Series-Regulator Supply Overheating at Full Load
A shore station's 13.8 V, 25 A linear supply uses an 18 V RMS transformer secondary, a bridge rectifier, a 40,000 µF filter capacitor, and four parallel pass transistors bolted to one unfanned heatsink of 1.1 °C/W in a 45 °C radio room. It powers a 100 W PEP HF transceiver drawing about 8 A average and 20 A on voice peaks. The operator reports hum on the transmitted signal, 1.8 V peak-to-peak ripple on the 13.8 V rail at 8 A, a rail that sags to 12.4 V on peaks, and a heatsink too hot to touch after ten minutes. Mains is 60 Hz. The capacitor measures at its rated capacitance with normal ESR.
1. Is the 1.8 V ripple evidence of a failed capacitor? Show the calculation.
Read the worked answer
For a full-wave rectifier at 60 Hz, the ripple frequency is 120 Hz and the peak-to-peak ripple is approximately ΔV = I / (2fC).
ΔV = 8 A / (120 Hz × 0.040 F) = 8 / 4.8 = 1.67 V p-p, which matches the measured 1.8 V.
Conclusion: the capacitor is not the fault. The ripple is exactly what the design produces at 8 A. It is high because the reservoir is small for the load, not because it has dried out. That ripple amplitude explains the hum: the series regulator has only a few volts of headroom over the pass bank, so its ripple rejection collapses at high current, and the residual rides on the 13.8 V rail and modulates the SSB envelope.
Judgement: if the capacitor had failed, ripple would have risen without a load change and ESR would be abnormal - neither is true here. Symptom chasing would have replaced a good part.
2. Estimate the dissipation in the pass bank and the resulting junction temperature. Is the heatsink adequate?
Read the worked answer
DC input to the regulator: 18 V RMS × 1.414 = 25.5 V peak, minus about 1.4 V for the bridge = 24.1 V, and the capacitor sits near the top of the ripple, so allow roughly 23–24 V average at 8 A.
Dissipation: (23.5 − 13.8) V × 8 A ≈ 78 W at average load, and much more briefly on 20 A peaks. Split four ways that is ≈20 W per device.
Heatsink rise: 78 W × 1.1 °C/W = 86 °C above ambient → sink at ≈131 °C. That is consistent with "too hot to touch" and exceeds any reasonable derating.
Junction: with θjc ≈ 1.5 °C/W and θcs ≈ 0.5 °C/W, Tj = 131 + 20×(2.0) = 171 °C - under the 200 °C absolute maximum but far outside the safe operating area once second breakdown and thermal cycling are considered. The bank is being operated at the edge, and a single device carrying more than its share (unequal emitter resistors) will fail first.
Verdict: not adequate. A linear pass bank burning 78 W to make 110 W of output is a ~58 % efficient design.
3. The rail sags to 12.4 V on voice peaks. Diagnose what to measure and propose fixes ranked by effort.
Read the worked answer
What the sag is not: at 8 A the trough of the ripple (≈22 V) is far above 13.8 V + dropout, so this is not regulator dropout on average. Peak sag points to current limiting, transformer/rectifier regulation, or I²R in the wiring.
Measure in this order:
- Scope the raw DC (before the pass bank) during a voice peak - a large dip there means the transformer or rectifier is the limit, not the regulator.
- Measure DC volts across the pass bank and across the output leads under peak; 20 A through 0.05 Ω of wire and connectors is 1 V lost for free.
- Check the current-limit sense resistor and the emitter-ballast resistors for drift.
- Measure the capacitor's ESR under load, not just capacitance.
Fixes, cheapest first:
- Re-terminate and shorten the 13.8 V leads; clean and tighten every connection.
- Add forced air to the heatsink (1.1 → ~0.3 °C/W drops the sink to ~68 °C) and re-paste the devices.
- Fit a bigger reservoir (100,000 µF) to lift the trough and improve headroom.
- The real cure: replace the linear pass bank with a 13.8 V switching supply (≈90 % efficient, ~12 W dissipated) or use a transformer with a lower secondary and a buck-preregulator ahead of the pass bank. This also removes the hum.