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The number of silicon atoms per m^(3) is 5 xx 10^(28). This is doped simultaneously with 5 xx 10^(22) atoms per m^(3) of Arsenic and 5 xx 10^(20) per m^(3) atoms of Indium. Calculate the number of electrons and holes. Given that n_(i) = 1.5 xx 10^(16) m^(-3). ls the material n-type or p-type? |
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Answer» Solution :When a SEMICONDUCTOR has both donor (pentavalent) and acceptor(trivalent) atoms then for charge neutrality, following condition should be FULFILLED: `N_(D) - N_(A) - n_(e) - n_(h) ""`.... (1) Where `N_(D)` are donor atoms and `N_(A) ` are acceptor atoms . Also for a semiconductor having `n_(i)`intrinsic charge carriers, we have `n_(e).n_(h) = n_(i)^(2)""`....(2) From equation (1), `(n_(e) + n_(h) )^(2) = (n_(e) - n_(h))^(2) + 4 n_(e) n_(h) = (N_(D) - N_(A) )^(2) + 4 n_(i)^(2) ""`....(3) ` therefore ""n_(e) + n_(h) = sqrt((N_(D) - N_(A))^(2) + 4 n_(i)^(2)) ""`.... (4) adding (1) and (4) , we have `n_(e) = (1)/(2) [ N_(D) - N_(A) + sqrt((N_(D) - N_(A))^(2) + 4 n_(i)^(2)) ] ` Given, `N_(D) = 5 xx 10^(22) m^(-3), N_(A) = 5 xx 10^(20) m^(-3) = 0.50 xx 10^(22) m^(-3), and n_(i) = 1.5 xx 10^(16) m^(-3)` ` n_(e) = (1)/(2) [ (5 xx 10^(22) - 0.05 xx 10^(22) ) + sqrt((4.95 xx 10^(22))^(2) + 4 xx (1.5 xx 10^(16))^(2) )] ` = 4.95` xx 10^(22) m^(-3)` As `"" n_(e).n_(i) = n_(i)^(2)` ` rArr "" n_(h) = ((1.5 xx 10^(16))^(2))/(4.95 xx 10^(22)) = (2.25 xx 10^(32))/( 4.95 xx 10^(22)) = 4.5 xx 10^(9) m^(-3)` Since `n_(e) gt n_(h),` the CRYSTAL is N-type. |
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