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Suppose a pure Si crystal has 5xx10^(28) atoms m^(-3).It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that n_(i)=1.5xx10^(16)m^(-3). |
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Answer» Solution :Here, thermally generated electrons, `n_(i)=1.5xx10^(16)m^(-3)` 1 ppm = 1 part per million `=10^(6)` `therefore` Electron number density due to As atom, `n_(D)=(5xx10^(28))/(10^(6))=5xx10^(22)m^(-3)` Number of electron due to doping. `n_(i)` is neglected compare to `n_(D)` `therefore n_(e )=n_(D )=5xx10^(22)m^(-3)` Now `n_(i)^(2)=n_(e )n_(H)` `therefore n_(h)=(n_(i)^(2))/(n_(e ))` `=((1.5xx10^(16))^(2))/(5xx10^(22))` `therefore n_(h)=4.5xx10^(9) m^(-3)` |
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