Replacing a Reservoir Capacitor in a 19 V Power Supply
A student repairs a linear bench power supply. The original reservoir capacitor is 470 µF, 35 V, 105 °C, low ESR. The only local stock is 470 µF, 16 V, 85 °C, general purpose. The unregulated rail before the regulator measures 19.5 V with no load and 18.2 V at 1 A. The student fits the 16 V part because the capacitance and footprint match. The supply works for two days, then the capacitor bulges and leaks electrolyte. The schematic shows a bridge rectifier, the reservoir capacitor, and a 1 A series regulator; there is no bleeder resistor across the capacitor. The student asks whether the capacitor was defective or whether something else caused the failure, and what replacement to order. The regulator datasheet recommends a low-ESR input capacitor and warns that the input rail can rise before the regulator starts.
1. Explain why the 16 V capacitor failed even though its capacitance was correct. State the key ratings that were ignored.
Read the worked answer
The capacitance value sets the amount of charge stored, but the voltage rating is the maximum continuous voltage the dielectric can withstand. The unregulated rail is 19.5 V no-load and 18.2 V at 1 A, both above the 16 V rating. Even if the average voltage were slightly lower, the rectifier charges the capacitor to the peak of the AC waveform, and the no-load rail can rise above the normal operating value. Overvoltage causes dielectric breakdown, gas generation, bulging, and leakage. The 85 °C general-purpose part also has lower ripple-current capability and higher ESR than the original 105 °C low-ESR part, so ripple current adds internal heating. The capacitor was not defective; it was mis-specified. The replacement must meet or exceed the original voltage rating, temperature rating, ESR, and ripple-current rating.
2. Calculate the charge and stored energy in the 470 µF capacitor at 19.5 V. Design a safe discharge method and comment on a permanent bleeder.
Read the worked answer
Q = CV = 470 × 10⁻⁶ × 19.5 = 9.17 × 10⁻³ C = 9.17 mC. E = ½CV² = 0.5 × 470 × 10⁻⁶ × 19.5² = 0.089 J. The energy is small but still enough to damage a meter or startle the user. For service work, connect a 100 Ω, 5 W resistor across the capacitor for a few seconds; the time constant is τ = RC = 100 × 470 × 10⁻⁶ = 0.047 s, so it discharges in well under a second. Then measure the voltage before touching the board. A permanent bleeder of 2.2 kΩ, 1 W gives τ = 1.03 s and dissipates 19.5²/2200 = 0.17 W; it discharges to a safe level in about 5τ = 5 s. That standby waste may be acceptable on a bench supply, but it must be rated for the voltage and mounted safely.
3. Available replacements are 470 µF 25 V 105 °C, 1000 µF 35 V 105 °C, and two 470 µF 25 V capacitors. Recommend the best choice and explain the risks of series connection.
Read the worked answer
The preferred replacement is 470 µF 35 V, 105 °C, low ESR, with a ripple-current rating at least equal to the original. A 470 µF 25 V, 105 °C part is the minimum acceptable because 25 V gives 5.5 V margin over 19.5 V, but 35 V is safer if the rail can rise or if the supply is used in a hot environment. A 1000 µF 35 V part will usually work, but it increases inrush current through the bridge and may stress the regulator; check the bridge rating and space before using it. Connecting two 470 µF 25 V capacitors in series gives only 235 µF and does not guarantee equal voltage sharing; equalizing resistors, such as 10 kΩ 1 W across each capacitor, would be needed. Series connection also doubles ESR and is not equivalent to one correct capacitor. Do not use a 16 V part again.