Depletion region

The region around a p-n junction with reduced mobile charge carrier density.

An illustrative abrupt-junction depletion approximation assumes uniform, fully ionized donor impurities and acceptor impurities, negligible mobile space charge in the depleted layer, and permittivity ${\epsilon }_s$. For area $A$, doping number densities $N_A,N_D$ and built-in electric potential difference $V_{\mathrm{bi}}$, it givesHere $q>0$ is the elementary charge and $V$ is positive for forward bias. Increasing both doping densities reduces $W$ in this estimate. That explains the preparation strategy for a tunnel diode; it does not make this sharp-depletion approximation a precise microscopic model of a degenerate semiconductor.

Thermal generation of electron-hole pairs within a reverse-biased depletion region can supply a leakage electric current in addition to neutral-region diffusion. For a simple midgap trap with equal electron and hole lifetime parameters, its scale is $e\mathcal{A}W{n}_i/(2{\tau }_g)$, where $W$ is depletion width and ${\tau }_g$ the lifetime parameter. The electric field separates generated carriers. Unlike ideal diode minority-carrier diffusion current, this contribution scales approximately with $n_i$ rather than $n_i^2$ and can vary with reverse voltage through $W$. Consequently a measured activation energy need not equal the full band gap.

The depletion approximation replaces the junction's transition region by a region without mobile carriers, leaving only fixed ionized dopants. For a uniformly doped abrupt p-n junction, $\rho =-e{N}_A$ on the p side, $\rho =+e{N}_D$ on the n side and $\rho \simeq 0$ outside. Charge neutrality gives $N_A{x}_p=N_D{x}_n$. Integrating Poisson equation gives the electric field and the electrostatic band bending.

For a planar abrupt p-n junction with ionized densities $N_A,N_D$, p-side width $w_p$ and n-side width $w_n$, zero external field requires $N_A{w}_p=N_D{w}_n$. Integrating Poisson equation gives $V_{\mathrm{bi}}-V=e(N_A{w}_p^2+N_D{w}_n^2)/(2{\epsilon }_s)$, with positive $V$ denoting forward bias. Therefore $W=w_n+w_p$ has the displayed value, $w_n=N_{A}W/(N_A+N_D)$ and $w_p=N_{D}W/(N_A+N_D)$. The positive depletion capacitance per area is $C_A={\epsilon }_s/W$. It equals $d{Q}_p/dV=-d{Q}_n/dV$ when $Q_p=-e{N}_A{w}_p$ and $Q_n=e{N}_D{w}_n$ are signed space charges per area. Differentiating a positive charge magnitude with respect to forward voltage instead gives a negative derivative. The square-root law assumes partially depleted regions, negligible mobile charge and no substantial injected-carrier storage.

For a zero-bias abrupt p-n junction, choose the p side at $x<0$ and the n side at $x>0$. Ionized acceptors and donors remain fixed while mobile Electrons and electron holes diffuse and establish a built-in electric field. In the depletion approximation, mobile carriers are neglected on $-x_p<x<x_n$ and the net charge density isCharge neutrality gives $N_A{x}_p=N_D{x}_n$. The quasi-neutral bulk can have a large mobile carrier density while its net charge density is nearly zero, because fixed dopant charge compensates it. With complete ionization and nondegenerate carriers, equilibrium also gives $np=n_i^2$; real carrier profiles vary smoothly rather than forming exact step functions at depletion edges.

The incremental capacitance associated with changes of electric charge around a p-n junction. Its depletion contribution is often estimated by $C_j\simeq {\epsilon }_{s}A/W$, so an independent junction capacitance measurement can constrain the width $W$ of the depletion region under the model assumptions. Other stored electric charge, including injected minority carriers, can also contribute to a measured terminal capacitance.

Let each side of a symmetric abrupt p-n junction have thickness $d$, area $\mathcal{A}$ and doping density $N$. Partial depletion gives $w^2={\epsilon }_s(V_{\mathrm{bi}}-V)/(eN)$. Both sides become fully depleted at $V_{\mathrm{fd}}=V_{\mathrm{bi}}-eN{d}^2/{\epsilon }_s$. Beyond that point, the ionized dopant charges are fixed, but an incremental terminal voltage changes the electric field uniformly across the full thickness $2d$. Gauss's law then gives $d{Q}_{\mathrm{terminal}}={\epsilon }_s\mathcal{A}dV/(2d)$, up to terminal orientation. The measured positive capacitance approaches $C_{\mathrm{fd}}$ rather than vanishing. This geometrical result assumes ideal contacts, negligible mobile carriers, no breakdown and no other series capacitances.