Shunt Zener Regulator That Only Overheats When the Load Is Disconnected
A final-year project powers a sensor board from a 12-0-12 V, 500 mA transformer through a bridge rectifier, a 470 uF reservoir capacitor, a 220 ohm series resistor and a BZX55C5V1 zener (5.1 V, 500 mW). The board draws between 0 and 45 mA. On the bench with the sensor unplugged the zener becomes too hot to touch within a minute, and one diode has already failed short-circuit; with the sensor connected the rail sits at 5.1 V and the circuit behaves normally. DMM measurements: 32.4 V DC across the capacitor at no load and 31.3 V at full load. Mains is 230 V, 50 Hz. The student argues that the zener must be safe because 5.1 V x 45 mA = 0.23 W is well inside the 500 mW rating. Replacement parts of the same number are available, and the local shop stocks 1N4733A (5.1 V, 1 W) zeners, 330 ohm resistors and 7805 regulators.
1. Calculate the zener current and dissipation in both measured conditions. Is the student's 0.23 W reasoning correct, and why does the part survive only when the load is connected?
Read the worked answer
The student multiplied the load current, not the zener current.
With the sensor unplugged, the only path for the rectified DC is the series resistor and the zener:
- I_R = (32.4 V - 5.1 V) / 220 ohm = 124 mA, so I_Z = 124 mA
- P_Z = 5.1 V x 0.124 A = 0.63 W
With the board connected:
- I_R = (31.3 - 5.1) / 220 = 119 mA
- I_Z = 119 - 45 = 74 mA
- P_Z = 5.1 x 0.074 = 0.38 W
The worst case for a shunt regulator is always zero load, because the zener must then absorb the entire resistor current. A 500 mW glass zener is rated at 25 degrees C and derates by roughly 4-5 mW per degree, so at 40 degrees C inside a case its real capability is about 0.42 W. 0.63 W is far over that, 0.38 W is borderline. The observed failure sequence (hot, then short) is the expected overdissipation failure mode, not a faulty batch.
2. Select a fix from the available stock and justify it numerically, including a 10% low-mains condition.
Read the worked answer
Option A: 1N4733A (1 W) with the series resistor raised to 330 ohm.
- No load: I_R = (32.4 - 5.1)/330 = 82.7 mA, P_Z = 0.42 W, i.e. 42% of the 1 W rating, safe after derating.
- Full load: I_R = (31.3 - 5.1)/330 = 79.4 mA, so I_Z = 34 mA, comfortably above the 5-10 mA knee where regulation holds.
- Mains at -10%: V_DC falls to about 29 V and the trough of the ripple to about 28 V, so I_R = (28 - 5.1)/330 = 69 mA and I_Z = 24 mA. Regulation still holds.
Option B: 7805. Dropout is not a problem (input is 31 V), but dissipation would be (31.3 - 5) x 0.045 = 1.18 W, needing a heatsink for a TO-220 (theta_JA about 65 degrees C/W gives a 77 degree rise). More seriously, at no load the input is 32.4 V and at mains +10% it rises to about 35.6 V, which is above the 7805's 35 V absolute maximum input. The regulator is therefore the wrong tool unless a series dropper or pre-regulator is added.
Recommendation: fit the 1N4733A and change the resistor to 330 ohm; this fixes both the part rating and the current that the part has to swallow at no load.
3. The failed zener is found short-circuited. Explain what the sensor board experiences in that state and what single protective addition you would make to the design.
Read the worked answer
A shorted zener removes the shunt path, so the 5.1 V rail is pulled up towards the unregulated 31-32 V (limited only by the 220 ohm resistor and the load current). Every 5 V IC on the board then sees about 6 times its supply rating; the board usually dies before the resistor has time to heat.
Protection options, in order of effectiveness:
- Crowbar plus fuse. A 100 mA slow-blow fuse in the DC line and a 5.6 V thyristor (or a 5.6 V transient suppressor) across the rail. When the regulator fails the clamp conducts, the fuse opens within milliseconds and the ICs are spared. This is the standard approach and costs very little.
- Series resistor and 5.6 V TVS. A TVS alone will clamp but must survive the 124 mA continuously until something else fails; it is acceptable only as a backup.
- Accept the risk only if the regulator is massively derated (a 1 W zener running at 42% of rating, mounted away from heat), which makes a short unlikely in the first place.
Note the judgement point: the fuse alone does not protect the ICs, because 124 mA will not open a 100 mA fuse quickly; the clamp is what limits the voltage while the fuse clears.