Solution:
Formula: \(n_e n_h = n_i^2\). Since the semiconductor is heavily p-doped, \(n_h approx N_a = 10^{18}\text{ m}^{-3}\). Therefore, \(n_e = frac{n_i^2}{n_h} = frac{(10^{16})^2}{10^{18}} = 10^{14}\text{ m}^{-3}\).
A silicon (Si) specimen is doped with aluminium (Al). The concentration of acceptor atoms is \(10^{18}\text{ m}^{-3}\). Given that the intrinsic carrier concentration is \(10^{16}\text{ m}^{-3}\), the concentration of electrons in the specimen is
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