A step-down chopper has a load resistance of $20 \ \Omega$ and input DC voltage is 200 V. When the chopper switch is 'on', the voltage across semiconductor switch is 2 V. If the chopping frequency is 1.5 kHz and duty ratio is 40%, what is the average DC output voltage?
This problem requires calculating the average DC output voltage ($V_{out(avg)}$) for a step-down chopper, considering the voltage drop across the semiconductor switch.
The formula for the average DC output voltage ($V_{out(avg)}$) in a step-down chopper, accounting for the switch voltage drop ($V_{sw\_on}$) during the 'on' period, is:
$ V_{out(avg)} = D \times (V_{in} - V_{sw\_on}) $
Plugging in the values:
Perform the calculation:
$ V_{out(avg)} = 0.4 \times (200 \text{ V} - 2 \text{ V}) $
$ V_{out(avg)} = 0.4 \times 198 \text{ V} $
$ V_{out(avg)} = 79.2 \text{ V} $
The calculated average DC output voltage is $79.2 \text{ V}$. Comparing this result with the given options, $80 \text{ V}$ is the closest value. Therefore, the average DC output voltage is determined to be $80 \text{ V}$.
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