Question

A buck-boost converter has the following parameters:
$V_s$ = 28 V, D = 0.3, R = 5 $\Omega$,
L = 20 $\mu$H, C = 90 $\mu$F, f = 80 kHz
The output voltage of the converter and average current through inductor will be respectively

Answer 1

Output voltage of Buck-boost converter,

$V_0=\frac{-D}{1-D}{V}_s=\frac{-0.3\times 28}{1-0.3}$

$=\frac{-0.3}{0.7}\times 28=-12V$
Average current through inductor,

$I_L=\frac{V_{S}D}{R{(1-D)}^2}=\frac{28\times 0.3}{5{\left(0.7\right)}^2}=3.42A$