Diffusion process and Einstein relation

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.

  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.