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

# pn junction: potential, widths, field, Fermi levels

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2068 Baisakh, 2.5+2.5+1+2+2 marks
- Video: [Derivation of Width of Depletion Region in PN Junction | Electronics Devices and Circuits - EDC](https://www.youtube.com/watch?v=IPinLmxNC-0) — Engineering Funda
- 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-1e16-1e17-fermi-levels
- 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 10¹⁶ cm⁻³ and 10¹⁷ cm⁻³ respectively. Find: i) Built-in potential ii) Width of depletion layer iii) Maximum electric field iv) Width in n and p sides v) Fermi level n and p sides

## Short answer

Built-in potential 0.755 V, total depletion width 0.331 µm, maximum field 4.57×10^4 V/cm; n-side width 0.030 µm, p-side width 0.301 µm; Fermi level on n-side is 0.407 eV above E_Fi, on p-side 0.348 eV below E_Fi.

## 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. $N_a = 10^{16}\ \text{cm}^{-3}$ (p side), $N_d = 10^{17}\ \text{cm}^{-3}$ (n side).

#### i) Built-in potential

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

#### ii) Width of depletion layer

$$
\begin{aligned}
W_0 &= \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}}\\
&= \sqrt{\frac{2\times1.053\times10^{-12}\times1.1\times10^{17}\times0.755}{1.6\times10^{-19}\times10^{33}}}\\
&= 3.31\times10^{-5}\ \text{cm} = 0.331\ \mu\text{m}
\end{aligned}
$$

#### iii) Maximum electric field (at the junction)

$$
\begin{aligned}
\mathcal{E}_0 &= \frac{eN_dW_n}{\varepsilon} = \frac{2V_0}{W_0}\\
&= \frac{2\times0.755}{3.31\times10^{-5}} = 4.57\times10^{4}\ \text{V cm}^{-1}
\end{aligned}
$$

#### iv) Width on n and p sides

From $N_aW_p = N_dW_n$:

$$
\begin{aligned}
W_n &= W_0\frac{N_a}{N_a+N_d} = 3.31\times10^{-5}\times\frac{1}{11} = 3.01\times10^{-6}\ \text{cm}\\
W_p &= W_0\frac{N_d}{N_a+N_d} = 3.31\times10^{-5}\times\frac{10}{11} = 3.01\times10^{-5}\ \text{cm}
\end{aligned}
$$

The depletion region lies mostly (91%) in the lightly doped p side.

#### v) Fermi level on n and p sides

$$
\begin{aligned}
E_{Fn}-E_{Fi} &= kT\ln\frac{N_d}{n_i} = 0.02585\ln\frac{10^{17}}{1.45\times10^{10}} = 0.407\ \text{eV}\\
E_{Fi}-E_{Fp} &= kT\ln\frac{N_a}{n_i} = 0.02585\ln\frac{10^{16}}{1.45\times10^{10}} = 0.348\ \text{eV}
\end{aligned}
$$

Check: $0.407 + 0.348 = 0.755$ eV $= eV_0$. ✔

```text
 p side                     n side
 Ec ______
          \________________ Ec
 EF -------------------------- EF
 EFi_____ (0.348 above EF)
          \_______ EFi (0.407 below EF)
 Ev ______
          \________________ Ev
```

| Quantity | Value |
|---|---|
| $V_0$ | 0.755 V |
| $W_0$ | 0.331 µm |
| $\mathcal{E}_{max}$ | $4.57\times10^{4}$ V/cm |
| $W_n$ / $W_p$ | 0.030 µm / 0.301 µm |
| $E_F$ n side | 0.407 eV above $E_{Fi}$ |
| $E_F$ p side | 0.348 eV below $E_{Fi}$ |

## Check yourself

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

1. What is V_0 for N_a = 10^16, N_d = 10^17 in Si at 300 K?
   A. 0.348 V
   B. 1.51 V
   C. 0.407 V
   D. 0.755 V
   Answer: D. V_0 = 0.02585 ln(10^33 / 2.1 x 10^20) = 0.755 V. (Key point: Built-in potential)

2. How is the maximum field related to V_0 and W_0?
   A. E_0 = V_0 W_0
   B. E_0 = 2 V_0 / W_0
   C. E_0 = V_0 / (2 W_0)
   D. E_0 = W_0 / V_0
   Answer: B. V_0 is the area of the triangular field profile: V_0 = E_0 W_0 / 2. (Key point: Maximum field)

3. Which side holds most of the depletion region and why?
   A. n side, because it is more heavily doped
   B. Both equally
   C. p side, because it is lightly doped
   D. n side, because electrons are faster
   Answer: C. N_a W_p = N_d W_n, so the lightly doped p side has W_p = 10 W_n. (Key point: Which side is wider)

4. How do the Fermi level offsets relate to V_0?
   A. Their difference equals eV_0
   B. Each equals eV_0
   C. Their sum 0.407 + 0.348 equals eV_0
   D. They are unrelated
   Answer: C. The Fermi level is flat, so the band bending equals the sum of the two offsets from E_Fi: 0.755 eV. (Key point: Fermi level check)

## Close variants set in other exams

- [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 Electrical Engineering Material

- [pn junction 10¹⁷/10¹⁶: potential, widths, field](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-1e17-1e16-depletion) — Electrical Engineering Material, chapter 6, asked 3×
- [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 depletion layer](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-junction-builtin-depletion) — Electrical Engineering Material, chapter 6, asked 2×
- [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
- [Built-in potential and depletion width derivation](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-builtin-depletion-derivation) — Electrical Engineering Material, chapter 6, asked 4×
- [pn junction from 5 Ω cm resistivity](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-resistivity-5-ohm-cm) — Electrical Engineering Material, chapter 6, asked 2×