--- title: "Zener current: 18 V, 100 Ω, RL 1.2 kΩ" subject: "Basic Electronics Engineering" chapter: 2 key: "semester_2/basic_electronics_engineering:zener-current-18v-100" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/zener-current-18v-100 site: BCT Engineering --- # Zener current: 18 V, 100 Ω, RL 1.2 kΩ Basic Electronics Engineering · Chapter 2: Diodes - Asked in: 2074 Bhadra, 5 marks; 2071 Bhadra, 4 marks; 2067 Chaitra, 3 marks (asked 3 times) - Video: [Special Diode Circuits | Zener Diode (Fixed Vi and RL) | Solved Problem](https://www.youtube.com/watch?v=n5EbPnk6yXo) — Electronics Insight - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/zener-current-18v-100 - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2 ## Question Find the Zener current in the given circuit when RL = 1.2 kΩ. Assume VZ = 10 V. [Figure: V1 = 18 V source, series resistor R = 100 Ω, Zener diode in parallel with RL = 1.2 kΩ; output Vo = 10 V.] ## Short answer The Zener is ON because open-circuit voltage across RL (18×1200/(100+1200)=16.62 V) exceeds VZ=10 V. Series current IR = (18−10)/100 = 80 mA, load current IL = 10/1200 = 8.33 mA, so Zener current IZ = IR − IL = 71.67 mA. ## Answer **Zener regulator analysis (VZ = 10 V).** First check whether the Zener is ON. With the Zener removed, the voltage across RL would be V = 18 × 1200/(100 + 1200) = 16.62 V This is greater than VZ = 10 V, so the Zener is in breakdown and holds Vo = 10 V. - Current through the series resistor: IR = (Vin − VZ)/R = (18 − 10)/100 = 0.08 A = 80 mA - Load current: IL = Vo/RL = 10/1200 = 8.33 mA - Zener current (KCL): IZ = IR − IL = 80 − 8.33 = 71.67 mA ```text 18 V o--[100 ohm]--+--------+--o Vo = 10 V | | Z RL 1.2k | | 0 V o-------------+--------+-- ``` **Answer: IZ = 71.7 mA** (IR = 80 mA, IL = 8.33 mA). Power in the Zener = 10 × 71.67 mA = 0.717 W, so a 1 W Zener is needed. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is it decided that the Zener is in breakdown? A. The voltage with the Zener removed (16.62 V) exceeds VZ = 10 V B. The load is 1.2 kΩ C. The source is 18 V D. The series resistor is 100 Ω Answer: A. The open-circuit voltage across RL exceeds VZ, so the Zener conducts and holds Vo = 10 V. (Key point: Check if Zener is ON) 2. What is the current through the series resistor? A. (18 − 10)/100 = 80 mA B. 18/100 = 180 mA C. 10/100 = 100 mA D. 18/1300 = 13.8 mA Answer: A. IR = (Vin − VZ)/R = 8/100 = 80 mA. (Key point: Series current) 3. What is IL? A. 10/1200 = 8.33 mA B. 18/1200 = 15 mA C. 10/100 = 100 mA D. 80 mA Answer: A. IL = Vo/RL = 10/1200 = 8.33 mA. (Key point: Load current) 4. How is IZ found, and what is it? A. IZ = IR + IL = 88.3 mA B. IZ = IR − IL = 71.7 mA C. IZ = IL = 8.33 mA D. IZ = IR = 80 mA Answer: B. By KCL, IZ = 80 − 8.33 = 71.67 mA. 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