Hum and Ripple After a Reservoir Capacitor Replacement
A technician repairs a small audio preamp that is fed from a 12 V AC, 50 Hz transformer through a bridge rectifier and a 2200 µF reservoir capacitor. The preamp draws 0.4 A. The original capacitor fails, and the technician fits a 1000 µF, 25 V capacitor from stock. The DC output is about 15.8 V, but a 100 Hz hum appears and the oscilloscope shows 3.3 V peak-to-peak ripple. The transformer secondary measures 12.0 V RMS on a DMM. The preamp needs ripple below 1 V peak-to-peak to meet its hum specification. The technician has 1000 µF, 2200 µF and 4700 µF capacitors rated 25 V, and a 35 V 4700 µF. The bridge is rated 2 A, 50 V. The customer wants the repair completed same day.
1. Calculate the expected ripple with 1000 µF and 2200 µF, compare with the observation, and explain the 100 Hz hum.
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For a full-wave rectifier, ripple frequency is 2 × 50 = 100 Hz, which explains the hum. Using the estimate Vr(pp) ≈ I/(2fC):
1000 µF: Vr ≈ 0.4/(2 × 50 × 0.001) = 4.0 Vpp. 2200 µF: Vr ≈ 0.4/(2 × 50 × 0.0022) = 1.82 Vpp.
The measured 3.3 Vpp is lower than the 4.0 V estimate because ESR and the conduction angle affect the simple formula, but it is in the expected range and far above the 1 Vpp target. The 1000 µF capacitor is too small, and even the original 2200 µF would not meet the specification. Check capacitor polarity and voltage rating before blaming the bridge.
2. Choose a capacitor from stock that meets the ripple target and justify its voltage rating.
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Required capacitance: C ≥ I/(2f × Vr) = 0.4/(100 × 1) = 4000 µF. A 4700 µF capacitor gives Vr ≈ 0.4/(100 × 0.0047) = 0.85 Vpp, which meets the target.
Peak secondary voltage = 12 × √2 = 17.0 V. After two diode drops, the capacitor sees about 15.6 V DC, so a 25 V rating is nominally enough but leaves little margin for mains high line. The 35 V, 4700 µF capacitor is the better choice. Check that the bridge's 2 A rating covers the repetitive surge and that the larger capacitor fits the board. Verify the replacement capacitor before fitting it.
3. The technician asks whether adding a 12 V linear regulator after the filter would solve the hum. Comment on feasibility and the better same-day repair.
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A linear regulator can reduce hum because it has high ripple rejection, but it needs input headroom. With 15.8 V average and 0.85 Vpp ripple, the trough is about 15.4 V, so a 12 V regulator with about 2 V dropout is feasible. However, the regulator would dissipate (15.8 − 12) × 0.4 ≈ 1.5 W, so it needs a heatsink and thermal check. It would also hide, not correct, an undersized reservoir.
The better same-day repair is the 4700 µF, 35 V capacitor. Add the regulator only if hum remains after the reservoir is corrected and the thermal design is acceptable.