--- title: "Design zener shunt regulator (5.6 V, 15 V supply)" subject: "Electronics Devices and Circuit" chapter: 6 key: "semester_3/electronics_devices_and_circuit:zener-shunt-regulator-5v6-design" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/zener-shunt-regulator-5v6-design site: BCT Engineering --- # Design zener shunt regulator (5.6 V, 15 V supply) Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators - Asked in: 2079 Baisakh, 3+2 marks - Video: [Example: Zener Voltage Regulator Design](https://www.youtube.com/watch?v=pa7VfBXajf8) — Mateo Aboy - 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-shunt-regulator-5v6-design - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6 ## Question A zener diode exhibits a constant voltage of 5.6 V for currents greater than five times the knee current. Izk is specified to be 1mA. The zener is to be used in the design of a shunt regulator fed from a 15V supply. The load current varies over the range of 0 mA to 15 mA. Find a suitable value for the resistor R. What is the maximum power dissipation of the zener diode? ## Short answer With Iz(min)=5mA (5×Izk) and IL(max)=15mA, R = (15V–5.6V)/20mA = 470Ω. Maximum Zener power at no load is Pz(max) = 5.6V × 20mA = 112mW. ## Answer **Given:** Vz = 5.6 V (constant for Iz ≥ 5 Izk), Izk = 1 mA, Vs = 15 V, IL varies from 0 to 15 mA. #### Circuit ```text 15 V o----[ R ]-----+-------+---- Vo = 5.6 V | | (Dz) [RL] | | GND GND ``` #### Step 1: Minimum Zener current The Zener voltage is constant only for Iz ≥ 5 Izk = 5 × 1 mA = 5 mA. So the minimum Zener current must be Iz(min) = 5 mA. #### Step 2: Find R The worst case is the largest load current, IL = 15 mA. At this condition Iz should be at its minimum, 5 mA: IR = IL(max) + Iz(min) = 15 + 5 = 20 mA R = (Vs − Vz)/IR = (15 − 5.6)/20 mA = 9.4 V / 20 mA = **470 Ω** (470 Ω is a standard value; the supply is assumed to be fixed at 15 V.) #### Step 3: Maximum Zener power The Zener current is largest with no load (IL = 0), when the whole of IR flows in it: Iz(max) = IR = 20 mA **Pz(max) = Vz Iz(max) = 5.6 V × 20 mA = 112 mW** Hence a Zener rated at 250 mW (or 400 mW) is suitable. **Answer: R = 470 Ω; maximum Zener dissipation = 112 mW.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why must Iz(min) be 5 mA here? A. Because the load current is 5 mA B. The Zener voltage is constant only for Iz ≥ 5 Izk = 5 mA C. Because Izk equals 5 mA D. Because R equals 5 Ω Answer: B. Izk = 1 mA, and constant voltage needs five times that. (Key point: Minimum Zener current) 2. At which condition is R designed? A. Maximum load current (15 mA) with Zener at its minimum (5 mA), so IR = 20 mA B. No load with Zener at maximum C. Load 0 mA with Iz = 5 mA D. Load 15 mA with Iz = 0 Answer: A. This guarantees the Zener stays in regulation for all loads. (Key point: Worst case) 3. What is R? A. (15 - 5.6)/15 mA = 627 Ω B. 5.6/20 mA = 280 Ω C. 15/20 mA = 750 Ω D. (15 - 5.6)/20 mA = 470 Ω Answer: D. R = 9.4 V / 20 mA = 470 Ω. (Key point: R value) 4. What is the maximum Zener power and when? A. 84 mW at full load B. 300 mW at no load C. 112 mW, at no load when Iz = 20 mA D. 28 mW at no load Answer: C. At IL = 0 all 20 mA flows in the Zener: 5.6 x 20 mA = 112 mW. 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