Zener Shunt Regulator Overdissipating at High Line
A 24 V nominal industrial control bus varies from 21.6 V to 26.4 V. A technician must derive a 12 V rail for a sensor that draws 0 to 80 mA. The parts bin contains a 1N4742A 12 V zener rated 1 W at 25 °C, with IZT = 21 mA and ZZ = 9 Ω, plus a selection of 1/4 W resistors. A simple shunt regulator is proposed: a series resistor from the bus to the zener cathode, with the load in parallel with the zener. The zener must stay in regulation at minimum bus voltage and full load, while not exceeding its rating at maximum bus voltage and no load. Ambient inside the enclosure reaches 60 °C, and the zener derates linearly to 0 W at 150 °C junction. The resistor must also survive worst-case power. The technician must choose the series resistor and decide whether the 1N4742A is acceptable.
1. Choose a series resistor that keeps the zener biased at minimum bus and full load. State your minimum zener current assumption.
Read the worked answer
At minimum bus and full load, the zener still needs a minimum current for regulation. Choose IZ(min) = 21 mA, the zener's test current, to stay on the flat part of its characteristic. Then R ≤ (21.6 − 12)/(0.080 + 0.021) = 9.6/0.101 = 95.0 Ω. Select 91 Ω, a standard value. At 21.6 V and 80 mA, IR = 9.6/91 = 105.5 mA, so IZ = 25.5 mA, above 21 mA. Using 100 Ω would give IZ = 96 − 80 = 16 mA, below test current but still above the knee; regulation would be softer. The 91 Ω choice provides margin at the cost of higher worst-case dissipation.
2. Evaluate worst-case zener and resistor dissipation at 60 °C. Is the 1N4742A and a 1/4 W resistor acceptable?
Read the worked answer
Worst-case zener power occurs at maximum bus and no load: IZ = (26.4 − 12)/91 = 158 mA, so PZ = 12 × 0.158 = 1.90 W. The 1N4742A is rated 1 W at 25 °C and derates linearly to 0 W at 150 °C. At 60 °C ambient, assuming the junction is at ambient, Pmax = 1 W × (150 − 60)/(150 − 25) = 0.72 W. The 1.90 W worst case is far above 0.72 W, so the 1N4742A will overheat and fail. Resistor power at worst case is I²R = 0.158² × 91 = 2.27 W, so a 1/4 W resistor is completely inadequate; use at least a 5 W resistor. Even at full load and maximum bus, the resistor still dissipates 2.27 W because its current is set by the bus and R.
3. Recommend a practical fix and comment on the efficiency of the shunt regulator.
Read the worked answer
Replace the 1N4742A with a 5 W 12 V zener such as the 1N5349B. Its 60 °C derated rating is 5 × 90/125 = 3.6 W, so 1.90 W is acceptable with margin. Use a 5 W or 7 W series resistor, provide ventilation, and verify junction temperature using the actual thermal resistance and copper area. However, the shunt regulator is inefficient: at 26.4 V and no load it burns about 2.27 W in the resistor plus 1.90 W in the zener, roughly 4.2 W to deliver nothing. At 24 V and 80 mA, input power is about 24 × 0.132 = 3.17 W for 0.96 W output, or roughly 30% efficiency. A series-pass regulator or a small switching regulator would be thermally far better; keep the shunt only if simplicity and cost dominate and the enclosure can dissipate the heat.