Power Electronics GATE 2002

Problem: In the circuit shown in Figure, the source $I$ is a dc current source.The switch $S$ is operated with a time period $T$ and a duty ratio $D$. You may assume that the capacitance $C$ has a finite value which is large enough so that the voltage $V_{c}$ has negligible ripple, calculate the following under steady state conditions, in terms of $D$, $I$ and $R$.a) The voltage $V_{c}$, with the polarity shown in figure.b) The average output voltage $V_{o}$, with the polarity shown in figure.

The following solution is a little rigorous. A more intuitive and short solution is given at the end.

Solution 1 for (a):

Since the circuit is operating in steady state conditions, the amount of energy stored in its components has to be the same at the beginning and at the end of a commutation cycle. Capacitor voltage ripple is negligible.

 $\displaystyle\implies\boxed{\triangle {V_{c}}^{OFF}+ \triangle {V_{c}}^{ON}=0}$

When $S$ is OFF, the capacitor will be charged by the current source through the diode according to the equation $V_{c}(t)=V_{c}+\displaystyle\frac{I}{C}t$

 $\implies\boxed{\triangle{V_{c}}^{OFF}=\displaystyle\frac{I}{C}T_{OFF}}$

When $S$ is ON, the capacitor will be discharged through the resistor according to the equation $V_{c}(t)=V_{c}e^{\displaystyle\frac{-t}{RC}}$

 $\implies\boxed{\triangle{V_{c}}^{ON}=V_{c}e^{\displaystyle\frac{-T_{ON}}{RC}}-V_{c}}$

 But we have $\displaystyle\triangle {V_{c}}^{OFF}+ \triangle {V_{c}}^{ON}=0\implies\displaystyle\frac{I}{C}T_{OFF}+V_{c}e^{\displaystyle\frac{-T_{ON}}{RC}}-V_{c}=0$

$\implies\boxed{\displaystyle\frac{I}{C}T_{OFF}=V_{c}(1-e^{\displaystyle\frac{-T_{ON}}{RC}})}$

It is given that the value of $C$ is very large so we can neglect the higher order terms with degree $\geq2$ from the infinite series of  $e^{x}$. 

Therefore, $\boxed{\displaystyle\frac{I}{C}T_{OFF}=V_{c}\frac{T_{ON}}{RC}}$.
Substituting for $T_{ON}=DT$ and $T_{OFF}=(1-D)T$.

The capacitor voltage $\displaystyle\boxed{V_{c}=\displaystyle\frac{IR(1-D)}{D}}$
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 Solution 2 for (a):

The energy increase of $C$ during $OFF$ time $\approx V_{c}IT_{OFF}$

Energy decrease of $C$ during $ON$ time when it discharges through R $\approx\displaystyle\frac{V_{c}^2}{R}T_{ON}$

Both the above terms must be equal for steady state.

$\implies\boxed{V_{c}IT_{OFF}=\displaystyle\frac{V_{c}^2}{R}T_{ON}}$ leading to the same solution $\displaystyle\boxed{V_{c}=\displaystyle\frac{IR(1-D)}{D}}$
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Solution for (b):
Average value of $\boxed{V_{o}=\displaystyle\frac{-V_{c}T_{ON}+(0)T_{OFF}}{T}}$

Average $\boxed{V_{o}={-V_{c}T_{ON}}{T}\implies V_{o}=-V_{c}D} $

Therefore, average value of $\boxed{V_{o}=\displaystyle-IR(1-D)}$