Built-in potential and depletion width defined

When p-type and n-type regions meet, electrons diffuse from n to p and holes from p to n. They recombine near the junction and leave behind uncovered fixed ions: negative acceptor ions on the p side and positive donor ions on the n side. This region, empty of mobile carriers, is the depletion (space-charge) region, and its width $W_0 = W_p + W_n$ is the depletion width. The ions create an electric field and a potential difference $V_0$ across the region, called the built-in potential; it stops further diffusion.

   p side        |  W0  |       n side
  o o o o o  | - - | + + |  * * * * *
  holes      | - - | + + |  electrons
             |<Wp>|<-Wn->|
  field E0  <-----------
  potential: 0 ____/^^^^^ V0

Built-in potential

In equilibrium the Fermi level is flat. Far from the junction:

$$p_{p0} = N_a,\qquad p_{n0} = \frac{n_i^2}{N_d}$$

The hole concentrations on the two sides are related by the Boltzmann factor of the potential energy step $eV_0$:

$$\frac{p_{n0}}{p_{p0}} = \exp\left(-\frac{eV_0}{kT}\right)$$

Hence

$$V_0 = \frac{kT}{e}\ln\frac{p_{p0}}{p_{n0}} = \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2}$$

(The same result follows from equating the drift and diffusion currents of holes, using the Einstein relation.)

Depletion width

1. Charge neutrality: $eN_aW_p = eN_dW_n$.

2. Field from Gauss's law (Poisson's equation): the field rises linearly through each charged layer and peaks at the junction:

$$\mathcal{E}_0 = -\frac{eN_dW_n}{\varepsilon} = -\frac{eN_aW_p}{\varepsilon},\qquad \varepsilon = \varepsilon_0\varepsilon_r$$

3. Potential = area under the field triangle:

$$V_0 = \frac{1}{2}|\mathcal{E}_0|W_0 = \frac{eN_dW_nW_0}{2\varepsilon}$$

4. Eliminate $W_n$: from step 1, $W_n = W_0\dfrac{N_a}{N_a+N_d}$, so

$$V_0 = \frac{eW_0^2}{2\varepsilon}\cdot\frac{N_aN_d}{N_a+N_d}$$ $$W_0 = \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}}$$

with $W_n = W_0N_a/(N_a+N_d)$ and $W_p = W_0N_d/(N_a+N_d)$. The depletion layer extends mostly into the lightly doped side. Under bias, $V_0$ is replaced by $V_0 - V$.