---
title: "Diffusion process and Einstein relation"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:diffusion-einstein-relation"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/diffusion-einstein-relation
site: BCT Engineering
---

# Diffusion process and Einstein relation

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2082 Kartik (new course), 5 marks; 2081 Bhadra, 1+5 marks; 2081 Baisakh, 8 marks; 2079 Bhadra, 8 marks; 2076 Asoj, 10 marks; 2075 Asoj, 8 marks; 2073 Shrawan, 6 marks; 2068 Shrawan, 8 marks; 2068 Baisakh, 10 marks (asked 9 times)
- Video: [Einstein Relation in Semiconductor (Basics, Statement, Meaning & Proof) Explained](https://www.youtube.com/watch?v=t2sRYkVDZaQ) — 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/diffusion-einstein-relation
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Explain the diffusion process in semiconductor and derive the Einstein relationship.

## Short answer

Diffusion is the flow of carriers from high to low concentration, giving diffusion currents Jn = e Dn dn/dx and Jp = –e Dp dp/dx. In equilibrium under a concentration gradient, drift and diffusion cancel. Using n(x) ∝ exp(–(Ec–EF)/kT) and dEc/dx = eE yields D/μ = kT/e, the Einstein relation.

## Answer

**Diffusion** is the net flow of carriers from a region of high concentration to a region of low concentration, caused by their random thermal motion. It needs no electric field; only a concentration gradient.

#### Diffusion process
Consider electrons with concentration $n(x)$ falling with $x$. Each carrier moves randomly, but more carriers cross a plane from the crowded side than from the sparse side, so there is a net flux towards lower concentration.

```text
  n(x)
   |\
   | \      net electron flow -->
   |  \     electron current  <--
   |   `-._
   |       `--.___
   +--------------- x
```

By **Fick's law**, the particle flux is proportional to the gradient:

$$
\Gamma_e = -D_e \frac{dn}{dx}, \qquad \Gamma_h = -D_h \frac{dp}{dx}
$$

where $D_e$, $D_h$ are diffusion coefficients (cm$^2$/s). Since electrons carry charge $-e$ and holes $+e$, the diffusion current densities are

$$
J_{D,e} = e D_e \frac{dn}{dx}, \qquad J_{D,h} = -e D_h \frac{dp}{dx}
$$

With an electric field $E$ also present, the total currents are

$$
\begin{aligned} J_e &= e n \mu_e E + e D_e \frac{dn}{dx} \\ J_h &= e p \mu_h E - e D_h \frac{dp}{dx} \end{aligned}
$$

#### Einstein relation
Consider an n-type semiconductor with non-uniform doping, so $n$ varies with $x$, in equilibrium (no applied voltage).

1. Electrons diffuse from high to low $n$, leaving positive donor ions behind. This builds an internal field $E$ that drives a drift current opposing diffusion.
2. In equilibrium, the Fermi level $E_F$ is flat and the net current is zero:

$$
J_e = e n \mu_e E + e D_e \frac{dn}{dx} = 0 \quad (1)
$$

3. The bands bend, since $E_c - E_F$ changes with $x$. For a non-degenerate semiconductor:

$$
n(x) = N_c \exp\left[-\frac{E_c(x) - E_F}{kT}\right]
$$

$$
\frac{dn}{dx} = -\frac{n}{kT}\frac{dE_c}{dx} \quad (2)
$$

4. The electron potential energy is $E_c = \text{const} - eV(x)$, and $E = -dV/dx$, so

$$
\frac{dE_c}{dx} = -e\frac{dV}{dx} = eE \quad (3)
$$

5. From (2) and (3): $\dfrac{dn}{dx} = -\dfrac{neE}{kT}$. Put this in (1):

$$
\begin{aligned} e n \mu_e E - e D_e \frac{neE}{kT} &= 0 \\ \mu_e &= \frac{e D_e}{kT} \\ \frac{D_e}{\mu_e} &= \frac{kT}{e} \end{aligned}
$$

The same steps for holes give $D_h/\mu_h = kT/e$. So

$$
\frac{D_e}{\mu_e} = \frac{D_h}{\mu_h} = \frac{kT}{e}
$$

This is the **Einstein relation**. At 300 K, $kT/e = 0.0259$ V. For Si with $\mu_e = 1350$ cm$^2$/Vs, $D_e = 0.0259 \times 1350 = 35$ cm$^2$/s. It links diffusion (random motion) and drift (field-driven motion), since both arise from the same scattering processes.

## Check yourself

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

1. What is the electron diffusion current density?
   A. J = e n mu E only
   B. J = e De dn/dx
   C. J = -e De dn/dx
   D. J = e De n
   Answer: B. Flux is -De dn/dx and electrons carry -e, so J = +e De dn/dx. (Key point: Fick's law and current)

2. What condition is used to derive the Einstein relation?
   A. Drift current equals zero alone
   B. Diffusion current is maximum
   C. In equilibrium net current is zero: drift and diffusion currents cancel
   D. Applied voltage equals kT/e
   Answer: C. The built-in field's drift opposes diffusion, giving J = 0. (Key point: Equilibrium condition)

3. Which relation connects dn/dx to the field?
   A. dn/dx = n e E / (kT) with no sign
   B. dn/dx = kT E / (n e)
   C. dn/dx = -n kT/(eE)
   D. dn/dx = -n e E / (kT)
   Answer: D. From n = Nc exp(-(Ec - EF)/kT) and dEc/dx = eE. (Key point: Band bending step)

4. What is the Einstein relation?
   A. D/mu = kT/e
   B. D mu = kT/e
   C. D/mu = e/kT
   D. D/mu = kT e
   Answer: A. Substituting into J = 0 gives mu = eD/kT, i.e. D/mu = kT/e (0.0259 V at 300 K). (Key point: Einstein relation)

## Related in Electrical Engineering Material

- [Einstein relation between diffusion and mobility](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aeinstein-diffusion-mobility) — Electrical Engineering Material, chapter 6, asked 4×
- [Einstein relation; mobility, conductivity, resistivity](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aeinstein-mobility-conductivity) — Electrical Engineering Material, chapter 6
- [Band bending and Einstein relationship](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aband-bending-einstein) — Electrical Engineering Material, chapter 6, asked 2×

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