Sizing a Filter Capacitor for a Half-Wave Rectifier
A student builds a half-wave rectifier using a 12 V rms, 50 Hz transformer secondary, one silicon diode with a 0.70 V forward drop, a 1.0 kΩ load, and a filter capacitor. On an oscilloscope the output has a peak of about 16.3 V but a large sawtooth ripple. The student wants the ripple to be less than 1.0 V peak-to-peak. Available electrolytic capacitors are 100 µF, 470 µF, and 1000 µF, all rated 25 V. The diode is rated for a peak inverse voltage of 50 V and a forward current of 1 A. The transformer has no center tap. For an initial estimate, take the load current as 16 mA and the capacitor discharge time as approximately the full 20 ms period. Use V_r ≈ I_L/(fC), where f is the ripple frequency. The student also wonders whether using the largest capacitor is automatically the best choice.
1. Calculate the peak output voltage, the ripple frequency, and the capacitance needed for a 1.0 V peak-to-peak ripple. Which available capacitor should be chosen?
Read the worked answer
The secondary peak voltage is V_s,peak = 12√2 = 16.97 V. The diode drop reduces the output peak to V_o,peak = 16.97 − 0.70 = 16.27 V ≈ 16.3 V.
For a half-wave rectifier supplied at 50 Hz, the ripple frequency is f = 50 Hz. Using V_r ≈ I_L/(fC):
C = I_L/(fV_r) = (16×10⁻³)/(50 × 1.0) = 3.2×10⁻⁴ F = 320 µF.
A 100 µF capacitor would give V_r ≈ 0.016/[50(100×10⁻⁶)] = 3.2 V, too large. A 470 µF capacitor gives V_r ≈ 0.016/[50(470×10⁻⁶)] = 0.68 V, which meets the requirement with some margin. The 1000 µF also works, but the 470 µF is the smallest available capacitor that meets the specification.
2. Check the capacitor voltage rating and the diode peak inverse voltage. Is the diode safe? Explain how the PIV is estimated in a capacitor-input half-wave rectifier.
Read the worked answer
The capacitor charges to about the peak output, 16.3 V. A 25 V rating gives a margin of about 8.7 V, which is acceptable for a normal 12 V rms secondary, though a 35 V capacitor would be safer if the line voltage can rise.
In a capacitor-input half-wave rectifier, when the diode is reverse-biased, the secondary can be near its negative peak while the capacitor holds the positive peak voltage on the cathode. The reverse voltage across the diode is therefore approximately V_s,peak + V_C ≈ 2V_s,peak = 2(16.97) = 33.9 V, or about 33.2 V if the diode drop is included. This is below the 50 V PIV rating, so the diode is safe under normal operation.
3. If the 1000 µF capacitor is used, estimate the ripple and discuss one practical reason it may not be the best choice despite the lower ripple.
Read the worked answer
With 1000 µF, V_r ≈ I_L/(fC) = (16×10⁻³)/[50(1000×10⁻⁶)] = 0.32 V peak-to-peak. The ripple is lower, but a much larger capacitor draws a larger inrush current when the supply is switched on, because it presents nearly a short circuit until it charges. That surge can stress the diode, transformer, and fuse, and can exceed the diode’s repetitive surge rating even though the average forward current is only 1 A. A 470 µF capacitor gives adequate filtering with a gentler surge, so bigger is not automatically better.