Depletion Region Width in Unsymmetric PN Junction Diode

In a $p\text{-}n$ junction diode, charge neutrality across the depletion layer requires that the total uncompensated negative charge per unit area on the $p$-side is equal to the total uncompensated positive charge per unit area on the $n$-side.

Let:

• $N_A$ be the acceptor doping concentration on the $p$-side,
• $N_D$ be the donor doping concentration on the $n$-side,
• $x_p$ be the depletion region width on the $p$-side,
• $x_n$ be the depletion region width on the $n$-side,
• $q$ be the elementary charge.

The condition for charge neutrality in the depletion region is expressed as: $q \cdot N_A \cdot x_p = q \cdot N_D \cdot x_n$

Canceling the elementary charge $q$ from both sides gives: $N_A \cdot x_p = N_D \cdot x_n$

Rearranging the terms, we get the relationship between the depletion widths on both sides: $\frac{x_p}{x_n} = \frac{N_D}{N_A}$

Given values:

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

Substituting these values into the ratio: $\frac{x_p}{x_n} = \frac{10^{18}}{10^{15}} = 10^3 = 1000$

This simplifies to: $x_p = 1000 \cdot x_n$

Since $x_p \gg x_n$, the depletion region extends much farther into the lightly doped $p$-side compared to the heavily doped $n$-side.

Therefore, the depletion region width is more on the $p$-side compared to that in the $n$-side.

Correct Option: B