JEE Main 2026 — Semiconductors Question with Solution

In a p-n junction diode, the total charge on both sides of the depletion region must be equal to maintain charge neutrality.

Let $x_p$ and $x_n$ be the widths of the depletion region on the p-side and n-side, respectively. Let $N_A$ and $N_D$ be the doping concentrations on the p-side and n-side.

The charge neutrality condition is given by:
$N_A{x}_p=N_D{x}_n$

Given $N_A=10^{15}{\text{ atoms/cm}}^3$ and $N_D=10^{18}{\text{ atoms/cm}}^3$.

Substituting the values, we get:
$10^{15}{x}_p=10^{18}{x}_n$

$x_p=10^3{x}_n$

Since $x_p>x_n$, the depletion region penetrates more into the lightly doped side. Therefore, the depletion region width is more on the p-side compared to that in the n-side.

Answer: The depletion region width is more on p-side compared to that in n-side