12 V supply that loses regulation at full load
A student builds a 12 V, 0.5 A regulated supply using a 230 V/12 V transformer, a bridge rectifier, a 2200 µF filter capacitor and a 7812 regulator. At no load the output is 12.1 V. At 0.5 A it falls to 10.8 V and the regulator is warm. The ripple at the regulator input is 3.8 V peak-to-peak. Each bridge diode drops about 1.0 V at 0.5 A, and the transformer secondary resistance is 1.2 Ω. The 7812 needs a minimum input of 14.0 V to stay in regulation. Its junction-to-case thermal resistance is 5 °C/W; the heatsink is 8 °C/W and ambient is 35 °C. The student proposes either fitting a 4700 µF capacitor or changing to an 18 V transformer.
1. Estimate the peak and minimum voltage at the regulator input at full load. Explain why the output falls to 10.8 V.
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The 12 V rms secondary has a peak of 12 × 1.414 ≈ 16.97 V. At 0.5 A the secondary resistance drops 0.5 × 1.2 = 0.6 V. Two bridge diodes conduct at once, so the diode drop is about 2.0 V. The peak at the filter capacitor is therefore about 16.97 − 0.6 − 2.0 = 14.37 V. With 3.8 V peak-to-peak ripple, the average input is about 14.37 − 1.9 = 12.47 V and the trough is about 14.37 − 3.8 = 10.57 V. The 7812 needs at least 14.0 V at its input. The trough is well below that, so the regulator drops out on every ripple cycle and the output follows the sagging input; 10.8 V is consistent with this. The regulator is not necessarily faulty; the transformer/filter combination is inadequate.
2. Calculate the filter capacitance needed for 1.5 V peak-to-peak ripple. Would 4700 µF alone restore 12 V regulation?
Read the worked answer
For a full-wave rectifier, Vr ≈ I / (2fC). Rearranging, C = I / (2fVr) = 0.5 / (2 × 50 × 1.5) = 3333 µF. A 4700 µF capacitor would give Vr ≈ 0.5 / (100 × 4700 × 10⁻⁶) ≈ 1.06 V, which is better than requested. However, the peak input is still only about 14.37 V. With 1.06 V ripple, the trough is about 14.37 − 1.06 = 13.31 V, still below the 14.0 V minimum. Therefore 4700 µF alone improves ripple but does not guarantee regulation; the transformer voltage is too low.
3. Choose between a 15 V and an 18 V transformer, and check the regulator dissipation and junction temperature.
Read the worked answer
A 15 V transformer gives a peak of 21.2 V; subtracting 2.6 V gives about 18.6 V at the capacitor, and with 1.06 V ripple the average is about 18.1 V. The regulator dissipation is (18.1 − 12) × 0.5 ≈ 3.05 W. Junction temperature is 35 + 3.05 × (5 + 8) ≈ 75 °C, which is acceptable with the stated heatsink. An 18 V transformer gives about 22.9 V average input, dissipation (22.9 − 12) × 0.5 ≈ 5.45 W, and Tj ≈ 35 + 5.45 × 13 ≈ 106 °C, too close to the usual 125 °C limit. The 15 V transformer with 4700 µF is the better choice, provided the mains low-line condition is also checked.