Question

A buck boost converter is operating at 20 kHz with inductor L = 50 $\mu$H. The output capacitor is sufficiently large and source voltage $V_d=15V.$ The output is to be regulated at 10 V and the converter is supplying at load of 10 W. The duty ratio and mode of conduction of converter respectively will be

• A. 0.45, discontinuous
• B. 0.40, continuous
• C. 0.398, continuous
• D. 0.298, discontinuous

Answer 1

The correct option is D 0.298, discontinuous

Assuming continuous conduction mode of converter



$V_0=\frac{\alpha }{1-\alpha }{V}_d$

$10=\left(\frac{\alpha }{1-\alpha }\right)15$

$\frac{\alpha }{1-\alpha }=1.5$

$2.5\alpha =1$

$\alpha =0.4$

Checking for boundary conditions,


$I_{OB,max}=\frac{V_0}{2fL}=\frac{10}{2\times 20\times {10}^3\times 50\times {10}^{-6}}=5A$

$I_{OB}=I_{OB,max}(1-\alpha {)}^2=5(1-0.4{)}^2$

$=5\times 0.36=1.8A$

$P_0=V_0{I}_0$

$10=10I_{0}or{I}_0=1A$

As output current is less than calculated $I_{0B}$

$\therefore$ Mode of operation : Discontinuous mode and value of duty ratio,

${\alpha }^{\prime }=\frac{V_0}{V_d}\sqrt{\frac{I_0}{I_{OB,max}}}=\frac{10}{15}\sqrt{\frac{1}{5}}=0.298$