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The Number of silicon atoms per m^(3) is 5 xx 10^(28) . This is doped simultaneously with5 xx 10^(22) atoms per m^(3) of Arsenic and 5 xx 10^(20)per m^(3) atoms of Indium. Calculate the number of electron and holes. Given that n_(i)= 1.5 xx 10^(16) m^(-3). Is the material n-type or p-type? |
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Answer» Solution :We known that for each atomdoped of ARSENIC one free electron is received . Similarly , for each atom doped of indium a vacancyis created .So, the number of free electrons INTRODUCED by pantavalent impurity added. `n_( e) = N_( A s)= 5xx 10^(22) m^(3)`...(i) The number of holes introduced by trivalent impurity added. Now, `n_( e ) -n_(h) = 5 xx10^(22) - 5 xx 10^(20) =4.95 xx 10^(20)` ....(ii) So, ` ( n_( e ) +n_(h))^(2) = ( n_(e ) - n_(h))^(2) +4n_(e ) n_(h )` `n_( e ) + n_(h ) = sqrt((4.95xx10^(22))^(2)+( 4.15 xx 10^(16))^(2))`...(iii) AddingEqs (iii) and (ii) `2n_( e ) = 4.95 xx 10^(22)+ sqrt((4.95 xx10^(22))^(2) + 4( 1.5 xx 10^(10))^(2))` `n_(e ) = (1)/(2) ( 4.95 xx 10^(22) + sqrt(( 4.95 xx10^(22))^(2)))` `= n_( e ) = 4.95xx 10^(22) // m^(2) ` Now `n_(i) ^(2) = n_(h ) xx n _( e ) ` `n_( h ) =( n_(i)^(2))/( n_(e ))= ((1.5 xx 10^(16))^(2))/( 4.95 xx 10^(22))=4.54 xx 10^(9) //m^(3)` As number of electrons `n_(e )( = 4.95 xx 10^(22))` is greater than the number of holes `n_(h ) (=4.5 xx10^(9))` . So the MATERIAL is n-type semicoductor. |
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