Einstein relation between diffusion and mobility

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).

 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.