---
title: "pn junction 10¹⁸/10¹⁶: potential, width, field"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:pn-1e18-1e16-max-field"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/pn-1e18-1e16-max-field
site: BCT Engineering
---

# pn junction 10¹⁸/10¹⁶: potential, width, field

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2068 Chaitra, 8 marks
- Video: [PN Junction Example: Depletion Width, E-Field](https://www.youtube.com/watch?v=M3Hoa_C1Mms) — 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/pn-1e18-1e16-max-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

A pn junction is formed at 300k. The acceptor and donor concentration in p-side and n-side are 10¹⁸ cm⁻³ and 10¹⁶ cm⁻³ respectively. Calculate: i) Built in potential ii) Width of depletion layer iii) Maximum value of electric field

## Short answer

Built-in potential 0.814 V, depletion width 0.329 µm, maximum electric field 4.95×10^4 V/cm.

## Answer

**Data used:** $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$, $kT/e = 0.02585$ V at 300 K, $\varepsilon = \varepsilon_0\varepsilon_r = 8.85\times10^{-14}\times11.9 = 1.053\times10^{-12}\ \text{F cm}^{-1}$, $e = 1.6\times10^{-19}$ C. $N_a = 10^{18}\ \text{cm}^{-3}$ (p side), $N_d = 10^{16}\ \text{cm}^{-3}$ (n side) — a $p^+n$ junction.

#### i) Built-in potential

$$
\begin{aligned}
V_0 &= \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2} = 0.02585\ln\frac{10^{18}\times10^{16}}{(1.45\times10^{10})^2}\\
&= 0.02585\ln(4.756\times10^{13}) = 0.02585\times31.49\\
&= 0.814\ \text{V}
\end{aligned}
$$

#### ii) Width of depletion layer

$$
\begin{aligned}
W_0 &= \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}}\\
&= \sqrt{\frac{2\times1.053\times10^{-12}\times1.01\times10^{18}\times0.814}{1.6\times10^{-19}\times10^{34}}}\\
&= 3.29\times10^{-5}\ \text{cm} = 0.329\ \mu\text{m}
\end{aligned}
$$

Split between the sides:

$$
W_n = W_0\frac{N_a}{N_a+N_d} = 3.26\times10^{-5}\ \text{cm},\qquad W_p = W_0\frac{N_d}{N_a+N_d} = 3.26\times10^{-7}\ \text{cm}
$$

Almost all (99%) of the depletion layer is in the lightly doped n side.

#### iii) Maximum electric field

The field peaks at the metallurgical junction:

$$
\begin{aligned}
\mathcal{E}_0 &= \frac{eN_dW_n}{\varepsilon} = \frac{1.6\times10^{-19}\times10^{16}\times3.26\times10^{-5}}{1.053\times10^{-12}}\\
&= 4.95\times10^{4}\ \text{V cm}^{-1}
\end{aligned}
$$

Check: $\mathcal{E}_0 = 2V_0/W_0 = 2\times0.814/3.29\times10^{-5} = 4.95\times10^{4}$ V/cm. ✔

**Answer: $V_0 \approx 0.814$ V, $W_0 \approx 0.329\ \mu$m, $\mathcal{E}_{max} \approx 4.95\times10^{4}$ V/cm ($4.95\times10^{6}$ V/m).**

```text
   E
   |   p+ |      n
   |      |\
   |      | \
   |      |  \
   |      |   \
   +------+----\---- x
        -Wp  0   Wn
   Emax at x = 0
```

## Check yourself

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

1. What is V_0 for the p+n junction (10^18, 10^16)?
   A. 0.755 V
   B. 0.814 V
   C. 0.348 V
   D. 1.63 V
   Answer: B. V_0 = 0.02585 ln(4.756 x 10^13) = 0.814 V. (Key point: Built-in potential)

2. Where does most of the depletion layer lie for this junction?
   A. In the lightly doped n side (99%)
   B. In the heavily doped p side
   C. Equally in both
   D. Outside the junction
   Answer: A. W_n = W_0 N_a/(N_a+N_d) is about 99% of W_0. (Key point: Location of depletion layer)

3. What is W_0 approximately?
   A. 3.29 cm
   B. 0.329 mm
   C. 0.329 micrometre
   D. 3.29 nm
   Answer: C. W_0 = 3.29 x 10^-5 cm = 0.329 micrometre. (Key point: Depletion width)

4. Where is the electric field maximum and what is it?
   A. At the edge of the n side, 4.95 x 10^4 V/cm
   B. At the metallurgical junction, about 4.95 x 10^4 V/cm
   C. At the metallurgical junction, about 0.814 V/cm
   D. At the p contact, 2.5 x 10^4 V/cm
   Answer: B. The field rises linearly to a peak at x = 0, E_0 = e N_d W_n / eps = 2V_0/W_0 = 4.95 x 10^4 V/cm. (Key point: Maximum field location)

## Close variants set in other exams

- [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×

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