--- title: "Zener regulator design: Rs and IZM" subject: "Fundamental of Electrical and Electronics Engineering" chapter: 4 key: "semester_1/fundamental_of_electrical_and_electronics_engineering:zener-regulator-design" source: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/zener-regulator-design site: BCT Engineering --- # Zener regulator design: Rs and IZM Fundamental of Electrical and Electronics Engineering · Chapter 4: Diodes - Asked in: 2081 Baishakh (new course), 2+2 marks - Video: [Zener Diode Regulation - Varying load resistance, Varying input voltage and Varying input resistance](https://www.youtube.com/watch?v=CO18uJ_LxE8) — Electric Circuits - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/zener-regulator-design - Whole chapter: https://www.bctengineer.com/md/semester_1/fundamental_of_electrical_and_electronics_engineering/chapter-4 ## Question Design a Zener diode voltage regulator that will maintain an output voltage of 20 V across 1 kΩ load with an input that will vary between 30 V and 50 V. Determine the proper value of Rs and the maximum Zener current IZM. ## Short answer Load current IL = 20 mA. Rs = (30−20)/20 mA = 500 Ω. At max input, current through Rs = (50−20)/500 = 60 mA, so IZM = 60−20 = 40 mA. ## Answer A Zener regulator uses a series resistor Rs and a Zener diode in parallel with the load. The Zener keeps the load voltage at Vz; Rs drops the extra input voltage. ```text Vi o--[ Rs ]--+-------+ 30-50 V | | Zener RL = 1 kΩ Vz=20V | o-------------+-------+ ``` **Given:** Vz = VL = 20 V, RL = 1 kΩ, Vi = 30 V to 50 V. Assume an ideal Zener (Iz(min) ≈ 0, rz = 0), as in the usual textbook design (Boylestad). #### Load current (constant) ```text IL = VL/RL = 20/1000 = 20 mA ``` #### Choosing Rs (worst case: minimum input) At Vi(min) = 30 V the Zener must still be in breakdown, so the current through Rs must at least supply the load (Iz ≥ 0): ```text Rs = (Vi(min) − Vz)/IL = (30 − 20)/20 mA = 500 Ω ``` #### Maximum Zener current (maximum input) ```text I_Rs(max) = (Vi(max) − Vz)/Rs = (50 − 20)/500 = 60 mA IZM = I_Rs(max) − IL = 60 − 20 = 40 mA ``` **Answer: Rs = 500 Ω, IZM = 40 mA.** Zener power rating needed: PZM = Vz × IZM = 20 × 40 mA = 0.8 W, so select a 20 V Zener of at least 1 W. Rs dissipates (50 − 20)²/500 = 1.8 W at maximum input, so use a 2 W (or higher) resistor. In practice Rs is chosen a little below 500 Ω (e.g. 470 Ω standard value) so a small minimum Zener current Iz(min) flows at 30 V input and regulation is kept. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The load voltage is 20 V across 1 kohm. What is the load current? A. 20 mA B. 2 mA C. 200 mA D. 20 uA Answer: A. IL = VL/RL = 20/1000 = 20 mA. (Key point: Load current) 2. Why is Rs chosen using the minimum input voltage? A. The Zener gets maximum current then B. The Zener must stay in breakdown, so Rs current must at least supply the load C. To maximise power in Rs D. To protect the load from overvoltage Answer: B. At 30 V, (30-20)/Rs = IL for Iz = 0, so Rs = 10 V/20 mA = 500 ohm. (Key point: Choosing Rs at minimum input) 3. With Rs = 500 ohm, what is IZM at Vi = 50 V? A. 60 mA B. 20 mA C. 40 mA D. 30 mA Answer: C. I_Rs = 30/500 = 60 mA; IZM = 60 - 20 = 40 mA. (Key point: Maximum Zener current) 4. What Zener power rating is needed? A. 0.2 W B. 0.4 W C. 0.1 W D. At least 1 W (PZM = 20 V x 40 mA = 0.8 W) Answer: D. PZM = Vz x IZM = 0.8 W, so choose at least 1 W. 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