Question

A buck regulator has an input voltage of $V_s=12V$. The required average output voltage is $V_0=5V$ at R = 500 $\Omega$ and the peak-to-peak output ripple voltage is 20 mV and the switching frequency is 25 kHz. The peak-to-peak ripple current of inductor is limited to 0.8 A. The value of filter capacitor is

Answer 1

Output voltage, $V_0=D{V}_s$
5 = D ×12
D = 0.416
Peak-to-peak ripple current, $\Delta I=\frac{V_{s}D(1-D)}{Lf}$

$0.8=\frac{12\times 0.416(1-0.416)}{25\times {10}^3\times L}$

$L=0.145\times {10}^{-3}H$
We know that

Peak-to-peak ripple voltage, $\Delta V=\frac{V_{s}D(1-D)}{8LC{f}^2}$

$C=\frac{V_{s}D(1-D)}{8L{f}^2\times \Delta V}$

$=\frac{12\times 0.416(1-0.416)}{0.145\times {10}^{-3}\times {(25\times {10}^3)}^2\times 20\times {10}^{-3}\times 8}$

$=201.06\mu F$