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In the circuit shown in the figure, the switch is operated at a duty cycle of 0.5. A large capacitor is connected across the load. The inductor current is assumed to be continuous. What is the average voltage across the load and the average current through the diode ?

\$V_{load}=\frac{V_{source}}{1-D}\$

\$V_{load}=\frac{20}{1-0.5}\$

\$V_{load}=\frac{20}{0.5}\$

\$V_{load}=40 V\$

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For an ideal boost converter.

\$V_{s}*I_{s}=V_{o}*I_{o}\$

\$I_{o}=\frac{V_{s}*I_{s}}{V_{o}}\$

\$I_{o}=\frac{20*4}{40}\$

\$I_{o}=2 A\$

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    \$\begingroup\$ This looks like homework. Show your calculations. \$\endgroup\$ Apr 11 '13 at 12:01
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    \$\begingroup\$ Hello Rajesh. Homework (or homework-like) problems are welcome there, but you have to show that you tried to solve them and don't just dropped the assignment here. If you have no clue about how to proceed, then ask for directions. But maybe you would read something more about the topic. \$\endgroup\$
    – clabacchio
    Apr 11 '13 at 12:56
  • \$\begingroup\$ @ clabacchio : I'm looking for a better way to solve this. In nature there are atleast two ways to solve a problem at hand. \$\endgroup\$ Apr 11 '13 at 12:59
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You have a continuous mode boost converter. The point of the assignment is for you to understand how it operates. Simply giving you the answer circumvents that.

Note that the converter is operating in steady state and that the inductor current never goes to zero. Therefore, consider that the rise in inductor current when the switch is closed is exactly ballanced by the decrease in that current when the switch is open. What does that tell you?

If you are still stuck, use a calculator and plot out the inductor current over a switching cycle. Then also plot the inductor output voltage. Show those plots and we can discuss what is going on further.

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