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Zener input range, 91 Ω, 8 V, 400 mW

Short answer

With a 91Ω series resistor and a 220Ω load, the Zener remains ON for Vi from 11.31V (Zener just conducts) to 15.86V (Iz reaches 50mA at max power 400mW).

Determine the range of input voltage Vi that will maintain the Zener diode in an 'ON' state and not exceed the maximum power rating of the Zener diode. [Figure: source Vi, series resistor Rs = 91 Ω, Zener diode Vz = 8 V with Pzmax = 400 mW in parallel with load RL = 0.22 kΩ]

Given: Rs = 91 Ω, Vz = 8 V, Pzmax = 400 mW, RL = 0.22 kΩ = 220 Ω.

Load current and maximum Zener current

IL = Vz/RL = 8 V / 220 Ω = 36.36 mA (constant while the Zener is ON)

IZM = Pzmax/Vz = 400 mW / 8 V = 50 mA

Minimum Vi (Zener just ON, Iz ≈ 0)

Rs carries only the load current:

Vi(min) = Vz + IL Rs = 8 + 0.03636 × 91 = 8 + 3.31 = 11.31 V

(Check with the Zener removed: Vi × 220/(220 + 91) = 8 V gives Vi = 11.31 V ✓.)

Maximum Vi (Iz = IZM = 50 mA)

Is = IZM + IL = 50 + 36.36 = 86.36 mA

Vi(max) = Vz + Is Rs = 8 + 0.08636 × 91 = 8 + 7.86 = 15.86 V

Answer: 11.31 V ≤ Vi ≤ 15.86 V. Below 11.31 V the Zener turns OFF (no regulation); above 15.86 V its power exceeds 400 mW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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