--- title: "Design DC voltage regulator for 6 V" subject: "Electronics Devices and Circuit" chapter: 6 key: "semester_3/electronics_devices_and_circuit:zener-regulator-design-6v" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/zener-regulator-design-6v site: BCT Engineering --- # Design DC voltage regulator for 6 V Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators - Asked in: 2083 Baisakh, 6 marks; 2075 Chaitra, 5 marks; 2072 Chaitra, 4 marks (asked 3 times) - 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-regulator-design-6v - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6 ## Question Design DC voltage regulator for 6V output. Given data are Vz = 6V at Iz = 20mA, IZK = 2mA, Pzmax = 500mW and rz = 10Ω. The nominal input voltage is 15V±30% DC. Find the maximum current it can deliver to the load. ## Short answer Choose series resistor R = 180 Ω. At minimum input Vin = 10.5 V and Zener knee current IZK = 2 mA, total current IR = 26 mA; maximum load current IL(max) = IR − IZK = 24 mA. ## Answer **Given:** Vz = 6 V at Iz = 20 mA, rz = 10 Ω, IZK = 2 mA, Pzmax = 500 mW. Vin = 15 V ± 30 % → Vin(min) = 10.5 V, Vin(max) = 19.5 V. #### Circuit ```text Vin o----[ R ]-----+------+----o Vo = 6 V | | (Dz) [RL] | | GND GND ``` #### Step 1: Zener model The Zener is represented by Vz0 in series with rz: Vz0 = Vz − rz Iz = 6 − 10 × 0.02 = 5.8 V, so Vz = 5.8 + 10 Iz. #### Step 2: Maximum Zener current Pz = Vz0 Iz + rz Iz² = 0.5 W 10 Iz² + 5.8 Iz − 0.5 = 0 → Iz(max) = 76.2 mA, with Vz = 5.8 + 0.762 = 6.56 V. #### Step 3: Choose R The worst case for the Zener power is at Vin(max) with no load (RL open), when the entire current goes through the Zener. That current must not exceed 76.2 mA: R ≥ (Vin(max) − Vz)/Iz(max) = (19.5 − 6.56)/76.2 mA = 169.8 Ω Take the standard value **R = 180 Ω**. (Check: at Vin(max), Iz = (19.5 − 6.52)/180 = 72 mA, and Pz = 0.47 W < 0.5 W.) #### Step 4: Maximum load current The load current is largest when Vin is minimum, and the Zener is kept at the knee current IZK = 2 mA (the Zener must stay in breakdown): Vz at 2 mA = 5.8 + 10 × 0.002 = 5.82 V IR = (Vin(min) − Vz)/R = (10.5 − 5.82)/180 = 26.0 mA **IL(max) = IR − IZK = 26.0 − 2 = 24.0 mA** This corresponds to a minimum load resistance of about 6/24 mA = 250 Ω. Assumption: the design is based on R fixed from the Zener power rating at no-load with maximum input. **Answer: R = 180 Ω (≥ 170 Ω); the circuit can deliver a maximum load current of about 24 mA (at Vin = 10.5 V, with Vo ≈ 5.8 V to 6.5 V).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are Vin(min) and Vin(max) for 15 V ± 30 %? A. 10.5 V and 19.5 V B. 12 V and 18 V C. 10 V and 20 V D. 4.5 V and 19.5 V Answer: A. 15 x 0.7 = 10.5 V; 15 x 1.3 = 19.5 V. (Key point: Input range) 2. When is the Zener power highest, setting the lower limit of R? A. Vin(max) with no load B. Vin(min) with full load C. Vin(min) with no load D. Vin(max) with full load Answer: A. With no load, all current flows through the Zener. (Key point: Worst case for R) 3. When is load current maximum, and what Zener current is assumed? A. At Vin(min) with Zener at knee current IZK B. At Vin(max) with Zener at 76.2 mA C. At Vin(min) with Zener at 20 mA D. At Vin(max) with Zener at IZK Answer: A. The Zener must just stay in breakdown at IZK. (Key point: Max load current) 4. With R = 180 Ω, what is IL(max)? A. About 24 mA B. About 26 mA C. About 72 mA D. About 76 mA Answer: A. IR = (10.5 - 5.82)/180 = 26 mA, minus 2 mA = 24 mA. 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