Question

Problem 1: (40 points) A boost converter based power factor correction circuit needs to operate with universal input voltage, i.e the input voltage can be either 120V (RMS)+/- 10% at 60 Hz or 240 V(RMS) +/- 10% at 50 Hz. The output voltage is regulated at 400 V DC and the maximum (average) output power is 1kW. The switching frequency is 200 kHz. (a) Calculate the value of inductance needed to keep its peak peak ripple current to be less than 10% of the peak fundamental frequency line current at the maximum input voltage and full load condition. (b) Calculate the value of output capacitance needed to keep the peak-peak ripple (at twice the line frequency) to be less than 10% of the nominal output voltage at any operating condition. (c) Calculate the maximum RMS ripple current in the output capacitor (neglect switching frequency ripple). (d) Draw ū (1) over a complete fundamental period corresponding to 240V (RMS), 50 Hz input.

Answer 1

ee Given that A boost converter based power factor connection crven't needs to operate with universal input voltage ise The input voltage can be either 12ov. (RMS) I lo% at GoH2. Solution. Boost Converterin V is votac R vo Input voltage 120(RMS)+(-10% at. Goltz output voltage, 240v (RMS)+/-10% at sota L 400 VDC The switching frequency = 200kHz for Duty cycle No=vs a) 400=240 Hence D=0.4 an output output power = lkw current Io = f = 1000 400 = 2.5 Ampue vo= IOR 400 =2.5XR f=loor

for Inductance we have formula d=0(1-0)R 1 LF = 0.4 *0.6x160 =0.096 Mit 2x 200 x 103 ② output capacitane (c) D(1-0). Jo 2 vs F 0.4X0.6x2.5 24240x 200 x 162 =6.25nf c) output Cancent in capacitance) Dil = VS XD : 240 X0.4 45 I nxf 0.096x103 x 200 x 203 Hence AJL = Ic + Io 5= Ic+2.5 IC = 2.5 Ampere. D) F= 50H2 = 0.02 D=0.4

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