Question

In a Buck-Boost converter operating at 20 kHz, L = 0.05 mH. The output capacitor is sufficiently large and $V_{dc}=15V$. The output is to be regulated at 10 V and converter is supplying a load of 10 W. The duty ratio of converter is

• A. 0.3
• B. 0.4
• C. 0.2
• D. 0.1

Answer 1

The correct option is A 0.3



Assuming continuous conduction,

$L_{cr}=\frac{R(1-D{)}^2}{2f}=\frac{10(1-0.4{)}^2}{2\times 20\times {10}^3}=0.09mH$

$\because L<L_{cr},itisdiscontinuousconductionmode.$

$Hence,D\ne 0.4$



During ON :
$V_{L\left(ON\right)}=\frac{L\Delta I_L}{D_T}$

$\Rightarrow \Delta I_L=\frac{15\times D}{0.05\times {10}^{-3}\times 20\times {10}^3}=15D$
$P_{in}=P_{out}$
$V_s{I}_{s\omega }=10$
$\Rightarrow I_{s\omega }=\frac{10}{15}=\frac{2}{3}$

$I_{s\omega avg}=\frac{1}{2}\times \frac{\Delta I_L\times DT}{T}$
$\frac{D\Delta I_L}{2}=\frac{2}{3}$

$\Rightarrow 15D^2=\frac{4}{3}$

$\Rightarrow D=0.3$