Question

From a 240Vdc, 480Watt 48Vdc load needs to be fed by designing a buck-boost converter. The inductor current ripple must be %40 of the average inductor current (Example when the average ind. current is 1 A, ΔI becomes 0.4 Amp.). The switching frequency is 83 300 Hz. complete the design answer the followings? Show the solution with detail

a) D=? 10p b) L=? 15p

Answer 1

Civen buck Boost Converter D t SW Yii A va Vo Vs well •Ic 11. - Vs=2 =240v P = 480w Vo=480 oli = inducter Current ripple ILA = Average inducter current given I L= 0.4 I LA f= 83300 HZ -5 TE -3 83300 = = 12 x 10 1-2x18xio T= 0.0/2 m sec (0) Vos plus (ez duty ratio) x 240 48= D. 1-D 1-D 5D ID=5 60= 1 D = 1 | D = 0.166

also Iswa = I LAXTON T or ISA Iswa = switch ement Anerage supply ISA Current (Iswa = Isa) D ILA D = TON T also We that Know Pin = Pont (input power = output Power Vs. Isa = V.I. Vs Is A = 480 watt (given) IsA= = 2 A 480 240 IsA= 2 A (Average supply current) from equation o ISA = D. ILA ILA = Isa = 2 = 12 A 0.166 ILA= 12 A Average inducter Current) given that Also inducter cument ripple (AIL) = 40 I LA 100 AI = o. 4 ILA

so OIL= 0.4 x 12 4-8 A OL CON Also know that t AIL = Vs DT I Vse I , ILA I ot from DT Val di dt i- dt so DI L = Vs DT so La DT Vs DIL L = x 0.166 x 0.012X1-3 240 4.8 L = 0-099 X163 H La OomH so (a) D = 0.167 (6) L = 0.1 MH.