Bench Power Supply Overheating and Poor Regulation
During a practical test, a trainee assembles a 12 V, 1 A regulated supply using a 15 V transformer secondary, a bridge rectifier, a 2200 µF filter capacitor and a series pass transistor. The intended output is 12.0 V at 1.0 A with less than 1 V ripple. The bench ambient is 32 °C. Measurements are: transformer secondary 15.0 V AC; DC after the bridge 18.6 V at no load and 15.4 V at 1 A; regulated output 12.0 V at no load but 11.2 V at 1 A. The 2200 µF capacitor is rated 25 V, 85 °C and becomes warm. A 0.5 Ω current-sense resistor in series with the load smells hot; the trainee used a 1/4 W part. The mains fuse is 500 mA and has not blown. The load is a 12 V, 1 A lamp, and it flickers slightly.
1. Calculate the power dissipated in the 0.5 Ω sense resistor at 1 A and decide whether the 1/4 W part is acceptable. Recommend a suitable rating and explain the consequence of leaving it in place.
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At 1 A, P = I²R = (1)² × 0.5 = 0.5 W, not 0.25 W. A 1/4 W resistor is therefore dissipating twice its rated power. Even a 0.5 W part would run at its limit, and at 32 °C ambient there is little margin. The resistor will drift, smell, discolour and may fail open, causing intermittent output. Choose at least a 1 W metal-oxide or 2 W resistor mounted clear of the board, or reduce the sense resistance to 0.1 Ω if the circuit can tolerate the lower feedback. Remember that the resistor also drops 0.5 V at 1 A; this drop adds to the regulator dropout and reduces the voltage available to the lamp. A hot resistor is a symptom, not just a component-rating issue.
2. Use the no-load and loaded DC measurements to estimate the effective source resistance between the bridge and the regulator input, and explain why the output falls to 11.2 V at 1 A.
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The DC level falls from 18.6 V to 15.4 V when the current rises from 0 to 1 A, so the effective source resistance is ΔV/ΔI = 3.2 V / 1 A = 3.2 Ω. This is the sum of transformer winding resistance, bridge dynamic resistance, wiring and fuse resistance. At 1 A the average input to the regulator is only 15.4 V; the rectifier ripple makes the trough lower than the average, perhaps about 14 V. A 12 V regulator needs its input above about 14.5–15 V at the trough once dropout and the 0.5 V sense-resistor drop are included. Therefore the regulator drops out of regulation and the load sees 11.2 V. The fault is not the regulator alone; the transformer and filter are undersized for 1 A. Use a transformer with better regulation, a larger capacitor, or reduce the load current.
3. The filter capacitor is warm. Give two plausible causes and one corrective action, with justification.
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Two plausible causes are: (1) excessive ripple current because 2200 µF is too small for a 1 A load, so the capacitor heats internally; (2) the capacitor is a general-purpose 85 °C type with high ESR, or it is installed with incorrect polarity/overvoltage. The 25 V rating is not immediately unsafe at 18.6 V no-load, but derating is poor. Corrective action: replace it with a 4700 µF, 35 V, 105 °C low-ESR capacitor, observe polarity, and check ripple with a scope. Increased capacitance reduces ripple voltage and ripple current per capacitor. Also verify that the capacitor is not heated by the sense resistor or regulator. If the capacitor still warms after replacement, measure AC ripple current and consider paralleling capacitors or improving transformer regulation.