---
title: "pn junction built-in potential and diffusion coefficients"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:pn-builtin-diffusion-coefficient"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/pn-builtin-diffusion-coefficient
site: BCT Engineering
---

# pn junction built-in potential and diffusion coefficients

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2068 Chaitra, 6 marks
- 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-builtin-diffusion-coefficient
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

A p-n junction is made by silicon doped with 10¹⁷ donor atoms per cm³ with silicon doped 10¹⁶ acceptor atoms per cm³ at room temperature. Calculate built in potential across the junction and diffusion co-efficient in both parts.

## Short answer

Built-in potential 0.755 V. Diffusion coefficient for electrons in p-side D_e ≈ 34.9 cm²/s, for holes in n-side D_h ≈ 11.6 cm²/s, obtained via Einstein relation D = μ kT/e with μ_e=1350, μ_h=450 cm²/V·s.

## Answer

**Data used:** Si at 300 K, $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$, $kT/e = 0.02585$ V, $\mu_e = 1350$, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$ (standard values; doped-Si mobilities are somewhat lower). $N_d = 10^{17}$, $N_a = 10^{16}\ \text{cm}^{-3}$.

#### Built-in potential

$$
\begin{aligned}
V_0 &= \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2}\\
&= 0.02585\ln\frac{10^{16}\times10^{17}}{(1.45\times10^{10})^2}\\
&= 0.02585\ln(4.756\times10^{12})\\
&= 0.02585\times29.19 = 0.755\ \text{V}
\end{aligned}
$$

#### Diffusion coefficients (Einstein relation)

$$
D = \mu\frac{kT}{e}
$$

Diffusion across a junction is by **minority carriers**: electrons in the p part and holes in the n part.

**p part (electrons diffusing):**

$$
D_e = 1350\times0.02585 = 34.9\ \text{cm}^2\text{s}^{-1}
$$

**n part (holes diffusing):**

$$
D_h = 450\times0.02585 = 11.6\ \text{cm}^2\text{s}^{-1}
$$

**Answer: $V_0 \approx 0.755$ V; $D_e \approx 34.9\ \text{cm}^2\text{s}^{-1}$ (p side) and $D_h \approx 11.6\ \text{cm}^2\text{s}^{-1}$ (n side).**

The ratio $D/\mu = kT/e = 0.02585$ V is the same for both carriers at a given temperature; if mobilities read from a doping-dependent graph are used, D scales in the same proportion.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is V_0 for N_d = 10^17, N_a = 10^16 at 300 K?
   A. About 0.755 V
   B. About 0.348 V
   C. About 1.1 V
   D. About 0.0259 V
   Answer: A. V_0 = 0.02585 ln(4.756 x 10^12) = 0.755 V. (Key point: Built-in potential)

2. Diffusion coefficients across the junction are for which carriers?
   A. Majority carriers on both sides
   B. Holes in p part and electrons in n part
   C. Minority carriers: electrons in p part and holes in n part
   D. Only electrons
   Answer: C. Diffusion across a junction is by minority carriers. (Key point: Which carriers diffuse)

3. What are D_e (p part) and D_h (n part)?
   A. 11.6 and 34.9 cm^2/s
   B. 34.9 and 11.6 cm^2/s
   C. 34.9 and 34.9 cm^2/s
   D. 1350 and 450 cm^2/s
   Answer: B. D = mu kT/e: 1350 x 0.02585 = 34.9 and 450 x 0.02585 = 11.6. (Key point: Diffusion coefficients)

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