InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The number of silicon atoms per `m^(3) is 5xx10^(28)`. This is doped simultaneously with `5xx10^(22)` atoms per `m^(3)` of Arsenic and `5xx10^(20) per m^(3)` atoms of indium. Calculate the number of electrons and holes. Given that `n_(i)=1.5xx10^(16)m^(-3)`. Is the material n-type or p-type?A. `3.24xx10^(6)m^(-3)`B. `6.24xx10^(8)m^(-3)`C. `4.54xx10^(9)m^(-3)`D. None of these |
|
Answer» Correct Answer - C For each atom doped of arsenic, one free electron is received similarly of each atom doped of indium, a vecancy is created. So, number of free electrons introduced by pentavalent impurity is `N_(AS) = 5 xx 10^(22)m^(-3)` The number of holes introduced by trivalent impurity added is `N_(I)=5xx10^(20)m^(-3)` So, net number of electrons added is `n_(e)=N_(AS)-N_(I)` `=5 xx10^(22)-5xx10^(20)` `=4.95xx10^(22)m^(-3)` Now, by the law of mass action. `n_(e)n_(h) =n_(i)^(2)` So, `n_(h) =(n_(i)^(2))/(n_(e))=((1.5xx10^(16))^(2))/(4.95xx10^(22))` `rArr n_(h)=4.54xx10^(9)m^(-3)` As, `n_(e) gt n_(h)` ( number of hles ). So, the material is n-type semiconductor. |
|