--- title: "Dynamic resistance of diode, small signal" subject: "Electronics Devices and Circuit" chapter: 1 key: "semester_3/electronics_devices_and_circuit:diode-dynamic-resistance-small-signal" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-dynamic-resistance-small-signal site: BCT Engineering --- # Dynamic resistance of diode, small signal Electronics Devices and Circuit · Chapter 1: Diodes - Asked in: 2082 Shrawan, 6 marks; 2082 Baisakh, 1+5 marks; 2078 Kartik, 5 marks; 2075 Ashwin, 5 marks; 2074 Ashwin, 2+4 marks; 2069 Chaitra, 2+4 marks (asked 6 times) - Video: [Dynamic Resistance of Diode | AC Resistance of Diode | Derivation of Dynamic Resistance of Diode](https://www.youtube.com/watch?v=EmvfIiG_Xjg) — Engineering Funda - 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-dynamic-resistance-small-signal - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-1 ## Question Derive an expression for dynamic resistance of the diode using small signal analysis. ## Short answer Dynamic resistance rd = η VT / ID, where VT ≈ 25 mV at room temperature. It is the reciprocal slope of the I–V curve at the operating point, valid for small signals vd ≪ η VT. With η=1 and ID=1 mA, rd ≈ 25 Ω. ## Answer The **dynamic (incremental or ac) resistance** *r*d of a forward-biased diode is the reciprocal of the slope of its i–v curve at the operating point: *r*d = dv/di at the Q-point. It is the resistance the diode shows to a small ac signal riding on a dc bias. #### Derivation Shockley equation for forward bias (v ≫ ηV_T): ```text iD = IS [ exp(vD / (η VT)) - 1 ] ≈ IS exp(vD / (η VT)) ``` Here IS is the saturation current, η the emission coefficient (1 to 2) and VT = kT/q ≈ 25 mV at room temperature. Let the diode be biased at (VD, ID) and let a small signal vd be added: ```text vD = VD + vd iD = IS exp[(VD + vd)/(η VT)] = IS exp(VD/(η VT)) · exp(vd/(η VT)) = ID · exp(vd/(η VT)) ``` For a small signal (vd ≪ η VT) use exp(x) ≈ 1 + x: ```text iD = ID (1 + vd/(η VT)) = ID + (ID/(η VT)) vd ``` So the total current is the dc part ID plus an ac part id = (ID/(η VT)) · vd. Therefore ```text id / vd = ID / (η VT) rd = vd / id = η VT / ID ``` The same result comes from differentiating: 1/rd = diD/dvD = iD/(η VT) = ID/(η VT) at the Q-point. #### Conditions and meaning - Valid for vd ≲ 10 mV (η = 1) so the exponential can be linearised. - At room temperature with η = 1: rd = 25 mV / ID. For ID = 1 mA, rd = 25 Ω; for ID = 0.1 mA, rd = 250 Ω. - rd depends on the dc bias current, so the diode acts as a current-controlled resistor. The small-signal model is simply the resistor rd. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the dynamic resistance rd of a forward-biased diode represent? A. The bulk resistance of the semiconductor only B. The dc ratio VD/ID at the Q-point C. The reverse resistance of the junction D. The reciprocal of the slope of the i-v curve at the Q-point, seen by a small ac signal Answer: D. rd = dv/di at the operating point; it is the resistance the diode offers to a small ac signal on a dc bias. (Key point: Definition of rd) 2. Why is exp(vd/(eta VT)) replaced by 1 + vd/(eta VT) in the derivation? A. Because temperature is zero B. Because IS is negligible C. Because the diode is reverse biased D. Because vd is small compared with eta VT Answer: D. For a small signal (vd much less than eta VT) exp(x) is approximately 1 + x, giving an ac current proportional to vd. (Key point: Linearising the exponential) 3. What is the final expression for rd? A. rd = eta VT x ID B. rd = VT / (eta ID^2) C. rd = ID/(eta VT) D. rd = eta VT / ID Answer: D. The ac part id = (ID/(eta VT)) vd, so rd = vd/id = eta VT/ID. (Key point: Result) 4. At room temperature with eta = 1, what is rd for ID = 1 mA, and how does rd change with ID? A. 25 kohm; it is independent of ID B. 25 ohm; it falls as ID rises C. 25 ohm; it rises as ID rises D. 250 ohm; it falls as ID rises Answer: B. rd = 25 mV/1 mA = 25 ohm, and since rd is proportional to 1/ID it decreases as the bias current increases. 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