The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni= 1.5 × 1016 m−3. Is the material n-type or p-type?
Number of silicon atoms, N = 5 × 1028 atoms/m3
Number of arsenic atoms, nAs = 5 × 1022 atoms/m3
Number of indium atoms, nIn = 5 × 1020 atoms/m3
Number of thermally-generated electrons, ni = 1.5 × 1016 electrons/m3
Number of electrons, ne = 5 × 1022 − 1.5 × 1016 ≈ 4.99 × 1022
Number of holes = nh
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as:
$n_{e} n_{h}=n_{i}^{2}$
$\therefore n_{h}=\frac{n_{i}^{2}}{n_{e}}$
$=\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.99 \times 10^{22}} \approx 4.51 \times 10^{9}$
Therefore, the number of electrons is approximately 4.99 × 1022 and the number of holes is about 4.51 × 109. Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.