The Buck Resonant Converter

Transcription

1 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman The Buck Resnan Cnverer Replacg he swich by he resnan-ype swich, ba a quasi-resnan PWM buck cnverer can be shwn ha here are fur mdes f pera under he seady-sae cndi a b Fig 6.8 a Cnvenal buck cnverer wih -ype resnan swich. b Simplified equivalen circui.

2 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman Buck Cnverer: Equivalen Mdes a b c d Fig 6.9 a Equivalen circui fr Mde. b Equivalen circui fr mde. c Equivalen circui fr mde. d Equivalen circui fr mde.

3 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman Mde [ 0 < ] Buck Cnverer: Seady-Sae Analysis Mde sars a 0 when S is urned ON Assume fr > 0, bh S and D are ON The capacir vlage, v c, is zer and he pu vlage is equal he ducr vlage di d The ducr curren, i, is given by i As lng as he ducr curren is less han, he dide will cnue cnducg and he capacir vlage remas a zer. 6.3 Hence, he ime erval is given by This is he ducr curren chargg sae

4 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman Mde [ < ] Mde sars a, dide is pen resnan sage beween and C The firs-rder differenial equas ha represen his mde are dvc C i d di d nducr curren is given by d d i + i C v c C 0 Seady-Sae Analysis cn d The general slu fr i is given by i A s ω + A csω + A3 Resnan angular frequency and cnsans: ω A C ω A 0, A 3 6.5a 6.5b

5 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman Equas i and v c are given by i + ω Z s [ cs ] v c ω Z Seady-Sae Analysis cn d is knwn as he characerisic impedance C The ime erval his mde can be derived a by seg i 0, i l + ω 0 Z s 6.0 herefre, Z s π + s ω ω Z 6. Mde sars a, when he swich is urned OFF. 5

6 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman Mde [ < 3 ]: Seady-Sae Analysis cn d A, he ducr curren becmes zer, and he capacir learly discharges frm zer durg 3. v c The capacir curren equals as given by, i C dv C d c 6. The capacir vlage v c is baed frm Eq. 6. frm 3 wih c as he iial value, vc + c 6.3 C The iial value is baed frm previus mde, c [ ] cs c ω 6.4 Subsiug Eq. 6.4 Eq. 6.3, v c C [ ] + cs ω 6.5 A 3, he capacir vlage becmes zer, C 3 3 [ cs ω ] 6.6 A his p, he dide urns ON and he circui eners Mde. 6

7 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman Mde [ 3 < 4 ]: Seady-Sae Analysis cn d A his mde he swich remas OFF, bu he dide sars cnducg a 3. Mde cnues as lng as he swich is OFF, and he upu curren sars he free-wheelg sage hrugh he dide. niial cndis i 0 l v c 0 By urng ON he swich a ime 4 is given by, T 4 s 3 T s, he cycle repeas hese fur mdes. The 6.7 7

8 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman lage Ga The express fr he vlage ga, M Typical Seady-Sae Wavefrms Figure 6.0 shws he seady sae wavefrms fr and i fr he buck cnverer wih -ype swich. v c Fig 6.0 Seady-sae curren and vlage wavefrms f buck -ype 8

9 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman Subsiue fr v c frm ervals - and 3 -, yield, ] cs [ c s d C d T ω The vlage ga rai is given by, ] s [ 3 3 C T c s + ω ω Bk crrec Subsiue fr -, 3 - and C frm he abve mdes, a clsed frm express fr M erms f he circui parameers can be baed. lage Ga s mehd 9

10 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman lage Ga nd mehd The al pu energy ver ne swichg cycle, T s 0 E i d Sce i is equal i, Eq. 6.0 is rewrien as, E d + i 0 i d Subsiug fr i frm Eqs. 6. and 6.8 he abve egrals, Eq. 6. becmes, E Subsiug fr + [ csω ] Zω 3 csω C E + + Zω wih, Z ω, Eq. 6.3 becmes, E C C 3 frm Bk crrec

11 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman The upu energy ver ne swichg cycle is: Ts 0 E d T s Equag he pu and upu energy expresss T s lage Ga cn d Frm Eq. 6.6 he vlage ga is expressed by, T s Subsiug fr, - and 3 - frm previus equas, he vlage ga becmes Ts + s ω Z + C [ csω ] 6.8

12 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman Nrmaliza T simplify and generalize he ga equa, he fllwg nrmalized parameers are defed: M R Q Z R f ns f f s nrmalized upu vlage nrmalized lad average upu curren nrmalized swichg frequency 6.9a 6.9b 6.9c 6.9d By subsiug Eq. 6.9 Eq. 6.8, he fal vlage ga is simplified M where, α f ns M Q + α + α π Q M s M Q cs

13 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman Buck-Cnrl Characerisic Curve A pl f he cnrl characerisic curve f M vs. f ns nrmalized lads is given Fig. 6. under varius Fig 6. Cnrl characerisic curve f M vs. ƒ ns fr he ZCS buck cnverer. 3

14 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman Example 6. Cnsider he fllwg specificas fr a ZCS buck cnverer f Fig. 6.8a. Assume he parameers are: 5,, A, f s 50kHz Design fr he resnan ank parameers and C and calculae he peak ducr curren, and peak capacir vlage. Deerme he ime erval fr each mde. Slu: M 0.48 The vlage ga is. Selec. Deerme Q frm eiher he cnrl 5 f ns 0.4 characerisic curve f Fig. 6. r frm he ga equa f Eq This resuls Q apprximaely equalg. Sce R, he characerisic impedance is given by, R Z Ω Ω 6.3 Q C The secnd equa erms f and C is baed frm f. Frm he nrmalized swichg frequency, f may be given by, f f f f s ns fs kHz n erms f he angular frequency, ω, ω πf ZCS Buck Cnverer C 6.33 Slvg Eqs. 6.3 and 6.33 fr and C, we ba, Z Ω 3 ω π 65 0 rad /sec µ H C Z ω π µ F 3 4

15 EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman The peak ducr curren, is given by, l, peak 3 A + Z The peak capacir vlage is: v c,peak 50 Example 6. cn d The ime ervals are calculaed frm he fllwg expresss: A H 6 0.µ s + s ω 0. + πf µ s Z s 5 C csω cs 0.67 Fr max we have A π π ω0 + ω 0 ω.79 µ s Exercise 6. fr pracice 0.4µ s + 0.µ s 0. 5µ s 4µ s 4 T s 5