--- title: "Forward biased pn junction band diagram" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:forward-bias-band-diagram" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/forward-bias-band-diagram site: BCT Engineering --- # Forward biased pn junction band diagram Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2069 Asar, 8 marks - Video: [PN Junction Band Diagram](https://www.youtube.com/watch?v=6VSUshhCpYg) — Jordan Edmunds Chetty - 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-bias-band-diagram - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question Explain with energy band diagram, the forward biased P-N junction and derive mathematical expressions for the same. ## Short answer Forward bias raises the n‑side bands, lowering the barrier to e(V₀−V). The junction law p_n(0)=p_n0 exp(eV/kT) describes minority injection. The total current is I = I₀(e^(eV/ηkT)−1) with η=1 for ideal diffusion; depletion width shrinks as √(V₀−V). ## Answer **Forward bias** means connecting the p side to the positive terminal and the n side to the negative terminal. The applied voltage $V$ opposes the built-in potential $V_0$, lowers the barrier to $e(V_0 - V)$, and allows a large diffusion current of majority carriers. #### Energy band diagram In equilibrium $E_F$ is flat and the bands bend by $eV_0$. Under forward bias the n-side bands are raised by $eV$ relative to the p side (the n side is at lower potential), so the Fermi levels separate by $eV$ and the bending is reduced to $e(V_0 - V)$. ```text Equilibrium Forward bias V p n p n Ec_ Ec__ \___ Ec \__ Ec EF ------- EF EFp --- -.- EFn Ev_ barrier: eV gap \___ Ev eV0 Ev__ \__ Ev barrier e(V0-V) ``` #### Mathematical expressions **1. Barrier and width** $$ W = \sqrt{\frac{2\varepsilon(N_a+N_d)(V_0-V)}{eN_aN_d}} $$ **2. Minority carrier injection (law of the junction).** Using the Boltzmann relation across the lowered barrier: $$ p_n(0) = p_{n0}\exp\left(\frac{eV}{kT}\right),\qquad n_p(0) = n_{p0}\exp\left(\frac{eV}{kT}\right) $$ with $p_{n0} = n_i^2/N_d$ and $n_{p0} = n_i^2/N_a$. **3. Excess carriers decay by recombination** in the neutral n region: $$ \Delta p_n(x) = \Delta p_n(0)\,e^{-x/L_h},\qquad L_h = \sqrt{D_h\tau_h} $$ **4. Diffusion current** of holes at the edge $x = 0$: $$ J_h = -eD_h\frac{d\Delta p_n}{dx}\Big|_{0} = \frac{eD_h}{L_h}\,\frac{n_i^2}{N_d}\left(e^{eV/kT}-1\right) $$ Similarly for electrons in the p region: $$ J_e = \frac{eD_e}{L_e}\,\frac{n_i^2}{N_a}\left(e^{eV/kT}-1\right) $$ **5. Total current — Shockley diode equation** $$ I = I_0\left(e^{eV/\eta kT}-1\right),\qquad I_0 = Aen_i^2\left(\frac{D_h}{L_hN_d}+\frac{D_e}{L_eN_a}\right) $$ where $\eta = 1$ for ideal diffusion current (Ge) and $\eta \approx 2$ when recombination in the depletion layer dominates (Si at low current). #### Key results - The current rises exponentially with V; for $V \gg kT/e$, $I \approx I_0e^{eV/kT}$. - Each 60 mV increase in V raises the current about ten times at 300 K. - Noticeable conduction starts at the cut-in voltage (~0.7 V Si, ~0.3 V Ge). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What happens to the band bending and Fermi levels under forward bias V? A. Bending increases to e(V0 + V); Fermi levels merge B. Bending unchanged; Fermi levels flat C. Bending zero; Fermi levels separate by eV0 D. Bending reduces to e(V0 - V); Fermi levels separate by eV Answer: D. Forward bias lowers the barrier by eV and splits the Fermi levels by eV. (Key point: Band diagram under forward bias) 2. What does the law of the junction state at the edge of the depletion region? A. pn(0) = pn0 exp(eV/kT) B. pn(0) = pn0 exp(-eV/kT) C. pn(0) = pn0 + V D. pn(0) = ni Answer: A. Excess minority density at the edge rises by exp(eV/kT). (Key point: Law of the junction) 3. How do excess holes vary with distance in the n region? A. Increase linearly B. Decay as exp(-x/Lh), with Lh = sqrt(Dh tau_h) C. Stay constant D. Decay as exp(+x/Lh) Answer: B. Recombination makes them decay with the diffusion length. (Key point: Diffusion of excess carriers) 4. Which is the diode equation? A. I = I0 exp(-eV/kT) B. I = I0 (1 - exp(eV/kT)) V C. I = I0 (exp(eV/ηkT) - 1) D. I = I0 V Answer: C. Summing electron and hole diffusion currents gives the Shockley equation. 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