Semiconductors: Common Mistakes and Fixes (3)

Question

In an n-type semiconductor, the number density of electrons is $n_e = 5 \times 10^{22}\,\text{m}^{-3}$ and the intrinsic carrier concentration at room temperature is $n_i = 1.5 \times 10^{16}\,\text{m}^{-3}$. Find the hole concentration $n_h$. Many students answer $n_h = n_i$. Why is that wrong?

Solution — Step by Step

The hole concentration is $n_h = 4.5\times 10^{9}\,\text{m}^{-3}$, far smaller than $n_i$.

Why This Works

Doping adds majority carriers (electrons in n-type) but also suppresses minority carriers via recombination. The product $n_e n_h$ stays equal to $n_i^2$ at a given temperature, no matter how heavily doped. So when $n_e$ shoots up, $n_h$ must drop proportionally.

This is the critical idea students miss: doping doesn’t conserve carrier counts; it conserves the product. The intrinsic value $n_i$ is a property of the material at a given temperature, not a floor for either carrier type.

Alternative Method

Take logs. $\log n_h = 2\log n_i - \log n_e$. With $n_i \sim 10^{16}$ and $n_e \sim 10^{22}$, $\log n_h \sim 32 - 22 = 10$, so $n_h \sim 10^{10}$. The order of magnitude check confirms the answer.

Common Mistake

Confusing $n_h$ with $n_i$. The fix is automatic if you write $n_e n_h = n_i^2$ as your first line, every time. JEE Main 2023 Shift 2 had this exact problem with different numbers.