Zener Shunt Regulator Failure After Load Disconnect
A trainee repairs a small 6 V mains adapter built from a 12 V rms transformer, a bridge rectifier, a 1000 µF smoothing capacitor, a 100 Ω series resistor and a 6.2 V, 400 mW zener diode. The load normally draws 80 mA. The adapter worked for months, but after the load was left disconnected overnight the zener was found short-circuited. With the load reconnected the secondary measures 12.0 V rms; with the load disconnected it rises to 13.2 V rms. The bridge diodes drop about 1.4 V in total. The trainee proposes fitting an identical zener and adding a 1 kΩ resistor across the output to 'keep some load on it'. A second suggestion is to change the 100 Ω resistor to 47 Ω so the zener 'gets more current and regulates harder'.
1. Estimate the no-load zener current and dissipation using 13.2 V rms, and decide whether the failure is explained.
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No-load secondary is 13.2 V rms, peak = 13.2 × 1.414 = 18.67 V. After the bridge, about 18.67 − 1.4 = 17.27 V. With no load the capacitor holds near peak, so I_Z = (17.27 − 6.2) / 100 = 110.7 mA. P_Z = 6.2 × 0.1107 = 0.686 W, well above the 400 mW rating.
The failure is explained: the zener was designed on the assumption that the 80 mA load was always present. When the load was removed, the series resistor passed almost all its current through the zener.
2. Evaluate the 1 kΩ bleeder and the 47 Ω proposal.
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A 1 kΩ bleeder at 6.2 V draws 6.2 / 1000 = 6.2 mA and dissipates 38 mW. That reduces zener current from about 111 mA to 105 mA, still about 0.65 W; it does not protect the zener. To absorb the excess 46 mA the bleeder would need to be about 6.2 / 0.046 = 135 Ω, wasting 0.28 W continuously, and it would still not fix low-mains operation.
The 47 Ω proposal is much worse. No-load current would be (17.27 − 6.2) / 47 = 235 mA and P = 1.46 W. That destroys the zener faster and overloads the transformer and bridge.
3. Is the design capable of supplying 80 mA with a 400 mW zener over ±10% mains? Show the conflict and recommend a direction.
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At high mains and no load, allowable zener current for 400 mW is 0.4 / 6.2 = 64.5 mA, so R must be at least (17.27 − 6.2) / 0.0645 = 172 Ω.
At low mains, 10.8 V rms, peak after the bridge = 10.8 × 1.414 − 1.4 = 13.87 V. Even with an ideal capacitor and R = 172 Ω, maximum current is (13.87 − 6.2) / 172 = 44.6 mA, less than the 80 mA load. No single series resistor can satisfy both ends with this zener.
The design needs a higher-wattage zener with a guaranteed minimum load, a larger secondary, or preferably a series-pass regulator. A shunt zener alone is the wrong topology for a load that can be disconnected.