--- title: "PN junction diode I-V characteristics" subject: "Basic Electronics Engineering" chapter: 2 key: "semester_2/basic_electronics_engineering:pn-diode-iv-characteristics" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/pn-diode-iv-characteristics site: BCT Engineering --- # PN junction diode I-V characteristics Basic Electronics Engineering · Chapter 2: Diodes - Asked in: 2073 Magh, 1+4 marks; 2070 Magh, 6 marks; 2069 Poush, 6 marks; 2067 Mangsir, 7 marks; 2067 Chaitra, 3 marks (asked 5 times) - Video: [V-I Characteristics of PN Junction Diode](https://www.youtube.com/watch?v=_vKeaPHXF9U) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/pn-diode-iv-characteristics - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2 ## Question What is a diode? Explain the working of a PN junction diode and its I-V characteristics for forward and reverse bias (with circuit diagram and graph). ## Short answer A PN junction diode conducts when forward biased beyond its knee voltage (~0.7 V Si). In forward bias, the depletion region narrows and current rises exponentially; in reverse bias, only a small saturation current I0 flows until breakdown voltage VBR, where current surges. The diode obeys I = I0(e^(V/ηVT)−1). ## Answer A **diode** is a two-terminal semiconductor device that allows current to flow easily in one direction (forward) and blocks it in the opposite direction (reverse). A PN junction diode is formed by joining p-type and n-type semiconductors; the terminals are anode (P) and cathode (N). #### Working of a PN junction diode Near the junction, electrons from the n-side diffuse to the p-side and holes diffuse the other way. This leaves fixed ions, forming a **depletion region** with a built-in barrier potential of about 0.7 V for silicon (0.3 V for germanium). **Forward bias** (P to positive, N to negative) - The external voltage opposes the barrier. The depletion region becomes narrow. - Until the applied voltage reaches the cut-in (knee) voltage, only a very small current flows. - Above about 0.7 V (Si), the barrier is overcome and current rises rapidly (exponentially). The majority carriers carry this current. **Reverse bias** (P to negative, N to positive) - The external voltage adds to the barrier. The depletion region widens. - Only the minority carriers move, giving a very small **reverse saturation current** I0 (nA to µA), almost constant with voltage. - If the reverse voltage is increased to the breakdown voltage VBR, the current increases sharply (avalanche or Zener breakdown). #### Test circuit ```text Forward bias: Reverse bias: +--[R]--(mA)--|>|--+ +--[R]--(uA)--|<|--+ | | | | +-----|E (V)|------+ +-----|E (V)|------+ ``` #### Diode equation I = I0 (e^(V/ηVT) − 1) where VT = 25 mV at room temperature, η = 1 (Ge) or about 2 (Si). #### I-V characteristic ```text I (mA) | / | / | / Reverse | / ---+----+--------+-/------- V (volt) VBR | 0.7V | | I0 (uA) | ``` - **Forward region:** negligible current below the knee voltage (Ge about 0.3 V, Si about 0.7 V), then a steep rise. - **Reverse region:** very small constant current I0. - **Breakdown region:** at VBR the reverse current rises sharply. The diode therefore acts as a closed switch in forward bias and an open switch in reverse bias, which is the basis of rectifiers. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What forms the barrier potential in a PN junction? A. External voltage B. Fixed ions left after diffusion, forming the depletion region C. Free electrons only D. The terminals Answer: B. Diffusion leaves fixed ions, which set a barrier of about 0.7 V (Si) or 0.3 V (Ge). (Key point: Depletion region) 2. What happens to the depletion region in forward bias and when does current rise? A. It widens; current rises at once B. It narrows; current rises rapidly above the knee voltage (about 0.7 V for Si) C. It widens; current rises above breakdown D. No change; current is constant Answer: B. Forward bias opposes the barrier, narrowing the depletion region; current rises steeply after the knee. (Key point: Forward bias) 3. What causes the small reverse saturation current? A. Majority carriers B. Minority carriers C. Breakdown D. The external resistor Answer: B. Only minority carriers move in reverse bias, giving a small, almost constant I0. (Key point: Reverse current) 4. What happens when reverse voltage reaches VBR? A. Current stays at I0 B. Reverse current rises sharply (avalanche or Zener) C. The diode becomes forward biased D. Current drops to zero Answer: B. At breakdown the reverse current increases sharply. 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