Module 2. AC to DC Converters. Version 2 EE IIT, Kharagpur 1
|
|
- Steven Butler
- 7 years ago
- Views:
Transcription
1 Mdule AC t DC Cnverters Versn EE IIT, Kharagpur 1
2 Lessn 9 Sngle Phase Uncntrlled Rectfer Versn EE IIT, Kharagpur
3 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
4 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
5 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
6 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
7 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
8 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
9 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
10 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
11 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 Versn EE IIT, Kharagpur 11
12 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
13 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
14 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 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
15 . 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 ˆv pp = V (1-sn ϕ) =. 1 =.4 V. V max = V = = 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 ( 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 ( ) dωt = 1+ω RC cs(ωt-θ)dωt π ϕ R π β -ϕ 1 1 = 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
16 9.4.1 Splt supply sngle phase uncntrlled full wave rectfer. Versn EE IIT, Kharagpur 16
17 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
18 Versn EE IIT, Kharagpur 18
19 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 = 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
20 Exercse 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 ( ϕ) ϕ Or sn θ = e = e Frm whch φ = V = 9.33 vlts. ˆv pp = V (1-sn ϕ) = 1. vlts. ˆv pp V AV = VMax - = V -.6V = V = 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 = sn( ϕ -θ).53 π = 4 3 Amps. Versn EE IIT, Kharagpur
21 9.4. Sngle phase uncntrlled full brdge rectfer Versn EE IIT, Kharagpur 1
22 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
23 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
24 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
25 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
26 Versn EE IIT, Kharagpur 6
27 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
28 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 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 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
29 ) ) 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 = 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 = V Frm whch φ = 86.5 tan ϕ = ωl = r ωl = hms R L = 5. mh. References [1] P.C. Sen, Pwer Electrncs, Tata McGraw Hll Publshng Cmpany Lmted [] Muhammad H. Rashd, Pwer Electrncs, crcuts, Devces and applcatns Prentce Hall f Inda Prvate Lmted, Secnd Edtn, 1994 Versn EE IIT, Kharagpur 9
30 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
31 Practce Prblems and Answers Versn EE IIT, Kharagpur 31
32 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
33 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
34 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
35 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 = = 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 = = = R 1 csφ =.635 φ = 1.57 rad. and snφ = = [.635 sn( 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 = 164 RPM. 185 Versn EE IIT, Kharagpur 35
36 Average armature current s Trque = = 3.6% 15 V R = π f rated trque. Amps. Versn EE IIT, Kharagpur 36
Module 2. AC to DC Converters. Version 2 EE IIT, Kharagpur 1
Module 2 AC to DC Converters erson 2 EE IIT, Kharagpur 1 Lesson 1 Sngle Phase Fully Controlled Rectfer erson 2 EE IIT, Kharagpur 2 Operaton and Analyss of sngle phase fully controlled converter. Instructonal
More informations s f h s s SPH3UW Unit 7.7 Concave Lens Page 1 of 7 Notes Properties of a Converging Lens
SPH3UW Unt 7.7 Cncave Lens Page 1 f 7 Ntes Physcs Tl bx Thn Lens s an ptcal system wth tw refractng surfaces. The mst smplest thn lens cntan tw sphercal surfaces that are clse enugh tgether that we can
More informationSOLID MECHANICS TUTORIAL FRICTION CLUTCHES
SOLI MECHANICS TUTORIAL FRICTION CLUTCHES Ths wrk cvers elements f the syllabus fr the Edexcel mdule 17P HNC/ Mechancal Prncples OUTCOME. On cmpletn f ths shrt tutral yu shuld be able t d the fllwng. escrbe
More informationTUTORIAL 7 STABILITY ANALYSIS
TUTORIAL 7 STABILITY ANALYSIS Ths tutral s specfcally wrtten fr students studyng the EC mdule D7 Cntrl System Engneerng but s als useful fr any student studyng cntrl. On cmpletn f ths tutral, yu shuld
More informationChapter 14, Problem 1. Find the transfer function V o. /V i. of the RC circuit in Fig. 14.68. Express it using ω = 1/RC. Figure 14.68 For Prob. 14.1.
Chapter 4, Prblem. Fnd the traner unctn V /V the C crcut n Fg. 4.68. Expre t ung /C. Fgure 4.68 Fr Prb. 4.. Chapter 4, Slutn. V H ( ) V H() jc jc jc j 0 j 0, where 0 C H 0 π - H ( ) φ H ( ) tan ( 0 ) 0
More informationSOLUTIONS ELECTRICAL ENGINEERING (EE) PRACTICE PROBLEMS FOR TECHNICAL MAJORS
SOUTIONS EECTRICA ENGINEERING (EE) PRACTICE PROBEMS FOR TECHNICA MAJORS Nte: The text entitled Applied Engineering Principles Manual is used as the reference fr the questins and prblems belw. Althugh nly
More informationSo far circuit analysis has been performed on single-
Three phase systems ntrdctn S far crct analyss has been perfrmed n sngle- phase crcts,.e. there has been ne crct wth a nmber f dfferent vltage and crrent srces whch were nt synchrnsed n any prpsefl way.
More informationChapter 25 The Reflection of Light: Mirrors. The content contained in all sections of chapter 25 of the textbook is included on the AP Physics B exam.
Chapter 25 The Reflectn f Lght: Mrrrs Chapter 25 THE REFLECTION OF LIGHT: MIRRORS PREVIEW The ray mel f lght states that lght may be represente by a straght lne alng the rectn f mtn, an ray ptcs s the
More informationBipolar-Junction (BJT) transistors
Bplar-Junctn (BJT) transstrs References: Barbw (Chapter 7), Hayes & Hrwtz (pp 84-4), Rzzn (Chapters 8 & 9) A bplar junctn transstr s frmed by jnng three sectns f semcnductrs wth alternately dfferent dpngs.
More informationSinusoidal Steady State Response of Linear Circuits. The circuit shown on Figure 1 is driven by a sinusoidal voltage source v s (t) of the form
Sinusidal Steady State espnse f inear Circuits The circuit shwn n Figure 1 is driven by a sinusidal ltage surce v s (t) f the frm v () t = v cs( ωt) (1.1) s i(t) + v (t) - + v (t) s v c (t) - C Figure
More informationHEAT TRANSFER HEAT EXCHANGERS
HET EXCHNGER Types f Heat Exchangers Heat exchangers are classfed accrdng t flw arrangement and type f cnstructn.. Duble-ppe heat exchanger ne flud flws thrugh the smaller ppe whle the ther flud flws thrugh
More informationNanoscale Heat Transfer using Phonon Boltzmann Transport Equation
Presented at the COMSOL Cnference 9 Bstn Nanscale Heat Transfer usng Phnn Bltzmann Transprt Equatn Sangwk Shn Ar Frce Research Labratry, Materals and Manufacturng Drectrate, AFRL/RX, Wrght-Pattersn AFB,
More informationChapter 31B - Transient Currents and Inductance
Chapter 31B - Transent Currents and Inductance A PowerPont Presentaton by Paul E. Tppens, Professor of Physcs Southern Polytechnc State Unversty 007 Objectves: After completng ths module, you should be
More informationA Web Application Framework for Reservation Systems and its Reusability Evaluation
A Web Applcatn wrk fr Reservatn Systems and ts Reusablty Evaluatn Feng Zhu and Takesh Chush Abstract Web applcatn s d n varus busness felds n the Internet and ntranets. It s an effcent way t develp Web
More informationMotor Calculations. Calculating Mechanical Power Requirements Torque - Speed Curves Numerical Calculation Sample Calculation Thermal Calculations
Mtr Calculatins Calculating Mechanical Pwer Requirements Trque - Speed Curves Numerical Calculatin Sample Calculatin Thermal Calculatins Calculating Mechanical Pwer Requirements Physically, pwer is defined
More informationELECTRICAL ENGINEERING Vol. II - Electronic Voltmeters and Ammeters - Alessandro Ferrero, Halit Eren
ELECTRICAL ENGINEERING Vl. II - Electrnc Vltmeters and Ammeters - Alessandr Ferrer, Halt Eren ELECTRONIC VOLTMETERS AND AMMETERS Alessandr Ferrer Dpartment d Elettrtecnca, Pltecnc d Mlan, Italy Halt Eren
More informationYOU ARE RECEIVING THIS NOTICE AS REQUIRED BY THE NEW NATIONAL HEALTH REFORM LAW (ALSO KNOWN AS THE AFFORDABLE CARE ACT OR ACA)
YOU ARE RECEIVING THIS NOTICE AS REQUIRED BY THE NEW NATIONAL HEALTH REFORM LAW (ALSO KNOWN AS THE AFFORDABLE CARE ACT OR ACA) On January 1, 2014, the Affrdable Care Act (ACA) wll be mplemented n Massachusetts
More informationHigh-frequency response of a CG amplifier
Hgh-requency respnse a G apler between surce & grund between dran & grund w-pass lter Md-band p w-pass lter db Input Ple Output Ple n sb Hgh- respnse a G apler Exact Slutn ( db sb ( / : Nde ( / : Nde n
More information= 6degrees of freedom, if the test statistic value f = 4.53, then P-value =.
Sectn 9.5 4. In testng H : σ = σ versus Ha : σ > σ wth ν = 4 and ν = 6degrees f freedm, f the test statstc value f = 4.53, then P-value =..05 75. The sample standard devatn f sdum cncentratn n whle bld
More informationThe circuit shown on Figure 1 is called the common emitter amplifier circuit. The important subsystems of this circuit are:
polar Juncton Transstor rcuts Voltage and Power Amplfer rcuts ommon mtter Amplfer The crcut shown on Fgure 1 s called the common emtter amplfer crcut. The mportant subsystems of ths crcut are: 1. The basng
More informationOblique incidence: Interface between dielectric media
lecrmagnec Felds Oblque ncdence: Inerface beween delecrc meda Cnsder a planar nerface beween w delecrc meda. A plane wave s ncden a an angle frm medum. The nerface plane defnes he bundary beween he meda.
More informationComparisons between CRM and CCM PFC *
Energy and Pwer Engineering, 13, 5, 864-868 di:1.436/epe.13.54b165 Published Online July 13 (http://www.scirp.rg/jurnal/epe Cmparisns between CRM and CCM PFC * Weiping Zhang,Wei Zhang,Jianb Yang 1,Faris
More informationOperational Amplifier Circuits Comparators and Positive Feedback
Operatinal Amplifier Circuits Cmparatrs and Psitive Feedback Cmparatrs: Open Lp Cnfiguratin The basic cmparatr circuit is an p-amp arranged in the pen-lp cnfiguratin as shwn n the circuit f Figure. The
More informationMultiple stage amplifiers
Multple stage amplfers Ams: Examne a few common 2-transstor amplfers: -- Dfferental amplfers -- Cascode amplfers -- Darlngton pars -- current mrrors Introduce formal methods for exactly analysng multple
More information1.3. The Mean Temperature Difference
1.3. The Mean Temperature Difference 1.3.1. The Lgarithmic Mean Temperature Difference 1. Basic Assumptins. In the previus sectin, we bserved that the design equatin culd be slved much easier if we culd
More informationComparison Of A One- Dimensional Model Of A High-Temperature Solid- Oxide Electrolysis Stack With CFD And Experimental Results
INL/EXT-05-00398 EINT Cmparsn Of A One- Dmensnal Mdel Of A Hgh-Temperature Sld- Oxde Electrlyss Stack Wth CFD And Expermental esults 005 ASME Internatnal Mechancal Engneerng Cngress And Expstn J. E. O
More informationLuby s Alg. for Maximal Independent Sets using Pairwise Independence
Lecture Notes for Randomzed Algorthms Luby s Alg. for Maxmal Independent Sets usng Parwse Independence Last Updated by Erc Vgoda on February, 006 8. Maxmal Independent Sets For a graph G = (V, E), an ndependent
More informationWatlington and Chalgrove GP Practice - Patient Satisfaction Survey 2011
Watlingtn and Chalgrve GP - Patient Satisfactin Survey 2011 Backgrund During ne week in Nvember last year patients attending either the Chalgrve r the Watlingtn surgeries were asked t cmplete a survey
More informationLinear Circuits Analysis. Superposition, Thevenin /Norton Equivalent circuits
Lnear Crcuts Analyss. Superposton, Theenn /Norton Equalent crcuts So far we hae explored tmendependent (resste) elements that are also lnear. A tmendependent elements s one for whch we can plot an / cure.
More informationChapter 12 Inductors and AC Circuits
hapter Inductors and A rcuts awrence B. ees 6. You may make a sngle copy of ths document for personal use wthout wrtten permsson. Hstory oncepts from prevous physcs and math courses that you wll need for
More informationTimes Table Activities: Multiplication
Tny Attwd, 2012 Times Table Activities: Multiplicatin Times tables can be taught t many children simply as a cncept that is there with n explanatin as t hw r why it is there. And mst children will find
More informationNetwork Theorems - Alternating Current examples - J. R. Lucas
Netwrk Therems - lternating urrent examples - J. R. Lucas n the previus chapter, we have been dealing mainly with direct current resistive circuits in rder t the principles f the varius therems clear.
More informationRotation Kinematics, Moment of Inertia, and Torque
Rotaton Knematcs, Moment of Inerta, and Torque Mathematcally, rotaton of a rgd body about a fxed axs s analogous to a lnear moton n one dmenson. Although the physcal quanttes nvolved n rotaton are qute
More informationDescription of the Force Method Procedure. Indeterminate Analysis Force Method 1. Force Method con t. Force Method con t
Indeternate Analyss Force Method The force (flexblty) ethod expresses the relatonshps between dsplaceents and forces that exst n a structure. Prary objectve of the force ethod s to deterne the chosen set
More information+ + + - - This circuit than can be reduced to a planar circuit
MeshCurrent Method The meshcurrent s analog of the nodeoltage method. We sole for a new set of arables, mesh currents, that automatcally satsfy KCLs. As such, meshcurrent method reduces crcut soluton to
More informationSPEE Recommended Evaluation Practice #6 Definition of Decline Curve Parameters Background:
SPEE Recommended Evaluaton Practce #6 efnton of eclne Curve Parameters Background: The producton hstores of ol and gas wells can be analyzed to estmate reserves and future ol and gas producton rates and
More informationLecture 26. Dielectric Slab Waveguides
Lectue 6 Delectc Slab Wavegudes In ths lectue yu wll lean: Delectc slab wavegudes T and TM guded mdes n delectc slab wavegudes C 303 Fall 005 Fahan Rana Cnell Unvesty T Guded Mdes n Paallel-Plate Metal
More information2. Before we answer the question, here are four important terms relating to redox reactions and galvanic cells.
CHAPTER SEVENTEEN ELECTROCHEMISTRY Fr Review 1. Electrchemistry is the study f the interchange f chemical and electrical energy. A redx (xidatin-reductin) reactin is a reactin in which ne r mre electrns
More informationCalculation of Sampling Weights
Perre Foy Statstcs Canada 4 Calculaton of Samplng Weghts 4.1 OVERVIEW The basc sample desgn used n TIMSS Populatons 1 and 2 was a two-stage stratfed cluster desgn. 1 The frst stage conssted of a sample
More informationGoals Rotational quantities as vectors. Math: Cross Product. Angular momentum
Physcs 106 Week 5 Torque and Angular Momentum as Vectors SJ 7thEd.: Chap 11.2 to 3 Rotatonal quanttes as vectors Cross product Torque expressed as a vector Angular momentum defned Angular momentum as a
More informationGetting Your Fingers In On the Action
Rry Garfrth Getting Yur Fingers In On the Actin Once yu are able t strum with yur fingers, yu can begin fingerpicking! The first task is t learn yur hand psitin and t learn which fingers are used n which
More informationCalculating the high frequency transmission line parameters of power cables
< ' Calculatng the hgh frequency transmsson lne parameters of power cables Authors: Dr. John Dcknson, Laboratory Servces Manager, N 0 RW E B Communcatons Mr. Peter J. Ncholson, Project Assgnment Manager,
More informationDerivative Markets and Instruments
1 Mdule # 6 Cmpnent # 1 This Cmpnent: fcuses n the basics f as part f ur examinatin f Derivatives. assumes a base level f financial thery, but attempts t add a level f practical applicatin. We attempt
More informationIntroduction to Fractions and Mixed Numbers
Name Date Intrductin t Fractins and Mixed Numbers Please read the Learn prtin f this lessn while cmpleting this handut. The parts f a fractin are (shwn belw): 5 8 Fractins are used t indicate f a whle.
More informationPhi Kappa Sigma International Fraternity Insurance Billing Methodology
Phi Kappa Sigma Internatinal Fraternity Insurance Billing Methdlgy The Phi Kappa Sigma Internatinal Fraternity Executive Bard implres each chapter t thrughly review the attached methdlgy and plan nw t
More informationBusiness Plan Overview
Business Plan Overview Organizatin and Cntent Summary A business plan is a descriptin f yur business, including yur prduct yur market, yur peple and yur financing needs. Yu shuld cnsider that a well prepared
More informationImplementation of Doppler Radar
Implementatin f Dppler Radar Applicatin Nte Justin Erskine ECE 480 Design Team 5 Executive Summary Dppler radar is an easy and effective way t measure relative speed. With the ability t make smaller antennas,
More informationAn Alternative Way to Measure Private Equity Performance
An Alternatve Way to Measure Prvate Equty Performance Peter Todd Parlux Investment Technology LLC Summary Internal Rate of Return (IRR) s probably the most common way to measure the performance of prvate
More informationHow do I evaluate the quality of my wireless connection?
Hw d I evaluate the quality f my wireless cnnectin? Enterprise Cmputing & Service Management A number f factrs can affect the quality f wireless cnnectins at UCB. These include signal strength, pssible
More informationWHITE PAPER. Vendor Managed Inventory (VMI) is Not Just for A Items
WHITE PAPER Vendr Managed Inventry (VMI) is Nt Just fr A Items Why it s Critical fr Plumbing Manufacturers t als Manage Whlesalers B & C Items Executive Summary Prven Results fr VMI-managed SKUs*: Stck-uts
More informationElectrochemical cells
Electrchemical cells In this chapter, we turn ur attentin t electrn transfer reactins. T identify an electrn transfer reactins, we must assign xidatin states review rules in Chapter 3. e.g. Zn(s) Cu(NO
More informationOPF Based CRM Using Supply Side and Demand Side Management
I (Prnt) : 30 3765 I (Onlne): 78 8875 Internatnal Jurnal f Adanced Research n Electrcal, Electrncs and Instrumentatn Enneern (An IO 397: 007 Certfed Oranzatn) Vl. 4, Issue 1, January 015 OPF Based CRM
More informationThe Gibbs Free Energy and Cell Voltage
The Gibbs Free Energy and Cell Vltage When an amunt f charge, Q, mves thrugh a ptential difference, E w = - Q E b/c wrk dne by the system E > 0 fr galvanic (vltaic) cells Recall, G = H TS = E + PV TS Fr
More informationPersonal Selling. Lesson 22. 22.1 Objectives. 22.2 Meaning of Personal Selling
Persnal Selling Lessn 22 Persnal Selling When yu want t buy smething yu usually g t a cncerned shp and purchase it frm there. But, smetimes yu find peple bring certain gds r prducts and make them available
More informationInformation Guide Booklet. Home Loans
Infrmatin Guide Bklet Hme Lans This Infrmatin Guide bklet prvides yu with general infrmatin nly. It will als help yu t better understand any recmmendatins we have made fr yu. Infrmatin Guide Hme Lans January
More informationHelpdesk Support Tickets & Knowledgebase
Helpdesk Supprt Tickets & Knwledgebase User Guide Versin 1.0 Website: http://www.mag-extensin.cm Supprt: http://www.mag-extensin.cm/supprt Please read this user guide carefully, it will help yu eliminate
More informationExperiment 1: Freezing Point Depression
Experiment 1: Freezing Pint Depressin Purpse: The purpse f this lab is t experimentally determine the freezing pint f tw slutins and cmpare the effect f slute type and cncentratin fr each slutin. Intrductin:
More informationGain vs. Proportional Band
Gain vs. Prprtinal Band The end result f the analysis f a typial lp tuning predure, whether pen lp r lsed lp, is a set f parameters with whih t adjust the ntrller. One f these parameters is ntrller gain.
More informationJet Engine. Figure 1 Jet engine
Jet Engne Prof. Dr. Mustafa Cavcar Anadolu Unversty, School of Cvl Avaton Esksehr, urkey GROSS HRUS INAKE MOMENUM DRAG NE HRUS Fgure 1 Jet engne he thrust for a turboet engne can be derved from Newton
More informationTHE UNEARNED NO CLAIM BONUS. C. P. WELTEN Amsterdam
THE UNEARNED NO CLAIM BONUS C. P. WELTEN Amsterdam I. The claims experience f a mtrcar insurance is assumed t give sme indicatin abut the risk (basic claim frequency) f that insurance. The experience rating
More informationHough-Domain Image Registration By Metaheuristics
Hugh-Dman mage Regstratn By etaheurstcs Shubn Zha Jangsu Autmatn Research nsttute Lanyungang, Jangsu, P. R. Chna 6 Emal: zha_shubn@63.cm Abstract mage regstratn s the prcess f regsterng tw r mre mages,
More informationCrnwall Partners in Care
Crnwall Partners in Care Mving Frward Versin 2.0 8 th January 2014 By Richard Mnk Crnwall Partners in Care August 2013 Page 1 f 6 CPIC mving frward This dcument has been created t help prvide a little
More informationUSING ARTIFICIAL NEURAL NETWORK TO MONITOR AND PREDICT INDUCTION MOTOR BEARING (IMB) FAILURE.
Internatnal Engneerng Cnventn, Jedda, Saud Araba, 10-14 Marc 2007 001 USING ARTIFICIAL NEURAL NETWORK TO MONITOR AND PREDICT INDUCTION MOTOR BEARING (IMB) FAILURE. A.K Maamad 1, S. San 1, M.H Abd Waab
More information8.5 UNITARY AND HERMITIAN MATRICES. The conjugate transpose of a complex matrix A, denoted by A*, is given by
6 CHAPTER 8 COMPLEX VECTOR SPACES 5. Fnd the kernel of the lnear transformaton gven n Exercse 5. In Exercses 55 and 56, fnd the mage of v, for the ndcated composton, where and are gven by the followng
More informationMerchant Management System. New User Guide CARDSAVE
Merchant Management System New User Guide CARDSAVE Table f Cntents Lgging-In... 2 Saving the MMS website link... 2 Lgging-in and changing yur passwrd... 3 Prcessing Transactins... 4 Security Settings...
More informationHow to put together a Workforce Development Fund (WDF) claim 2015/16
Index Page 2 Hw t put tgether a Wrkfrce Develpment Fund (WDF) claim 2015/16 Intrductin What eligibility criteria d my establishment/s need t meet? Natinal Minimum Data Set fr Scial Care (NMDS-SC) and WDF
More informationHow much life insurance do I need? Wrong question!
Hw much life insurance d I need? Wrng questin! We are ften asked this questin r sme variatin f it. We believe it is NOT the right questin t ask. What yu REALLY need is mney, cash. S the questin shuld be
More informationJune 29, 2009 Incident Review Dallas Fort Worth Data Center Review Dated: July 8, 2009
The purpse f this dcument is t capture the events and subsequent respnse t the incident that tk place in the DFW datacenter n 29 June, 2009. I. Executive Summary On 29 June, an area f the Rackspace DFW
More informationOperation- and Assembly Instructions
Operatn- and Assembly Instructns Master Cntrl Order-N. 7574 00 1X Cntent 1 Safety nstructns... 3 2 Devce cmpnents... 3 3 Functn... 3 4 Operatn... 4 5 Infrmatn fr qualfed electrcans... 13 5.1 Fttng and
More informationSolution: Let i = 10% and d = 5%. By definition, the respective forces of interest on funds A and B are. i 1 + it. S A (t) = d (1 dt) 2 1. = d 1 dt.
Chapter 9 Revew problems 9.1 Interest rate measurement Example 9.1. Fund A accumulates at a smple nterest rate of 10%. Fund B accumulates at a smple dscount rate of 5%. Fnd the pont n tme at whch the forces
More informationSoftware Quality Assurance Plan
Sftware Quality Assurance Plan fr AnthrpdEST pipeline System Versin 1.0 Submitted in partial fulfillment f the requirements f the degree f Master f Sftware Engineering Prepared by Luis Fernand Carranc
More informationHow To Install Fcus Service Management Software On A Pc Or Macbook
FOCUS Service Management Sftware Versin 8.4 fr Passprt Business Slutins Installatin Instructins Thank yu fr purchasing Fcus Service Management Sftware frm RTM Cmputer Slutins. This bklet f installatin
More informationState Variable Model for Considering the Parasitic Inductor Resistance on the Open Loop Performance of DC to DC Converters
Jurnal f Cmputer and Cmmunicats, 04,, 4-48 Published Onle Nvember 04 SciRes. http://www.scirp.rg/jurnal/jcc http://dx.di.rg/0.436/jcc.04.3006 State ariable del fr Cnsiderg the Parasitic Inductr Resistance
More informationTRAINING GUIDE. Crystal Reports for Work
TRAINING GUIDE Crystal Reprts fr Wrk Crystal Reprts fr Wrk Orders This guide ges ver particular steps and challenges in created reprts fr wrk rders. Mst f the fllwing items can be issues fund in creating
More informationBERNSTEIN POLYNOMIALS
On-Lne Geometrc Modelng Notes BERNSTEIN POLYNOMIALS Kenneth I. Joy Vsualzaton and Graphcs Research Group Department of Computer Scence Unversty of Calforna, Davs Overvew Polynomals are ncredbly useful
More informationApplied Radio Labs. Group Delay Explanations and Applications. Introduction. The Delay of Signals. Definition of Group Delay.
Grup Delay - Explanatins and Applicatins Applied Radi Labs Grup Delay Explanatins and Applicatins www.radi-labs.cm Design File: DN4 5 Nvember 999 Intrductin Grup delay is a cncept that radi engineers are
More informationChemical Engineering Thermodynamics II
Chemcal Engneerng hermdynamcs II (CHE 303 Curse Ntes).K. Nguyen Chemcal and Materals Engneerng Cal ly mna (Wnter 009) Cntents Chapter : Intrductn. Basc Defntns -. rperty -.3 Unts -3.4 ressure -4.5 emperature
More informationChapter 6 Inductance, Capacitance, and Mutual Inductance
Chapter 6 Inductance Capactance and Mutual Inductance 6. The nductor 6. The capactor 6.3 Seres-parallel combnatons of nductance and capactance 6.4 Mutual nductance 6.5 Closer look at mutual nductance Oerew
More informationbenefit is 2, paid if the policyholder dies within the year, and probability of death within the year is ).
REVIEW OF RISK MANAGEMENT CONCEPTS LOSS DISTRIBUTIONS AND INSURANCE Loss and nsurance: When someone s subject to the rsk of ncurrng a fnancal loss, the loss s generally modeled usng a random varable or
More informationELEC 204 Digital System Design LABORATORY MANUAL
ELEC 204 Digital System Design LABORATORY MANUAL : Design and Implementatin f a 3-bit Up/Dwn Jhnsn Cunter Cllege f Engineering Kç University Imprtant Nte: In rder t effectively utilize the labratry sessins,
More informationPeak Inverse Voltage
9/13/2005 Peak Inerse Voltage.doc 1/6 Peak Inerse Voltage Q: I m so confused! The brdge rectfer and the fullwae rectfer both prode full-wae rectfcaton. Yet, the brdge rectfer use 4 juncton dodes, whereas
More informationFOCUS Service Management Software Version 8.5 for Passport Business Solutions Installation Instructions
FOCUS Service Management Sftware fr Passprt Business Slutins Installatin Instructins Thank yu fr purchasing Fcus Service Management Sftware frm RTM Cmputer Slutins. This bklet f installatin instructins
More informationWireless Light-Level Monitoring
Wireless Light-Level Mnitring ILT1000 ILT1000 Applicatin Nte Wireless Light-Level Mnitring 1 Wireless Light-Level Mnitring ILT1000 The affrdability, accessibility, and ease f use f wireless technlgy cmbined
More informationFlexible PCB Antenna with Cable Integration Application Note
Flexible PCB Antenna with Cable Integratin Applicatin Nte APN-11-8-004/C/JC Page 1 f 10 1. BASICS A Flexible PCB antenna with cable is a very flexible, lw prfile, highly reliable, and ecnmical slutin widely
More informationFOCUS Service Management Software Version 8.5 for CounterPoint Installation Instructions
FOCUS Service Management Sftware Versin 8.5 fr CunterPint Installatin Instructins Thank yu fr purchasing Fcus Service Management Sftware frm RTM Cmputer Slutins. This bklet f installatin instructins will
More informationGauss Law. AP Physics C
Gauss Law AP Physics C lectric Flux A Let's start be defining an area n the surface f an bject. The magnitude is A and the directin is directed perpendicular t the area like a frce nrmal. Flux ( r FLOW)
More informationConduction in the Cylindrical Geometry
Cnductin in the Cylindrical Gemetry R. Shankar Subramanian Department f Chemical and Bimlecular Engineering Clarksn University Chemical engineers encunter cnductin in the cylindrical gemetry when they
More informationSMA5101 Thermodynamics of Materials Ceder 2001
SMA5101 Thermdynamcs f Materals Ceder 2001 Chapter 4 Slutn Thery In the frst chapters we dealt prmarly wth clsed systems fr whch nly heat and wrk s transferred between the system and the envrnment. In
More informationA class of Petri nets for modeling and analyzing business processes
A class f Petr nets fr mdelng and analyzng busness prcesses W.M.P. van der Aalst Department f Mathematcs and Cmputng Scence, Endhven Unversty f Technlgy, P.O. Bx 513, NL-5600 MB, Endhven, The Netherlands,
More informationCHECKING ACCOUNTS AND ATM TRANSACTIONS
1 Grades 6-8 Lessn 1 CHECKING ACCOUNTS AND ATM TRANSACTIONS Tpic t Teach: This lessn is intended fr middle schl students in sixth thrugh eighth grades during a frty minute time perid. The lessn teaches
More informationOtomasyon, Danışmanlık Ltd. Şti. INDILOAD DATACARD M AN UAL. Antetli03.doc / Page 9 of 1. Kibele Hedef mükemmele ulasmaksa.
Otmasyn, Danışmanlık Ltd. Şti. INDILOAD DATACARD M AN UAL Antetli03.dc / Page 9 f 1 Otmasyn, Danışmanlık Ltd. Şti. Chapter 1 : General Chapter 2 : Cnnectins Chapter 3 : Operatin Chapter 4 : Technical sheet
More informationFaraday's Law of Induction
Introducton Faraday's Law o Inducton In ths lab, you wll study Faraday's Law o nducton usng a wand wth col whch swngs through a magnetc eld. You wll also examne converson o mechanc energy nto electrc energy
More informationRetirement Planning Options Annuities
Retirement Planning Optins Annuities Everyne wants a glden retirement. But saving fr retirement is n easy task. The baby bmer generatin is graying. Mre and mre peple are appraching retirement age. With
More informationVerification of Workflow Nets
Verfcatn f Wrkflw Nets W.M.P. van der Aalst Endhven Unversty f Technlgy, P.O. Bx 513, NL-5600 MB, Endhven, The Netherlands. wsnwa@wn.tue.nl Abstract. Wrkflw management systems wll change the archtecture
More informationefusion Table of Contents
efusin Cst Centers, Partner Funding, VAT/GST and ERP Link Table f Cntents Cst Centers... 2 Admin Setup... 2 Cst Center Step in Create Prgram... 2 Allcatin Types... 3 Assciate Payments with Cst Centers...
More informationRecurrence. 1 Definitions and main statements
Recurrence 1 Defntons and man statements Let X n, n = 0, 1, 2,... be a MC wth the state space S = (1, 2,...), transton probabltes p j = P {X n+1 = j X n = }, and the transton matrx P = (p j ),j S def.
More informationHow To Understand The Power Control System
THREE-PHASE CIRCUITS PART I AC GENERATOR Single-phase AC generatr - designed t generate a single sinusidal vltage fr each rtatin f the shaft (rtr). Plyphase AC generatr - designed t generate multiple utf-phase
More informationCopyrights and Trademarks
Cpyrights and Trademarks Sage One Accunting Cnversin Manual 1 Cpyrights and Trademarks Cpyrights and Trademarks Cpyrights and Trademarks Cpyright 2002-2014 by Us. We hereby acknwledge the cpyrights and
More informationDesign for securability Applying engineering principles to the design of security architectures
Design fr securability Applying engineering principles t the design f security architectures Amund Hunstad Phne number: + 46 13 37 81 18 Fax: + 46 13 37 85 50 Email: amund@fi.se Jnas Hallberg Phne number:
More informationGetting started with Android
Getting started with Andrid Befre we begin, there is a prerequisite, which is t plug the Andrid device int yur cmputer, and lad the drivers fr the OS. In writing this article, I was using Windws XP, 7
More information