McGraw Hll Publshng Cmpany Lmted. 1995 [] Muhammad H. Rashd, Pwer Electrncs, crcuts, Devces and applcatns Prentce Hall f Inda Prvate Lmted, Secnd Edtn, 1994 Versn EE IIT, Kharagpur 9
Mdule Summary A rectfer s a pwer electrnc cnverter whch cnverts ac vltage r current surces t dc vltage and current. In a rectfer, electrcal pwer flws frm the ac nput t the dc utput. In many rectfer crcuts, pwer can als flw frm the dc sde t the ac sde, where upn, the rectfer s sad t be peratng n the nverter mde. Rectfers can be classfed based n the type f devce they use, the cnverter crcut tplgy, number f phases and the cntrl mechansm. All rectfers prduce unwanted harmnes bth at the ut put and the nput. Perfrmance f a rectfer s judged by the relatve magntudes f these harmnes wth respect t the desred utput. Fr a gven nput vltage and lad, the utput vltage (current) f an uncntrlled rectfer can nt be vared. Hwever, the utput vltage may vary cnsderably wth lad. Sngle phase uncntrlled half wave rectfer wth resstve r nductve lad have lw average utput vltage, hgh frm factr and pr rpple factr f the utput vltage wavefrm. Sngle phase uncntrlled full wave rectfer have hgher average utput vltage and mprved rpple factr cmpared t a half wave rectfer wth resstve and nductve lad. Wth hghly nductve lad the utput vltage wavefrm f a full wave rectfer may be ndependent f the lad parameters. Wth a capactve lad the utput vltage frm factr appraches unty wth ncreasng capactance value fr bth the half wave and the full wave rectfers. Hwever, THD f the nput current als ncreases. A full wave brdge rectfer generates hgher average dc vltage cmpared t a splt supply full wave rectfer. Hwever t als uses mre number f ddes. Versn EE IIT, Kharagpur 3
Practce Prblems and Answers Versn EE IIT, Kharagpur 31
Q1. What wll be the lad vltage and current wavefrm when a sngle phase half wave uncntrlled rectfer supples a purely nductve lad? Explan yur answer wth wavefrms. Q. The splt supply f a sngle phase full wave rectfer s btaned frm a sngle phase transfrmer wth a sngle prmary and a center tapped secndary. The rectfer supples a purely resstve lad. Assumng the transfrmer t be deal fnd ut the, dsplacement factr, dstrtn factr and the pwer factr at the prmary sde f the transfrmer. Q3. A sngle phase splt supply full wave rectfer s desgned t supply an nductve lad. The average lad current s A, and the rpple current s neglgble. Can the same rectfer be used wth a capactve lad drawng the same Amps average current? Justfy yur answer. Q4. A V, 15 Amps, 15 rpm separately excted dc mtr has an armature resstance f 1 Ω and nductance f 5 mh. The mtr s suppled frm a sngle phase full wave brdge rectfer wth nput vltage f 3 V, 5 HZ. Neglectng all n lad lsses, fnd ut the n lad speed f the machne. Als fnd ut the trque and speed at the bundary between cntnuus and dscntnuus cnductn. Versn EE IIT, Kharagpur 3
Answer 1 Wthut lss f generalty t can be assumed that S s turned ON at ωt =. Snce, f t s turned ON anytme after ωt =, the vlt-sec. acrss the nductr wll dctate that the current thrugh t becmes zer befre the next pstve gng zer crssng f v. In the regn ωt < π D s frward based and v = v d L = Vsnωt () = dt d r ωl = Vsnωt = dωt ωt = V V =I- csωt I = ωl ωl V = (1- csωt) ωl V at ωt = π, = ωl > D cnducts beynd ωt = π untl s zer agan. Let the extnctn angle be ωt = β > π. ` Then fr ωt β V = (1- csωt) ωl V = (1-csβ) fr π β π the nly slutn s β = π ωt =β ωl Versn EE IIT, Kharagpur 33
V v = v fr ωt π and = (1- csωt) ωl ωt π Answer Fgure shws the secndary vltage and current wavefrms f the rectfer. Frm the gven data NS v S1 = VPsnωt NP NS VP S1 = snωt fr ωt π NP R S1 = therwse. NS v S =- VPsnωt NP NS VP S =- snωt fr π ωt π NP R S = therwse Frm the MMF balance f an deal transfrmer N P P-N S S1+N S S= NS VP r P = (S1- S)= snωt NP NP R N S At the nput Dsplacement factr = Dstrtn factr = Pwer factr = 1. Versn EE IIT, Kharagpur 34
Answer 3 If the lad current s A wth neglgble rpple. The requred RMS current ratng f the rectfer dde, wth reference t Fg 9.6 (b) wll be I D1RMS =I DRMS = Amps. Hwever frm Fg 9.7 (b) and Prblem f Exercse 9.3 the requred RMS current fr a capactve lad wll be much larger than Amps. Therefre the same rectfer can nt be used. Answer 4 Snce all n lad lsses are neglected the develped pwer at n lad and hence the n lad trque wll be zer. Therefre, the average armature current wll als be zer. Hwever, snce a dde rectfer can nt cnduct nstantaneus negatve lad current, zer average current wll mply that the nstantaneus value f the armature current at all tme wll be zer at n lad.wth reference t Fg 9.1 ths cndtn wll requre the rectfer ddes t reman reverse based at all tme. Hence at n lad E V Hwever E wll nt exceed V, snce nce a becmes zer when E= V there wll be n develped trque t accelerate the mtr. Hence the mtr speed and E wll nt ncrease any further. Thus at n lad E = V = 35.7 vlts. Under the rated cndtn at 15 rpm E rated = 15 1. = 185 vlts. E N Nw = Erated Nrated E 35.7 N = N rated = 15 = 637 rpm. Erated 185 At the bundary between the cntnuus and dscntnuus mde f cnductn. snφ -θ tanφ snθ e =sn(φ - θ)+ -π tanφ 1-e csφ snφ r [ ] θ tanφ -π tanφ = cs ϕ sn(φ - θ) + snθ e 1-e -3 ωl 1π 5 1 where tanφ = = = 15.78 R 1 csφ =.635 φ = 1.57 rad. and snφ =.168.6995 = [.635 sn(1.57 - θ) + snθ] e.6366 θ -1 E frm whch θ = sn = 38.5 E =.48 V V but E at 15 RPM = 185 vlts. Speed at the junctn f cntnuus and dscntnuus cndtn s.48 15 = 164 RPM. 185 Versn EE IIT, Kharagpur 35
Average armature current s Trque = 4.593 1 = 3.6% 15 V R = 4.593 π f rated trque. Amps. Versn EE IIT, Kharagpur 36