--- title: "Three-diode string voltage regulation" subject: "Electronics Devices and Circuit" chapter: 1 key: "semester_3/electronics_devices_and_circuit:diode-string-regulation" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-string-regulation site: BCT Engineering --- # Three-diode string voltage regulation Electronics Devices and Circuit · Chapter 1: Diodes - Asked in: 2079 Baisakh, 3+2 marks - Video: [Electronics I: Diodes: Consider the circuit shown in Fi. 4.15. A string of three diodes is used t...](https://www.youtube.com/watch?v=JYDRWVxzkNA) — Computer Engineering life - 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/diode-string-regulation - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-1 ## Question A string of three diodes is used to provide a constant voltage of about 2.1 V. Calculate the change in this regulated voltage caused by (i) a ±10% change in the power supply voltage; (ii) connection of a 1 kΩ load resistance. Assume η = 2. [Figure: supply 10 ± 1 V, series resistor R = 1 kΩ, three diodes in series to ground giving output Vo, with load RL = 1 kΩ connectable across the output] ## Short answer Bias current I=7.9 mA; small-signal string resistance 3rd=19 Ω. ±10% supply change causes ΔVo≈±19 mV. Connecting 1 kΩ load draws 2.1 mA, dropping output by −40 mV. ## Answer A forward-biased diode drops almost constant voltage, so three diodes in series with a resistor form a simple voltage regulator. Changes are found with the small-signal (incremental) resistance of the string. #### Given V+ = 10 V ± 10 % (±1 V), R = 1 kΩ, three diodes (0.7 V each, Vo ≈ 2.1 V), η = 2, VT = 25 mV. #### Bias current and diode resistance ```text I = (10 - 2.1)/1 kΩ = 7.9 mA rd (each) = η VT / I = 2 × 25 mV / 7.9 mA = 6.33 Ω 3 rd = 19.0 Ω ``` #### (i) Supply change of ±10 % The signal ΔV+ = ±1 V divides between R and the string resistance: ```text ΔVo = ΔV+ × 3rd / (R + 3rd) = ±1 × 19.0 / (1000 + 19.0) = ±18.6 mV ``` **Output changes by about ±19 mV** (±0.9 % of 2.1 V), against a ±10 % change in supply: line regulation = 18.6 mV/V. #### (ii) Connecting RL = 1 kΩ The load draws about 2.1 V / 1 kΩ = 2.1 mA from the string, so the string current drops by 2.1 mA: ```text ΔVo = -ΔI × 3rd = -2.1 mA × 19.0 Ω = -40 mV ``` **Output falls by about 40 mV** (≈ 2 %). (More exactly, the diodes' rd rises a little as I falls, so the true drop is slightly larger.) Load regulation = ΔVo/ΔIL = -19 mV/mA. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can three forward diodes in series act as a voltage regulator? A. They block dc B. They have infinite resistance C. Each drops a nearly constant voltage regardless of current D. They amplify the supply Answer: C. The forward drop changes very little with current, so the output stays near 2.1 V. (Key point: Why a diode string regulates) 2. What is the bias current and total 3rd? A. 7.9 mA and 19 ohm B. 7.9 mA and 6.3 kohm C. 2.1 mA and 19 ohm D. 10 mA and 50 ohm Answer: A. I = (10 - 2.1)/1 kohm = 7.9 mA; rd = 50 mV/7.9 mA = 6.33 ohm each, 19 ohm total. (Key point: Bias current and rd) 3. What is the output change for a +/-1 V supply change? A. About +/-0.5 V B. About +/-190 mV C. About +/-1 V D. About +/-19 mV Answer: D. The signal divides between R and 3rd: 1 x 19/1019 = 18.6 mV. (Key point: Line regulation) 4. Why does connecting a 1 kohm load reduce the output about 40 mV? A. The supply voltage drops B. The load draws about 2.1 mA from the string, so the string current falls and DeltaVo = -2.1 mA x 3rd C. The load adds diodes D. R increases Answer: B. The reduced string current lowers VO by DeltaI x 3rd = 2.1 mA x 19 ohm = 40 mV. 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