--- title: "Forward and reverse biased pn junction" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:forward-reverse-bias" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/forward-reverse-bias site: BCT Engineering --- # Forward and reverse biased pn junction Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2070 Asar, 8 marks - Video: [PN junction Diode Explained | Forward Bias and Reverse Bias](https://www.youtube.com/watch?v=USrY0JspDEg) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/forward-reverse-bias - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question With the help of P-N junction, explain the phenomena of forward and reverse biased. ## Short answer Forward bias (p positive) reduces the barrier to V₀−V, injects majority carriers and gives a large diffusion current I = I₀(e^(eV/kT)−1). Reverse bias raises the barrier to V₀+Vr, only minority carriers are swept across, resulting in a tiny saturation current I₀. ## Answer A **pn junction** in equilibrium has a depletion region of width $W_0$ and a built-in barrier $V_0$ that stops majority carriers from diffusing across. Applying an external voltage changes this barrier: lowering it is **forward bias**, raising it is **reverse bias**. #### Forward bias (p to +, n to −) ```text + | p |-|+| n | - | --> holes | | electrons<-- W small, barrier V0 - V ``` - The applied voltage $V$ opposes the built-in field. Barrier becomes $V_0 - V$; depletion width shrinks: $$ W = \sqrt{\frac{2\varepsilon(N_a+N_d)(V_0-V)}{eN_aN_d}} $$ - Many more majority carriers can now cross. Holes are **injected** into the n side and electrons into the p side, where they become excess minority carriers: $$ p_n(0) = p_{n0}e^{eV/kT},\qquad n_p(0) = n_{p0}e^{eV/kT} $$ - These carriers diffuse away from the junction and recombine, giving a **diffusion current** that rises exponentially with V. - Large current (mA) flows once V exceeds the cut-in voltage (~0.6–0.7 V Si, ~0.2–0.3 V Ge). #### Reverse bias (p to −, n to +) ```text - | p |--|++| n | + | majority pulled away | | minority swept across | W large, barrier V0 + Vr ``` - The applied voltage adds to the built-in field; barrier becomes $V_0 + V_r$ and the depletion region widens. - Majority carriers cannot cross. Only **minority carriers** generated thermally near the junction are swept across by the field. - A tiny, almost constant **reverse saturation current** $I_0$ (nA in Si, µA in Ge) flows. It increases with temperature. - At a large reverse voltage, **breakdown** (Zener or avalanche) occurs and the current rises sharply. #### Diode equation and characteristic $$ I = I_0\left(e^{eV/\eta kT} - 1\right) $$ ```text I (mA) | / | / forward | / Vbr ____|___/______ V | |-I0 (uA/nA) | reverse ``` | Point | Forward bias | Reverse bias | |---|---|---| | Connection | p to +, n to − | p to −, n to + | | Barrier | $V_0 - V$ (lower) | $V_0 + V_r$ (higher) | | Depletion width | Decreases | Increases | | Current carriers | Majority (injected) | Minority | | Current | Large, exponential | Small, saturates at $-I_0$ | | Resistance | Low | Very high | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Under forward bias, what happens to the barrier and depletion width? A. Barrier V0 + V, width decreases B. Barrier V0 - V, width increases C. Barrier unchanged, width decreases D. Barrier V0 - V, width decreases Answer: D. The applied voltage opposes the built-in field, lowering the barrier and narrowing W. (Key point: Forward bias barrier) 2. What carries the large forward current? A. Injected majority carriers that become minority carriers diffusing across B. Thermally generated minority carriers only C. Ions drifting across the junction D. Carriers tunnelling through the barrier Answer: A. Majority carriers cross, are injected as excess minority carriers and diffuse and recombine. (Key point: Forward current) 3. What is the reverse saturation current due to? A. Majority carriers crossing the barrier B. Thermally generated minority carriers swept across C. Avalanche breakdown D. Recombination in the p region Answer: B. In reverse bias majority carriers are blocked; only minority carriers near the junction are swept across. (Key point: Reverse current) 4. How does the reverse saturation current vary with temperature? A. It decreases B. It is exactly constant C. It increases D. It becomes zero Answer: C. More thermally generated minority carriers at higher T raise I0. 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