question_answer

The number of silicon atoms per \[{{m}^{3}}\] is \[5\times {{10}^{28}}.\] This is doped simultaneously with \[5\times {{10}^{22}}\] atoms per \[{{m}^{3}}\] Arsenic and \[5\times {{10}^{22}}\] of atoms of indium. Calculate the number of electrons and holes, Given that \[=1.5\times {{10}^{16}}\,{{m}^{-3}}\] Is the material n type or p-type?

Answer:

\[{{n}_{e}}=4.95\times {{10}^{22}}{{m}^{-3}}\] \[{{n}_{h}}=4.5\times {{10}^{9}}{{m}^{-3}}\] material is n-type. Given, \[{{N}_{D}}=5\times {{10}^{22}}{{m}^{-3}},{{N}_{A}}=5\times {{10}^{20}}{{m}^{-3}}\] \[{{n}_{i}}=1.5\times {{10}^{16}}{{m}^{-3}}\] For the semiconductor to remain electrically neutral, \[{{N}_{D}}-{{N}_{A}}={{n}_{e}}-{{n}_{h}}\]                       ?(i) Also      \[{{n}_{e}}{{n}_{h}}=n_{i}^{2}\] \[\therefore \]      \[{{({{n}_{e}}+{{n}_{h}})}^{2}}={{({{n}_{e}}-{{n}_{h}})}^{2}}+2{{n}_{e}}{{n}_{h}}\]                                     \[={{({{N}_{D}}-{{N}_{A}})}^{2}}=4n_{i}^{2}\]             \[{{n}_{e}}+{{n}_{h}}=\sqrt{{{({{N}_{D}}-{{N}_{A}})}^{2}}=4n_{i}^{2}}\]         ?(ii) On adding Eqs. (i) and (ii), we get             \[{{n}_{e}}=\frac{1}{2}({{N}_{D}}-{{N}_{A}})+\sqrt{{{({{N}_{D}}-{{N}_{A}})}^{2}}=4n_{i}^{2}}\] \[=\frac{1}{2}\left[ (5\times {{10}^{22}}-0.05\times {{10}^{22}}) \right.\] \[\left. +\sqrt{{{(4.95\times {{10}^{22}})}^{2}}+{{(1.5\times {{10}^{16}})}^{2}}} \right]\] \[\approx \frac{1}{2}\left[ 4.95\times {{10}^{22}}+\sqrt{{{(4.95\times {{10}^{22}})}^{2}}} \right]\]             \[=4.95\times {{10}^{22}}{{m}^{-3}}\] Also      \[{{n}_{h}}=\frac{n_{i}^{2}}{{{n}_{e}}}=\frac{{{(1.5\times {{10}^{16}})}^{2}}}{4.95\times {{10}^{22}}}=\frac{2.25\times {{10}^{32}}}{4.95\times {{10}^{22}}}\]             \[=4.5\times {{10}^{9}}{{m}^{-3}}\] Since \[{{n}_{e}}>{{n}_{h}},\] the material is of n-type.