Question

In an intrinsic semiconductor, carrier’s concentration is \( 5 \times 10^8 \ \text{m}^{-3} \). On doping with impurity atoms, the hole concentration becomes \( 8 \times 10^{12} \ \text{m}^{-3} \).

[(a)] Identify (i) the type of dopant and (ii) the extrinsic semiconductor so formed.

[(b)] Calculate the electron concentration in the extrinsic semiconductor.

Hint:

In a doped semiconductor, the product of electron and hole concentrations remains equal to the square of intrinsic carrier concentration: \( n_e \cdot n_h = n_i^2 \).

Answer 1

(a) (i) Since the hole concentration increases due to doping, this means electrons are the minority carriers and holes are the majority carriers.
This indicates a p-type semiconductor.
(ii) The dopant used must be an acceptor impurity (i.e., a trivalent atom) to increase the hole concentration.
(b) For an intrinsic semiconductor, the product of the electron and hole concentrations remains constant: \[ n_i^2 = n_e \times n_h \] Given: \[ n_i = 5 \times 10^8 \ \text{m}^{-3}, \quad n_h = 8 \times 10^{12} \ \text{m}^{-3} \] Now compute electron concentration \( n_e \) using: \[ n_e = \frac{n_i^2}{n_h} = \frac{(5 \times 10^8)^2}{8 \times 10^{12}} = \frac{25 \times 10^{16}}{8 \times 10^{12}} = 3.125 \times 10^4 \ \text{m}^{-3} \] Final Answer: \( 3.125 \times 10^4 \ \text{m}^{-3} \)