Zener Shunt Regulator Losing Regulation at Heavy Load
A student builds a simple zener shunt regulator to provide 6.0 V to a load from an unregulated 10 V supply. The zener has Vz = 6.0 V at 5 mA, a rated power of 0.50 W, and a dynamic resistance small enough to ignore. The series resistor is 120 Ω. The load is a resistor that can be varied from 1.0 kΩ down to 150 Ω. The input remains at 10 V unless stated. The student first tests with no load, then with 1.0 kΩ, then with 150 Ω, and finally replaces the 120 Ω resistor with 47 Ω to get more current for a heavier load. The zener is connected with the correct polarity.
1. Calculate the current through the series resistor, the load current, the zener current and the zener power for the no-load test and for the 1.0 kΩ load. Is the zener safe in both cases?
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The voltage across the series resistor is 10 − 6 = 4 V, so IR = 4/120 = 33.3 mA. No load: all of this current passes through the zener. Iz = 33.3 mA and Pz = VzIz = 6 × 0.0333 = 0.20 W, below the 0.50 W rating. With RL = 1.0 kΩ: IL = 6/1000 = 6.0 mA. The zener carries Iz = IR − IL = 33.3 − 6.0 = 27.3 mA, and Pz = 6 × 0.0273 = 0.164 W. Both cases are safe, and the zener current remains well above the 5 mA knee current, so the output stays near 6.0 V.
2. Determine the minimum load resistance for which regulation is maintained and evaluate the 150 Ω test. What actually happens at 150 Ω?
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Regulation fails when the zener current falls to its knee value of 5 mA. The largest load current that still leaves 5 mA for the zener is IL,max = 33.3 − 5.0 = 28.3 mA. Thus RL,min = 6.0/0.0283 = 212 Ω, to two significant figures about 210 Ω. At 150 Ω the load would demand 6.0/150 = 40 mA if the output stayed at 6 V, which exceeds the 33.3 mA available through the 120 Ω resistor. The zener therefore turns off and the circuit becomes a simple potential divider: Vout = 10 × 150/(150 + 120) = 5.56 V, with about 37 mA flowing. The output has dropped below 6 V, so regulation has been lost even though nothing has been damaged.
3. Evaluate the change from 120 Ω to 47 Ω. Consider the no-load test and a possible rise of the input to 12 V.
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With 47 Ω and 10 V input, IR = (10 − 6)/47 = 85.1 mA. With no load, all of this flows through the zener, giving Pz = 6 × 0.0851 = 0.511 W, which already exceeds the 0.50 W rating, if only marginally. The no-load test is therefore unsafe. If the input rises to 12 V, the voltage across 47 Ω becomes 6 V and IR = 6/47 = 128 mA. The no-load zener power would be 6 × 0.128 = 0.77 W, well above the rating, and the zener could be destroyed. Even with a load, the higher input leaves little margin. A better fix is to keep a series resistance that limits the worst-case current, or to use a higher-power zener or a different regulator topology, rather than lowering the series resistance.