Built-in potential and depletion in forward bias

When p and n regions are joined, electrons diffuse to the p-side and holes to the n-side, leaving behind fixed ionized donors ($+$) and acceptors ($-$). This depletion (space charge) region creates an electric field and a potential barrier, the built-in potential $V_o$.

        p-side   |  W  |   n-side
   o o o o o o |- - | + +| e e e e e
   o o o o o o |- - | + +| e e e e e
              -Wp   0   Wn
         E-field points n -> p

Built-in potential

In equilibrium the Fermi level is flat across the junction. On the p-side, far from the junction, $p_{p0} = N_a$; on the n-side, $n_{n0} = N_d$ and the minority hole density is $p_{n0} = n_i^2/N_d$. The Boltzmann relation between hole densities on the two sides of the barrier is

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

(the same result follows from setting hole drift current = hole diffusion current). Therefore

$$V_o = \frac{kT}{e}\ln\frac{p_{p0}}{p_{n0}} = \frac{kT}{e}\ln\left(\frac{N_a N_d}{n_i^2}\right)$$

Depletion width at equilibrium

Assume an abrupt junction and full depletion. Charge neutrality: $N_a W_p = N_d W_n$. Poisson's equation $\dfrac{d^2V}{dx^2} = -\dfrac{\rho(x)}{\varepsilon}$, with $\rho = -eN_a$ for $-W_p<x<0$ and $\rho = eN_d$ for $0<x<W_n$. Integrating once gives a triangular field with maximum at $x = 0$:

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

Integrating again, the potential drop equals the area of the field triangle:

$$\begin{aligned} V_o &= \frac{1}{2}|E_o| W = \frac{e N_a N_d W^2}{2\varepsilon (N_a + N_d)} \\ W &= W_p + W_n = \sqrt{\frac{2\varepsilon (N_a + N_d) V_o}{e N_a N_d}} \end{aligned}$$

where $\varepsilon = \varepsilon_0\varepsilon_r$.

Forward bias

With forward voltage $V$ (p positive), the applied voltage opposes the built-in field. Almost all of $V$ drops across the high-resistance depletion region, so the barrier falls from $V_o$ to $V_o - V$:

$$W_{fwd} = \sqrt{\frac{2\varepsilon (N_a + N_d)(V_o - V)}{e N_a N_d}}$$

• The depletion layer becomes narrower as $V$ increases ($W \propto \sqrt{V_o - V}$).
• The lower barrier lets majority carriers cross; minority carrier density at the edges rises by $e^{eV/kT}$, giving the diode current $I = I_o(e^{eV/kT} - 1)$.
• For reverse bias, $V$ is replaced by $-V_r$, and $W$ widens.

For a one-sided junction ($N_a \gg N_d$), $W \approx \sqrt{2\varepsilon (V_o - V)/(eN_d)}$, so the depletion region lies almost entirely in the lightly doped side.