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NCERT Class XII Chapter
Semiconductor Electronics
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Question : 12 of 19
Marks: +1, -0
The number of silicon atoms per m3\mathrm{m}^3 is 5 × 102810^{28}. This is doped simultaneously with 5 × 102210^{22} atoms per m3\mathrm{m}^3 of Arsenic and 5 × 102010^{20} per m3\mathrm{m}^3 atoms of Indium. Calculate the number of electrons and holes. Given that ni = 1.5 × 1016 m−310^{16}\,\mathrm{m}^{-3}. Is the material n-type or p-type?
Solution:  
We know that for each atom doped of Arsenic, one free electron is received. Similarly, for each atom doped of indium, a vacancy is created. So, the number of free electrons introduced by pentavalent impurity added
nen_e = NAsN_{\mathrm{As}} = 5 × 1022 m−310^{22}\,\mathrm{m}^{-3}
The number of holes introduced by trivalent impurity added
nhn_h = NInN_{\mathrm{In}} = 5 × 1020 m−310^{20}\,\mathrm{m}^{-3}
We know the relation, nenhn_e n_h = ni2n_i^2 ...(i)
Now net electrons,
n′e = ne−nhn_e - n_h = 5 × 102210^{22} – 5 × 102010^{20}
= 4.95 × 1022 m−310^{22}\,\mathrm{m}^{-3} ...(ii)
Now using equation (i), net holes
nh′n'_h = ni2ne′\frac{n_i^2}{n'_e} = (1.5×1016)24.95×1022\frac{(1.5\times 10^{16})^2}{4.95\times 10^{22}} = 4.5 × 109 m−310^{9}\,\mathrm{m}^{-3}
So, ne′n'_e ≫ nh′n'_h , the material is of n-type.
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