Built-in potential of a pn junction

The built-in potential $V_0$ is the potential difference across the depletion region of a pn junction in equilibrium. It balances the diffusion of carriers.

Assumptions

1. Abrupt (step) junction: $N_a$ on the p-side, $N_d$ on the n-side, uniform.
2. All dopants ionised: $p_{p0} = N_a$, $n_{n0} = N_d$.
3. Non-degenerate doping, so Boltzmann statistics and $np = n_i^2$ hold.
4. Thermal equilibrium, no applied voltage: $E_F$ is constant through the device.
5. Depletion region free of mobile carriers.

Derivation (using the Fermi level)

Before contact, measured from the intrinsic level:

$$E_{Fn} - E_{Fi} = kT\ln\frac{N_d}{n_i}, \qquad E_{Fi} - E_{Fp} = kT\ln\frac{N_a}{n_i}$$

   before contact             after contact
  p          n              p    |W|    n
 --Ec      --Ec           --Ec__
           ..EFn                 \___  Ec
 - Ei -    - Ei -     EF ...............
 ..EFp                     Ev__     eV0
 --Ev      --Ev                \___  Ev

On contact, electrons flow from n to p (and holes from p to n) until the Fermi levels line up. The bands on the n-side drop by $eV_0$ relative to the p-side, where

$$\begin{aligned} eV_0 &= E_{Fn} - E_{Fp} = kT\ln\frac{N_d}{n_i} + kT\ln\frac{N_a}{n_i} \\ V_0 &= \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2} \end{aligned}$$

Check using the Boltzmann relation

Hole concentrations on the two sides differ by the barrier $eV_0$:

$$\frac{p_{p0}}{p_{n0}} = \exp\left(\frac{eV_0}{kT}\right), \quad p_{n0} = \frac{n_i^2}{N_d}$$ $$\frac{N_aN_d}{n_i^2} = \exp\left(\frac{eV_0}{kT}\right) \;\Rightarrow\; V_0 = \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2}$$

Example: Si, $N_a = 10^{17}$, $N_d = 10^{16}$ cm$^{-3}$, $n_i = 1.45\times10^{10}$ cm$^{-3}$: $V_0 = 0.02586\ln(4.76\times10^{12}) = 0.755$ V.