Question

In a buck-boost converter operating at 20 kHz, L = 0.05 mH. The output capacitor C is sufficiently large and input voltage $V_s=15V$. The output is to be regulated at 10 V and the converter is supplying a load of 10 W. The duty ratio D is

• A. 0.3
• B. 0.5
• C. 0.7
• D. 0.4

Answer 1

The correct option is A 0.3

$I_0=\frac{10}{10}=1A$

Mode of conduction is not known. Assume current to be at the edge of continuous conduction. For Buck-boost converter,

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

$\frac{D}{1-D}=\frac{10}{15}$

D = 0.4

$I_{OB,max}=\frac{T_s{V}_0}{2L}$

$I_{OB,max}=\frac{0.05\times 10}{2\times 0.05}=5A$

$I_{OB}=I_{OB,max}(1-D{)}^2$

$=5(1-0.4{)}^2=1.8A$

Since the output current $I_0=1A$ is less than $I_{OB}$, the current conduction is discontinuous

Duty ratio, $D=\frac{V_0}{V_s}\sqrt{\frac{I_0}{I_{OB,max}}}$

$=\frac{10}{15}\sqrt{\frac{1}{5}}=0.3$