Question

A boost converter has an input voltage of $V_s=5V$. The average output voltage $V_0=15V$ and the average load current $I_0=0.5A$. The switching frequency is 25 kHz. If $L=150\mu H$ and $C=220\mu F$, then the peak inductor current and the ripple voltage of the filter capacitor $\Delta V_c$ respectively, are

• A. 1.945 A, 60.61 mV
• B. 3.5 A, 20.61 mV
• C. 2.79 A, 50.61 mV
• D. 3 A, 30 .61 mV

Answer 1

The correct option is A 1.945 A, 60.61 mV

Peak inductor current $I_2=I_s+\frac{\Delta I}{2}$

$\Delta I=\frac{k{V}_s}{fL}=\frac{V_s(V_0-V_s)}{fL{V}_0}$

$V_0=\frac{V_s}{(1-k)}$

$k=1-\frac{5}{15}=\frac{2}{3}$

$\Delta I=\frac{2}{3}\times \frac{5}{25\times {10}^3\times 150\times {10}^{-6}}=0.89A$

$I_s=\frac{0.5}{(1-\frac{2}{3})}=1.5A$

$I_2=I_s+\frac{\Delta I}{2}=1.5+\frac{0.89}{2}=1.945A$

$\Delta V_c=\frac{k{I}_0}{fC}=\frac{2}{3}\times \frac{0.5}{25\times {10}^3\times 220\times {10}^{-6}}=60.61mV$