---
title: "Built-in potential and depletion width defined"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:builtin-depletion-definition"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/builtin-depletion-definition
site: BCT Engineering
---

# Built-in potential and depletion width defined

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2074 Chaitra, 6 marks
- Video: [Width of Depletion Region](https://www.youtube.com/watch?v=d3VIDO0S7xk) — 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/builtin-depletion-definition
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

What is Built-in potential and depletion width? Derive the expression of these with necessary diagram.

## Short answer

Built-in potential V0 = (kT/e) ln(N_aN_d/n_i^2) arises from diffusion of carriers across a pn junction, leaving fixed ion charges. Depletion width W0 = sqrt(2ε(N_a+N_d)V0/(e N_a N_d)) spans the space-charge region devoid of mobile carriers.

## Answer

When p-type and n-type regions meet, electrons diffuse from n to p and holes from p to n. They recombine near the junction and leave behind uncovered **fixed ions**: negative acceptor ions on the p side and positive donor ions on the n side. This region, empty of mobile carriers, is the **depletion (space-charge) region**, and its width $W_0 = W_p + W_n$ is the **depletion width**. The ions create an electric field and a potential difference $V_0$ across the region, called the **built-in potential**; it stops further diffusion.

```text
   p side        |  W0  |       n side
  o o o o o  | - - | + + |  * * * * *
  holes      | - - | + + |  electrons
             |<Wp>|<-Wn->|
  field E0  <-----------
  potential: 0 ____/^^^^^ V0
```

#### Built-in potential

In equilibrium the Fermi level is flat. Far from the junction:

$$
p_{p0} = N_a,\qquad p_{n0} = \frac{n_i^2}{N_d}
$$

The hole concentrations on the two sides are related by the Boltzmann factor of the potential energy step $eV_0$:

$$
\frac{p_{n0}}{p_{p0}} = \exp\left(-\frac{eV_0}{kT}\right)
$$

Hence

$$
V_0 = \frac{kT}{e}\ln\frac{p_{p0}}{p_{n0}} = \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2}
$$

(The same result follows from equating the drift and diffusion currents of holes, using the Einstein relation.)

#### Depletion width

**1. Charge neutrality:** $eN_aW_p = eN_dW_n$.

**2. Field from Gauss's law (Poisson's equation):** the field rises linearly through each charged layer and peaks at the junction:

$$
\mathcal{E}_0 = -\frac{eN_dW_n}{\varepsilon} = -\frac{eN_aW_p}{\varepsilon},\qquad \varepsilon = \varepsilon_0\varepsilon_r
$$

**3. Potential = area under the field triangle:**

$$
V_0 = \frac{1}{2}|\mathcal{E}_0|W_0 = \frac{eN_dW_nW_0}{2\varepsilon}
$$

**4. Eliminate $W_n$:** from step 1, $W_n = W_0\dfrac{N_a}{N_a+N_d}$, so

$$
V_0 = \frac{eW_0^2}{2\varepsilon}\cdot\frac{N_aN_d}{N_a+N_d}
$$

$$
W_0 = \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}}
$$

with $W_n = W_0N_a/(N_a+N_d)$ and $W_p = W_0N_d/(N_a+N_d)$. The depletion layer extends mostly into the **lightly doped** side. Under bias, $V_0$ is replaced by $V_0 - V$.

## Check yourself

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

1. What makes up the depletion region of a pn junction?
   A. Mobile electrons and holes in equal numbers
   B. A region of intrinsic silicon
   C. Uncovered fixed acceptor and donor ions with no mobile carriers
   D. Only the metallurgical junction plane
   Answer: C. Diffusing carriers recombine near the junction, leaving fixed negative acceptor and positive donor ions. (Key point: Depletion region)

2. Which expression gives the built-in potential?
   A. (kT/e) ln(N_a N_d / n_i^2)
   B. (kT/e) ln(N_a / N_d)
   C. (kT/e) ln(n_i^2 / (N_a N_d))
   D. kT ln(N_a + N_d)
   Answer: A. From the Boltzmann relation p_n0/p_p0 = exp(-eV_0/kT) with p_p0 = N_a and p_n0 = n_i^2/N_d. (Key point: Built-in potential expression)

3. What is the charge-neutrality condition used in the depletion width derivation?
   A. N_a W_n = N_d W_p
   B. N_a W_p = N_d W_n
   C. W_p = W_n always
   D. N_a + N_d = W_0
   Answer: B. Total negative charge on the p side equals total positive charge on the n side: eN_aW_p = eN_dW_n. (Key point: Charge neutrality)

4. Which expression gives the depletion width W_0?
   A. sqrt(2 eps N_a N_d V_0 / (e (N_a + N_d)))
   B. sqrt(eps (N_a + N_d) V_0 / (2 e N_a N_d))
   C. 2 eps (N_a + N_d) V_0 / (e N_a N_d)
   D. sqrt(2 eps (N_a + N_d) V_0 / (e N_a N_d))
   Answer: D. Setting V_0 as the area under the triangular field and eliminating W_n gives W_0 = sqrt(2 eps (N_a+N_d) V_0 / (e N_a N_d)). (Key point: Depletion width formula)

## Close variants set in other exams

- [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×

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