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) An increase in hole concentration due to doping signifies that holes are majority carriers and electrons are minority carriers, indicating a p-type semiconductor.
(ii) To elevate the hole concentration, the dopant must be an acceptor impurity, such as a trivalent atom.

(b) In an intrinsic semiconductor, the product of electron and hole concentrations is constant: \[n_i^2 = n_e \times n_h\] Given: \[n_i = 5 \times 10^8 \ \text{m}^{-3}, \quad n_h = 8 \times 10^{12} \ \text{m}^{-3}\] The electron concentration \( n_e \) is computed as: \[n_e = \frac{n_i^2}{n_h} = \frac{(5 \times 10^8)^2}{8 \times 10^{12}} = \frac{25 \times 10^{16}}{8 \times 10^{12}} = 3.125 \times 10^4 \ \text{m}^{-3}\]

Final Answer: \( 3.125 \times 10^4 \ \text{m}^{-3} \)