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Zener input range, 220 Ω, 20 V, 60 mA

Short answer

Load current IL = 20 V / 1.2 kΩ = 16.67 mA. For Zener ON, Vi(min) = 20 + 16.67mA × 220 = 23.67 V (Iz = 0). Vi(max) = 20 + (16.67+60)mA × 220 = 36.87 V (Izmax = 60 mA). Vi range: 23.67 V to 36.87 V.

Determine the range of values of Vi that will maintain the Zener diode of the figure below in ON state. [Figure: source Vi, series resistor R = 220 Ω, Zener diode Vz = 20 V with izmax = 60 mA in parallel with load RL = 1.2 kΩ]

Given: R = 220 Ω, Vz = 20 V, IZmax = 60 mA, RL = 1.2 kΩ. The Zener is ON (in breakdown) when Iz is between 0 and IZmax, and Vo = Vz = 20 V.

Load current

IL = Vz/RL = 20 V / 1.2 kΩ = 16.67 mA (constant, since Vz is fixed)

Minimum Vi

Zener just turns ON: Iz ≈ 0, so the series current IR = IL.

Vi(min) = Vz + IR R = 20 + (16.67 mA)(220 Ω) = 20 + 3.67 = 23.67 V

Maximum Vi

The Zener current is at its maximum, Iz = 60 mA, so

IR(max) = IZmax + IL = 60 + 16.67 = 76.67 mA

Vi(max) = Vz + IR(max) R = 20 + (76.67 mA)(220 Ω) = 20 + 16.87 = 36.87 V

Check

At Vi = 23.67 V: IR = 3.67/220 = 16.67 mA = IL, so Iz = 0 ✓. At Vi = 36.87 V: IR = 16.87/220 = 76.7 mA, Iz = 76.7 − 16.7 = 60 mA ✓.

Answer: 23.67 V ≤ Vi ≤ 36.87 V keeps the Zener in the ON state.

✓ Answer checked by Claude Opus 5.5 (high effort)

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