---
title: "pn junction 10¹⁸/10¹⁶: built-in potential and widths"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:pn-1e18-1e16-depletion"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/pn-1e18-1e16-depletion
site: BCT Engineering
---

# pn junction 10¹⁸/10¹⁶: built-in potential and widths

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2081 Bhadra, 6 marks; 2081 Baisakh, 6 marks; 2069 Asar, 6 marks (asked 3 times)
- Video: [Barrier Potential](https://www.youtube.com/watch?v=itL2MQqZtqA) — Neso Academy
- 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-depletion
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

A heavily doped p side with acceptor concentration of 10¹⁸ cm⁻³ is connected to n-side with donor concentration of 10¹⁶ cm⁻³. Calculate the built-in potential and depletion width in n-side and p-side and overall depletion width of pn junction. The intrinsic concentration is 1.45×10¹⁰ cm⁻³ and temperature is 300 K.

## Short answer

V0 = (kT/e) ln(Na Nd / ni^2) = 0.814 V. Total depletion width W = sqrt[2ε(Na+Nd)V0/(e Na Nd)] = 0.329 μm, with Wn = 0.326 μm, Wp = 3.26 nm.

## 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:** $N_a = 10^{18}$ cm$^{-3}$, $N_d = 10^{16}$ cm$^{-3}$, $n_i = 1.45\times10^{10}$ cm$^{-3}$, $T = 300$ K.

#### Built-in potential
$$
\begin{aligned} \frac{N_aN_d}{n_i^2} &= \frac{10^{18}\times10^{16}}{(1.45\times10^{10})^2} = 4.76\times10^{13} \\ V_0 &= 0.02586 \times \ln(4.76\times10^{13}) = 0.02586 \times 31.49 \\ &= 0.814\ \text{V} \end{aligned}
$$

#### Total depletion width
$$
\begin{aligned} W &= \sqrt{\frac{2 \times 1.054\times10^{-12} \times (10^{18}+10^{16}) \times 0.814}{1.602\times10^{-19} \times 10^{18} \times 10^{16}}} \\ &= 3.29\times10^{-5}\ \text{cm} = 0.329\ \mu\text{m} \end{aligned}
$$

#### Width on each side
$$
\begin{aligned} W_n &= \frac{N_a}{N_a+N_d}W = \frac{10^{18}}{1.01\times10^{18}} \times 0.329 = 0.326\ \mu\text{m} \\ W_p &= \frac{N_d}{N_a+N_d}W = \frac{10^{16}}{1.01\times10^{18}} \times 0.329 = 3.26\times10^{-3}\ \mu\text{m} = 3.26\ \text{nm} \end{aligned}
$$

Check: $N_aW_p = 10^{18}\times3.26\times10^{-7} = 3.26\times10^{11}$ cm$^{-2}$ $= N_dW_n$. Since this is a p$^+$n junction, almost all the depletion region lies in the lightly doped n-side. (The peak field is $E_0 = eN_dW_n/\varepsilon = 4.95\times10^4$ V/cm.)

**Answer:** $V_0 = 0.814$ V; $W_n = 0.326$ $\mu$m; $W_p = 3.26$ nm; $W = 0.329$ $\mu$m.

## Check yourself

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

1. For Na = 10^18 and Nd = 10^16 cm^-3, which expression gives V0 and roughly what value?
   A. 0.02586 ln(Na/Nd), about 0.12 V
   B. 0.02586 ln(Na Nd / ni^2), about 0.814 V
   C. 0.02586 ln(ni^2 / Na Nd), about -0.81 V
   D. 0.02586 (Na Nd / ni^2), about 10^12 V
   Answer: B. Na Nd/ni^2 = 4.76e13, ln = 31.49, so V0 = 0.02586 x 31.49 = 0.814 V. (Key point: Built-in potential calculation)

2. In this p+n junction, where does almost all of the depletion width lie, and why?
   A. In the heavily doped p-side, because it has more charge
   B. Equally on both sides
   C. In the lightly doped n-side, because Na Wp = Nd Wn
   D. Outside the junction
   Answer: C. Charge neutrality forces the wider region on the lightly doped side: Wn = 0.326 um versus Wp = 3.26 nm. (Key point: Which side holds most of the depletion)

3. Which pair of formulas gives the widths on each side?
   A. Wn = Nd W/(Na+Nd), Wp = Na W/(Na+Nd)
   B. Wn = W/2, Wp = W/2
   C. Wn = Na W/(Na+Nd), Wp = Nd W/(Na+Nd)
   D. Wn = Na W/Nd, Wp = Nd W/Na
   Answer: C. From Na Wp = Nd Wn and W = Wn + Wp, Wn = Na W/(Na+Nd) and Wp = Nd W/(Na+Nd). (Key point: Wn and Wp formulas)

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