Zener Shunt Regulator That Drops Out at Full Load
A 12 V unregulated supply with ±10 % variation (10.8 V to 13.2 V) feeds a 6.2 V shunt regulator. The series resistor is 220 Ω and the Zener is a 1N4735A (Vz = 6.2 V at Izt = 41 mA, Zz = 10 Ω, Izk = 1 mA, 1 W at 25 °C with 8 mW/°C derating above 25 °C). The load is a sensor board that draws between 10 mA and 45 mA, is specified to work from 6.2 V ±5 %, and sits in an enclosure that reaches 45 °C inside. On the bench with a 10 mA dummy load the output reads 6.2 V. When the real board is connected the output collapses to about 2 V and the board resets intermittently. The technician has 100 Ω, 150 Ω and 220 Ω resistors rated 1 W in the drawer, and one spare 1N4735A.
1. From the given values, find the maximum load current for which the Zener still regulates, and explain the bench observation at full load.
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Current through the series resistor at nominal input:
I_R = (12 − 6.2) / 220 = 5.8 / 220 = 26.4 mA
The Zener must keep at least Izk = 1 mA:
IL(max) = 26.4 − 1 = 25.4 mA
The board needs 45 mA, so the design was sized for roughly half the real load. At 10 mA, Iz = 16.4 mA and Pz = 6.2 × 16.4 mA ≈ 102 mW, so the output sits correctly at 6.2 V. At 45 mA the Zener is starved below breakdown and simply stops conducting; the output then follows the resistive drop:
V = 12 − (45 mA × 220 Ω) = 12 − 9.9 = 2.1 V
That matches the observed collapse: the fault is a current-budget error in R_S, not a faulty Zener. The 220 Ω value must be replaced, not the device.
2. Select a resistor from the drawer that regulates over the full input and load range, and verify Zener and resistor dissipation at 45 °C ambient.
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Worst case for the series resistor is minimum input with maximum load, because that is when the Zener current is smallest:
R_S ≤ (10.8 − 6.2) / (45 mA + 1 mA) = 4.6 / 0.046 = 100 Ω
So choose 100 Ω. Check the opposite corner (13.2 V input, 10 mA load):
I_R = (13.2 − 6.2) / 100 = 70 mA → Iz = 60 mA → Pz = 6.2 × 0.060 = 372 mW
Derated rating at 45 °C: 1 W − 8 mW/°C × 20 °C = 840 mW. 372 mW is well inside that, so the spare Zener is safe. Resistor dissipation worst case:
P_R = (13.2 − 6.2)² / 100 = 0.49 W
A 1 W part runs at about 49 % of rating in a 45 °C enclosure, which is marginal; a 2 W part is the better engineering choice. Always check both corners, not just the nominal 12 V.
3. Estimate the load regulation caused by the Zener's dynamic resistance and say whether the ±5 % specification can be met.
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Across the full load range the Zener current swings from 60 mA (light load) down to 1 mA (full load), a change of 59 mA. With Zz = 10 Ω:
ΔVz = 59 mA × 10 Ω ≈ 0.59 V
That is about 9.5 % of 6.2 V, roughly double the ±5 % window that the sensor board allows, before even counting the Zener tolerance and temperature drift. Physically, a shunt regulator with a 60:1 current swing in a 10 Ω dynamic resistance cannot hold a tight window. Options: reduce the current swing (lower input variation, or bias the Zener from a current source), add a series-pass transistor or three-terminal regulator after the Zener, or use a reference with much lower dynamic impedance. The lesson is that line/load regulation must be computed from the Zener's dynamic resistance, not from its nominal voltage alone.