Zener Shunt Regulator at Start-Up
A student builds a zener shunt regulator to provide 6.2 V to a small radio from a 12 V supply. The series resistor is 220 Ω, and the zener diode is rated 6.2 V, 500 mW. The radio draws 20 mA in normal operation but 60 mA for a short time at start-up. With a 1 kΩ test load, the output measures close to 6.2 V, so the student concludes that the design is sound. The teacher warns that both the zener power dissipation and the start-up condition must be checked before the regulator is connected to the radio. The supply is a well-smoothed 12 V source with negligible internal resistance. The zener is ideal in the sense that it maintains 6.2 V only while it carries some current, and its power rating must not be exceeded. The student asks whether the circuit will regulate during start-up and whether any resistor value must be changed.
1. For the 1 kΩ test load, calculate the current through the series resistor, the load current, the zener current, and the zener power dissipation. Is the zener safe in this test?
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The voltage across the series resistor is 12 − 6.2 = 5.8 V, so I_R = 5.8/220 ≈ 26.4 mA. The test load current is I_L = 6.2/1000 = 6.2 mA. The zener current is I_Z = I_R − I_L ≈ 26.4 − 6.2 = 20.2 mA. The zener power is P_Z = 6.2 × 20.2 mA ≈ 125 mW, well below the 500 mW rating. The zener is safe under the test load, which explains why the student sees about 6.2 V.
2. Check the start-up condition when the radio draws 60 mA. Will the output remain regulated? Determine a suitable series resistance that keeps the zener conducting during start-up while respecting its power rating.
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At 60 mA, the resistor current is still about 26.4 mA, so the zener current would be 26.4 − 60 = −33.6 mA. A zener cannot supply a negative current; it simply turns off, and the output falls below 6.2 V. The regulator fails at start-up.
To keep the zener conducting, choose R so that I_R ≥ 60 mA plus a small zener current. With a minimum zener current of about 5 mA, require I_R ≥ 65 mA: R ≤ (12 − 6.2)/0.065 ≈ 89 Ω. A standard 82 Ω resistor gives I_R = 5.8/82 ≈ 70.7 mA. At 60 mA load, I_Z ≈ 10.7 mA, so regulation is maintained. At 20 mA load, I_Z ≈ 50.7 mA and P_Z ≈ 314 mW. With no load, I_Z ≈ 70.7 mA and P_Z ≈ 438 mW, below 500 mW but with little margin. A 1 W zener, or a slightly larger resistor such as 91 Ω, would give better safety margin.
3. With the original 220 Ω resistor, what happens if the radio is disconnected? Is no-load the worst case for the zener? Explain.
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With no load, all the resistor current flows through the zener: I_Z ≈ 26.4 mA. The zener power is P_Z ≈ 6.2 × 26.4 mA ≈ 164 mW, still well below 500 mW. No-load is therefore not the worst case for this particular design.
The worst case is the 60 mA start-up load, because the zener current would be driven negative and regulation would be lost. The design is limited by the minimum zener current under heavy load, not by the maximum zener current under light load. Checking only the no-load condition would miss the actual failure mode.