Buck-Boost Converter

Transcription

1 Buck - Bst Cnverter Buck-Bst cnverter s a dc-t-dc cnverter that has the capablty f steppng up r steppng dwn the utput vltage. n ther wrds, the utput vltage can be hgher r lwer than the nput (surce) vltage. t s als labeled as ndrect cnverter because the surce s never drectly cnnected t the lad. t s the nductr n the crcut that cntrls the flw f energy frm the nput sde t the utput sde. Fgure-1 shws the buck-bst cnverter n ts smplest frm. Pay attentn t the drectns f the currents thrugh the capactr and the lad resstr and the plarty f the utput vltage. Fgure 1: Bst Cnverter The swtch s usually an electrnc devce that perates ether n the cnductn mde (when clsed) r the cut-ff mde (when pen). The cnductn and cut-ff tmeperds are cntrlled by the sutably desgned gatng crcuts, whch are usually nt shwn. The cnductn (n) tme f the swtch s a fractn f ts tme perd T such that T ON T, where s the duty cycle. urng the cut-ff (ff) tme when the swtch s pen, (1 )T, the nductr current s drected tward the lad va dde. When T OFF the swtch s clsed, the dde blcks the drect current flw frm the surce tward the lad. urng ths tme, the current s channeled thrugh the nductr and the capactr supples the lad current. Only when the swtch s n ts pen pstn, the nductr current flws tward the lad and the capactr. The dde helps mantan the current cntnuty thrugh the nductr. et us make an assumptn that the crcut has been peratng fr a lng tme and the nductr current vares between ts mnmum and maxmum values durng each tme half-tme perd. Guru/CC/BuckBst/ February 8, Buck-Bst Cnverter

2 We begn ur analyss when the nductr current s at ts mnmum and the swtch s clsed. The dfferental equatn fr the nductr current, fr ts slutn are d (t) dt (t) 0 t T ON T, and t,mn (1) Accrdng t ths equatn, the nductr current ncreases lnearly and attans ts maxmum value as t T ON T such that. max,mn (),max T efnng the change n the current frm ts mnmum t maxmum value as the peak-t-peak current rpple, the abve equatn yelds an expressn fr as,max,mn T (3) As sn as the nductr current reaches ts maxmum value, the swtch s pened. The nductr current nw begns t supply the lad current and charge the capactr n accrdance wth the fllwng dfferental equatn. d (t) dt The slutn f ths equatn yelds (t) t A where A s the cnstant ntegratn and s determned by applyng the ntal cndtn at t T n. That s,max T A Frm ths equatn, we btan A and the nductr current as A,max T (t) t,max T (4) Guru/CC/BuckBst/ February 8, 006 Buck-Bst Cnverter

3 As per ths equatn, the nductr current decreases lnearly frm ts maxmum value and attans ts mnmum value when t T, such that,max (5),mn (1 )T The peak-t-peak current rpple nw s,max,mn (1 )T (6) The current rpple as gven by (3) must be the same as gven by (6). Therefre, equatng the tw equatns, we get T (1 )T Ths equatn upn smplfcatn yelds (7) 1 Equatn (7) states that the utput vltage f the buck cnverter s drectly prprtnal t and ndrectly prprtnal t (1-). When 0.5, the utput vltage s exactly equal t the appled vltage. The utput vltage s greater than the appled vltage as lng as > 0. 5, whch crrespnds t the bst peratn. On the ther hand, the buck-bst cnverter behaves as a buck cnverter wth utput vltage less than the appled vltage when < When the swtch, the nductr, and the capactr are treated as deal elements, the average pwer dsspated by them s zer. Cnsequently, the average pwer suppled by the surce must be equal t the average pwer delvered t the lad. That s, lad current as 1 Ths equatn helps us express the average surce current n terms f the average The utput current s (8) 1 (9) 1 Guru/CC/BuckBst/ February 8, Buck-Bst Cnverter

4 as Equatns (8) and (9) help us express the surce current n terms f the duty cycle (10) 1 Fgure : urce current n a buck-bst cnverter The surce current as a functn f tme s shwn n Fgure-. t s smply a plt f the nductr current as gven by (1) when the swtch s n ts clsed pstn. The average surce current s T 1 t,mn dt,mn T T (11) 0 Fgure 3: nductr current Frm Fgure-3, whch s a plt f nductr current, t s clear that,avg (1),mn T,mn Guru/CC/BuckBst/ February 8, Buck-Bst Cnverter

5 Cnsequently, the average values f the surce and the nductr currents are related as,avg (13) The average nductr current frm (10) and (13) s,avg (14) ( 1 ) The expressns fr the maxmum and mnmum currents thrugh the nductr may nw be wrtten as,avg (15) (1 ) f,max,mn,avg (16) (1 ) f The peak-t-peak current rpple can be expressed n terms f the nput and utput vltages, as (1 ) (17) f f Fgure 4: The dde current The current thrugh the dde s shwn n Fgure 4. ts average value s the same as the average lad current and can be cmputed as max,mn TOFF (18) T,avg nce the average current n the dde s equal t the average current thrugh the lad resstr, the average current n the capactr s zer. Guru/CC/BuckBst/ February 8, Buck-Bst Cnverter

6 Hence, frm When the swtch s n ts clsed pstn, the capactr supples the lad current. 0 t T ON T, the capactr current s C (t) (19) When the swtch s pened, the nductr current supples bth the capactr current and the lad current. Thus, durng the tme nterval frm 0 t TOFF (1 ) T, the capactr current s C (t) (t) (0) The maxmum and mnmum values f the capactr current when the swtch s n ts pen pstn as C,max C,mn t must be nted that,max (1) 1 f,mn () 1 f (3) C,max C,mn The capactr current wavefrm s shwn n Fgure 5. Fgure 5: Current thrugh the capactr The current wavefrm f Fgure-5 helps us determne the change n the vltage acrss the capactr. urng the tme the swtch s clsed, the charge n the capactr s decreasng because the capactr s supplyng the current t the lad. The change n the charge s Q TON T (4) Guru/CC/BuckBst/ February 8, Buck-Bst Cnverter

7 The decrease n the charge wll result n the decrease f the capactr vltage. Therefre, the magntude f the change n the capactr vltage s Q T (5) C C Nte that durng the tme the swtch s pen, the cmpnent f the nductr current that flws thrugh the capactr wll ncrease the capactr vltage by the same amunt. Hence, when we defne the capactr rpple as the rat f the ncrease n the capactr vltage t ts average value, t can then be expressed as T C Cf (6) Nte that the capactr rpple defne by (6) s nt the same as the peak-t-peak vltage rpple fr the rectfers. The peak-t-peak vltage rpple fr the buck-bst cnverter wll be twce f that gven by (6). Equatn (6) may be vewed as ne-sded vltage rpple. The buck-bst cnverter can perate ether n ts cntnuus cnductn mde r dscntnuus cnductn mde. When t perates n the cntnuus cnductn mde, there s always a current n the nductr. The mnmum current n the cntnuus cnductn mde can be zer at the tme f swtchng. Cnsequently, there s a mnmum value f the nductr that ensures ts cntnuus cnductn mde. t can be btaned frm (16) by settng Hence,, mn t zer as (1 ) 0 (1 ) f mn (7) f mn (1 ) Frm the peak-t-peak current rpple, we can als btan an expressn fr the percent current rpple as %C,avg f (1 ) 100 mn (8) Guru/CC/BuckBst/ February 8, Buck-Bst Cnverter

8 Example: A buck-bst cnverter peratng at a frequency f 0 khz s used t step-up a 10- dc supply t 480- s that t can prvde the rated vltage t a 480-, 1000-W heater. T ensure a relable peratn, nductr must at least be 0% greater than ts mnmum value and the vltage rpple shuld be wthn 1%. esgn the bst cnverter. lutn: Frm the gven data, the duty cycle, frm (7), s The tme perd, the n tme and ff tmes f the swtch are 1 1 T 50µ s f 0,000 The equvalent resstr f the heater s 6 TON T µ s 6 TOFF (1 )T (1 0.75)(50 10 ) 1.5µ s Ω P 1000 The mnmum value f the nductr fr the cntnuus cnductn mde, frm (), s 30.4 mn (1 ) (1 0.8) 30.4µ H f 0,000 The mnmum value f the nductr t satsfy the desgn requrement must be 6 1. mn µ H T allw fr nductr-t-nductr varatns, let us select 300µ H The average current thrugh the heater s A 30.4 The peak-t-peak current rpple frm (1) s 10 6 T A Guru/CC/BuckBst/ February 8, Buck-Bst Cnverter

9 The average surce current s A 10 The average nductr current, frm (13), s A 0.8,avg Hence, the maxmum and mnmum currents thrugh the nductr, frm (8) and (9), are 16,avg A,max 16,avg A,mn et us nw select the capactr based upn the desred vltage rpple f 1%. Usng (16), we determne C as C f µ F , Fr ths applcatn, let us select a standard 0-µF capactr. The percent current rpple f the nductr current can be cmputed as %C,avg % n sme applcatns, such a hgh percent rpple may nt be acceptable. We have t ncrease the value f the nductr t decrease the rpple. Guru/CC/BuckBst/ February 8, Buck-Bst Cnverter