Discussion
That chopper switches at 500 Hz. Its on-time in each period is
Answer: A
Correct answer: 1.2 ms
The switching period is the reciprocal of the frequency, and the on-time is the duty ratio of that period:
\[ T = \frac{1}{500} = 2 \text{ ms}, \qquad t_{on} = 0.6 \times 2 = 1.2 \text{ ms} \]The off-time is the remaining 0.8 ms. Raising the switching frequency shortens both and reduces the current ripple, at the cost of greater switching loss in the device.
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