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Question

A boost converter has an input voltage of VS=5V. The average output voltage V0=15 V and the average load current I0=0.5 A. The switching frequency is 25 kHz. If L=150 μH and C = 220 μF, then the peak inductor current will be ___________A

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Solution

Peak inductor current, I2=Is+ΔI2
ΔI=kVsfL=Vs(V0Vs)fLV0
V0=Vs(ak)
k=1515=23
ΔI=23×525×103×150×106=0.80 A
Is=0.5(123)=1.5A
I2=IS+ΔI2=1.5+0.892=1.945 A

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