Chapter 5: Discontinuous conduction mode. Introduction to Discontinuous Conduction Mode (DCM)

Transcription

1 haper 5. The isconinuous onducion Mode 5.. Origin of he disconinuous conducion mode, and mode boundary 5.. Analysis of he conversion raio M(,K) 5.3. Boos converer example 5.4. Summary of resuls and key poins Inroducion o isconinuous onducion Mode (M) Occurs because swiching ripple in inducor curren or capacior volage causes polariy of applied swich curren or volage o reverse, such ha he curren- or volage-unidirecional assumpions made in realizing he swich are violaed. ommonly occurs in dc-dc converers and recifiers, having singlequadran swiches. May also occur in converers having wo-quadran swiches. Typical example: dc-dc converer operaing a ligh load (small load curren). Someimes, dc-dc converers and recifiers are purposely designed o operae in M a all loads. Properies of converers change radically when M is enered: M becomes load-dependen Oupu impedance is increased ynamics are alered onrol of oupu volage may be los when load is removed 5.. Origin of he disconinuous conducion mode, and mode boundary Buck converer example, wih single-quadran swiches Q i () coninuous conducion mode (M) i () I i i () Minimum diode curren is (I i ) c componen I = / urren ripple is i = ( ) T s = ' Noe ha I depends on load, bu i does no. conducing devices: Q Q i () I i 3

2 educion of load curren Increase, unil I = i Q i () M-M boundary i () i () Minimum diode curren is (I i ) c componen I = / urren ripple is i = ( ) T s = ' I conducing devices: Q Q i () i Noe ha I depends on load, bu i does no. I i 4 Furher reduce load curren Increase some more, such ha I < i Q i () i () isconinuous conducion mode i () Minimum diode curren is (I i ) c componen I = / urren ripple is i = ( ) T s = ' I conducing 3 devices: Q X Q i () Noe ha I depends on load, bu i does no. The load curren coninues o be posiive and non-zero. 5 Mode boundary I > i I < i for M for M Inser buck converer expressions for I and i : < ' Simplify: < ' This expression is of he form K < K cri () for M where K = T and K cri ()=' s 6

3 K and K cri vs. for K < : for K > : K < K cri : M K > K cri : M K > K cri : M K = / K cri () = K cri () = K = / 7 riical load resisance cri Solve K cri equaion for load resisance : where < cri () for M > cri () for M cri ()= 'T s 8 Summary: mode boundary K > K cri () or < cri () for M K < K cri () or > cri () for M Table 5.. M-M mode boundaries for he buck, boos, and buck-boos converers onverer K max cri() ( K ) cri min cri() ( cri ) Buck ( ) Boos ( ) 7 4 Buck-boos ( ) ( ) T s ( ) 7 Ts ( ) T s 9

4 5.. Analysis of he conversion raio M(,K) Analysis echniques for he disconinuous conducion mode: Inducor vol-second balance v = v () d apacior charge balance = i = i () d = Small ripple approximaion someimes applies: because v << i() I is a poor approximaion when i > I onverer seady-sae equaions obained via charge balance on each capacior and vol-second balance on each inducor. Use care in applying small ripple approximaion. Example: Analysis of M buck converer M(,K) i () v () i () subinerval Q i () i () subinerval v () i () i () i () subinerval 3 v () i () Subinerval v ()= i ()=i ()/ i () v () i () Small ripple approximaion for (bu no for i()!): v () i () i () /

5 Subinerval v ()= i ()=i ()/ i () v () i () Small ripple approximaion for bu no for i(): v () i () i () / 3 Subinerval 3 v =, i = i ()=i ()/ i () v () i () Small ripple approximaion: v ()= i ()= / 4 Inducor vol-second balance v () 3 ol-second balance: Solve for : v () = ( ) () 3 () = = noe ha is unknown 5

6 apacior charge balance node equaion: i ()=i () / i () i () / capacior charge balance: i = hence i = / i () mus compue dc componen of inducor curren and equae o load curren (for his buck converer example) i pk <i > = I 3 6 Inducor curren waveform peak curren: i () i ( )=i pk = average curren: <i > = I i pk i = i () d i riangle area formula: i () d = i pk ( ) =( ) ( ) 3 equae dc componen o dc load curren: = ( )( ) 7 Soluion for Two equaions and wo unknowns ( and ): = (from inducor vol-second balance) = ( )( ) (from capacior charge balance) Eliminae, solve for : = 4K / where K = / valid for K < K cri 8

7 Buck converer M(,K). M(,K) K =..8 K = K =.5 K M = for K > K cri for K < K cri 4K / Boos converer example i() v () i () i () Q Mode boundary: Previous M soln: I > i I < i for M for M I = ' i = Mode boundary ' > for M K cri ().5 K cri( 3 ) = 4 7 > ' for M. where K > K cri () for M K < K cri () for M K = T s and K cri ()='

8 Mode boundary.5 where K > K cri () for M. K < K cri () for M K = T s and K cri ()=' M M K < K cri M K > K cri K.5 Kcri () onversion raio: M boos i() v () i () subinerval i() i () i() v () Q i () subinerval v () i () i() v () i () subinerval 3 3 Subinerval i() v ()= i ()=/ v () i () Small ripple approximaion for (bu no for i()!): v () i () / < < 4

9 Subinerval v ()= i ()=i()/ i() v () i () Small ripple approximaion for bu no for i(): v () i () i() / < < ( ) 5 Subinerval 3 v =, i = i ()=/ i() v () i () Small ripple approximaion: v ()= i ()= / ( ) < < 6 Inducor vol-second balance v () 3 ol-second balance: ( ) 3() = Solve for : = g noe ha is unknown 7

10 apacior charge balance node equaion: i ()=i ()/ capacior charge balance: i = hence i = / i () i () mus compue dc componen of diode curren and equae o load curren (for his boos converer example) 8 Inducor and diode curren waveforms peak curren: i pk = average diode curren: i() i pk riangle area formula: i = i () d i () d = i pk 3 i () i pk <i > 3 9 Equae diode curren o load curren average diode curren: i = i pk = equae o dc load curren: = 3

11 Soluion for Two equaions and wo unknowns ( and ): = g = (from inducor vol-second balance) (from capacior charge balance) Eliminae, solve for. From vol-sec balance eqn: = Subsiue ino charge balance eqn, rearrange erms: K = 3 Soluion for K = Use quadraic formula: = ± 4 / K Noe ha one roo leads o posiive, while oher leads o negaive. Selec posiive roo: = M(,K)= 4 / K g where valid for K = / K < K cri () Transisor duy cycle = inerval duy cycle 3 Boos converer characerisics 5 M(,K) 4 3 K =. K =.5 K =. K 4/7 M = 4 / K for K > K cri for K < K cri Approximae M in M: M K 33

12 Summary of M characerisics Table 5.. Summary of M-M characerisics for he buck, boos, and buck-boos converers onverer K cri() M M(,K) M (,K) M M() Buck ( ) 4K / K M(,K) Boos ( ) 4 / K K M(,K) Buck-boos ( ) K K wih K = /. M occurs for K < K cri. 34 Summary of M characerisics M M(,K) Boos Buck-boos ( ) Buck K M buck and boos characerisics are asympoic o M = and o he M buck-boos characerisic M buck-boos characerisic is linear M and M characerisics inersec a mode boundary. Acual M follows characerisic having larger magniude M boos characerisic is nearly linear 35 Summary of key poins. The disconinuous conducion mode occurs in converers conaining curren- or volage-unidirecional swiches, when he inducor curren or capacior volage ripple is large enough o cause he swich curren or volage o reverse polariy.. ondiions for operaion in he disconinuous conducion mode can be found by deermining when he inducor curren or capacior volage ripples and dc componens cause he swich on-sae curren or off-sae volage o reverse polariy. 3. The dc conversion raio M of converers operaing in he disconinuous conducion mode can be found by applicaion of he principles of inducor vol-second and capacior charge balance. 36

13 Summary of key poins 4. Exra care is required when applying he small-ripple approximaion. Some waveforms, such as he oupu volage, should have small ripple which can be negleced. Oher waveforms, such as one or more inducor currens, may have large ripple ha canno be ignored. 5. The characerisics of a converer changes significanly when he converer eners M. The oupu volage becomes loaddependen, resuling in an increase in he converer oupu impedance. 37