25 Jun 2025

p-n Junction Formation, Depletion Region, and Barrier

Diffusion, space charge, built-in potential, electric field, and depletion width of an abrupt p-n junction.

bsc semester-iv mj-7 semiconductor-devices pn-junction depletion-region

Consider a one-dimensional abrupt homojunction with uniformly doped p material at $x<0$ and n material at $x>0$. Let $q=1.602\times10^{-19}\,\mathrm C$ denote the positive elementary-charge magnitude. The depletion approximation used below assumes complete dopant ionization and neglects mobile carriers inside the depleted layer.

Formation of the space-charge region

Before contact, the n side contains many electrons and the p side many holes. After contact, electrons diffuse from n to p and holes diffuse from p to n. Recombination near the interface leaves fixed ionized acceptors $-qN_A$ on the p side and fixed ionized donors $+qN_D$ on the n side. Their electric field points from the positive donor charge toward the negative acceptor charge, namely from n to p.

If the depletion edges are $-x_p$ and $x_n$, the charge density is

\[\rho(x)= \begin{cases} -qN_A,&-x_p<x<0,\\ +qN_D,&0<x<x_n,\\ 0,&\text{elsewhere}. \end{cases}\]

Charge neutrality of the depletion layer requires

\[\boxed{N_Ax_p=N_Dx_n}.\]

Poisson’s equation and the electric-field sign convention are

\[\frac{dE}{dx}=\frac{\rho}{\varepsilon_s}, \qquad E=-\frac{d\phi}{dx},\]

where $\varepsilon_s$ is the semiconductor permittivity in $\mathrm{F\,m^{-1}}$. With $E(-x_p)=E(x_n)=0$,

\[E(x)= \begin{cases} -\dfrac{qN_A}{\varepsilon_s}(x+x_p),&-x_p\le x\le0,\\[6pt] \dfrac{qN_D}{\varepsilon_s}(x-x_n),&0\le x\le x_n. \end{cases}\]

Thus $E<0$ throughout the depletion region: it opposes further majority-carrier diffusion. Equilibrium is reached when drift and diffusion currents cancel separately for electrons and holes.

Depletion approximation for an illustrative asymmetric junction. Equal charge areas enforce $N_Ax_p=N_Dx_n$; the field is directed from n to p and the potential rises from p to n.

The editable source is pn-junction-depletion.tex.

Built-in potential and depletion width

At thermal equilibrium the Fermi level is constant. For non-degenerate material with $n_n\simeq N_D$ and $p_p\simeq N_A$, the electrostatic potential difference is

\[\boxed{V_{bi}=\phi_n-\phi_p =V_T\ln\!\left(\frac{N_AN_D}{n_i^2}\right)}, \qquad V_T=\frac{k_BT}{q}.\]

$V_T$ is the thermal voltage, equal to about $25.9\,\mathrm{mV}$ at $300\,\mathrm K$. The formula assumes $N_AN_D>n_i^2$, complete ionization, and Maxwell-Boltzmann statistics. Integrating $-E=d\phi/dx$ gives

\[V_{bi}=\frac{q}{2\varepsilon_s} \left(N_Ax_p^2+N_Dx_n^2\right).\]

Writing $W_0=x_p+x_n$,

\[\boxed{W_0=\sqrt{\frac{2\varepsilon_sV_{bi}}q \left(\frac1{N_A}+\frac1{N_D}\right)}},\] \[x_p=\frac{N_D}{N_A+N_D}W_0, \qquad x_n=\frac{N_A}{N_A+N_D}W_0.\]

The depletion region extends farther into the more lightly doped side. The peak field magnitude is

\[\lvert E_{\max}\rvert=\frac{qN_Ax_p}{\varepsilon_s} =\frac{qN_Dx_n}{\varepsilon_s} =\frac{2V_{bi}}{W_0},\]

with units $\mathrm{V\,m^{-1}}$. Since an electron’s band energy changes as $-q\phi$, the conduction and valence band edges fall by $qV_{bi}$ from the p side to the n side; this band bending is the equilibrium barrier to majority-carrier diffusion.

The charge-neutrality and built-in-potential identities are checked in junction-and-mos-check.mac; every displayed residual is zero.