---
title: "pn junction 10¹⁷/10¹⁶: potential, widths, field"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:pn-1e17-1e16-depletion"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/pn-1e17-1e16-depletion
site: BCT Engineering
---

# pn junction 10¹⁷/10¹⁶: potential, widths, field

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2080 Baisakh, 8 marks; 2073 Shrawan, 6 marks; 2070 Asar, 8 marks (asked 3 times)
- Video: [PN Junction Electric Field Profile](https://www.youtube.com/watch?v=h5qLpyraT9M) — 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-1e17-1e16-depletion
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

A heavily doped N-side with donor concentration of 10¹⁷ cm⁻³ and P-side with acceptor concentration of 10¹⁶ cm⁻³ are connected. Find (i) Built in potential (V₀) (ii) Depletion width (W₀, Wn and Wp) (iii) Electric field at metallurgical junction (E₀)

## Short answer

V0 = 0.755 V. W0 = 0.330 μm, Wn = 0.030 μm, Wp = 0.300 μm. Peak electric field |E0| = e Nd Wn / ε = 4.57×10^4 V/cm.

## Answer

**Formulas** (abrupt junction, depletion approximation):

$$
V_0 = \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2}, \quad W = \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}}, \quad W_n = \frac{N_a W}{N_a+N_d}, \quad W_p = \frac{N_d W}{N_a+N_d}
$$

**Constants:** $kT/e = 0.02586$ V at 300 K, $e = 1.602\times10^{-19}$ C, $\varepsilon = \varepsilon_r\varepsilon_0 = 11.9 \times 8.854\times10^{-14} = 1.054\times10^{-12}$ F/cm (silicon assumed).

**Given:** $N_d = 10^{17}$ cm$^{-3}$ (n-side), $N_a = 10^{16}$ cm$^{-3}$ (p-side). $n_i$ and $T$ are not given; assume silicon at 300 K with $n_i = 1.45\times10^{10}$ cm$^{-3}$.

#### (i) Built-in potential
$$
\begin{aligned} \frac{N_aN_d}{n_i^2} &= \frac{10^{16}\times10^{17}}{(1.45\times10^{10})^2} = 4.76\times10^{12} \\ V_0 &= 0.02586 \times \ln(4.76\times10^{12}) = 0.02586 \times 29.19 \\ &= 0.755\ \text{V} \end{aligned}
$$

#### (ii) Depletion widths
$$
\begin{aligned} W_0 &= \sqrt{\frac{2 \times 1.054\times10^{-12} \times (10^{16}+10^{17}) \times 0.755}{1.602\times10^{-19} \times 10^{16} \times 10^{17}}} \\ &= 3.30\times10^{-5}\ \text{cm} = 0.330\ \mu\text{m} \\ W_n &= \frac{N_a}{N_a+N_d}W_0 = \frac{10^{16}}{1.1\times10^{17}} \times 0.3305 = 0.0300\ \mu\text{m} \\ W_p &= \frac{N_d}{N_a+N_d}W_0 = \frac{10^{17}}{1.1\times10^{17}} \times 0.3305 = 0.300\ \mu\text{m} \end{aligned}
$$

The depletion region lies mostly in the lightly doped p-side.

#### (iii) Field at the metallurgical junction
$$
\begin{aligned} |E_0| &= \frac{eN_dW_n}{\varepsilon} = \frac{1.602\times10^{-19} \times 10^{17} \times 3.005\times10^{-6}}{1.054\times10^{-12}} \\ &= 4.57\times10^{4}\ \text{V/cm} = 4.57\ \text{MV/m} \end{aligned}
$$

It points from the n-side to the p-side ($E_0 = -4.57\times10^4$ V/cm if $x$ is measured from p to n). Check: $\frac{1}{2}|E_0|W_0 = 0.5 \times 4.57\times10^4 \times 3.305\times10^{-5} = 0.755$ V $= V_0$.

**Answer:** $V_0 = 0.755$ V; $W_0 = 0.330$ $\mu$m, $W_n = 0.030$ $\mu$m, $W_p = 0.300$ $\mu$m; $|E_0| = 4.57\times10^4$ V/cm.

## Check yourself

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

1. For Nd = 10^17 and Na = 10^16 cm^-3 with ni = 1.45 x 10^10, V0 is about
   A. 0.755 V
   B. 0.814 V
   C. 0.30 V
   D. 1.2 V
   Answer: A. ln(1e33/2.1e20) = ln(4.76e12) = 29.19, and 0.02586 x 29.19 = 0.755 V. (Key point: Built-in potential)

2. Which statement about the depletion widths is right?
   A. Wn is larger because the n-side is more heavily doped
   B. Wn = Wp
   C. Wp = 0.300 um is larger than Wn = 0.030 um because the p-side is lightly doped
   D. Wp is smaller since the p-side is lightly doped
   Answer: C. Wp/Wn = Nd/Na = 10, so the depletion region extends mostly into the lightly doped p-side. (Key point: Depletion on lightly doped side)

3. How is the peak field at the metallurgical junction found?
   A. E0 = e Nd Wn / epsilon
   B. E0 = V0 x W
   C. E0 = epsilon Nd Wn / e
   D. E0 = e Nd W / V0
   Answer: A. Integrating Poisson's equation across the n-side gives |E0| = e Nd Wn/epsilon (equal to e Na Wp/epsilon). (Key point: Field at the junction)

4. Which check confirms the result is consistent?
   A. E0 W equals V0
   B. (1/2)|E0| W equals V0
   C. E0 / W equals V0
   D. |E0| Wn equals Wp
   Answer: B. The potential is the area of the triangular field profile, (1/2)|E0|W = 0.755 V. (Key point: Check using triangular field)

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