ECE 320 Energy Conversion and Power Electronics Dr. Tim Hogan

Transcription

1 ECE 3 Energy Cnverin an Pwer Elecrnic Dr. Tim Hgan Chaper 8: Pwer Elecrnic (Texbk Chaper 1, an Secin 11., 11.3, an reerve bk: Pwer Elecrnic) Chaper Objecive A we aw in he la chaper, cnrl ver he rque an pee f he mr can be gaine hrugh vlage an frequency cnrl he mr. Thi can be accmplihe by cnvering he inpu AC urce pwer a DC urce (recifying i), hen filering i reuce harmnic, an finally cnvering i back an AC urce having he eire frequency an ampliue (inverer). 8.1 ine Cnrlle Recifier We ar wih a ecripin f hw raw pwer frm a 1-phae r 3-phae yem prvie DC a la. The characeriic f he yem here inclue ha he evice ue will urn hemelve ff (cmmuae) an ha he yem raw reacive pwer frm he la One-Phae an Three-Phae Circui wih Die If he urce i 1-phae, a ie i ue an he la i purely reiive, a hwn in Figure 1 hen i i a relaively imple cnfigurain. When he urce vlage i piive, he curren flw hrugh he ie an he vlage f he urce equal he vlage f he la. v, v v ie i i v v R v ie v, v ie Figure 1. Simple circui wih ie an reiive la. If he la inclue an inucance an a urce (uch a a baery we wih charge), a in Figure, hen he ie will cninue cnuc even when he la vlage becme negaive a lng a he curren i mainaine. Thi cme frm he characeriic f he inucr:

2 1 3 v = i( ) i( ) = Thu, he hae area A in Figure mu equal he hae area B. 3 (8.1) v ie v i v v E E v i 1 3 v ie 5m 1m 15m m 5m A v B Figure. Simple circui wih ie an inucive la wih vlage urce One-Phae Full Wave Recifier Mre cmmn i a ingle phae ie brige recifier uch a hwn in Figure 3. The la can be mele wih ne f w exreme: eiher a a cnan curren urce, repreening he cae f a large

3 inucance ha keep he curren hrugh i alm cnan, r a a reir, repreening he cae f minimum line inucance. We will uy he fir cae wih AC an DC ie curren an vlage wavefrm hwn in Figure 4 fr he ieal cae f =. i i v C v Figure 3. One-phae full wave recifier. v i I v i = I Figure 4. Wavefrm fr a ne-phae full wave recifier wih inucive la.

4 If we analyze hee wavefrm, he upu vlage will have a DC cmpnen, (where he ubcrip repreen ha hi i he ieal cae wih = ): 9 =. (8.) π where i he RMS value f he inpu AC vlage. On he her han he RMS value f he upu vlage will be = (8.3) cnaining cmpnen f higher frequency. Similarly n he AC ie he curren i n inuial, raher i change abruply beween I an I. an again, he RMS value are he ame I 9 The al harmnic irin, THD, can hen be fun a: 1 = I. I (8.4) π I = I (8.5) I I THD = 48.43% (8.6) I 1 1 I i impran ne here ha if he urce ha me inucance (an i uually e), hen cmmuain will be elaye afer he vlage reache zer, unil he curren ha rppe zer a hwn in Figure 5. Thi will lea a ecreae f he upu DC vlage belw wha i expece frm (8.)

5 i i v v I v v v i A Figure 5. One-phae full wave recifier wih inucive la an urce inucance.

6 8.1.3 Three-Phae Full Die Recifier The circui f Figure 3 can be mifie hanle hree phae, by uing 6 ie a hwn in Figure 6. Figure 7 hw he AC ie curren an DC ie vlage fr he cae f high la inucance. Similar analyi a befre hw ha n he DC ie, he vlage i: 3 = l l = π l l Frm Figure 6, we ee ha n he AC ie, he RMS curren, I i: I while he funamenal curren, i.e. he curren a pwer frequency i: (8.7) = I =. I (8.8) 1 I 78 1 = 6I =. I (8.9) π Again, inucance n he AC ie will elay cmmuain, cauing a vlage l, i.e. he DC vlage will be le han ha preice by equain (8.7). i a i a n b C v R c Figure 6. Three-phae full-wave recifier wih ie.

7 v v an v bn v cn 1m m 3m 4m A i a D 1 D 4 1º 6º 1º A D D 5 i b A D 3 D 6 i c Figure 7. Wavefrm f a hree-phae full-wave recifier wih ie an inucive la Cnrlle Recifier wih Tyrir Thyrir give u he abiliy vary he DC vlage. Remember ha make a hyrir ar cnucing, he hyrir mu be frwar biae an a gae pule prvie i gae. Al, urn ff he hyrir he curren hrugh i mu revere irecin fr a hr peri f ime, rr, an reurn zer One-Phae Cnrlle Recifier Fig. hw he ame 1-phae brige we have alreay uie, nw wih hyrir inea f ie, an fig hw he upu vlage an inpu curren wavefrm. In hi figure α i he elay angle, crrepning he ime we elay riggering he hyrir afer hey became frwar biae.

8 T 1 T 3 i v v I T 4 T Figure 8. One-phae full wave cnverer wih hyrir. Thyrir 1 an are riggere geher an f cure are 3 an 4. Each pair f hyrir i urne ff immeiaely (r hrly) afer he her pair i urne n by gaing. Analyi imilar ha fr ie circui will give: an he relain fr he curren i he ame = c( α ) =.9 c( α ) (8.1) π I 1 = I c( α ) =.9I c( α ) (8.11) π I hul al be pine u ha in Figure 9 he curren wavefrm n he AC ie i ffe in ime wih repec he crrepning vlage by he ame angle α, hence i he funamenal f he curren, reuling in a lagging pwer facr.

9 v i ω α = v v ω α α > v i ω -1 v Figure 9. Wavefrm f ne-phae full wave cnverer wih hyrir. On he DC ie, nly he DC cmpnen f he vlage carrie pwer, ince here i n harmnic cnen in he curren. On he AC ie he pwer i carrie nly by he funamenal, ince here are n harmnic in he vlage. ( ) I P = I c α = 1 (8.1) Inverer Me If he curren n he DC ie i uaine even if he vlage revere plariy, hen pwer will be ranferre frm he DC he AC ie. The vlage n he DC ie can revere plariy when he elay angle excee 9º, a lng a he curren i mainaine. Thi can nly happen when he la vlage i a hwn in Figure 1, e.g. a baery.

10 i T 1 T 3 i v v E T 4 T Figure 1. Operain f a ne-phae cnrlle cnverer a an inverer Three-Phae Cnrlle Cnverer A wih ie, nly 6 hyrir are neee accmmae hree phae. Figure 11 hw he chemaic f he yem, an Figure 1 hw he upu vlage wavefrm. i T 1 T 3 T 5 a i a n b i a v I c i a T 4 T 6 T Figure 11. Schemaic f a hree-phae full-wave cnverer bae n hyrir.

11 v v an v bn v cn i a ω α = 15m m 5m 3m 35m v v an v bn v cn i a α ω α > 16m m 4m 8m 3m Figure 1. Wavefrm f a hree-phae full-wave cnverer bae n hyrir. The elay angle α i again meaure frm he pin ha a hyrir becme frwar biae, bu in hi cae he pin i a he inerecin f he vlage wavefrm f w ifferen phae. The vlage n he DC ie i hen (he ubcrip here again meaning = ): while he pwer fr bh he AC an DC ie i which lea : 3 = l l c( α ) = 1.35l l c( α ) (8.13) π ( α ) = 3 I c( α ) P = I = 1.35l li c l l 1 (8.14) I. 78I an he relainhip beween an α in Figure 1 i: 1 = (8.15) ( α ) α = c (8.16) Again, if he elay angle α i exene beyn 9º, he cnverer ranfer pwer frm he DC ie he AC ie, becming an inverer. We hul keep in min, hugh ha even in hi cae he cnverer i rawing reacive pwer frm he AC ie.

12 8.1.7 Ne Fr bh 1-phae an 3-phae cnrlle recifier, a elay in α creae a phae iplacemen f he phae curren wih repec he phae vlage, equal α. The cine f hi angle i he pwer facr f he funamenal harmnic. Fr bh mr an generar me he cnrlle recifier abrb reacive pwer frm he hreephae AC yem, alhugh i can eiher abrb r pruce real pwer. I al nee he pwer line cmmuae he hyrir. Thi mean ha inverer perain i pible nly wih he recifier i cnnece a pwer line. When a DC mr r a baery i cnnece he erminal f a cnrlle recifier an α becme greaer han 9º, he erminal DC vlage change plariy, bu he irecin f he curren ay he ame. Thi mean ha in rer fr he recifier raw pwer frm he baery r a mr ha perae a a generar urning in he ame irecin, he erminal have be wiche. 8. Inverer Here we uy yem ha can cnver DC AC hrugh he ue f evice ha can be urne n an ff uch a GTO, BJT, IGBT, an MOSFET, which allw he ranfer f pwer frm he DC urce any AC la, an give cnierable cnrl ver he reuling AC ignal. The general blck iagram f he cmplee yem i hwn in Figure (Hz) AC _ AC Mr Recifier Filer Capacir Swich-me inverer Figure 13. Typical variable vlage an variable frequency yem One-Phae Inverer Figure 14 hw he perain f ne leg f an inverer regarle f he number f phae. T illurae he pin beer, he inpu DC vlage i ivie in w equal par. When he upper wich, S1, i cle while S i pen, he upu vlage A will be cle while S1 i pen, he upu vlage will be, an when he lwer wich, S, i v A S1 S v AN Figure 14. One leg f an inverer.

13 T cnrl he upu wavefrm he wiche can be cnrlle by pule wih mulain, PWM, where he ime each wich i cle can be eermine by he ifference beween a cnrl wavefrm, an a carrier (r riangular) wavefrm a hwn in Figure 15. When he cnrl wave i greaer han he riangular wave, S1 i cle, an S i pen. When he cnrl wave i le han he riangular wave, S1 i pen, an S i cle. In hi way, he wih f he upu i mulae (hence he name) m 18m m m 4m 6 m 8m 3m 3m 34 m 1(ri) 1(cn) Time Figure 15. Cnrl (re inewave) an carrier r riangular wavefrm (green) eermine when he wiche are cle. Fr he ime when he cnrl wave i greaer han he riangular wave, S1 i cle, an S i pen. When he cnrl wave i le han he riangular wave, S1 i pen, an S i cle m 18m m m 4m 6m 8m 3m 3m 34m (3)- () Time Figure 16. Oupu vlage A crrepning he cnrl an carrier wave hwn in Figure 15.

14 The frequency f he riangular wave i f c, an he frequency f he cnrl wave i f. We efine he rai f hee a he frequency mulain inex, m f. f m c f = (8.17) f ikewie, he ampliue mulain inex, m a, i efine a he rai f he cnrl vlage he riangular wave vlage. m cnrl a = (8.18) riangular A ne phae, bull wave inverer i hwn in Figure 17. I ha fur cnrlle wiche, each wih an aniparallel ie. A S1 S B S4 S3 v AB = v A- Figure 17. One-phae full wave inverer. The iagnal wiche perae geher uch ha S1 an S3 pen an cle geher, an S an S4 pen an cle geher. The upu will cillae beween an. If he Furier ranfrmain f he pule wih mulae quare wave hwn in Figure 18 i aken, he ampliue f he funamenal will be a linear funcin f he ampliue inex = m a / a lng a m a 1. Then he RMS value f he upu vlage will be: m 353 a 1 = =. ma (8.19) v B

15 m 18m m m 4m 6m 8m 3 m 3 m 34m ()- (3) Time Figure 18. Oupu vlage AB fr he ne-phae, full wave inverer f Figure 17. When m a increae beyn 1, he upu vlage increae al, bu n linearly wih m a. The upu 4 ampliue can reach a peak value f when he reference ignal becme infinie an he upu i a π quare wave. Uner hi cniin, he RMS value i: 45 Equaing he pwer f he DC ie wih ha f he AC ie give Thu fr nrmal perain: an in he limi fr a quare wave: 1 = =. (8.) π P = I = 1 I 1 pf (8.1) I = a 1 I.353m I pf (8.) = 1.45I pf (8.3)

16 8.. Three-Phae Inverer Fr hree-phae la i make mre ene ue a hree-phae inverer hwn in Fig., raher han uing hree ne-phae inverer. S1 S S3 S4 S5 S6 A B C Figure 19. Three-phae, full wave inverer. The baic PWM cheme fr a hree-phae inverer ha ne cmmn carrier an hree eparae cnrl wavefrm. If he wavefrm we wan achieve are inuial he frequency mulain inex, m f, i lw we ue a ynchrnize carrier ignal wih m f an ineger muliple f m 17m 18m 19m m 1m m 3m 4m 1(cn1) 1(cn) 1(cn3) 1(ri) Time (a)

17 m 17 m 1 8m 19m m 1m m 3m 4m () Time (b) m 17 m 1 8m 19m m 1m m 3m 4m (3) Time (c) m 17m 18m 19m m 1m m 3m 4m ()- (3) Time () Figure. Three-phae, full wave inverer hwing (a) cnrl an riangular wave, (b) phae A line-neural, (c) phae B line--neural, an () line--line ignal frm phae A B.

18 m 17.4m 17. 8m 17. 1m m 17. m 17.4m 17.8m 17.3m 17.36m 17.4m () Time (a) m 17.4m 17. 8m 17. 1m m 17. m 17.4m 17.8m 17.3m 17.36m 17.4m (3) Time (b) m 17.4m 1 7.8m 17.1m m 17.m 17.4 m 17.8m 17.3m 17.36m 17.4m ()- ( 3) Tim e (c) Figure 1. Shrer ime pan fr he (a) phae A line--neural, (b) phae B line--neural, an (c) line-line ignal frm phae A B f he upu in Figure.

19 The PSPICE neli fr hi hree phae PWM circui i given belw: 3 phae FU-BRIDGE INERTER ************ OUTPUT IS (,3) **************** **********INPUT PARAMETERS ****************.PARAM urce = 1 ; DC inpu inverer.param F = ; funamenal frequency.param Mf = 1 ; carrier, muliple f F.PARAM Ma =.8 ; ampliue rai.param Fc = {Mf*F} ; carrier frequency S 1 DC {urce} ; c urce ******* OTAGE-CONTROED SWITCHES ***** S SWITCH S SWITCH S SWITCH S4 3 4 SWITCH S SWITCH S SWITCH ************** FEEDBACK DIODES ************* D1 1 DMOD D 3 1 DMOD D3 4 1 DMOD D4 DMOD D5 3 DMOD D6 4 DMOD ******************** OAD ****************** R1 5 1 ; la beween ne an 1 5 6MH R ; la beween ne 3 an 6 6MH R ; la beween ne 4 an 3 7 6MH *************** TRIANGE CARRIER ************** ri 3 PUSE (1-1 {1/(*Fc)} {1/(*Fc)} 1n {1/Fc}) ******************** REFERENCE ***************** cn1 4 SIN( {Ma} {F} {-9/Mf}) cn 5 SIN( {Ma} {F} {-9/Mf - 1}) cn3 6 SIN( {Ma} {F} {-9/Mf - 4}) ************ MODES AND COMMANDS *************.MODE SWITCH SWITCH(RON=.1 ON=.7 OFF=-.7).MODE DMOD D ; efaul ie.probe.tran.5ms 33.33MS 16MS.1MS.FOUR 5 I(R1) ; Furier ranfrm.options NOPAGE IT5=.END

20 A lng a m a i le han 1, he RMS value f he funamenal f he upu vlage i a linear funcin f i: 3 = m. 61m 1 l l a a (8.4) In he limi, when he cnrl vlage becme infinie, he RMS value f he funamenal f he upu i hen: 3 4 l l = π Again, equaing he pwer n he DC an AC ie we bain: r fr nrmal PWM perain: an in he limi fr he quare wave: (8.5) P = I = 3 l l1i1 pf (8.6) I = a 1 I 1.6m I pf (8.7) = I pf (8.8) Finally, here are her way cnrl he perain f an inverer. If i i n he upu vlage wavefrm we wan cnrl, bu raher he curren, we can eiher impe a fa cnrller n he vlage wavefrm, riven by he errr beween he curren ignal an he reference, r we can apply a hyerei ban cnrller, hwn fr ne leg f he inverer in Figure.

21 v A S1 S v AN 4.A.A A -.A -4.A 16 m 18 m m m 4 m 6 m 8 m 3 m 3 m 34m I( R) Ti me Cmparar lerance ban * A i _ Σ i ε i ε Swich- Me Inverer i A Figure. Curren cnrl wih hyerei ban Inverer Ne Wih a ine-riangle PWM he harmnic f he upu vlage are f frequency arun nf n, where n i an ineger an f n i he frequency f he carrier (riangle) wavefrm. The higher hi frequency i he eaier filer u hee harmnic. On he her han, increaing he wiching frequency al increae prprinally he wiching le. Fr 6-ep perain f a 3-phae inverer he harmnic are even, excep he riple ne, i.e. hey are f rer 5, 7, 11, 13, 17 ec. When he la f an inverer i inucive he curren in each phae remain piive afer he vlage in ha phae became negaive, i.e. afer he p wich ha been urne ff. The curren hen flw hrugh he aniparallel ie f he bm wich, reurning pwer he DC link. The ame

22 happen f cure when he bm wich i urne ff an he curren flw hrugh he aniparallel ie f he p wich Example 1 A hree-phae cnrlle recifier i upplying a DC mr ha ha k = 1 ( ) an R = 1 (Ω). The recifier i fe frm a 8 ( l-l ) urce. 8 () Filer Figure 3. Figure fr example prblem a) Calculae he maximum n-la pee f he DC mr: Wihu a la he curren i zer, herefre: = kω IR = kω The maximum pee i hen fun by he maximum DC vlage max = kω max The maximum DC vlage i prvie by he cnrlle recifier fr α = max = l l = 81.8 () Therefre, ω max = 8.8 (ra/) b) The mr nw i prucing rque f (N m). Wha i he maximum pee he mr can achieve? Nw ha here i la rque, here i al curren: Then T = ki I = (A) IR ω = = = 6.8 (ra/) k 1 c) Fr he cae in b) calculae he al RMS curren f he funamenal an he pwer facr a he AC ie A he maximum, he funamenal f he AC curren i:

23 The pwer facr i hen 1. I 1.78I = 15.6 (A) ) If he mr i nw cnnece a a generar wih a cuner rque f (N m) a 15 (rpm). Wha hul be he elay angle an AC curren? Fr a DC generar: T = kω IR = kω k π R = () 6 1 Since hi i a generar, he vlage i negaive fr he inverer () = c α = 119.º ( α ) 8..5 Example Fr he yem hwn in Figure 4, he AC urce i cnan, an he la vlage i 15 ( l-l ), (A), 5 (Hz),.85 pf lagging. 8 () Filer 6-ep invere α Figure 4. Figure fr example prblem a) Calculae he vlage n he DC ie an he DC cmpnen f he curren. Fr he 6-ep inverer P l l, 1 =.78 = 19 () pf = I I = 19 = 3 l lil = b) Calculae he AC urce ie RMS an funamenal curren an pwer facr. Fr a 3-phae recifier = 1.35 c α 19 = c α c α =. l l 3 (A) ( ) ( ) ( ) I I 1 = = pf = c α = (A) ( ) lagging

24 8.3 DC-DC Cnverin DC DC cnverin i fen aciae wih abilizing he upu while he inpu varie, hwever he cnvere i al require in me applicain, which i pruce a variable DC frm a fixe r variable urce. The iue f elecing cmpnen parameer an calculaing he perfrmance f he yem i he fcu f hi ecin. Since hee cnverer are wiche me yem, hey are fen referre a chpper Sep-Dwn r Buck Cnverer The baic circui f hi cnverer i hwn in Figure 5 cnnece a purely reiive la. If we remve he lw pa filer hwn an he ie, he upu vlage v () i equal he inpu vlage when he wich i cle, an zer when he wich i pen. The average upu,, i hen: 1 n T = n = T n T (8.9) w-pa filer i v C v = R (la) v n ff T = 1 f Figure 5. Tplgy f he buck chpper.

25 The rai f n /T = D, he uy rai. The lw pa filer aenuae he high frequencie (muliple f he wiching frequency) an leave alm nly he DC cmpnen. The energy re in he filer inucr (r he la inucr) ha be abrbe mewhere her han he wich, hence he ie, which cnuc when he wich i pen. Here we uy hi cnverer in he cninuu me f perain uch ha he curren hrugh he inucr never becme zer. A he wich pen an cle he circui aume ne f he plgie f Figure 6. v - A B - T = 1 f = I n ff v v C R C R Figure 6. Operain f he buck chpper. We will ue he fac ha he average vlage acr he inucr i zer, an aume a perfec filer uch ha he vlage acr he inucr i ( ) uring n, an fr he remainer f he cycle. n = ( ) ( ) = T n (8.3)

26 ( ) ( T ) = (8.31) n n = n D (8.3) T = Al uing he fac ha he inpu an upu pwer are he ame give: I = I (8.33) I I = D (8.34) = We analyze hi uner he aumpin f cninuu me f perain. In he icninuu me, he upu DC vlage i le han wha i given here, an he chpper i le eay cnrl. A he bunary beween cninuu an icninuu me, he inucr curren reache zer fr ne inan every cycle, a hwn in Figure 7. v - I, avg = I I,max = T D n ff T = 1 f Figure 7. Operain f he buck cnverer a he bunary f cninuu cnucin. Frm hi figure we can ee ha a hi peraing pin, he average inucr curren i I = ½, an: I 1 = n ( ) = ( ) DT (8.35) Since he average inucr curren i he average upu curren (he average capacir curren i zer), equain (8.35) efine he minimum la curren ha will uain cninuu cniin. Finally, a cnierain i he upu vlage ripple. We aume he ripple curren i abrbe by he capacir, i.e. he vlage ripple i mall. The ripple vlage i hen ue he eviain frm he average f he inucr curren a hw in fig. Uner hee cniin:

27 where ΔQ 1 1 ΔI T Δ = = (8.36) C ( D) T ΔI = 1 (8.37) Δ = 1 8 T C ( 1 D) (8.38) v - A B - T = 1 f = I n ff v Δ Figure 8. Analyi f he upu vlage ripple f he buck cnverer.

28 Anher way view hi i efine he wiching frequency f = 1/T an ue he crner frequency f he = 1 π C : filer ( ) f crn Δ π f ( 1 ) = D crn f (8.39) 8.3. Sep-Up r B Cnverer Here he upu vlage i alway higher han he inpu. The chemaic i hwn in Figure 9. v C R Figure 9. Schemaic iagram f a b cnverer. Bae n he cniin f he wich, here are w pible plgie a hwn in Figure 3. Again, he way calculae he relainhip beween inpu an upu vlage we ake he average curren f he inucr be zer, an he upu pwer equal he inpu pwer giving: ( )( T ) = (8.4) n n 1 = 1 D (8.41) I = 1 D (8.4) I

29 v - A B - T = 1 f = I n ff v v C R C R Figure 3. Tw circui plgie f he b cnverer. T eermine he value f inucance an capaciance we uy he bunary f cninuu cnucin like befre an he upu vlage ripple. A he bunary f he cninuu cnucin, he gemery f he curren wavefrm give: I T = D( 1 D) (8.43) The upu curren mu excee hi value fr cninuu cnucin. Uing he ripple analyi a hwn in Figure 31 we fin: Δ = DT RC (8.44)

30 I i impran ne ha he perain f a b cnverer epen n paraiic cmpnen, epecially fr uy cycle appraching uniy. Thee cmpnen will limi he upu vlage level well belw he given by equain (8.41). i D ΔQ ΔQ i D = I v Δ DT (1-D)T Figure 31. Calculaing he upu vlage ripple fr a b inverer Buck-B Cnverer Thi cnverer ha a chemaic hwn in fig. an can prvie upu vlage ha can be lwer r higher han he inpu vlage. Again he perain f he cnverer can be analyze uing he w plgie reuling frm he perain f he wich a hwn in fig. By equaing he inegral f he inucr vlage zer we ge: ( )( 1 D) T = DT (8.45) D = 1 D A he bunary beween cninuu an icninuu cnucin we fin I (8.46) T = ( 1 D) (8.47)

31 The upu ripple, a calculae frm Fig. i Δ = DT RC (8.48) i D v C R Figure 3. Baic Buck-B cnverer. i

32 v A B - T = 1 f I = (I I ) DT (1-D)T i D v C R v C R i i Figure 33. Operain f a Buck-B chpper. 8.3 Example The inpu f a ep wn cnverer varie frm 3 () 4 () an he upu vlage i be cnan a (), wih upu pwer varying beween 1 (W) an (W). The wich i peraing a 1 (khz). Wha i he inucr neee keep he inucr curren cninuu? Wha i hen he filer capacir neee keep he upu ripple belw %? The uy cycle will vary beween D 1 = /3 =.667 an D = /4 =.5. The la curren will range beween I 1 = 1/ = 5 (A), an I = / = 1 (A). The minimum curren neee keep he inucr curren cninuu i: I, min DT = ( )

33 Since he cnan i he upu vlage,, an he minimum la curren mu be greaer han I,min, we can expre i a a funcin f an make i le han r equal 5 (A), r: DT T ( A) I = ( ) = ( 1 D) 5, min T = 1/1(kHz), = (), an he maximum value i achieve fr D =.5, leaing min = 5 (µh). Fr he ripple, he highe will ccur a 1-D =.5, hu: π f. =.5 crn f 3 crn 1 1 = 9 (Hz) π C = 9 (Hz) C = 65 (µf)