Question

The semi-converter switch as shown in figure is operated at a frequency of 25 kHz with duty cycle ratio of 0.5. The circuit operates in steady state and waveform of inductor is shown below. If $V_1$ = 75 V and $V_2$ = 200 V , then the energy transferred to $V_2$ during five cycles of steady state is _________(mJ). (Assume converter to be ideal ) .

Answer 1

Since inductor is connected to source side, so it is boost converter.
Given that , $V_1=75V,V_2=200V,D=0.5,T=25kHz$
From 0<t<DT
Apply KVL ,

$V_L=V_1$

L.$\frac{di}{dt}=V_1$

$\frac{di}{dt}=\frac{V_1}{L}$

${\int }_0^{I_P}di={\int }_0^{DT}\frac{V_1}{L}.dt$

$I_p=\frac{V_{1}D}{fL}$
$I_p=\frac{75\times 0.5}{25\times {10}^3\times 75\times {10}^{-6}}$

$=\frac{500}{25}$ = 20 A

From DT < t < $\beta T$

$-V_1+V_L+V_2=0$

$V_L=V_1-V_2$

$L.\frac{di}{dt}=V_1-V_2$

${\int }_{I_p}^{0}di={\int }_{DT}^{\beta T}\frac{(V_1-V_1)(\beta -D)}{L}$

$I_p=\frac{(V_2-V_1)(\beta -D)}{f.L}$
$20=\frac{(200-75)(\beta -D)}{25\times 75\times {10}^{-3}}$

$\frac{20\times 25\times 75}{125\times {10}^3}=\beta -D$

$\beta -D=\frac{3}{10}$

$\beta =\frac{3}{10}+\frac{5}{10}=\frac{8}{10}$

Energy is transferred to $V_2$ during DT to $\beta T$
Energy transferred in one cycle,
$E_1$ = $V_2.\frac{I_p}{2}\times (\beta -D)T$
$=200\times \frac{20}{2}\times 0.3\times 4\times {10}^{-5}=\frac{3}{125}J$

Energy transferred in 5 cycles

$E_5=5\times E_1=\frac{15}{125}$ = 120 mJ

Alternate Solution :

For DCD,
$V_0=\frac{D}{B}{V}_s\left(buck\right)$

$=\left(\frac{\beta }{\beta -D}\right)V_s\left(boost\right)$

$=\left(\frac{D}{\beta -D}\right)V_s(buck-boost)$
For DCD mode (boost ) shown in question

$\left(\frac{\beta }{\beta -D}\right)V_s=V_0$

$\left(\frac{\beta }{\beta -0.5}\right)\times 75=200$
$\beta =0.8$
$\beta -D=0.3$

$\Delta I_L=\frac{D{V}_s}{fL}$
$=\frac{0.5\times 75}{25\times 75}\times {10}^3$
= 20A
$I_0=\frac{1}{2}\times \Delta I_L\times (\beta -D)=\frac{1}{2}\times 20\times 0.3=3A$

Then energy transferred $=V_o{I}_{o}t$
$=200\times 3\times 5\times \frac{1}{25000}=120mJ$