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Suppose a pure Si crystal has5 xx 10^(28) atoms m^(-3) . It is droped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that n_(i) = 1.5 xx 10^(16) m^(-3). |
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Answer» Solution :Note that thermally GENERATED electrons`( n_(i) ~ 10^(16)m^(-3))` are negligibly SMALL as compared to those produced by doping. Therefore, `n_(e ) = N_(D) ` Since `n_( e) n_(h ) = n_(i )^(2) `,The number of holes ,`n_(h) = ( 1.5 xx 10^(16))^(2) // 5 xx 10^(28) xx 16^(-6)` `n_(h) =( 2.25 xx 10^(32)) //( 5xx 10^(22)) ~ 4.5xx 10^(9) m^(-3)` |
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