Zener Regulator Failure on a 28 V Bus
A line-replaceable unit takes power from a 28 V DC bus (25.2 V to 30.8 V in service) and derives a 12 V rail for a 20 mA sensor load using a series resistor and a 1N4742A zener. The zener is rated 12 V, 1 W at 25 °C, derating linearly at 10 mW/°C above 50 °C, and its knee current is 5 mA. The series resistor is 220 Ω, 0.25 W. Units returned from service show the zener short-circuited and the sensor destroyed. Bench tests reproduce the fault when the load is disconnected while the bus is at its upper limit. The internal ambient of the unit is measured at 70 °C. The design authority asks whether the 220 Ω / 1 W combination was ever valid, and what to do about it.
1. Calculate the zener dissipation at nominal bus with the load connected, and at maximum bus with the load disconnected. Is the 1N4742A adequate at 70 °C internal ambient?
Read the worked answer
Nominal bus, load connected
- Resistor current: I = (28 − 12) / 220 = 72.7 mA
- Zener current: 72.7 − 20 = 52.7 mA
- Zener dissipation: 12 V × 52.7 mA = 0.63 W
Maximum bus (30.8 V), load disconnected - the true worst case, because none of the current is diverted to the load:
- I = (30.8 − 12) / 220 = 85.5 mA
- P = 12 V × 85.5 mA = 1.03 W
Derating: the 1 W rating applies only up to 50 °C. At 70 °C the permissible dissipation is 1 − (20 × 0.010) = 0.80 W.
1.03 W exceeds 0.80 W, so the part is over its rating in the worst case, and even the nominal-bus figures (0.63 W now, 0.85 W if the load is ever unplugged) leave almost nothing in hand. The 220 Ω choice is only defensible if the load is guaranteed present forever, which a line-replaceable unit cannot guarantee. The correct judgement is that the resistor value, not the zener batch, is the root cause.
2. The zener has failed short-circuit. Explain the sequence that destroyed the sensor and say what the series resistor is doing in that state.
Read the worked answer
Once the junction melts, the zener behaves as a low-value resistance, so the 12 V node is no longer clamped.
- The sensor (12 V, 20 mA → roughly 600 Ω equivalent) now sits in a divider with the 220 Ω resistor across the bus.
- Worst-case node voltage: 30.8 × 600 / (600 + 220) = 22.5 V across a 12 V part - immediate destruction.
- The resistor sees the whole bus: P = 30.8² / 220 = 4.3 W in a 0.25 W body, so it is also destroyed or opened.
So a single zener overload takes out the load as a secondary failure. That is why a shunt regulator feeding expensive downstream electronics is normally backed by a crowbar or a fused/current-limited arrangement, and why the failure mode of the pass element matters as much as its steady-state rating. It also explains why the returned units show two failed parts: the technician who replaces only the zener will see the fault return.
3. Choose a corrected series resistor and verify it at both supply extremes with the load connected and disconnected. State the resistor's power rating.
Read the worked answer
Two constraints define a window for R.
Lower limit - protect the zener at maximum bus. Allowable zener current at 70 °C with a little margin, say P ≤ 0.68 W: I_Z ≤ 0.68 / 12 = 57 mA. With the load disconnected, I_R = I_Z, so
R ≥ (30.8 − 12) / 0.057 = 217 Ω
Upper limit - keep the zener above its 5 mA knee at minimum bus with full load.
I_R ≥ 20 + 5 = 25 mA → R ≤ (25.2 − 12) / 0.025 = 528 Ω
A 330 Ω standard value sits comfortably inside the window:
- Max bus, no load: I = 18.8 / 330 = 57.0 mA → P_Z = 684 mW (< 800 mW rating) ✔
- Nominal, full load: I = 16 / 330 = 48.5 mA → I_Z = 28.5 mA ✔
- Min bus, full load: I = 13.2 / 330 = 40.0 mA → I_Z = 20 mA ✔ (well above the knee)
Resistor rating: worst-case dissipation is 0.057² × 330 = 1.07 W, so specify at least a 2 W part (3 W preferred for derating in a 70 °C enclosure). If the envelope will not take a 2 W resistor, the honest answer is that a shunt regulator is the wrong topology here and a series pass regulator should be used.