Depletion region width

The depletion region width $W$ is the total thickness of the Depletion region of a PN junction, split into a part $x_n$ on the n-side and a part $x_p$ on the p-side. Doping concentrations and the bias set its value.

$W=x_n+x_p=\sqrt{\frac{2{\epsilon }_s}{q}\left(\frac{1}{N_A}+\frac{1}{N_D}\right)V_0}$

where:

• $x_n$ = depletion depth into the n-side, $x_p$ = depletion depth into the p-side.
• ${\epsilon }_s$ = permittivity of the semiconductor (for silicon $\approx 1.04\times 10^{-12}\text{F/cm}$).
• $q$ = electron charge, $1.602\times 10^{-19}\text{C}$.
• $N_A$, $N_D$ = acceptor and donor doping concentrations.
• $V_0$ = the Built-in voltage (in equilibrium; under bias, replace it with the total junction potential, see below).

Where it comes from (Poisson’s equation)

Poisson’s equation relates the electric field to charge density: $d^2\varphi /d{x}^2=-\rho /{\epsilon }_s$. Inside the depletion region the only charge is the fixed dopant ions, so on the n-side $\rho =+q{N}_D$ and on the p-side $\rho =-q{N}_A$. Overall charge neutrality of the exposed charge requires the positive charge on the n-side to balance the negative charge on the p-side:

$q{N}_D{x}_n=q{N}_A{x}_p⟹N_D{x}_n=N_A{x}_p$

Integrating Poisson’s equation once gives the (triangular) field profile, peaking at the junction; integrating again and evaluating the total potential drop across both sides sets that drop equal to $V_0$. Solving for the total width $x_n+x_p$ produces the boxed expression above. The combination $(\frac{1}{N_A}+\frac{1}{N_D})$ falls out of eliminating $x_n$ and $x_p$ with the neutrality condition.

Two consequences worth remembering

1. The lightly-doped side gets the wider depletion. From $N_D{x}_n=N_A{x}_p$, the side with more doping has the smaller depletion depth, so the depletion mostly extends into the lightly-doped side. A region needs a certain amount of fixed charge exposed to support the field; if the doping density is low, you have to deplete a wider slab to expose that much charge. So in a $p^{+}n$ junction (heavily doped p, lightly doped n) almost all of $W$ lies on the n-side.

Width $W$ splits into $x_n$ and $x_p$; the heavier-doped side has the narrower depletion.

2. Bias changes the width. An applied voltage changes the total potential the depletion region must support. Replace $V_0$ in the formula with the net junction potential:

• Reverse bias (applied $V$ adds to the barrier): total potential $\to V_0+V_R$, so $W$ increases, the region widens.
• Forward bias (applied $V$ opposes the barrier): total potential $\to V_0-V_F$, so $W$ decreases, the region narrows.

This voltage-controllable width is why a reverse-biased junction behaves like a voltage-variable capacitor (a varactor), and it’s central to how field-effect and bipolar transistors modulate current.