• $summary_{type}$: concept
• $word_{count}$: 170

When p-type meets n-type, electrons diffuse from n to p and holes from p to n (driven by concentration gradients). This exposes immobile ions: positive donors on n-side, negative acceptors on p-side. The resulting charge distribution creates the depletion region — a thin zone (~0.1-1 um) depleted of free carriers. The built-in electric field E (from n to p, opposing diffusion) establishes equilibrium where diffusion current equals drift current. The contact potential: $V_0$ ≈ 0.7 V (Si) or 0.3 V (Ge). Depletion width: W = sqrt(2epsilon($V_0$)$\frac{N_A+N_D}{(e*N_A*N_D)}$). The depletion region is wider on the lightly doped side (charge neutrality: $N_A$$x_p$ = $N_D$*$x_n$). Higher doping = thinner depletion, stronger field. The depletion region has no free carriers and acts like an insulator — it is the key to all p-n junction behavior. JEE asks about how bias changes depletion width and barrier height.