---
title: "PN junction formation and built-in potential"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:pn-formation-builtin"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/pn-formation-builtin
site: BCT Engineering
---

# PN junction formation and built-in potential

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2070 Chaitra, 6 marks
- Video: [PN Junction Introduction](https://www.youtube.com/watch?v=BHA4teZmwT0) — 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-formation-builtin
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Explain how PN junction is formed when n-type and p-type semiconductor are brought together. Derive the relation of built-in-potential of a PN junction.

## Short answer

When p and n regions are joined, carrier diffusion creates a depletion layer with ionized acceptors and donors, setting up a built-in field. At equilibrium, drift balances diffusion. Integrating the hole current equation yields built-in potential V0=(kT/e) ln(Na Nd/ni²).

## Answer

#### Formation of a pn junction

A pn junction is formed when one region of a single crystal is made p-type and the adjacent region n-type (e.g. by diffusion or ion implantation). On contact:

1. **Diffusion:** the p side has many holes and the n side many electrons. Holes diffuse into the n side and electrons into the p side, where they recombine with majority carriers.
2. **Space charge:** near the junction the p side loses holes and is left with fixed negative acceptor ions $N_a^-$; the n side is left with fixed positive donor ions $N_d^+$. This carrier-free layer is the **depletion region** of width $W_0 = W_p + W_n$.
3. **Built-in field:** the ions set up a field $\mathcal{E}_0$ from n to p. It drives a drift current opposite to the diffusion current.
4. **Equilibrium:** diffusion continues until drift exactly balances it. The net current is zero, the Fermi level is flat across the device, and a potential difference $V_0$ (the **built-in potential**) exists between the n and p sides.

```text
    p              W0             n
 o o o o |- - - - |+ + + + | * * * *
 o o o o |- - - - |+ + + + | * * * *
         |<- Wp ->|<- Wn ->|
          E0 <---------------
 V:  0 ______/^^^^^^^^^^^^ V0
```

#### Derivation of built-in potential

Consider holes. In equilibrium the hole drift current balances the hole diffusion current:

$$
ep\mu_h\mathcal{E} - eD_h\frac{dp}{dx} = 0
$$

With $\mathcal{E} = -dV/dx$ and the Einstein relation $D_h/\mu_h = kT/e$:

$$
-p\frac{dV}{dx} = \frac{kT}{e}\frac{dp}{dx}\;\Rightarrow\; dV = -\frac{kT}{e}\frac{dp}{p}
$$

Integrate from the p side ($V = 0$, $p = p_{p0}$) to the n side ($V = V_0$, $p = p_{n0}$):

$$
V_0 = \frac{kT}{e}\ln\frac{p_{p0}}{p_{n0}}
$$

With $p_{p0} = N_a$ and $p_{n0} = n_i^2/N_d$:

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

Equivalently, $p_{n0}/p_{p0} = e^{-eV_0/kT}$: the Boltzmann relation across the barrier.

**Example:** Si, $N_a = 10^{18}$, $N_d = 10^{16}\ \text{cm}^{-3}$, 300 K: $V_0 = 0.02585\ln(10^{34}/2.1\times10^{20}) \approx 0.81$ V.

## Check yourself

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

1. What happens first when p and n regions are joined?
   A. Holes diffuse to n and electrons to p, where they recombine
   B. Ions drift across
   C. Electrons only move to p side
   D. Nothing, since there is no field
   Answer: A. Concentration differences cause majority carriers to diffuse across and recombine. (Key point: Initial diffusion)

2. What defines equilibrium in the junction?
   A. Diffusion stops completely
   B. Drift current exactly balances diffusion, net current zero, flat Fermi level
   C. Drift exceeds diffusion
   D. The depletion region vanishes
   Answer: B. The built-in field drives drift opposite to diffusion until they cancel. (Key point: Equilibrium condition)

3. The built-in potential derivation starts from which equation?
   A. Hole drift current plus hole diffusion current add to I_0
   B. J = sigma E for electrons only
   C. Hole drift equals hole diffusion: e p mu_h E - e D_h dp/dx = 0
   D. np = N_d N_a
   Answer: C. Setting net hole current to zero and using E = -dV/dx with the Einstein relation integrates to V_0. (Key point: Derivation starting point)

4. What is the final V_0 and its value for Na = 10^18, Nd = 10^16 Si at 300 K?
   A. (kT/e) ln(N_a/N_d), 0.12 V
   B. (kT/e) ln(N_a N_d / n_i^2), about 0.81 V
   C. (kT/e) ln(N_a N_d / n_i), about 1.6 V
   D. (kT/e) ln(n_i^2 / (N_a N_d)), negative
   Answer: B. V_0 = 0.02585 ln(10^34 / 2.1 x 10^20) = about 0.81 V. (Key point: Result)

## Related in Electrical Engineering Material

- [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×
- [Built-in potential of a pn junction](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-builtin-potential-derivation) — Electrical Engineering Material, chapter 6, asked 2×
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- [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
- [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×
- [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×

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