--- title: "Zener: range of load for VL = 5 V" subject: "Electronics Devices and Circuit" chapter: 6 key: "semester_3/electronics_devices_and_circuit:zener-load-range-5v" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/zener-load-range-5v site: BCT Engineering --- # Zener: range of load for VL = 5 V Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators - Asked in: 2078 Kartik, 5 marks - Video: [Solved Problems on the Zener Diode](https://www.youtube.com/watch?v=kYtebfx-yHc) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/zener-load-range-5v - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6 ## Question 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] ## 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. ## Answer **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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the current in Rs constant? A. Because RL is fixed B. Vs and Vz are fixed so Is = (10 - 5)/100 = 50 mA while the Zener is ON C. Because Is changes with Iz D. Because Vz changes with load Answer: B. Is depends only on Vs, Vz and Rs. (Key point: Series current) 2. What is RLmin and what limits it? A. 100 Ω; at IL = 50 mA, Iz = 0 and the Zener would cut off B. 250 Ω; Zener power exceeded C. 50 Ω; Zener at IZM D. 100 Ω; Zener at IZM Answer: A. RLmin = 5 V / 50 mA. (Key point: Minimum RL) 3. What is RLmax? A. 500 Ω B. 100 Ω C. 167 Ω D. 250 Ω; IL = 50 - 30 = 20 mA Answer: D. RLmax = 5 V / 20 mA = 250 Ω, where Iz = IZM. (Key point: Maximum RL) 4. Why does too-large RL damage the Zener? A. Load current would become negative B. Vs would rise C. Less load current means more Zener current, exceeding IZM D. Rs would burn out first Answer: C. Is = Iz + IL is constant, so lower IL raises Iz. 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