• question_answer The resistivity of pure silicon of $3000\,\Omega \,m.$ and the electron and hole mobilities are $0.12\,{{m}^{2}}{{V}^{-1}}{{s}^{-1}}$ and 0$0.045\,{{m}^{2}}{{V}^{-1}}{{s}^{-1}}$ respectively, determine (i) the resistivity of a specimen of the material when ${{10}^{19}}$ atoms of phosphorous are added per ${{m}^{3}}.$ (ii) the resistivity of the specimen if further $2\times {{10}^{19}}$ boron atoms perm3 are also added.

The resistivity of pure silicon is given by $\rho =\frac{1}{\sigma }=\frac{1}{e{{n}_{i}}({{\mu }_{e}}+{{\mu }_{h}})}$ $\therefore$ Intrinsic carrier concentration ${{n}_{i}}=\frac{1}{e\rho ({{\mu }_{e}}+{{\mu }_{h}})}$ $=\frac{1}{1.6\times {{10}^{-19}}\times 3000\times (0.12+0.045)}{{m}^{-3}}$ $=1.26\times {{10}^{16}}{{m}^{-3}}$ (i) When ${{10}^{19}}$ donor atoms of phosphorous are added per${{m}^{3}}$ ${{n}_{e}}{{n}_{h}}=n_{i}^{2}={{(1.437\times {{10}^{16}})}^{2}}=2.066\times {{10}^{32}}$ and       ${{n}_{e}}-{{n}_{h}}={{n}_{d}}-{{n}_{a}}={{10}^{19}}$ As         ${{n}_{e}}>>{{n}_{h}},$ therefore ${{n}_{e}}\approx {{10}^{19}}$ Hence, $\rho =\frac{1}{e{{n}_{e}}{{\mu }_{e}}}=\frac{1}{1.6\times {{10}^{-19}}\times {{10}^{19}}\times 0.12}$             $=25\,\Omega \,\,\text{-}\,\,m$ (ii) When $2\times {{10}^{19}}$ acceptor atoms of boron are further added : ${{n}_{h}}-{{n}_{e}}={{n}_{a}}-{{n}_{d}}=2\times {{10}^{19}}-{{10}^{19}}={{10}^{19}}$ As ${{n}_{e}}>>{{n}_{h}},$ therefore ${{n}_{h}}\simeq {{10}^{19}}$ Hence, $\rho =\frac{1}{e{{n}_{h}}{{\mu }_{h}}}=\frac{1}{1.6\times {{10}^{-19}}\times {{10}^{19}}\times 0.045}$             $=13.9\,\Omega \,\,\text{-}\,\,m$