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Zener: range of load for VL = 5 V

Short answer

Series resistor current Is=(10-5)/100=50 mA. Zener current must stay 0–30 mA, so load current IL=50–Iz. IL max 50 mA gives RLmin=5/0.05=100 Ω; IL min 20 mA gives RLmax=5/0.02=250 Ω. Range: 100 Ω ≤ RL ≤ 250 Ω.

Determine the Range of load RL that will maintain the zener diode load voltage VL at 5V. Given Vs = 10 V, Rs = 100 Ω, IZM = 30 mA. [Figure: source Vs in series with Rs, Zener diode Vz in parallel with load RL, with currents Is, Iz, IL]

Given: Vs = 10 V, Rs = 100 Ω, Vz = VL = 5 V, IZM = 30 mA.

Current in the series resistor

The current through Rs is constant while the Zener is ON:

Is = (Vs − Vz)/Rs = (10 − 5)/100 = 50 mA

Is = Iz + IL, so IL = 50 mA − Iz. The Zener must carry Iz between 0 (just ON) and IZM = 30 mA.

Minimum load resistance (maximum load current)

Iz is smallest, ≈ 0:

IL(max) = 50 − 0 = 50 mA

RLmin = VL / IL(max) = 5 V / 50 mA = 100 Ω

(Below this value the Zener cuts off and VL drops below 5 V.)

Maximum load resistance (minimum load current)

Iz is highest, = IZM = 30 mA:

IL(min) = 50 − 30 = 20 mA

RLmax = 5 V / 20 mA = 250 Ω

(Above this value Iz exceeds 30 mA and the Zener can be damaged.)

Answer: 100 Ω ≤ RL ≤ 250 Ω (load current from 20 mA to 50 mA).

✓ Answer checked by Claude Opus 5.5 (high effort)

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