Zener Shunt Regulator That Collapses Under Load
A trainee builds a 5.1 V Zener shunt regulator for a sensor board. The unregulated input is 12 V nominal but varies from 10.8 V to 13.2 V. The load draws between 20 mA and 60 mA. The chosen Zener is a 5.1 V, 500 mW type with a specified minimum knee current of 5 mA. The series resistor is 100 Ω, 1/4 W. On test, with no load, the output is 5.1 V and the resistor becomes too hot to touch; with a 60 mA load and the input at 10.8 V, the output falls to 4.6 V. The trainee has E12 resistors from 10 Ω to 1 kΩ in 1/4 W and 2 W, and 1 W Zeners in 5.1 V. The board must work over the full input and load range.
1. Show why the 100 Ω resistor fails at full load. Calculate the allowable series-resistance range and choose an E12 value with suitable power ratings.
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At low line and full load, the 100 Ω resistor passes: I = (10.8 − 5.1)/100 = 57 mA. The load needs 60 mA, so the Zener cannot supply the deficit; the output collapses to about 4.6 V.
Design limits: Rs ≤ (Vin(min) − Vz)/(Iz(min) + IL(max)) = 5.7/0.065 = 87.7 Ω. For the 500 mW Zener at high line and no load: Rs ≥ Vz(Vin(max) − Vz)/Pz = 5.1 × 8.1/0.5 = 82.6 Ω. No standard value lies between 82.6 Ω and 87.7 Ω, so the 500 mW Zener is marginal. Use a 1 W Zener and the E12 value 82 Ω.
Check 82 Ω: At 10.8 V and 60 mA: I = 5.7/82 = 69.5 mA, Iz = 9.5 mA, above the 5 mA knee. At 13.2 V and no load: I = 8.1/82 = 98.8 mA, Pz = 5.1 × 0.0988 = 0.504 W, within 1 W. Resistor worst case: P = I²R = 0.0988² × 82 = 0.80 W, so use a 2 W resistor. Choose 82 Ω, 2 W and a 5.1 V, 1 W Zener.
2. The trainee suggests simply fitting a 1 W Zener and keeping the 100 Ω resistor. Evaluate this suggestion.
Read the worked answer
It does not solve the problem. The failure at full load is caused by insufficient current through the 100 Ω resistor, not by the Zener's power rating. At 10.8 V input and 60 mA load, the resistor passes only 57 mA, so the Zener is below its knee and the output falls. A 1 W Zener can survive more dissipation, but it cannot create the missing 3 mA.
The 100 Ω value is too high for the low-line, full-load condition. Either reduce the resistor to 82 Ω and use the 1 W Zener, or reduce the maximum load current or input range. Keeping 100 Ω would require accepting a lower full-load current or a higher minimum input voltage.
3. What happens if the load is disconnected while the input is at 13.2 V, and what protection or design change would you add?
Read the worked answer
With 82 Ω and no load, the Zener current is (13.2 − 5.1)/82 = 98.8 mA, so it dissipates about 0.50 W. A 1 W Zener has margin but will run hot; a 500 mW type would be overstressed.
Fit the 1 W Zener with short leads and some thermal copper, add a fuse or series resistor on the input, and consider a minimum bleeder load, such as 1 kΩ drawing about 5 mA, so the Zener never carries the full no-load current. Confirm the sensor board's minimum load and specify the regulator for the disconnected-load condition. The resistor should be rated for the worst-case high-line current, not the nominal current.