Power Electronics Circuits
A sinusoidal AC voltage of 100 V RMS is full-wave rectified with an ideal diode bridge. Without a smoothing capacitor, which of the following is closest to the average output voltage?
- AAbout 90 V
- B100 V
- CAbout 141 V
Show answer and explanation
Answer: A. About 90 V
The peak of the sine wave is Vm = √2 × 100 ≈ 141 V. The average of the full-wave rectified waveform is (2/π) × Vm = (2√2/π) × V ≈ 0.90 × 100 ≈ 90 V. 100 V is the RMS value of the full-wave rectified waveform, the same as that of the input. With half-wave rectification the negative half cycles are missing, so the average is half of this, about 45 V.
- A. Correct. The average is (2√2/π) × 100 ≈ 90 V.
- B. 100 V is the RMS value of the full-wave rectified waveform. The average is smaller than the RMS value.
- C. About 141 V is the peak of the sine wave, the value the output approaches with a smoothing capacitor and a very light load.