Built-in potential and depletion width derivation

Built-in potential

Consider an abrupt pn junction with acceptor concentration $N_a$ on the p-side and donor concentration $N_d$ on the n-side. Assumptions: abrupt junction, full ionisation, non-degenerate doping, depletion approximation (no free carriers in the depletion region), equilibrium.

When the two sides are joined, holes diffuse from p to n and electrons from n to p. They leave behind uncovered negative acceptor ions on the p-side and positive donor ions on the n-side. These form a space charge (depletion) region whose field opposes further diffusion. In equilibrium the Fermi level is constant through the junction and the bands bend by $eV_0$.

        p-side  |  depletion  |  n-side
                | -Wp  0  +Wn |
  Ec ___________
                \_____________
  EF ............................. (flat)
  Ev ___________   eV0
                \_____________

In equilibrium, carrier concentrations at the two ends of the depletion region follow the Boltzmann factor:

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

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

$$\begin{aligned} \frac{N_a N_d}{n_i^2} &= \exp\left(\frac{eV_0}{kT}\right) \\ V_0 &= \frac{kT}{e}\ln\left(\frac{N_a N_d}{n_i^2}\right) \end{aligned}$$

(The same result follows from $eV_0 = E_{Fn} - E_{Fp}$ before contact, using $E_{Fn} - E_{Fi} = kT\ln(N_d/n_i)$ and $E_{Fi} - E_{Fp} = kT\ln(N_a/n_i)$.)

Depletion width

Charge density: $\rho = -eN_a$ for $-W_p < x < 0$ and $\rho = +eN_d$ for $0 < x < W_n$.

  rho
  +eNd     +-----+
           |     |
 ---+------+-----+--- x
 -Wp|   0        Wn
    +------+
  -eNa

1. Charge neutrality:

$$N_a W_p = N_d W_n$$

2. Poisson's equation: $\dfrac{dE}{dx} = \dfrac{\rho}{\varepsilon}$, with $\varepsilon = \varepsilon_0\varepsilon_r$. Integrating from $-W_p$ (where $E = 0$) gives a field that is maximum at the junction:

$$E_0 = -\frac{eN_d W_n}{\varepsilon} = -\frac{eN_a W_p}{\varepsilon}$$

3. Potential: $V_0$ is the area under the triangular field plot:

$$V_0 = -\frac{1}{2}E_0 W = \frac{eN_d W_n W}{2\varepsilon}, \quad W = W_n + W_p$$

4. From neutrality, $W_n = \dfrac{N_a W}{N_a + N_d}$. Substituting:

$$\begin{aligned} V_0 &= \frac{e N_a N_d W^2}{2\varepsilon (N_a + N_d)} \\ W &= \sqrt{\frac{2\varepsilon (N_a + N_d) V_0}{e N_a N_d}} \end{aligned}$$

   E(x)
  ---+-----------+---- x
  -Wp \         / Wn
       \       /
        \     /
         \   /
          \ /
          -E0

The depletion region extends mostly into the lightly doped side ($W_n/W_p = N_a/N_d$). With reverse bias $V_r$, replace $V_0$ by $V_0 + V_r$.