Power Electronics Circuits
Estimate on the assumption that the capacitor supplies all of the load current for one full ripple period. When 50 Hz AC is full-wave rectified, smoothed with a 10,000 µF capacitor, and a constant 2 A flows to the load, which of the following is the approximate peak-to-peak output voltage ripple?
- AAbout 1 V
- BAbout 2 V
- CAbout 4 V
Show answer and explanation
Answer: B. About 2 V
With full-wave rectification the ripple period is half a supply cycle, 1/(2 × 50 Hz) = 10 ms. If the capacitor supplies 2 A during that time, it gives up a charge of 2 A × 10 ms = 20 mC, and the voltage drops by ΔV ≈ I/(2fC) = 20 mC ÷ 10,000 µF = 2 V. In reality the next charging starts slightly before the peak, so the ripple is a little smaller than this estimate. With half-wave rectification the ripple period is 20 ms, giving about 4 V under the same conditions.
- A. This assumes discharge for only half the ripple period (5 ms). Even with full-wave rectification, the capacitor discharges for almost 10 ms.
- B. Correct. ΔV ≈ I/(2fC) = 2 ÷ (2 × 50 × 0.01) = 2 V.
- C. This is the value for half-wave rectification. With full-wave rectification the capacitor is charged twice per cycle, so the ripple is half as large.