Designing a Zener Regulator for a Fluctuating Supply
A student is building a 10 V regulated supply for a project. The raw DC input from a transformer, rectifier and filter varies between 18 V and 24 V as the mains fluctuates. The load draws between 0 and 30 mA. The only Zener diode available is rated 10 V, 1 W, and the manufacturer specifies a minimum Zener current of 5 mA for good regulation. The student first tries a 100 Ω series resistor and finds that at high mains voltage the Zener becomes very hot and eventually fails. He then tries 470 Ω, but with the load connected the output falls well below 10 V. The teacher asks the group to choose a proper series resistor, to check the worst-case power dissipation in both the resistor and the Zener, and to explain why each of the two trials failed. The Zener is to be operated without a heatsink, so its dissipation must stay below 1 W.
1. Choose a suitable series resistor and verify the circuit at the worst-case conditions. What power ratings are needed for the resistor and the Zener?
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The Zener must stay in regulation when the input is lowest and the load current is highest. Then the series resistor must still supply I_load + I_Z(min): I_R ≥ 30 mA + 5 mA = 35 mA. Rs ≤ (V_in(min) − V_Z)/I_R = (18 − 10)/0.035 = 229 Ω. The nearest standard value below this is 220 Ω.
Check 1 - low line, full load (18 V, 30 mA): I_R = (18 − 10)/220 = 36.4 mA; I_Z = 36.4 − 30 = 6.4 mA ≥ 5 mA. Regulation holds.
Check 2 - high line, no load (24 V, 0 mA): I_R = (24 − 10)/220 = 63.6 mA; all of it flows in the Zener: P_Z = 10 × 0.0636 = 0.64 W < 1 W, but only a 36% margin. In a warm box this is marginal; a 1.3 W or 2 W Zener, or a small heatsink, would be safer. P_R = (24 − 10) × 0.0636 = 0.89 W, so use a 2 W resistor, not a ¼ W one.
Check 3 - high line, full load: I_Z = 63.6 − 30 = 33.6 mA; P_Z = 0.34 W. Fine.
So Rs = 220 Ω, 2 W, with the Zener preferably rated above 1 W for thermal margin.
2. Explain why the 100 Ω trial destroyed the Zener and why the 470 Ω trial failed to regulate.
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100 Ω trial: at 24 V with no load, I_R = (24 − 10)/100 = 140 mA, P_Z = 10 × 0.140 = 1.4 W. This exceeds the 1 W rating, so the junction temperature rises until the device fails. Even with a load, the current is too high at high line; the resistor is only safe if the input never rises and the load never falls.
470 Ω trial: at 18 V with a 30 mA load, the resistor can supply only I_R = (18 − 10)/470 = 17 mA. That is less than the 30 mA the load demands, so the Zener cannot stay at 10 V. The Zener effectively turns off, and the output falls toward V_in − I_load Rs (the supply sags). Regulation requires the resistor current to be at least I_load(max) + I_Z(min) at the lowest input.
The two failures illustrate the design window: Rs must be small enough for the worst case of low input and full load, but large enough to limit Zener current at high input and no load. If that window is too narrow or empty, a simple Zener regulator cannot meet the specification and a better regulator topology is needed.
3. The student now wants the same supply to deliver up to 50 mA. Using the same 1 W, 10 V Zener and the same 18–24 V input, is this possible? Justify your answer with worst-case calculations and suggest alternatives.
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For 50 mA load, the resistor must pass at least 50 + 5 = 55 mA at 18 V: Rs ≤ (18 − 10)/0.055 = 145 Ω. Take Rs = 145 Ω (or a standard 140 Ω). Now check the high-line, no-load condition: I_R = (24 − 10)/145 = 96.6 mA, P_Z = 10 × 0.0966 = 0.97 W. This is just under the 1 W rating, with essentially no margin. Any rise in ambient temperature, any increase in mains voltage, or any unit with a low thermal resistance will push the junction over its limit. The resistor dissipation is also (14)²/145 = 1.35 W, so a 3 W or 5 W resistor is needed.
So the design is not acceptable with a 1 W Zener without a heatsink. Options:
- Use a higher-power Zener (e.g. 5 W) mounted on a heatsink and recalculate the resistor for the larger I_Z(min) and the thermal derating.
- Add a series pass transistor (emitter follower) so that the Zener only supplies base current; the transistor carries the load current and the dissipation is shared.
- Use a three-terminal regulator (e.g. 7810) with a heatsink, which also gives better line and load regulation and current limiting.
- Reduce the input voltage range with a better transformer or preregulator, which widens the allowable Rs window.
The key judgement is that a Zener regulator's worst case is not full load but maximum input with minimum load, and the design must keep the Zener below its rated dissipation there with a real thermal margin.