Question

A semiconductor has equal electron and hole concentration of 6 x 10⁸ m^-3. On doping with certain impurity, electron concentration increases to 9 x 10¹² m^-3.

(i) Identify the new semiconductor obtained after doping.

(ii) Calculate the new hole concentration.

Answer 1

(i) Given n_i = 6 x 10⁸ m^-3

After doping with certain impurity, the electron concentration becomes ne = 9 x 10¹² m^-3. Hence the doping is of pentavalent impurity and the new semiconductor is of n-type.

(ii) n_h = ?

Since n_eh_n = n_i²

∴ n_h = \(\frac{n_i^2}{n_e} = \frac{(6 \times 10^8)}{9 \times 10^2}\)

or n_h = 4 x 10⁴ m^-3.