Question 12

The number of silicon atoms per m^{ 3} is 5 × 10^{ 28} . This is doped simultaneously with 5 × 10^{ 22} atoms per m^{ 3} of Arsenic and 5 × 10^{ 20} per m^{ 3} atoms of Indium. Calculate the number of electrons and holes. Given that n_{ i} = 1.5 × 10^{ 16} m^{ −3} . Is the material n-type or p-type?

Answer

Number of silicon atoms, N = 5 × 10^{ 28} atoms/m^{ 3}

Number of arsenic atoms, n_{ As} = 5 × 10^{ 22} atoms/m^{ 3}

Number of indium atoms, n _{In} = 5 × 10^{ 20} atoms/m^{ 3}

Number of thermally-generated electrons, n _{i} = 1.5 × 10^{ 16} electrons/m^{ 3}

Number of electrons, n _{e} = 5 × 10 ^{22} − 1.5 × 10^{ 16} ≈ 4.99 × 10 ^{22}

Number of holes = n _{h}

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} = n _{i}^{2 }/ n_{e }= (1.5x10^{16})^{2 }/ 4.99 x 10^{22 }≈ 4.51 x 10^{9}

Therefore, the number of electrons is approximately 4.99 × 10 ^{22} and the number of holes is about 4.51 × 10 ^{9} . Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.

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