Question

A dc-dc buck converter has input voltage of $400V$ and a load in form of resistor $R=40\Omega$ is connected accordingly. The switching frequency of converter is $2kHz$ and the converter switch has voltage drop of $2V$. When switch is conducting and duty cycle is $0.5$, then value of chopper efficiency will be

Answer 1

Given, $R=40\Omega ;V_s=400V$

Switch voltage drop, $V_{st}=2V,f_s=2kHz$

Average output voltage considering switch voltage drop is

$V_0=\alpha (V_s-2)$

$=0.5(400-2)=199V$

rms value of output voltage

$V_{rms}=\sqrt{\alpha }[V_s-2]$

$=\sqrt{0.5}\left(398\right)=281.428V$

Power delivered to the load,

$P_0=\frac{V_{rms}^2}{R}=\frac{(281.428{)}^2}{40}=1980.042W$

Input $P_1=V_s{I}_0$

$=400\left(\frac{199}{40}\right)=1990W$

$\therefore$ Chopper efficiency, $\eta =\frac{Outputpower}{Inputpower}\times 100=\frac{1980.042}{1990}\times 100=99.499\%$