For a boost converter with switching frequency of 500 Hz, input of 50 V dc, output of
Question:
For a boost converter with switching frequency of 500 Hz, input of 50 V dc, output of 75 V dc, inductor of 2 mH, and load resistance of 2.5 Ω, we determine the following:
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period T = 1/500 = 0.002 s= 2 ms Vout/Vin = 75/50 = 1/(1-D), which gives D = 0.333 Ton = 0.333 x 2 ms = 0.666 ms Toff = 2-0.666 = 1.334 ms lout = 75/2.5 = 30A in=30=(1-0.333) = 45 A = Vin Ton L=50 V x 0.666 ms + 2 mH = 16.65 A (a large ripple). A/Ripple.pk-pk
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Related Book For
Introduction To Electrical Power And Power Electronics
ISBN: 9781466556607
1st Edition
Authors: Mukund R. Patel
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