Small-Signal Modeling and Analytical Analysis of Power Converters

Transcription

1 Smll-Signl Modeling nd Anlytil Anlysis of Power Converters Christophe Bsso EEE Senior Member Chris Bsso APEC 03

2 Course Agend Liner nd Non-Liner Funtions Wht is Smll-Signl Model? Fst Anlytil Tehniques t Work From Swithed to Linerized Model The CCM M Smll-Signl PWM Swith Model The DCM M Smll-Signl PWM Swith Model Pek Current Mode Control in Lrge Signl The CCM CM Smll-Signl PWM Swith Model The DCM CM Smll-Signl PWM Swith Model The PWM Swith in Boundry Mode Chris Bsso APEC 03

3 Course Agend Liner nd Non-Liner Funtions Wht is Smll-Signl Model? Fst Anlytil Tehniques t Work From Swithed to Linerized Model The CCM M Smll-Signl PWM Swith Model The DCM M Smll-Signl PWM Swith Model Pek Current Mode Control in Lrge Signl The CCM CM Smll-Signl PWM Swith Model The DCM CM Smll-Signl PWM Swith Model The PWM Swith in Boundry Mode 3 Chris Bsso APEC 03

4 Liner nd Nonliner Funtions Funtions or systems n be liner or non-liner y f f y x x b y y x b The slope is onstnt long the onsidered setion: The slope is not onstnt long the onsidered setion: the funtion f is liner the funtion f is non liner x 4 Chris Bsso APEC 03

5 The Definition of Liner A liner system must fulfill the superposition priniple f x y f x f y Additivity u system y u y system u u Liner system y y f k x k f x Proportionlity of homogeneity u system y k u Liner system k y 5 Chris Bsso APEC 03

6 Time nvrine System properties do not hnge s time elpses u t y t t t t t u t y t t t t t t t Superposition Time invrine Liner Time nvrint (LT) 6 Chris Bsso APEC 03

7 Who is Liner? Chek properties of the liner funtion with n offset y u system u b f u system u b b y x b x u u system u u b u u b u u b Nonliner! k u k u b k u kb system k u b 7 Chris Bsso APEC 03

8 How to Linerize? Split the funtion in two prts, offset (d) nd urve () y y y f f b b + y x b x y0 d b x y x x u u u u k u system k u k u k u y = x is liner! 8 Chris Bsso APEC 03

9 How Do You Clulte The Slope? Look t the vertil vrition brought by horizontl step y y x x y x x x y y y y x x As y is lrger thn y, the slope in this se is positive 9 Chris Bsso APEC 03

10 Linerity in Non-Liner Funtion A non-liner funtion n be liner t n observtion point y y f Zoom y f y y x x x x x x - x must be smll enough to isolte liner setion Zooming further mens hving x x lose to eh other y x 0 Chris Bsso APEC 03

11 From Slope to Differentition To see liner zone y nd x must be very smll quntities x f x x x x When x is lose to x, we hve lulted the slope t x The slope t this point is lled the derivtive of f t x f f ( x x) f ( x ) x ' lim x0 ( x x ) x Also noted x nfinitesiml vritions re noted dy nd dx (Leibniz) df dx Chris Bsso APEC 03

12 Tngent nd Differentition A tngent is grphil representtion of differentition y B A C D E mx F min G f 'G x Piee Wise Liner y B A C D E F G x When the slope is zero, the funtion is minimum or mximum A PWL is liner pproximtion of non-liner funtion Chris Bsso APEC 03

13 Differentition nd Continuity "A smll input hnge gives smll output hnge" y y lim tt t f t f t f (t) f ' t t ontinuous t t lim f? tt f (t) ontinuous f ' t t Funtion f is disontinuous t point t : no finite differentition 3 Chris Bsso APEC 03

14 Why Do We Need Liner Equtions? f the funtion is liner, we n extrpolte points positions y y x x f y y x x Linerity implies no gin hnge during modultion y y x f x y? y y y G y G t t y y G y? Frequeny response explortion requires liner bloks t t 4 Chris Bsso APEC 03

15 Wht Does Smll-Signl Opertion Men? A smll mplitude omponent strddles d level Superimposed omponent d vlue bis point quiesent point v t d ˆv Full response u 79u 465u 65u 837u t s The signl mplitude is smll to mintin linerity 5 Chris Bsso APEC 03

16 Bis Point nd Liner Zone Where does the devie operte in its hrteristis? A 8.0m F 8.0m A F 4.0m 4.0m 0.0m F 0.0m F 6.00m.00m b 00m 300m 500m 700m 900m 7 ma 70m 6.00m.00m b 00m 300m 500m 700m 900m.8 ma 60m Bis point tells where the devie stnds in its d response Plese note pitl letters for d vlues 6 Chris Bsso APEC 03

17 8.0m 4.0m 0.0m 6.00m A Modultion t Bis Point The modultion linerly moves the bis point A F 4.9 ma b F 700m i f t t The modultion mplitude is smll enough: system is in liner mode.00m 598m 639m 680m 7m 76m v f t Smll-signl exittion t 7 Chris Bsso APEC 03

18 A Modultion t Bis Point Bis point hnge brings distortion to the modulted signls A 3.4m 0.0m 6.64m 3.6m F 650 m.9 ma b F i f t t The modultion mplitude is too big: system is in nonliner mode 0 399m 499m 599m 699m 799m v f t t distortion Lrge-signl exittion 8 Chris Bsso APEC 03

19 Course Agend Liner nd Non-Liner Funtions Wht is Smll-Signl Model? Fst Anlytil Tehniques t Work From Swithed to Linerized Model The CCM M Smll-Signl PWM Swith Model The DCM M Smll-Signl PWM Swith Model Pek Current Mode Control in Lrge Signl The CCM CM Smll-Signl PWM Swith Model The DCM CM Smll-Signl PWM Swith Model The PWM Swith in Boundry Mode 9 Chris Bsso APEC 03

20 A Smll-Signl Model How n we predit the response from the diode?.we need model ounting for the operting point.t must reflet the reltionship between nd t this point. We need smll-signl model f f d lineriztion r d Non-liner opertion 0 Chris Bsso APEC 03

21 Methods to Derive Smll-Signl Model Strt from the non-liner eqution.perturb the eqution: dd smll vrition to ll vribles.re-rrnge, sort d nd terms The urrent in the diode obeys Shokley eqution: F nt F S e the diode forwrd urrent n F F T S the diode forwrd drop therml voltge, 6 m t 5 C emission oeffiient, vries between nd the reverse bis sturtion urrent, few na for N448 Chris Bsso APEC 03

22 Method, Eqution is Perturbed Considering the exponentil muh lrger thn, we hve: F F nt nt F S e Se Now, eh vrible is perturbed by smll vrition The vrition is onsidered smll, the system remins liner, iˆ f vˆ, vˆ f in out perturb vˆ vˆ vˆ F F F F F iˆ e e e e n n n n F F S s F F T T T T in in out out S, n nd T re onstnt, perturb only F nd F : Chris Bsso APEC 03

23 Method, Eqution is Perturbed x Use the Pdé pproximnt of order : e x ˆ vˆ ˆ F vf F if F F F nt n Full response d T solte the term iˆ F F vˆ F n T r d vˆ n ˆF T i F F r d is the smll-signl model of the diode t urrent F 3 Chris Bsso APEC 03

24 A 3.4m Method, Eqution is Perturbed The dynmi resistne r d is the tngent slope 0.0m 6.64m F r d r d F F F F 3.6m 0 F r d F 399m 499m 599m 699m 799m r d F F F t is evluted t given operting point 4 Chris Bsso APEC 03

25 Reple the Nonliner Model For the trnsfer funtion, ompute the smll-signl model.30.3 AC = 0. in s 3 R k F out s 64m D N448 F F 64m.66 ma From the urve rd 00Ω Plug it into the iruit nd solve the equtions in AC = 0. s 3 R k out rd 00 s out in d d s r s r R 9m 5 Chris Bsso APEC 03

26 Method, Use Prtil Differentition Sometimes, sorting out nd d equtions is triky Prtil derivtive is n esy wy to get -only terms There is n nlogy with the superposition theorem vˆin ˆv ˆd G vˆout You ompute the output vlue while inputs re zero vˆ G vˆ G vˆ G dˆ out,, d in,, d, d, out in out out in in The totl response is the sum of the individul responses 6 Chris Bsso APEC 03

27 Prtil Differentition A funtion depends on multiple vribles We n lulte the rte of hnge of this funtion when: df x x vries while y nd z re onstnt y vries while x nd z re onstnt z vries while x nd y re onstnt dx df y dy df z f x, y, z Mthemtilly, we perform prtil differentition, noted: Totl rte of hnge f f f df dx dy dz x y z Prtil rtes of hnge dz 7 Chris Bsso APEC 03

28 Prtil Differentition With one vrible, prtil nd totl derivtives re similr f ' x df x f x dx x Prtil derivtive is fster with omplex equtions out L p Ri out F sw You hve three vribles, you n perturb nd ollet out iˆ out L ˆ v Ri F fˆ vˆ p sw sw out out out... iˆ... out d eqution eqution 8 Chris Bsso APEC 03

29 Prtil Differentition or you n use Mthd nd utomte the proess,,,,,, ˆ out Fsw out ˆ out Fsw out out Fsw out i f vˆ vˆ F out sw out sw out You hve the result in twinkling of n eye L F L F L iˆ f vˆ vˆ p ˆ sw p sw p out sw out Ri out Ri out Ri out We n pply this tehnique to our diode eqution F nt Se F ˆ n vˆ T F F i vˆ e vˆ n n F F S F F T T r d n F T 9 Chris Bsso APEC 03

30 Course Agend Liner nd Non-Liner Funtions Wht is Smll-Signl Model? Fst Anlytil Tehniques t Work From Swithed to Linerized Model The CCM M Smll-Signl PWM Swith Model The DCM M Smll-Signl PWM Swith Model Pek Current Mode Control in Lrge Signl The CCM CM Smll-Signl PWM Swith Model The DCM CM Smll-Signl PWM Swith Model The PWM Swith in Boundry Mode 30 Chris Bsso APEC 03

31 The Bipolr Smll-Signl Model A bipolr trnsistor is highly non-liner system Reple it by its smll-signl model to get the response R b R b b g in Q out r e b R b Re Ce e Remember bipolrs Ebers-Moll model, sonny! 3 Chris Bsso APEC 03

32 A Liner System The new iruit is smll-signl liner rhiteture s b s in b s r s b s out R b R b e e s R Re Ce Lple nottion pplies to get the trnsfer funtion 3 Chris Bsso APEC 03

33 The Trnsfer Funtion A trnsfer funtion links response to n exittion H s out in s s response exittion t must be written in so-lled low entropy form b b s 0 H s s 0 b 0 b s b s ftor b z G0 ftor 0 0 s s n this st -order expression you n identify the terms G 0 b 0 0 z b b 0 H s d gin, s = 0 zero pole p N s D s 0 p zeros poles 33 Chris Bsso APEC 03

34 Finding the Zeros G 0 is found by shorting indutors nd opening pitors s s s s in out in out R r C C R d R r C R G 0 R R R Zeros prevent the exittion from rehing the output s s in Z s out response r C r C C sc sr C sc exittion Z s 0 R C Z s 0 sr C 0 C z rc C 34 Chris Bsso APEC 03

35 Finding the Poles The poles re linked to the time onstnts of the system These time onstnts solely depend on the struture Remove the exittion signl to isolte the struture voltge soure short iruit urrent soure open iruit The denomintor order depends on the storge elements storge element L storge elements st -order C C nd -order 35 Chris Bsso APEC 03

36 The Finl Answer Short the voltge soure nd lulte the time onstnt R r C C R R R? r C R R r R R p C rc R R C r R R C C The omplete trnsfer funtion is obtined in 3 steps H s G 0 s z R src C s R R s r R R C p C We derived the trnsfer funtion by inspetion! Fst Anlytil Tehniques for Eletril nd Eletroni Ciruits,. orpérin, Cmbridge Press, Chris Bsso APEC 03

37 The Trnsistor Amplifier We hve just one pitor, this is st -order system s s out z G0 s s in G 0 is found by opening the pitor b s out s in s b s Not involved r b r b r R e b s e e p s R s s R s R s out b b e s s s R be in e e r b s r in s s r R s s b b G 0 s s R in b e R R r R R e e r e 37 Chris Bsso APEC 03

38 Looking for the Zeros Wht ould prevent the exittion from rehing the output? s b s in b s r s b s out R b R b e e s R Re Ce Z s When this impedne is infinite, there is no bse urrent 38 Chris Bsso APEC 03

39 Looking for the Zeros The frequeny where the impedne is infinite is our zero Re Ce Z s Re Re sc sr C e e e When the denomintor is equl to zero, Z is infinite srece 0 z We re lmost there R C e s R sr C s R D s out e e in e e 39 Chris Bsso APEC 03

40 Looking for the Poles For the system time onstnt, short the exittion b r b b b R R? b R b R e e R R r R e R Write two simple equtions R b r b R R r Re 40 Chris Bsso APEC 03

41 The Finl Trnsfer Funtion We hve our trnsfer funtion in few steps only out s R srece in s Re r scere The next step is to hek mths versus simultion AC = 4 3 rp 000 Re.k Ce u B Current ()*00 R 0k out R e.k r P k R 0k C e F 00 ( xy ) z R z f R e C z 7.343Hz G e 0 r P ( ) R e p p f r p P R e C e 6.47kHz x y x y 4.48 s z H ( s) G 0 s p Mthd 4 Chris Bsso APEC 03

42 The Finl Trnsfer Funtion The urves superimpose, lultions re good! H f db H f f Hz Alwys run this snity hek to verify your results 4 Chris Bsso APEC 03

43 Agend Liner nd Non-Liner Funtions Wht is Smll-Signl Model? Fst Anlytil Tehniques t Work From Swithed to Linerized Model The CCM M Smll-Signl PWM Swith Model The DCM M Smll-Signl PWM Swith Model Pek Current Mode Control in Lrge Signl The CCM CM Smll-Signl PWM Swith Model The DCM CM Smll-Signl PWM Swith Model The PWM Swith in Boundry Mode 43 Chris Bsso APEC 03

44 Non-Linerity in Swithing Converter A swithing onverter is ruled by liner equtions x L SW L y u x C R during DT sw u on off C R x x L C R during DT sw ombining so-lled stte vribles 44 Chris Bsso APEC 03

45 Stte ribles Stte vribles desribe the mthemtil stte of system n stte vribles for n independent storge elements knowing vribles stte t t 0 helps ompute outputs for t > t 0 u u u t t t 3 nput vetor u System desribed by stte vribles x, x, x3..., xn y y y t t t 3 Output vetor y x is the indutor urrent nd x is the pitor voltge Differentition gives the stte vrible rte of hnge x i t L x dil dt t Predit future system stte x v t C x dvc dt t 45 Chris Bsso APEC 03

46 Desribe the System During the On-Time Observe the system during the on-time durtion or dt SW : u x L ic t x R C x R t dvc ic t C Cx x x dt R t di u L v t Lx x dt L C x x u L L x x x C RC x 0 L x L 0 u x C RC x 0 0 u Stte oeffiients Soure oeffiients 46 Chris Bsso APEC 03

47 Desribe the System During the Off-Time Repet the exerise during the off-time durtion or -dt SW vl t L x ic t C R ir t x L v t x R x L Lx v t Lx i t x i t x Cx R C x i t R x R x Cx x 0 x L x x x C RC x 0 L x 0 0 u x C RC x 0 0 u Stte oeffiients Soure oeffiients 47 Chris Bsso APEC 03

48 Mke it Fit the Stte Eqution Formt Arrnge expressions to mke them fit the formt: x Ax t Bu t Stte eqution on-time network x 0 L x L 0 u x C RC x 0 0 u A 0 L L 0 C RC B 0 0 off-time network x 0 L x 0 0 u x C RC x 0 0 u A 0 L 0 0 C RC B 0 0 How do we link mtrixes A nd A, B nd B? We smooth the disontinuity by weighting them by D nd -D x A D A D x t B D B D u t 48 Chris Bsso APEC 03

49 The Stte-Spe Averging Method (SSA) We now hve ontinuous lrge-signl eqution We need to linerize it vi perturbtions D D dˆ x x0 xˆ u u0 uˆ 0 x A D A D x t B d B D u t u ˆ 0d L ˆ ˆ D d x xˆ uˆ u L L L ˆ x xˆ xˆ C RC 0 0 u D0 ˆd ˆx C ˆx Cnonil smllsignl model 49 Chris Bsso APEC 03

50 The Stte-Spe Averging Method (SSA) SSA ws pplied to swithing onverters by Dr Ćuk in 976 t is long, pinful proess, mnipulting numerous terms Wht if you dd n EM filter for instne? L SW L y u L L C C R u C on off C R L u C L C R 4 stte vribles nd you hve to re-derive ll equtions! 50 Chris Bsso APEC 03

51 The PWM Swith Model in oltge Mode We know tht non-linerity is brought by the swithing ell L u C R p Why don't we linerize the ell lone? : tive : ommon p: pssive d PWM swith M Swithing ell p Smll-signl model (CCM voltge-mode)... orpérin, "Simplified Anlysis of PWM Converters using Model of PWM Swith, prts nd " EEE Trnstions on Aerospe nd Eletroni Systems, ol. 6, NO. 3, 990 p 5 Chris Bsso APEC 03

52 Reple the Swithes by the Model Like in the bipolr iruit, reple the swithing ell L u C R p.. nd solve set of liner equtions!.. L u C R 5 Chris Bsso APEC 03

53 An nvrint Model The swithing ell mde of two swithes is everywhere! buk d PWM swith M p d PWM swith M p boost buk-boost d PWM swith M p d PWM swith M p d PWM swith M Ćuk p 53 Chris Bsso APEC 03

54 Smoothing the Disontinuity A CRT T displys frmes t ertin rte, 50 per seond t The opti nerve time onstnt is lrger thn n intervl A suession of disrete events is seen s ontinuous Low-frequeny filtering ntegrtion See "phi phenomen", 54 Chris Bsso APEC 03

55 Averging Wveforms The keyword in the PWM swith is verging pek vt A t v t T sw DT sw T sw T sw sw vt vt dt pek dt pek D pek D ' Tsw T T sw 0 T sw DT The resulting funtion is ontinuous in time sw 55 Chris Bsso APEC 03

56 From Steps to Continuous Funtion Some funtions require mthemtil bstrtion: duty rtio T sw vt t t t 3 t t 4 5 D D D3 D 4 Disrete vlues of D Averge nd ontinuous evolution of d(t) D n t T n sw D 5 f mod F sw At the modultion frequeny sle, points look ontiguous Link them through ontinuous-time ripple-free funtion d(t) 56 Chris Bsso APEC 03

57 The Common Pssive Configurtion The PWM swith is single-pole double-throw model d i t i t d ' v t v t p nstll it in buk onverter nd drw the wveforms i t d i t L d ' in C R vp t vp t p p p out CCM 57 Chris Bsso APEC 03

58 The Common Pssive Configurtion Averge the urrent wveforms ross the PWM swith i t i t T sw i 0 t 0 t i t t T sw D Averged vribles DT sw dt sw i t i t dt D i t D T sw T T sw sw 0 CCM 58 Chris Bsso APEC 03

59 The Common Pssive Configurtion Averge the voltge wveforms ross the PWM swith vp t v p t T sw vp t 0 t p D p 0 DT sw dt v t p t sw v t v t dt D v t D p T p p p p sw T Tsw sw 0 T sw Averged vribles CCM 59 Chris Bsso APEC 03

60 A Two-Port Representtion We hve link between input nd output vribles p D p d Two-port ell p D p t n further be illustrted with urrent nd voltge soures d p p D D p p p CCM 60 Chris Bsso APEC 03

61 A Trnsformer Representtion The PWM swith lrge-signl model is d "trnsformer"!.. D p p D t n be immeditely plugged into ny -swith onverter D p p p D D D r L L. p. in C R d equtions! CCM 6 Chris Bsso APEC 03

62 Simulte mmeditely with this Model SPCE n get you the d bis point 0.0 R 00m L 00u 9.80 C (,p)*(d) 5 p p 4.0 out 7 g 0 300m AC = d Rdum u (d)*(c) C 470u R 0 but lso the response s it linerizes the iruit db rg H f 0 00 k 0k 00k H Hz f CCM 6 Chris Bsso APEC 03

63 We Wnt Trnsfer Funtions Derive the d trnsfer funtion: open ps., short indutors in r L p D.. p out R out R out out R out D D R out in rl p out in rl Dout out out D in r L out in out in D rl D D ' r L RD ' R CCM 63 Chris Bsso APEC 03

64 Plotting Trnsfer Funtions Plot the lossy boost trnsfer funtion in snpshot f( d0.) 5 R 0Ω rl 0.Ω out in f( d0.) 4 f( d0.3) f( d0.4) f( d0.5) 3 f( d0.6) f( d0.7) f( d0.8) f( d0.9) rl 0.Ω f( d) rl Ω Duty rtio Above ertin onversion rtio, lth-up ours d CCM 64 Chris Bsso APEC 03

65 Course Agend Liner nd Non-Liner Funtions Wht is Smll-Signl Model? Fst Anlytil Tehniques t Work From Swithed to Linerized Model The CCM M Smll-Signl PWM Swith Model The DCM M Smll-Signl PWM Swith Model Pek Current Mode Control in Lrge Signl The CCM CM Smll-Signl PWM Swith Model The DCM CM Smll-Signl PWM Swith Model The PWM Swith in Boundry Mode 65 Chris Bsso APEC 03

66 A Smll-Signl Model We need smll-signl version to get the response We n pply the lineriztion tehnique we lerned D vribles,, f D ˆ ˆ f D i ˆ d i D iˆ dˆ Diˆ p dˆ D p vribles Diˆ, p, p f D f D vˆ ˆ ˆ p d v D dˆ p Dvˆp p dˆ p dˆ p D vˆ dˆ Dvˆ p p p.. D iˆ p vˆp p Smll-signl model CCM 66 Chris Bsso APEC 03

67 Plug this Smll-Signl Model in the Boost You now hve ompletely liner nd ontinuous model = 0 r L ˆ pd D dˆ R out in L D.. p C r C Re-rrnge the iruit with the PWM swith r L L dˆ p D.. p dˆ C R r C out 67 Chris Bsso APEC 03

68 Reflet on the Other Side if Needed Write d equtions to define the d trnsfer funtion in D.. p p out in p out p Dout D in out out out D D ' in Apply similr tehnique to reflet impednes R eq?.. D R p out P out Pin R R out in D ' R R out out eq eq Req RD ' 68 Chris Bsso APEC 03

69 p D s Simplify the Ciruit nd Solve it! The iruit n now be updted with Lple nottions s rl L D ' D s s.. D s p r C C R out s We hve two storge elements, it is seond-order system sz sz... out s N s H s H H D s D s s s 0Q Chris Bsso APEC 03

70 dentify the Zeros A zero prevents the exittion from rehing the output s s s s p s D s rl L D ' D s.. D p C s R p r C out s No response mens: s s 0 r C sc 0 rc C is short or iˆ p 0 0 D s s sl r p L src C Two zeros 0 sc 70 Chris Bsso APEC 03

71 dentify the Zeros The first zero is strightforwrd src C 0 sz We know s s D s D s r C C This is LHP zero from the smll-signl model Sine p s 0, we hve s s s D s D s We n rerrnge the equtions 0 D s s sl r p L s D s D s s p D s sl r s s D D s L D ' D s 7 Chris Bsso APEC 03

72 in Equte nd solve for s D s pd s D ' sl r L dentify the Zeros Wht is the vlue of?.. D R D ' p rl p sz D ' rl L L out P in out in P out in out out out D ' out D ' p This is RHP zero L sz D ' R rl out p D ' R 7 Chris Bsso APEC 03

73 dentify the Poles Let's reflet the lod nd pitor on the other side rl D s p L D ' D s C D ' RD '.. D out s rc D ' The denomintor is solely dependent on the struture t is independent from the exittion: set it to zero! dentify the time onstnts 73 Chris Bsso APEC 03

74 dentify the Poles We redue the exittion to zero: Ds 0 r L C D ' L r D ' C RD ' D.. out D must be dimensionless thus: s D s s s This is two-storge elements network 0Q 0 Hz Hz s s The two possible terms for re The two possible terms for re ' ' L or R RC B. Erikson, "The n Extr Element Theorem", 74 Chris Bsso APEC 03

75 dentify the Poles For look t the resistne R driving L nd C Look t the driving impedne t L while C is in its d stte Look t the driving impedne t C while L is in its d stte R? R? rl r ' CD RD ' r L rc D ' RD ' L C... s rl RD ' r D ' rl RD ' D ' C R r RD L ' R rl RD ' r D ' C 75 Chris Bsso APEC 03

76 dentify the Poles how (involving L) ombines with ' (involving C)? how (involving C) ombines with ' (involving L)? Look t the driving impedne t C while L is in its HF stte Look t the driving impedne t L while C is in its HF stte r L rc D R? ' RD ' C D ' r R ' D ' f we hose R? r ' L RD L Sme result ' ' r L rc D ' RD ' ' rl RD ' rc D ' f we hose C rl RD ' rc D ' D ' L 76 Chris Bsso APEC 03

77 dentify the Poles We hve our denomintor! L rl R rc R D s s C r C s LC rl RD ' rl RD ' rl RD ' We n identify the terms 0 rl LC RD ' r R C Q 0 r L L C r R The d term H 0 in the trnsfer funtion is: H 0 dout D D dd out C in in in H D 0 D D ' 77 Chris Bsso APEC 03

78 The CCM Boost Trnsfer Funtion The finl trnsfer funtion n be written s: s s out s z z H 0 D s s s 0Q 0 z z D ' R rl rc C L 0 rl LC r C RD ' R Q 0 r L L C r R C H 0 in D ' Time to hek the response versus simultion! 78 Chris Bsso APEC 03

79 Cheking A Responses Cpture simple boost onverter shemti R L 00u X PWMCCMM 9.73 d PWM swith M p m R 0. out R3 0 in 0 D 0.4 in out D D D 0.6 L 00H C 0F r L 0. r C 0. R 0 z r C C z D L R r z L f 7.343kHz 0 L C 0 f 0 r L D R r C R Hz z f Hz ESR zero RHP zero 0.4 AC = C 0u in Q.646 H R r L L 0 0 C r C D R D r L R D r L s s z H ( s) H z 0 s s 0 Q 0 Chek the result versus Mthd spredsheet 79 Chris Bsso APEC 03

80 Cheking the A Responses The urves must perfetly superimpose db H f rg H f 0 00 k 0k 00k f not, go bk to the sheet nd fix the equtions! f Hz 80 Chris Bsso APEC 03

81 Agend Liner nd Non-Liner Funtions Wht is Smll-Signl Model? Fst Anlytil Tehniques t Work From Swithed to Linerized Model The CCM M Smll-Signl PWM Swith Model The DCM M Smll-Signl PWM Swith Model Pek Current Mode Control in Lrge Signl The CCM CM Smll-Signl PWM Swith Model The DCM CM Smll-Signl PWM Swith Model The PWM Swith in Boundry Mode 8 Chris Bsso APEC 03

82 The Disontinuous Cse n DCM, third timing event exists when i L (t) = 0 il t L D D SW L in C R during D T sw D 3 in il t on off C R il t L C R during D T sw 0 DT sw T sw DTsw il t 0 D3Tsw t il t 0 during C R D T 3 sw 8 Chris Bsso APEC 03

83 The Sme Configurtion s in CCM Drw the wveforms in the "ommon pssive" onfigurtion i vp t t pek in t in p D D D 3 on off L C out R i vp t t DT sw in T sw pek DTsw D3Tsw out t t t Averge the wveforms: pek D D D D D pek pek pek D D D D D D 83 Chris Bsso APEC 03

84 Derive p to Unveil the New Model The ddition of the third event omplites the equtions vp t D D p p p 3 D D D3 p DT sw p DTsw p D3Tsw.. N t p D D D p p p D p p D D N N p p Np N Control input N D D D f D 84 Chris Bsso APEC 03

85 Finlly, Get the D lue n DCM the indutor verge voltge per yle is lwys 0 p out Wht is the verged indutor pek urrent? pek v t L DT sw L v t L D T sw pek D D D T sw L D T The pek urrent uses previous expression D pek D pek D LFsw D D sw 85 Chris Bsso APEC 03

86 CCM to DCM Auto-Toggling is Possible We hve two equtions for the DCM model D D D D p p D D When D 3 shrinks nd finlly disppers D D D3 D D f you substitute D in the two bove equtions D D D D p p D D p D D p CCM equtions 86 Chris Bsso APEC 03

87 Clmp the Eqution to Auto Toggle D in DCM is smller thn D in CCM D, DCM D D3 D, CCM D Clmping D eqution to D offers uto-toggling LF sw F D D D THEN D D LFsw ELSE D D D (DCM) SPCE offers severl mens to lmp (CCM) Ed d 0 vlue = { F ( ((*{L}*{Fs}*(M)/((d)*(,)+u)) - (d))>(-(d)),-(d), + *{L}*{Fs}*(M)/((d)*(,)+u)) - (d) ) } Bd d 0 =((*{L}*{Fs}*(M)/((d)*(,)+u)) - (d))>(-(d))? -(D) : + (*{L}*{Fs}*(M)/((d)*(,)+u)) - (d) PSpie sspie C. Bsso,"SMPS: SPCE Simultions nd Prtil Designs", MGrw-Hill Chris Bsso APEC 03

88 Run Simultion mmeditely! Re-use the CCM boost exmple nd plug the DCM model 0.0 R 00m L {L} C (,p)*(n) 5 p p 3.5 out 7 g 0 50m AC = d 3.n Rdum u (N)*(C) C 470u R 50 d Bd oltge prmeters Fsw=00k L=00u (*{L}*{Fsw}*(C)/((d)*(,)))-(d) N BN oltge (d)/((d)+(d)) mode 79m 58m 0 DCM Bmode oltge (*{L}*{Fsw}*(C)/((d)*(,)))-(d) > (-(d))? : 0 88 Chris Bsso APEC 03

89 db A Response in Snpshot The response is tht of seond-order system! H f Low-frequeny pole High-frequeny pole rg H f RHP zero! 0 00 k 0k 00k Meg - f Hz 89 Chris Bsso APEC 03

90 g 0 R 00m 0.0 Clulte the D Operting Point Let's rework the DCM model to fit tht of the CCM N D D D D LFsw D D (p)*()*(d)^/(*{fsw}*{l}*(c)) C p out m AC = d N F B Current out D L sw Short indutors, open ps. nd verify bis point 3.5 -()*(d)^/(*{fsw}*{l}) R 50 out prmeters Fsw=00k L=00u R in D out F L R sw Equtions: out out D LF sw Chris Bsso APEC 03

91 dentify the Duty Rtio nd M Expressions Solve nd from the first equtions: R 0 in D R LF R R out sw in oltge t node R D LF D R R R D F L in sw out out in R sw Lout Fsw Substitute these vribles in the third eqution in in LFsw out in LFsw out R D out in R D LFsw out LF M M sw D M out LFsw out in R in R D.348 LFsw 0.5 ok ok 9 Chris Bsso APEC 03

92 Deriving the Smll-Signl Model Lrge-signl expressions require lineriztion Apply prtil derivtives to the equtions N 3 vribles,,,,,, f D ˆ ˆ f D f D i d iˆ vˆ D Mthd utomtes the proess esily ˆ sw ˆ D D i d vˆ F L F L sw k D F sw L k D F sw L iˆ k dˆ k vˆ 9 Chris Bsso APEC 03

93 Smll-Signl Soures in DCM Smll-signl perturbtions would be more diffiult! p N p 4 vribles,,, p,,, p,,, p,,, p f D f D f D f D vˆ dˆ vˆ iˆ vˆ p p D p p D ˆ D p D ˆ pd p p FswL FswL F F sw L swl vˆ d vˆ i vˆ pd D pd p D sw sw F sw L sw k k k k F L F L F L vˆ k dˆ k vˆ k iˆ k vˆ p 3 4 p 5 6 A snity hek is reommended before proeeding 93 Chris Bsso APEC 03

94 erifying the ntermedite Step A SPCE simultion will let us know if the derivtion is ok prmeters Fsw=00k L=00u d=50m =-9.99 p=-3.5 =-3.m =-0.956m 7 R 00m g AC = L {L} d k=*d/(fsw*l) k=d^/(*fsw*l) k3=*p*d/(fsw*l*) k4=*d^/(*fsw*l*) k5=-*p*d^/(*fsw*^*l) k6=p*d^/(*fsw**l) Rdum u Sme results s with lrge-signl model! C {k3}*(d)+{k4}*(,p)+{k5}*(c)+{k6}*(,) 5 {k}*(d)+{k}*(,) db H p p f rg H f C 470u R 50 out -360 f 0 00 k 0k 00k Meg Hz 94 Chris Bsso APEC 03

95 Crefully Deriving the Expressions The iruit n be further simplified s vˆ 0 in is zero C {k3}*(d)-{k4}*(out)+{k5}*(c)-{k6}*() out s L {L} 5 s D s {k}*(d)-{k}*() B Current C 470u R 50 out sc R Z s s s out There re three equtions s s sl out s Z s out s s s k D s k s k s k s out out s k D s k s 95 Chris Bsso APEC 03

96 Rerrnging the Expression Further to few mnipultions, you should find H s k kk3 kk6 sl out s R k R k k D s k R R k R k C L R k k k R k R R k 3 5 k3 kk s s CL k5 R Rk 4 R k5 Rk 4 This is seond-order system ffeted by RHP zero s s z G H s 0 s s 0Q 0 z k k k k k L k3 kk5 3 6 Q k k R k L k kk4 Ck5 R R G 0 R k R k k k R R k LC k R 5 k k Chris Bsso APEC 03

97 A Low-Q System A seond-order system fetures two poles These poles re the hrteristi eqution roots D s s s 0Q 0 Ds 0 s p, s p 4 Q 0 Q When Q is low, use MLurin series to simplify n x x nx x x s s p p 4Q Q Q Q Q 4Q Q 0 0 Q0 Q Q D s s s s s p p A nd -order low-q system is equivlent to sded poles 97 Chris Bsso APEC 03

98 A Simpler Expression We n thus rewrite simpler trnsfer funtion H s G 0 s s s s z z s s s s p p Further to rerrnging, simplifying nd hving fun G 0 R M M M F sw L p Adding the ESR M M R C R z F sw z M L p sw sw M F F D The DCM model is of the sme order s the CCM model The differene is in the dmping C r C. orpérin,"anlytil Methods in Power Eletronis", n-house lss, Toulouse, 004 r C C src C 0 sc 98 Chris Bsso APEC 03

99 H ( s) Full Mthd Expression versus SPCE The urves perfetly superimpose, lultions re orret! 50 0 H i f k 0 log 0 0 rgh i f k 50 R k 3 R k k 5 k 5 R R k sl k k k 3 k k 6 k 3 k k 5 R k 5 C L R k k 4 k 6 R k s s C k 5 R R k L 4 db H f rg H f f k R R k 6 k 5 R R k 4 f 80 Hz Yes! 99 Chris Bsso APEC 03

100 Why RHP Zero in DCM? The RHP Zero is present in CCM nd is still there in DCM! d (t) i L,pek D 3 T sw t D T sw D T sw T sw When D inreses, [D,D ] stys onstnt but D 3 shrinks 00 Chris Bsso APEC 03

101 Why RHP Zero in DCM? The tringle is simply shifted to the right by ˆd d (t) i L,pek D 3 T sw ˆd t T sw D T sw D T sw The pitor refueling time is delyed nd drop ours 0 Chris Bsso APEC 03

102 The Refueling Time is Shifted f D inreses, the diode urrent is delyed by ˆd d (t) D(t) ˆd.0m.04m.06m.09m.m.85m.99m.3m.7m.4m out (t) out (t).0m.04m.06m.09m.m Simultion, no pek inrese 0 Chris Bsso APEC 03

103 Course Agend Liner nd Non-Liner Funtions Wht is Smll-Signl Model? Fst Anlytil Tehniques t Work From Swithed to Linerized Model The CCM M Smll-Signl PWM Swith Model The DCM M Smll-Signl PWM Swith Model Pek Current Mode Control in Lrge Signl The CCM CM Smll-Signl PWM Swith Model The DCM CM Smll-Signl PWM Swith Model The PWM Swith in Boundry Mode 03 Chris Bsso APEC 03

104 Pek Current Mode Control n voltge-mode, the loop ontrols the duty rtio n urrent-mode, the indutor pek urrent is ontrolled lok PWM lth Q R vr t Slope omp. i v L t t R i i t R v t L i r D D ' R i p An rtifiil rmp is dded for stbiliztion purposes 04 Chris Bsso APEC 03

105 pek How Does Pertubtion Propgte? The perturbtion is rried yle by yle in CCM t t il S S 0 i T L R i L sw L t L pek il S b t Current setpoint S L T sw R t i T sw DT sw T sw DT sw T 0 S DT S D ' T L sw L sw sw S DT sw ' S D T sw S S =0 D D ' L 0 pek St b pek St L Tsw pek pek b S S L ntsw L 0 generlized D D ' n 05 Chris Bsso APEC 03

106 il t nstbility in CCM for D > 50% pek R i L 0 pek il t D 50% Asymptotilly stble T d sw Tsw dtsw d3tsw d4tsw vllley t R i L 0 D 50% Asymptotilly unstble t 06 Chris Bsso APEC 03

107 Add n Artifiil Rmp An externl rmp helps to stbilize the urrent L L pek 0 0 il 0 t S Current setpoint S DT sw T sw S R i L T sw R t i L( ntsw ) L(0) S S S 0 S 0 for 00% duty rtio S S S D ' S D S 50% S n Considering 00% duty rtio overompenstes the design 07 Chris Bsso APEC 03

108 Wht is the Control Lw in CCM? Drw the primry urrent with stbiliztion rmp S pek i t Current setpoint S R i R i Theoretil vlue S = 0 T sw S S Effetive vlue 0 DT sw t T sw The urrent t point is defined by: R sw i S DT R i 08 Chris Bsso APEC 03

109 We Wnt the Averge Current Definition The vlue is the indutor urrent t hlf the ripple t i pek T sw S R i b L S 0 DT sw T sw D Tsw t Express urrent t point b nd substitute the definition L S D ' T b T sw sw T sw S S D T R ' sw DTsw i Ri 09 Chris Bsso APEC 03

110 Define the Converter off-slope Use buk onfigurtion to see voltges t ply i in t d p d ' i v t v t p p t L C R out The downslope depends on the output voltge out : S out L The indutor verge voltge is 0 t stedy-stte p out S p L 0 Chris Bsso APEC 03

111 A Current Mode Genertor Updte the previous eqution to obtin finl definition Tsw S D DT R L R p sw i i Pek urrent setpoint Hlf indutor ripple Compenstion rmp Group nd nd 3 rd terms Tsw S p D DT L R i sw ndutor ripple nd ompenstion rmp lter pek vlue R i L SDTsw R i R i p p Chris Bsso APEC 03

112 CM or M Led to Similr nput Currents Averge the urrent wveforms ross the PWM swith i t i t T sw i 0 t t i t T sw D 0 dt sw t p dt sw i t i t dt D i t D T sw T T sw sw 0 D p p CCM Chris Bsso APEC 03

113 The PWM Swith Model in Current Mode The finl model ssoites three urrent soures in p p R i L C R p This is the lrge-signl urrent-mode PWM swith model Cn you think of something simpler?! No!. orpérin,"anlysis of Current-Controlled PWM Converters using the Model of the PWM Swith", PCM Conferene,990 3 Chris Bsso APEC 03

114 How to Model Subhrmoni nstbilities? The model, s it is, nnot predit instbilities Let's observe smll-signl perturbtion in v i t vˆ vˆp L 0 t T sw The off-slope does not hnge s vˆp keeps onstnt This "memory" effet is modeled with pitor 4 Chris Bsso APEC 03

115 Finl Model nludes Subhrmoni Effets A simple pitor is enough to mimi instbility in p p R i C s resonnt tnk L C R p As the instbility is pled t hlf the swithing frequeny: Fsw LC s C s L F sw 5 Chris Bsso APEC 03

116 The Lrge-Signl Model t Work A urrent-mode 5-/5-A buk onverter s n exmple p B3 Current B ()/{Ri} Current (D)*(C) R3 m p {Se}*(D)/({Ri}*{Fsw}) + v(,p)*(-(d))*({/fsw}/(*{l})) C L uH B4 Current -.50m C {Cs} C 00uF 0 v D.8 500m stim B R 00m.8 oltge v(,p)/v(,p) AC = 4 R out prmeters Fsw=00kHz L=00u Cs=/(L*(Fsw*3.4)^) Ri=50m Se=0 Duty rtio lultion Rmp ompenstion is djusted to see its effets 6 Chris Bsso APEC 03

117 nstbilities Show up s Expeted Peking is tmed by inresing slope ompenstion db Se 0 Subhrmoni peking out f f Se 0 k s Se out f f 0 00 k 0k 00k f Hz Se 0 k s 7 Chris Bsso APEC 03

118 Another Wy of Modeling We n build duty rtio ftory nd use the M model pek i t Current setpoint S R i R i T sw S S 0 t DT sw T sw S DT DT R R L sw sw i i D F R R i S L sw i 8 Chris Bsso APEC 03

119 Build Duty Rtio Ftory An in-line eqution builds D from the ontrol voltge p B Current (D)*(C) R3 m -.50m p 4.99 C B3 oltge (D)*(,p).8 v stim.8 AC = PWM M 500m D 4.99 B oltge L 00uH C 00uF 0 R 00m R out prmeters Fsw=00kHz L=00u Cs=/(L*(Fsw*3.4)^) Ri=50m Se=0 ((()-(C)*{Ri})*{Fsw})/({Se}+{Ri}*(,)/(*{L})) Duty rtio ftory 9 Chris Bsso APEC 03

120 No Sub-Hrmoni Osilltions Predition Low-frequeny gins re identil but no peking t F sw / db PWM CM out f f Duty rtio ftory Duty rtio ftory out f f PWM CM k 0k 00k f Hz Se 0 0 Chris Bsso APEC 03

121 Agend Liner nd Non-Liner Funtions Wht is Smll-Signl Model? Fst Anlytil Tehniques t Work From Swithed to Linerized Model The CCM M Smll-Signl PWM Swith Model The DCM M Smll-Signl PWM Swith Model Pek Current Mode Control in Lrge Signl The CCM CM Smll-Signl PWM Swith Model The DCM CM Smll-Signl PWM Swith Model The PWM Swith in Boundry Mode Chris Bsso APEC 03

122 Smll-Signl Modeling of PWM CM There re two equtions to linerize: p Tsw S p p Tsw R i p L Ri p p p p Tsw S p T p p R i p L Ri p p sw Apply prtil differentition: 3 vribles,,,,,, p p p p p p iˆ vˆ vˆ vˆ p p p p,,,,,, p p p p p p iˆ vˆ vˆ vˆ p p p p 3 vribles Chris Bsso APEC 03

123 Smll-Signl Modeling of PWM CM Rerrnge the oeffiients nd rewrite equtions: iˆ k vˆ g vˆ g vˆ iˆ k vˆ g vˆ g vˆ o f p o p i i p r p Where: k k k k o i o i R i D R i R i D R i g f g g f g i i STswD TswD R L i p D g f p DD ' T Dgo L D g f p sw g g r o Tsw D TswD STswD L L R p Tswp D L p Using orpérin nottion nd T sw S go D ' D L Sn gr god p iˆ i p... g vˆ o p S n, on-slope S f, off-slope S, externl rmp 3 Chris Bsso APEC 03

124 A Smll Signl Model The model inludes urrent soures nd ondutnes g i v k vˆp g r vˆp g f vˆ k o ˆ i g o C s p k k o i R i D R i g f g i Dg o D g f DD ' T L p sw T sw S go D ' D L Sn gr god p 4 Chris Bsso APEC 03

125 The Model t Work Compre responses of the linerized lrge-signl model 0.0 prmeters Fsw=00kHz L=00u Cs=/(L*(Fsw*3.4)^) Ri=50m Se=.5k 0 p B3 Current B ()/{Ri} Current (D)*(C) R3 m p {Se}*(D)/({Ri}*{Fsw}) + v(,p)*(-(d))*({/fsw}/(*{l})) C L 4.99 {L} B4 Current -.50m C {Cs} C 00uF 0 v D.8 500m stim B R 00m.8 oltge v(,p)/v(,p) AC = 4 R out with tht of the smll-signl model prmeters Fsw=00kHz Tsw=/Fsw L=00u Cs=/(L*(Fsw*3.4)^) Ri=50m Se=.5k p R4 {/gi} R3 m B5 Current (v)*{ki} p B6 Current (,p)*{gr} B3 Current (,p)*{gf} 9 B4 Current (v)*{ko} v D.8 500m stim {} AC = 4.99 R5 {/go} C C {Cs} -.50m B oltge v(,p)/v(,p) L 4.99 {L} 4.99 C 00uF 0 R 00m 5 4 R out 5 Chris Bsso APEC 03

126 Results re dentil dentil results prove tht our smll-signl modeling is ok! db out f f out f f 0 00 k 0k 00k f Hz 6 Chris Bsso APEC 03

127 Current Loop nstbilities Let's ssume input nd output voltges re onstnt onstnt vˆ 0, onstnt onstnt vˆ 0 in out p p The buk implementtion simplifies to: 0 0 vˆp g f vˆ k o g o C s L C 0 R g o C s L p Fsw Ciruit pproximtes urrent loop dynmis ner 7 Chris Bsso APEC 03

128 This is Resonting Tnk For look t the resistne R driving L nd C Look t the driving impedne t L while C is in its d stte Look t the driving impedne t C while L is in its d stte C L R? C R? L g o g o Lgo 0 D s slg s o 8 Chris Bsso APEC 03

129 Find the Seond-Order Coeffiient how (involving L) ombines with ' (involving C)? Look t the driving impedne t C while L is in its HF stte g o C R? L ' C Lgo g o LC ' C g o The denomintor n thus be expressed s: Ds D s slg s LC o 0 sw Lg o s s 0Q 0 Q Q 0 sw Q Lg sw o 9 Chris Bsso APEC 03

130 Wht Brings Q to nfinity? We now substitute the definition of g o Q swlgo T sw S S L D ' D D ' D Tsw L Sn Sn This definition mthes the one derived by Ry Ridley Q m D ' 0.5 m S Wht hppens if S = 0 with 50% duty rtio? S n Q D D = 50% Q R. B. Ridley, "A New Continuous-Time Model for Current-Mode Control", EEE Trnstions of Power Eletronis, April Chris Bsso APEC 03

131 3 Chris Bsso APEC 03 How to Keep the Loop Stble? To prevent overompenstion, dmp Q below ' n S D D S 0.5 ' n S D S D Keeps the denomintor wy from 0 up to D = 00% ' 0 n S D D S ' 0.5 n S D D S 0.5 n f n S S D S S f S S ' f n S D S D D Using Ry Ridley's nottions, we hve: 0.5 ' m D 0.5 ' n S S D 0.5 ' n S D S D

132 Do Not Over Compenste the Loop Adding more rmp shifts the ontrol towrds voltge mode d t S v t + Sn vout t The modultor trnsfer funtion hnges to F pek m pek v t sense S S T n sw S sense Sn F m S T sw S T Smll signl pek sw Ds F m s + G PWM pek H s s R e L i k s k s f in r out 3 Chris Bsso APEC 03

133 Externl Rmp Effets on Power Stge The double pole splits nd joins the low-frequeny one db S 0 P S S n oltge-mode LC peking strts to pper P, P P P out f f 0 00 k 0k 00k P 3 f Hz 33 Chris Bsso APEC 03

134 Agend Liner nd Non-Liner Funtions Wht is Smll-Signl Model? Fst Anlytil Tehniques t Work From Swithed to Linerized Model The CCM M Smll-Signl PWM Swith Model The DCM M Smll-Signl PWM Swith Model Pek Current Mode Control in Lrge Signl The CCM CM Smll-Signl PWM Swith Model The DCM CM Smll-Signl PWM Swith Model The PWM Swith in Boundry Mode 34 Chris Bsso APEC 03

135 The PWM Swith in DCM The PWM CM n work in disontinuous mode pek i t Current setpoint S S S R i R i Theoretil vlue S = 0 T sw D3T sw 0 DT sw DTsw t T sw The verge urrent is somewhere in the downslope S pek D T S R sw i D T S R sw i D T S sw 35 Chris Bsso APEC 03

136 Derive the ndutor Averge Current We must now obtin the vlue of to get pek S pek T sw DTsw is the re under the tringle divided by the swithing period t D D pek D D pek pek pek pek pek D D D D 36 Chris Bsso APEC 03

137 Adopt the CCM Struture for DCM Substitute nd rerrnge to get the indutor urrent D T S D D R R L sw p DTsw i i f we stik to the originl CCM rhiteture with Ri D T S D D sw p DTsw Ri L R i D T S D D sw p DTsw Ri L p 37 Chris Bsso APEC 03

138 Disontinuous Wveforms Let's hve look t the PWM swith voltges in DCM i t pek T sw vp t in t i t DCM t vp t p D3T sw t 0 DT sw DTsw out t 38 Chris Bsso APEC 03

139 Derive the Duty Rtios From the DCM voltge-mode PWM swith we hve: D p p D D D D p p p From the operting DCM wveforms pek D pek D D D pek D D D Almost there, just need to express D pek L D T sw nd DT sw L D D pek D D D LFsw D D 39 Chris Bsso APEC 03

140 The DCM Model is Complete! in We n use this model for DCM simultions p R i L C R D D D D T S D D sw p DTsw Ri L 40 Chris Bsso APEC 03

141 Another Wy of Modeling We n build duty rtio ftory nd use the M model pek i t Current setpoint S R i pek R i pek R i S R i DT sw S S pek L DT sw 0 t DT sw T sw D Fsw R S L R i i 4 Chris Bsso APEC 03

142 Build Duty Rtio Ftory An in-line eqution builds D from the ontrol voltge PWM M 4.93 C 4.93 L {L} 4.93 out prmeters Fsw=00kHz L=00u Ri= Se= (N)*(C) R3 m -5.0u 57m p v p stim AC = 57m (,p)*(n) D N D 30m 493m 38m B Bd oltge C 00uF 0 R 00m 4 R 00 (/((bs(v(,))/{l})+({se}/{ri})))*{fsw}*(v)/{ri} BN (*{L}*{Fsw}*(C)/(((d)*(,)))-(d)+u) oltge (d)/((d)+(d)) Duty rtio ftory 4 Chris Bsso APEC 03

143 Build Duty Rtio Ftory You n pply the tehnique to nother M model 0.0 L {L} out prmeters Fsw=00kHz L=00u Ri= Se= 0 57m X CCM-DCM FS = Fsw L = L v stim AC = 57m d s k 304m B C 00uF 0 R 00m 4 R 00 (/((bs(v(,))/{l})+({se}/{ri})))*{fsw}*(v)/{ri} This is the uto-toggling CoPEC model 43 Chris Bsso APEC 03

144 Chek the DCM Model Results Chek simultion results with Ridley's models AC model prmeters Fsw=00kHz L=00u Ri= in D D Gnd 57m Ctrl 3 out stim AC = 57m X BUCKDCM R = Ri L = L RS = 0M FS = Fsw OUT = 5 RL = 00 N = 0 MC = C 00uF 0 R 00m 4 R 00 out 44 Chris Bsso APEC 03

145 Chek the DCM Model Results Bode plots re similr long the frequeny xis db out f f out f f k 0k 00k Meg 0 45 Chris Bsso APEC 03

146 Agend Liner nd Non-Liner Funtions Wht is Smll-Signl Model? Fst Anlytil Tehniques t Work From Swithed to Linerized Model The CCM M Smll-Signl PWM Swith Model The DCM M Smll-Signl PWM Swith Model Pek Current Mode Control in Lrge Signl The CCM CM Smll-Signl PWM Swith Model The DCM CM Smll-Signl PWM Swith Model The PWM Swith in Boundry Mode 46 Chris Bsso APEC 03

147 Why QR Opertion? More onverters re using vrible-frequeny opertion This is known s Qusi-Squre Wve Resonnt mode: QR lley swithing ensures extremely low pitive losses DCM opertion sves losses in the seondry-side diode Esier synhronous retifition The Right Hlf-Plne Zero is pushed to high frequenies i t i t D Less noise d Smooth signls vds t Low C² losses vds t 47 Chris Bsso APEC 03

148 Wht is the Priniple of Opertion? The drin-soure signl is mde of peks nd vlleys A vlley presene mens: The drin is t minimum level, pitors re nturlly dishrged The onverter is operting in the disontinuous ondution mode in vlley 0 0 t on t off DT vds t.06m.064m.066m.069m.07m BCM = Borderline or Boundry Condution Mode Flybk struture 48 Chris Bsso APEC 03

149 A QR Ciruit Does not Need Clok The system is self-osillting urrent-mode onverter Demg detetor bulk m L p N p :N s... C out R lod out R esr S Q Q dd R + FB R pullup - CTR G FB R sense 49 Chris Bsso Huwei Tehnil Seminr 0

150 vux t A Winding is Used to Detet Core Reset When the flux returns to zero, the ux. voltge drops Disontinuous mode is lwys mintined dely bulk Core is reset vds t il p t 00 0 dely d vux t N dt t vds t A 800m 400m 0 il p t vux t m -800m 0.5m.55m.60m.64m.69m set 50 Chris Bsso Huwei Tehnil Seminr 0

151 rible-frequeny Current-Mode Observing the wveforms helps us deriving n verge model i t i t DT sw pek pek The struture is similr to tht of the urrent-mode model t D R i T sw t t T sw pek p 5 Chris Bsso Huwei Tehnil Seminr 0

152 Derive the Operting Point nsert the model in simple buk onverter pplition in D R i L C out R p Derive the d operting point: short indutors, open ps 5 Chris Bsso APEC 03

153 Write d Equtions The indutor pek urrent is disontinuous i t S L L DT sw pek t pek R i D S t DT L pek L on sw L R T i sw T sw Derive the swithing frequeny expression t on L R i t on L R i p L Tsw ton toff R i p S on S off 53 Chris Bsso APEC 03

154 A Lrge Signl Model A Qusi Resonnt model is built with the PWM swith model D p R i D L R T i sw L R T i sw T sw L R i p These re lrge-signl equtions tht need lineriztion f ˆ i vˆ ˆ i vˆ vˆ k k R i R vrible i R,,,,,, p p p iˆ vˆ iˆ vˆ p p,,, p p i p 3 vribles 54 Chris Bsso APEC 03

155 Lrge to Smll-Signl Finl steps before the smll-signl model ˆ i vˆ iˆ vˆ k 0 0 p0 p0 0 p p p 0 0 p0 p 0 0 p0 0 p0 0 p0 p k i p0 p00 k 0 p0 p0 0 A smll-signl model n now be ssembled k k i k k ondutne ondutne ondutne p 55 Chris Bsso APEC 03

156 The Model t Work in n solted Converter The model n be inserted into flybk onfigurtion in i Lp v.k v(,p).kp i.ki v(,).k p Linerized QR PWM swith model X5 XFMR RATO = N Resr Cout Rlod out p0 k k k p i out N 0 0 Ri 0 in 0N R N i out in out out Nin out 0N R N in i out in 56 Chris Bsso APEC 03

157 Strt the Study with the d Gin in is onstnt in smll-signl, hene v = 0 ˆin R eq R N lod k p X4 XFMR RATO = N out B8 Current ()*{k}*({ki}-) R8 {/kp} RlodN {Rlod/N^} G k k R N 0 i eq 57 Chris Bsso APEC 03

158 Put ll Elements Bk in Ple The nlysis requires the omplete model p B6 Current ()*{k} ESR {ESR/N^} X5 XFMR RATO = N out B5 Current (,p)*{kp}+()*{k}*{ki}+()*{k} Lp {Lp} Cout {Cout*N^} Rlod {Rlod/N^} Kirhhoff's Current nd oltge Lws now pply 58 Chris Bsso APEC 03

159 Finl Trnsfer Funtion Equtions The plnt fetures one Right Hlf-Plne Zero kp k s s sr ESRCout p ˆ sl v s Nk k k s z s z G v s N R s out i i 0 ˆ N k ESR p N kpresr Rlod R lod sc sp out G Nk k i 0 k p N R lod s p s z k p R N R ESR C out N C lod out lod s N RESR kpr Rlod N kp R z ESR ki kp k L p Lp R i in 59 Chris Bsso APEC 03

160 Chek Anlytil Results ersus Simultion A snity hek is importnt to verify the derivtion prmeters in=00 Rlod=0 N=-0.5 ESR= Cout=00u Lp=m =.7 Ri= Fsw =5.6k =/(*Ri) =00 p=76.9 k=/(*ri) kp=*/(p+)^ ki=p/(p+) k=p*/(p+)^ Lrge-signl model in {in} 7.0 ton 5.6 Fsw 850m 00 ton Fsw (khz) 7 0 B3 oltge (Lp) 5 Lp {Lp} v 9 PWM swith BCM p B oltge ()+{} X PWMBCMCM L = Lp Ri = Ri 0 AC = X XFMR RATO = N ESR {ESR} Cout {Cout} out Rlod {Rlod} 60 Chris Bsso APEC 03

161 Finl Lp! Compre simulted results with Mthd plots 7.9 db 0 5 f p =0 Hz log G ( i f ) 0 rg ( G ( i f ) ) f z =.6 khz 80 f z =8.7 khz f They perfetly superimpose 6 Chris Bsso APEC 03

162 Conlusion Smll-signl modeling is n importnt prt of design Understnding the tehnique is key to quikly deriving equtions Stte-spe verging is n option but it is tedious nd long Smll-signl modeling using the PWM swith is simple nd fst Avilble tools help to perform intermedite snity heks Anlytil nlysis does not shield you ginst lb. experiments Anlytil nlysis, simultion nd benh: the best ombintion! Meri! Thnk you! Xiè-xie! 6 Chris Bsso APEC 03