--- title: "PN junction built-in potential and depletion layer" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:pn-junction-builtin-depletion" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/pn-junction-builtin-depletion site: BCT Engineering --- # PN junction built-in potential and depletion layer Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2076 Chaitra, 8 marks; 2072 Kartik, 10 marks (asked 2 times) - Video: [Basic Electronics Part 3: PN Junction Explained | Depletion Region, Equilibrium & Built-in Potential](https://www.youtube.com/watch?v=v5Sz8gxEND4) — YOMS 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/pn-junction-builtin-depletion - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question What is PN junction? Derive the relation for built in potential and depletion layer of a PN junction. ## Short answer A pn junction is the interface between p-type and n-type regions. In equilibrium, diffusion forms a depletion layer with built-in potential V0 = (kT/e) ln(NaNd/ni²). The total width is W = √[2ε(Na+Nd)V0/(eNaNd)], extending more into the lighter-doped side. ## Answer A **pn junction** is the boundary formed inside a single semiconductor crystal when one region is doped p-type (acceptors, $N_a$) and the adjacent region n-type (donors, $N_d$). It is the basic structure of diodes, BJTs, solar cells and LEDs. ```text p-type | n-type + + + + + - - |+ + - - - - - + + + + + - - |+ + - - - - - holes (+) depletion electrons (-) region W <----E---- ``` When formed, holes diffuse from p to n and electrons from n to p and recombine near the junction. This leaves a **depletion region** of fixed negative acceptor ions (p-side) and positive donor ions (n-side) with a built-in field from n to p, which stops further net diffusion. #### Built-in potential Consider an abrupt pn junction with acceptor concentration $N_a$ on the p-side and donor concentration $N_d$ on the n-side. Assumptions: abrupt junction, full ionisation, non-degenerate doping, depletion approximation (no free carriers in the depletion region), equilibrium. In equilibrium the Fermi level is constant through the junction, so the bands bend by $eV_0$ across the depletion region. ```text p-side | depletion | n-side | -Wp 0 +Wn | Ec ___________ \_____________ EF ............................. (flat) Ev ___________ eV0 \_____________ ``` In equilibrium, carrier concentrations at the two ends of the depletion region follow the Boltzmann factor: $$ \frac{p_{p0}}{p_{n0}} = \exp\left(\frac{eV_0}{kT}\right) $$ With $p_{p0} = N_a$ and $p_{n0} = n_i^2/N_d$: $$ \begin{aligned} \frac{N_a N_d}{n_i^2} &= \exp\left(\frac{eV_0}{kT}\right) \\ V_0 &= \frac{kT}{e}\ln\left(\frac{N_a N_d}{n_i^2}\right) \end{aligned} $$ (The same result follows from $eV_0 = E_{Fn} - E_{Fp}$ before contact, using $E_{Fn} - E_{Fi} = kT\ln(N_d/n_i)$ and $E_{Fi} - E_{Fp} = kT\ln(N_a/n_i)$.) #### Depletion width Charge density: $\rho = -eN_a$ for $-W_p < x < 0$ and $\rho = +eN_d$ for $0 < x < W_n$. ```text rho +eNd +-----+ | | ---+------+-----+--- x -Wp| 0 Wn +------+ -eNa ``` 1. **Charge neutrality:** $$ N_a W_p = N_d W_n $$ 2. **Poisson's equation:** $\dfrac{dE}{dx} = \dfrac{\rho}{\varepsilon}$, with $\varepsilon = \varepsilon_0\varepsilon_r$. Integrating from $-W_p$ (where $E = 0$) gives a field that is maximum at the junction: $$ E_0 = -\frac{eN_d W_n}{\varepsilon} = -\frac{eN_a W_p}{\varepsilon} $$ 3. **Potential:** $V_0$ is the area under the triangular field plot: $$ V_0 = -\frac{1}{2}E_0 W = \frac{eN_d W_n W}{2\varepsilon}, \quad W = W_n + W_p $$ 4. From neutrality, $W_n = \dfrac{N_a W}{N_a + N_d}$. Substituting: $$ \begin{aligned} V_0 &= \frac{e N_a N_d W^2}{2\varepsilon (N_a + N_d)} \\ W &= \sqrt{\frac{2\varepsilon (N_a + N_d) V_0}{e N_a N_d}} \end{aligned} $$ ```text E(x) ---+-----------+---- x -Wp \ / Wn \ / \ / \ / \ / -E0 ``` The depletion region extends mostly into the lightly doped side ($W_n/W_p = N_a/N_d$). With reverse bias $V_r$, replace $V_0$ by $V_0 + V_r$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is a pn junction? A. A contact between a metal and an n-type semiconductor B. Two separate crystals pressed together without doping change C. A boundary inside one crystal between p-type (acceptor) and n-type (donor) regions D. A region of intrinsic silicon only Answer: C. A pn junction is formed where p-doped and n-doped regions meet in one crystal. (Key point: Definition) 2. How does the depletion region form? A. Holes and electrons diffuse across and recombine, leaving fixed ions that create a field opposing further diffusion B. An external battery removes carriers C. Donors and acceptors annihilate D. Light generates carriers Answer: A. Diffusion leaves uncovered negative acceptor and positive donor ions whose built-in field stops further net diffusion. (Key point: Origin of depletion region) 3. Which is the built-in potential? A. V0 = (kT/e) ln(ni^2 / Na Nd) B. V0 = (kT/e) ln(Na Nd / ni^2) C. V0 = kT ln(Na/Nd) D. V0 = e ln(Na Nd / kT) Answer: B. Using the Boltzmann relation p_p0/p_n0 = exp(eV0/kT) with p_n0 = ni^2/Nd. (Key point: Built-in potential) 4. What is the depletion width, and how does it change with reverse bias Vr? A. W = sqrt(2 eps (Na+Nd) V0 / (e Na Nd)); replace V0 with V0 + Vr B. W = sqrt(2 eps (Na+Nd) V0 / (e Na Nd)); replace V0 with V0 - Vr C. W = 2 eps V0 / (e Na); independent of Vr D. W = sqrt(e Na Nd / (2 eps V0)); replace V0 with V0 + Vr Answer: A. Derived from V0 = e Na Nd W^2 / (2 eps (Na+Nd)); reverse bias increases the total barrier to V0 + Vr, so W grows. (Key point: Depletion width) ## Related in Electrical Engineering Material - [Built-in potential of a pn junction](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-builtin-potential-derivation) — Electrical Engineering Material, chapter 6, asked 2× - [PN junction formation and built-in potential](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-formation-builtin) — Electrical Engineering Material, chapter 6 - [pn junction 10¹⁸/10¹⁶: built-in potential and widths](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-1e18-1e16-depletion) — Electrical Engineering Material, chapter 6, asked 3× - [pn junction built-in potential and diffusion coefficients](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-builtin-diffusion-coefficient) — Electrical Engineering Material, chapter 6 - [pn junction: potential, widths, field, Fermi levels](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-1e16-1e17-fermi-levels) — Electrical Engineering Material, chapter 6, asked 2× - [pn junction 10¹⁸/10¹⁶: potential, width, field](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-1e18-1e16-max-field) — Electrical Engineering Material, chapter 6, asked 2× ## Related in other subjects - [PN junction diode I-V characteristics](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Apn-diode-iv-characteristics) — Basic Electronics Engineering, chapter 2, asked 5× - [Large signal models of PN junction diode](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alarge-signal-models-diode) — Electronics Devices and Circuit, chapter 1, asked 4× - [Photochemistry behind ozone layer depletion](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Aozone-photochemistry) — Engineering Chemistry, chapter 3