Zener regulator worst-case power
A student builds a simple zener regulator. The unregulated input is nominally 12 V but can rise to 15 V. The zener is rated 6.2 V, 250 mW. The series resistor is 220 Ω, and the load is 1.0 kΩ. At 12 V input and with the load connected, the output is a steady 6.2 V. The student then removes the load and notices that the zener becomes noticeably hot. The student wants to know whether the 250 mW rating is sufficient and whether the circuit will survive the input rising to 15 V with the load disconnected. A 1/2 W zener and a 330 Ω resistor are available as alternatives. The zener needs at least 5 mA to regulate.
1. Calculate the zener current and power at 12 V with the load, and at 15 V with the load removed.
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With the load connected at 12 V: I_R = (12 - 6.2)/220 = 0.0264 A = 26.4 mA. I_L = 6.2/1000 = 6.2 mA. I_Z = 26.4 - 6.2 = 20.2 mA. P_Z = 6.2 × 0.0202 = 0.125 W = 125 mW. With the load removed at 15 V: I_R = (15 - 6.2)/220 = 0.0400 A = 40.0 mA, and all of it passes through the zener, so P_Z = 6.2 × 0.0400 = 0.248 W = 248 mW.
2. Is the 250 mW zener adequate? What is the worst case if the input can be 15 V and the load can be disconnected?
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248 mW is essentially the full 250 mW rating, with no margin for zener tolerance, temperature derating, or a slightly higher input. The worst case is maximum input with no load, because then the zener carries the whole series current. A 1/2 W zener would be safer. Alternatively, the 330 Ω resistor reduces the worst-case current to (15 - 6.2)/330 = 26.7 mA and P_Z = 165 mW. At 12 V with the 1 kΩ load, I_R = (12 - 6.2)/330 = 17.6 mA, I_Z = 17.6 - 6.2 = 11.4 mA, still above the 5 mA minimum. The 330 Ω resistor plus a 1/2 W zener is the better choice.
3. If the load is 1.0 kΩ and the input falls to 10 V, will regulation hold? What is the minimum input for regulation?
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At 10 V input, I_R = (10 - 6.2)/220 = 17.3 mA. With the 1 kΩ load drawing 6.2 mA, I_Z = 11.1 mA, which exceeds 5 mA, so regulation holds. The minimum input is V_in,min = 6.2 + (I_L + I_Z,min)R = 6.2 + (0.0062 + 0.005)(220) = 6.2 + 2.46 = 8.66 V. Below about 8.7 V the zener drops out of regulation.