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.

Answer:

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, \[\rho =\frac{1}{e{{n}_{h}}{{\mu }_{h}}}=\frac{1}{1.6\times {{10}^{-19}}\times {{10}^{19}}\times 0.045}\]             \[=13.9\,\Omega \,\,\text{-}\,\,m\]