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

# Einstein relation between diffusion and mobility

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2081 Baisakh, 6 marks; 2070 Chaitra, 8 marks; 2070 Asar, 6 marks; 2069 Chaitra, 6 marks (asked 4 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/einstein-diffusion-mobility
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Derive the Einstein relationship showing the relation between electron diffusion co-efficient in n-type semiconductor and electron mobility.

## Short answer

In equilibrium, the electron drift current e n μn E and diffusion current e Dn dn/dx cancel. With n(x) = Nc exp[–(Ec–EF)/kT] and dEc/dx = eE, we get dn/dx = –(n eE)/kT. Substituting into zero net current gives Dn/μn = kT/e, the Einstein relation.

## Answer

The **Einstein relation** links the diffusion coefficient and drift mobility of a carrier: $D_e/\mu_e = kT/e$. It is derived by noting that, in equilibrium, drift and diffusion currents in a non-uniformly doped sample cancel.

#### Setup
Take an n-type semiconductor whose donor concentration falls with $x$, in thermal equilibrium (no applied voltage).

```text
 Ec  \__
        \___
            \_______         (band bends)
 EF ---------------------   (flat in equilibrium)

 n high              n low
   diffusion of e- ---->
   drift of e-     <----  (due to built-in E)
```

- Electrons diffuse towards low $n$, leaving positive donor ions behind.
- This creates an internal electric field $E$, which drives electrons back by drift.
- In equilibrium the net electron current is zero:

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

#### Electron concentration and band bending
For a non-degenerate semiconductor,

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

Since $E_F$ is constant,

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

The conduction band edge follows the electron potential energy, $E_c(x) = \text{const} - eV(x)$. With $E = -dV/dx$:

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

From (2) and (3):

$$
\frac{dn}{dx} = -\frac{n e E}{kT} \quad (4)
$$

#### Result
Substitute (4) in (1):

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

This is the Einstein relation for electrons. A similar result holds for holes: $D_h/\mu_h = kT/e$.

At 300 K, $kT/e \approx 25.9$ mV. For Si, $\mu_e = 1350$ cm$^2$V$^{-1}$s$^{-1}$ gives $D_e \approx 35$ cm$^2$/s. Thus, once mobility is measured (e.g. from Hall and conductivity data), the diffusion coefficient follows directly.

## Check yourself

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

1. Why do the drift and diffusion currents cancel in the derivation?
   A. An external voltage balances them
   B. Diffusion leaves positive donors, building an internal field whose drift opposes diffusion; equilibrium gives net J = 0
   C. Holes cancel the electron current
   D. Temperature is zero
   Answer: B. Non-uniform doping gives diffusion and a built-in field; at equilibrium they balance. (Key point: Setup)

2. What does constant EF in equilibrium give?
   A. dn/dx = 0
   B. dn/dx = n kT dEc/dx
   C. dn/dx = -(n/kT) dEc/dx
   D. dn/dx = dEc/dx
   Answer: C. Differentiating n = Nc exp(-(Ec - EF)/kT) with EF constant. (Key point: Flat Fermi level)

3. How is dEc/dx related to the field E?
   A. dEc/dx = -eE
   B. dEc/dx = E/e
   C. dEc/dx = 0
   D. dEc/dx = eE, since Ec = const - eV and E = -dV/dx
   Answer: D. Differentiating Ec = const - eV(x) with E = -dV/dx. (Key point: Band edge slope)

4. What follows, and what is De for Si with mu = 1350 cm^2/Vs at 300 K?
   A. De/mu = kT/e, De ~ 35 cm^2/s
   B. De mu = kT/e, De ~ 0.03 cm^2/s
   C. De/mu = e/kT, De ~ 5.2e4 cm^2/s
   D. De/mu = kT/e, De ~ 3.5 cm^2/s
   Answer: A. 0.0259 V x 1350 = 35 cm^2/s. (Key point: Result and number)

## Related in Electrical Engineering Material

- [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
- [Diffusion process and Einstein relation](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Adiffusion-einstein-relation) — Electrical Engineering Material, chapter 6, asked 9×
- [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×
- [Electron concentration in an extrinsic semiconductor](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aextrinsic-electron-concentration) — Electrical Engineering Material, chapter 6
- [Compensated semiconductor: electron and hole concentrations](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Acompensated-carrier-concentration) — Electrical Engineering Material, chapter 6, asked 4×