--- title: "PN and Zener I-V graphs; diode ac resistance" subject: "Electronics Devices and Circuit" chapter: 1 key: "semester_3/electronics_devices_and_circuit:diode-iv-ac-equivalent-resistance" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-iv-ac-equivalent-resistance site: BCT Engineering --- # PN and Zener I-V graphs; diode ac resistance Electronics Devices and Circuit · Chapter 1: Diodes - Asked in: 2068 Baisakh, 7 marks; 2071 Chaitra, 2+4 marks (asked 2 times) - Video: [AC or Dynamic Resistance of a Diode](https://www.youtube.com/watch?v=6QlFxif-JfQ) — Neso Academy - 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-iv-ac-equivalent-resistance - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-1 ## Question Draw graphs of IV characteristics of ordinary PN junction diode and zener diode. Draw ac equivalent model for PN junction diode and derive its ac resistance. ## Short answer PN diode I–V: exponential forward, small reverse leakage. Zener adds a sharp reverse breakdown at VZ. AC equivalent model is a resistor rd = ηVT/ID, derived from the slope of the Shockley equation. At room temperature with η=1, rd = 25 mV/ID. ## Answer #### I–V characteristics **Ordinary PN junction diode:** in forward bias almost no current flows until the cut-in voltage (about 0.7 V for Si, 0.3 V for Ge); after that the current rises exponentially. In reverse bias only a very small saturation current IS flows, up to the breakdown voltage. An ordinary diode is not designed to work in breakdown and is usually damaged there by the large current and power. **Zener diode:** the forward part is the same as for a normal diode. In reverse bias it blocks until the Zener voltage VZ; then the current rises sharply while the voltage stays almost constant at VZ. The device is designed to work safely in this region. ```text ID (mA) | / | / forward region | / (same for both diodes) | _/ -VZ (Zener) | _-' ------+-----------------+---''-------------- VD (V) | -IS (µA) 0 0.7 V (Si) | reverse leakage | | breakdown: very sharp knee for the Zener; | current rises steeply at almost constant VZ | -IZ ``` (The first quadrant is the same for both diodes; the Zener's special feature is the sharp reverse knee at -VZ. An ordinary diode breaks down at a much larger reverse voltage, -VBR.) #### AC equivalent model For a small signal around a forward bias point (VD, ID), the diode behaves like a resistor rd: ```text o----/\/\/----o rd ``` (In a more complete model, a cut-in voltage battery and the diode capacitance are added.) #### Derivation of ac resistance ```text iD = IS exp(vD/(η VT)) (v ≫ η VT) 1/rd = diD/dvD at Q = IS exp(VD/(η VT))/(η VT) = ID/(η VT) rd = η VT / ID ``` At 300 K, VT = 25 mV, so for η = 1: rd = 25 mV / ID(mA) ohms. Example: ID = 2.5 mA gives rd = 10 Ω. For a given diode and temperature, rd depends only on the dc current, and it is small at high currents. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the Zener I-V characteristic differ from an ordinary diode? A. It has no breakdown B. It has a sharp reverse knee at VZ with near-constant voltage in breakdown C. It conducts only in forward bias D. Its forward part is different Answer: B. The Zener is designed to operate safely in reverse breakdown where voltage stays nearly VZ. (Key point: Zener vs normal diode) 2. What is the approximate cut-in voltage of Si and Ge? A. 1.2 V Si, 0.7 V Ge B. 0.7 V for both C. 0.3 V Si, 0.7 V Ge D. 0.7 V Si, 0.3 V Ge Answer: D. Si is about 0.7 V and Ge about 0.3 V. (Key point: Cut-in voltage) 3. What is the basic ac equivalent model of a forward-biased diode? A. An open circuit B. A resistor rd C. A capacitor only D. A battery of 0.7 V Answer: B. For small signals the diode behaves as the resistor rd. (Key point: AC model) 4. What is rd for eta = 1 and ID = 2.5 mA at 300 K? A. 62.5 ohm B. 25 ohm C. 10 ohm D. 100 ohm Answer: C. rd = 25 mV/2.5 mA = 10 ohm. 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