Question

The Buck boost regulator shown in figure has an input voltage of $V_s=12V.$ The duty cycle is 0.25 and the switching frequency is 25 kHz. The inductance $L=150\mu H$ and filter capacitance $C=220\mu F.$ The average load current $I_a=1.25A$. Average output voltage $\left(V_0\right)$ and peak current of the transistor, $I_p$(A) is

• A. -4V, 2.47 A
• B. 4 V, 1.67 A
• C. -4V, 1.67 A
• D. 4V, 2.07 A

Answer 1

The correct option is D 4V, 2.07 A

For Buck-boost regulator,
Average output voltage, $V_0=\frac{V_s\left(\alpha \right)}{(1-\alpha )}=\frac{12\times 0.25}{0.75}=4V$
Peak current, $I_P=\frac{I_s}{\alpha }+\left(\frac{\Delta I}{2}\right)$
Now, $V_0{I}_a=V_s{I}_s$
$I_s=\frac{4\times 1.25}{12}=0.4167A$
Ripple current, $\Delta I=\frac{\alpha V_s}{fL}=\frac{0.25\times 12}{25\times {10}^3\times 150\times {10}^{-6}}=0.8$
Peak current, $I_P=\frac{0.4167}{0.25}+\frac{0.8}{2}=2.067A$