Buck Boost Converter
Consider the following circuit:
The voltage source is a DC supply. The positive terminal of the supply is connected to the switch. The inductance is . The capacitance is . The diode is ideal.
Initially, the switch is off, and the inductor and the capacitor are both unenergized. The switch is turned on for , and then turned off. Determine the magnitude of the final voltage on the capacitor. Express your answer in volts and round off to the nearest integer.
The answer is 316.
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1 solution
Just after the switch is turned on, the capacitor voltage is zero, so the diode will be reverse biased. No current will pass through the diode and the capacitor will remain uncharged while the switch stays on.
The following equation for the inductor current holds while the switch is on:
1 0 V = 1 m H d t d i
The initial inductor current is zero, so after 1 m s , the inductor current is 1 0 A .
Just after the switch is turned off, the inductor current is still 1 0 A , so the diode will be forward biased. Current will pass through the diode and the capacitor will start a charging process.
Note that we now have an LC circuit. As the capacitor is charging, the inductor current decreases. Eventually, the inductor current reaches zero and will not start to flow in the opposite direction because of the diode. The capacitor voltage will now remain at a constant level.
All of the energy that the inductor had just after the switch was turned off, has been transferred to the capacitor:
( 1 m H ) ( 1 0 A ) 2 = ( 1 μ F ) V c a p 2
V c a p = 3 1 6 V