Shunt-Regulated 12 V Supply That Dies at No Load
A student assembles a 12 V, 100 mA regulated supply for a small project. The transformer is rated 230 V/12 V–0–12 V at 500 mA and feeds a bridge of four 1N4007 diodes; the filter capacitor is 470 µF/25 V. A 1N4742 zener (12 V, 1 W) is used as a shunt regulator with a 22 Ω series resistor, and the load draws 100 mA. On test, the no-load output sits at about 12.1 V and the full-load output falls to 10.9 V with roughly 2.2 V peak-to-peak ripple. After a few minutes the zener is warm, and when the load is disconnected briefly the zener is destroyed. The student insists the transformer and diode bridge are "fine because the meter shows 12 V" and wants to know which part of the chain is actually at fault. Mains frequency is 50 Hz, and the diodes may be treated as dropping 0.7 V each when conducting.
1. Calculate the peak voltage at the filter capacitor and the ripple voltage at 100 mA, then determine the maximum load current for which this zener branch can stay in regulation. Is the 100 mA load supportable?
Read the worked answer
Peak voltage. Two diodes conduct in series in a bridge: Vp = 12√2 − 2(0.7) = 16.97 − 1.40 = 15.57 V.
Ripple. For a full-wave rectifier with a capacitor-input filter, Vr ≈ IL/(2fC) = 0.100/(2 × 50 × 470 × 10⁻⁶) = 2.13 V p-p, which matches the measured 2.2 V and confirms the capacitor value and wiring are as intended.
Trough voltage. Vmin = 15.57 − 2.13 = 13.44 V.
Zener condition. At the trough the current through Rs must be at least the load current, otherwise the zener leaves breakdown and the output simply follows the ripple: (Vmin − VZ)/Rs ≥ IL, so IL,max = (13.44 − 12)/22 = 65 mA.
At 100 mA the zener is starved near the trough and the output sags to roughly 13.44 − 0.100 × 22 = 11.2 V, which is the observed 10.9 V. So the 100 mA load is not supportable by this shunt branch. In addition, the transformer RMS current is much larger than 100 mA because the charging pulses are narrow and tall, so the 500 mA rating is being stressed even when the DC meter looks healthy.
2. Estimate the zener current and dissipation the instant the load is removed, and explain the failure. What does this imply about selecting Rs and the zener power rating?
Read the worked answer
At no load the capacitor charges to Vp = 15.57 V and all of the current flows in the zener:
IZ = (15.57 − 12)/22 = 162 mA, so PZ = 12 × 0.162 = 1.95 W.
That is roughly twice the 1 W rating, and the rating is derated above 25 °C ambient, so destruction on load disconnection is exactly what the numbers predict. Even at full load the peak zener current is (15.57 − 12)/22 − 0.100 = 62 mA, i.e. 0.74 W, only about 26% margin. With a 10% high mains the peak rises to 17.1 V, giving (17.1 − 12)/22 = 232 mA and 2.8 W.
The real design finding is that the two requirements conflict. No-load survival needs Rs ≥ (Vp,max − VZ)/(PZ/VZ) = (17.1 − 12)/0.083 ≈ 61 Ω, while full-load regulation needs Rs ≤ (Vmin − VZ)/IL = 1.44/0.100 = 14.4 Ω. Those windows do not overlap with a 1 W zener and this ripple, so no single Rs fixes the circuit. The options are a much larger capacitor plus a 5 W zener (Rs around 27 Ω), or abandoning the shunt regulator for a series pass device.
3. Recommend a revised design that delivers 12 V at 100 mA from the same transformer and bridge, and justify the component values and the trade-off you accept.
Read the worked answer
Option A - series regulator (recommended). Fit a 7812 with a heatsink and raise the filter capacitor to 2200 µF. Then Vr = 0.100/(2 × 50 × 2200 × 10⁻⁶) = 0.45 V, so Vmin = 15.57 − 0.45 = 15.1 V, comfortably above the 7812's ≈2 V dropout. Dissipation in the regulator is (15.1 − 12) × 0.100 ≈ 0.31 W, easily handled by a small heatsink plus the tab. Add 0.33 µF at the input and 0.1 µF at the output for stability, and a 1N4007 from output to input for reverse-bias protection. The cost is a few extra components and about 2 V of headroom lost as heat.
Option B - keep the shunt. Use C = 2200 µF and Rs = 27 Ω, 2 W, with a 5 W zener. At full load the trough current is (15.1 − 12)/27 = 115 mA, leaving 15 mA in the zener; at no load the zener takes 115 mA and dissipates 1.38 W, which a 5 W part survives. The penalty is poor efficiency: the supply wastes roughly 1.4 W at all times and loads the transformer continuously.
Because the load is a fixed 100 mA and efficiency matters little here, Option A is the better engineering choice; Option B is only justified if the 7812 is unavailable or dropout headroom cannot be guaranteed at low mains.