A) \[2.25\times {{10}^{7}}/{{m}^{3}}\]
B) \[1.12\times {{10}^{3}}/{{m}^{3}}\]
C) \[3.11\times {{10}^{6}}/{{m}^{3}}\]
D) \[2.11\times {{10}^{5}}/{{m}^{3}}\]
Correct Answer: A
Solution :
[a] In 'n' region; number of \[{{e}^{-}}\] is due to As: \[{{n}_{e}}={{N}_{D}}=1\times {{10}^{-6}}\times 5\times {{10}^{28}}atoms/{{m}^{3}}\] \[{{n}_{e}}=5\times {{10}^{22}}/{{m}^{3}}\] The minority carriers (hole) is \[{{n}_{h}}=\frac{n_{i}^{2}}{{{n}_{e}}}=\frac{{{(1.5\times {{10}^{16}})}^{2}}}{5\times {{10}^{22}}}=\frac{2.25\times {{10}^{32}}}{5\times {{10}^{22}}}\] \[{{n}_{h}}=0.45\times 10/{{m}^{3}}\] Similarly, when Boron is implanted a ?p? type is created with holes \[{{n}_{h}}={{N}_{A}}=200\times {{10}^{-6}}\times 5\times {{10}^{28}}=1\times {{10}^{25}}/{{m}^{3}}\] This is far greater than \[{{e}^{-}}\] that existed in 'n' type wafer on which Boron was diffused. Therefore, minority carriers in created 'p' region \[{{n}_{e}}=\frac{n_{i}^{2}}{{{n}_{h}}}=\frac{2.25\times {{10}^{32}}}{1\times {{10}^{25}}}=2.25\times {{10}^{7}}/{{m}^{3}}\]
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