PN junction formation and built-in potential

Formation of a pn junction

A pn junction is formed when one region of a single crystal is made p-type and the adjacent region n-type (e.g. by diffusion or ion implantation). On contact:

1. Diffusion: the p side has many holes and the n side many electrons. Holes diffuse into the n side and electrons into the p side, where they recombine with majority carriers.
2. Space charge: near the junction the p side loses holes and is left with fixed negative acceptor ions $N_a^-$; the n side is left with fixed positive donor ions $N_d^+$. This carrier-free layer is the depletion region of width $W_0 = W_p + W_n$.
3. Built-in field: the ions set up a field $\mathcal{E}_0$ from n to p. It drives a drift current opposite to the diffusion current.
4. Equilibrium: diffusion continues until drift exactly balances it. The net current is zero, the Fermi level is flat across the device, and a potential difference $V_0$ (the built-in potential) exists between the n and p sides.

    p              W0             n
 o o o o |- - - - |+ + + + | * * * *
 o o o o |- - - - |+ + + + | * * * *
         |<- Wp ->|<- Wn ->|
          E0 <---------------
 V:  0 ______/^^^^^^^^^^^^ V0

Derivation of built-in potential

Consider holes. In equilibrium the hole drift current balances the hole diffusion current:

$$ep\mu_h\mathcal{E} - eD_h\frac{dp}{dx} = 0$$

With $\mathcal{E} = -dV/dx$ and the Einstein relation $D_h/\mu_h = kT/e$:

$$-p\frac{dV}{dx} = \frac{kT}{e}\frac{dp}{dx}\;\Rightarrow\; dV = -\frac{kT}{e}\frac{dp}{p}$$

Integrate from the p side ($V = 0$, $p = p_{p0}$) to the n side ($V = V_0$, $p = p_{n0}$):

$$V_0 = \frac{kT}{e}\ln\frac{p_{p0}}{p_{n0}}$$

With $p_{p0} = N_a$ and $p_{n0} = n_i^2/N_d$:

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

Equivalently, $p_{n0}/p_{p0} = e^{-eV_0/kT}$: the Boltzmann relation across the barrier.

Example: Si, $N_a = 10^{18}$, $N_d = 10^{16}\ \text{cm}^{-3}$, 300 K: $V_0 = 0.02585\ln(10^{34}/2.1\times10^{20}) \approx 0.81$ V.