1 / 10

Switch- Mode Regulators

Switch- Mode Regulators. Buck Regulator Boost Regulator Buck-Boost Regulator Cúk Regulator. Buck Regulator. L. i L. i 0. i s. v g. i c. R o. V s. C. v o. i D. v g. For 0 ≤ t ≤ KT v s – v o = L ∆ i L / (KT) assuming that v o ≈ constant

gray-burks
Télécharger la présentation

Switch- Mode Regulators

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Switch-Mode Regulators • Buck Regulator • Boost Regulator • Buck-Boost Regulator • Cúk Regulator Buck Regulator L iL i0 is vg ic Ro Vs C vo iD

  2. vg • For 0 ≤ t ≤ KT • vs – vo = L ∆ iL / (KT) assuming that vo ≈ constant • then, ∆ iL = K (vs – vo) / (f L) …..(1) • where ∆ iL = iLmax- iLmin • For KT ≤ t ≤ T • vo = L ( - ∆ iL /(T-KT) • ∆ iL = vo (1-K) / (f L) …..(2) • equating (1) and (2), • then, K (vs – vo) / (f L) = vo (1-K) / (f L) • then , Vo = K Vswhere Vo is the average value of vo • Another method; • Noting that is = iL in the interval 0 ≤ t ≤ KT Vo Vs t 2T T T+KT KT 0 iL iLmax io K iLmin 0 1 t 2T T T+KT KT 0 ic t T+KT KT 0 2T T is iLmax io vo iLmin ∆vo (exaggerated!) 2T T+KT KT 0 T t noting that Vs Is = Vo Io and, Vo = K Vs 2T T T+KT KT 0 then, Is = K io

  3. vg Ic is the average rectified value of ic Ic = ½ T/2 ∆IL/2 = ⅛ ∆IL /f ∆ Vo = 1/C . Ic = ∆IL / (8fC) = vo (1-K) / (8f2LC) t 2T T T+KT KT 0 Boost Regulator For 0 ≤ t ≤ KT mode 1 L iD i0 is ic Ro Vs C is vo vg Vs i0 vg ic Ro vo

  4. is ismax mode (2) For KT ≤ t ≤ T Is ismin t 2T T T+KT KT 0 iC is Is – Io For 0 ≤ t ≤ KT Vs = L ∆ is / (KT) ∆ is = KVs/(fL) …(1) For KT ≤ t ≤ T Vs – Vo = - L ∆ is /(T-KT) ∆ is = (1 – K) (Vo – Vs) / (fL) …(2) then, K Vs = (1 – K) (Vo – Vs) Vo = Vs / (1 – K) t Io . K T = (Is –Io) . ( T – KT) then, Is = Io / (1-K) ∆ Vo = 1/C . (Io . KT) ∆ Vo = K Io / f C 2T T+KT KT 0 T Io iD is i0 vo ic ∆vo Ro Vs C vo t 2T T T+KT KT 0 Vo ! 2Vs Vs K 0 0.5 1

  5. Buck-Boost Regulator is vg vg t 2T T T+KT KT 0 Ro Vs L iL C vo iL iLmax ic io iLmin i0 iD for 0 ≤ t ≤ KT is iLmax t is 2T T T+KT KT 0 iLmin t 2T 0 KT T T+KT iD Ro C vo Vs L iL iLmax iLmin ic io t 2T 0 i0 KT T T+KT Mode 1 Is K = Io (1 – K) Is = K / (1-K) . Io Vs = L ∆ iL / KT ∆ iL = K Vs / (f L)

  6. from mode 1 : Vs = L ∆ Is / (KT) ∆ Is = K Vs / (fL) …(1) from mode 2 : Vo = L ∆ Is/ (T - KT) ∆ Is = (1 – K) Vo /( fL) …(2) then, K Vs = ( 1-K ) Vo Vo = K Vs / (1-K) ∆ vC = 1/C . KTIo ∆ vC = K Io / (fC) Ro L iL C vo Vo ic ! i0 iD Vs K For KT ≤ t ≤ T mode 2 0 1 iC 0.5 t 2T 0 KT T T+KT IC vC ∆vC t 0 KT T 2T T+KT

  7. Ćuk Regulator vg L2 L1 C1 is i0 t 2T T T+KT KT 0 Vs Ro vo 2C vg ic iD iL2 L2 L1 C1 iL1 iL2 _ + VC1 Mode 1 For 0 ≤ t < KT Vs Ro vo 2C ic i0 iL2

  8. L2 L1 C1 iL2 Mode 2 For KT ≤ t < T _ + VC1 Vs Ro vo 2C ic i0 iL2 For 0 ≤ t < KT Vs = L1 ∆ IL1/(KT) (1) ∆ IL1 = K Vs /(fL1) For KT≤ t < T Vs – VC1 = - L1 ∆ IL1/(1-K)T (2) Substituting from (1) into (2), Vs – VC1 = - K /(1-K) Vs VC1 =[(1 + K/(1-K) ] Vs VC1 = Vs /(1-K) • For 0 ≤ t < KT • VC1 – Vo = L2 ∆ IL2/(KT) (3) • For KT≤ t < T • Vo = - L2 ∆ IL2/(1-K)T (4) • Substituting from (3) into (4), • – Vo = - K ( VC1 – Vo ) / (1-K) • Vo = K VC1 • Vo = K/(1-K) Vs iL1

  9. iL1 iL1max for 0 < t ≤ KT Vc1 – Vo = L2 ∆IL2 / (KT) ∆IL2 = K ( VC1 – Vo ) / (fL2) ∆IL2 = K [ Vs /(1-K) – KVs /(1-K) ] / (fL2) ∆IL2 = K Vs / fL2 iL1min t T KT iL2 iL2max Io iL2min iC1 iL2max Io iL2min t vc1 KT t T KT T ∆vC1 (exaggerated!) -iL1max t -iL1min KT T

  10. iC2 Vs Is = Vo Io Vs Is = K Vs /(1-K) . Io Is = K/(1-K) . Io ∆ vC1 = 1/C1 . ∫ iC1dt = Io T K/C1 ∆ vC1 = K Io/(fC1) ∆ vo= 1/C2. ∫ iC2dt = 1/C2 . ½. T/2. ½ ∆IL2 ∆ vo= K Vs / (8f2C2L2) t KT T T+KT 2T vo (exaggerated!) ∆Vo t

More Related