Question

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?

Answer 1

Arsenic is donor impurity No. of donor atoms added, `N_(D) = 5 xx 10^(22) m^(-3)`,
Indium is acceptor impurity, no. of accepter atoms added `N_(A) = 5 xx 10^(20) m^(-3)`
Therefore, no. of free electrons created `n_(e) = N_(D) = 5 xx 10^(22)`
Now, `n_(e) gt n_(h)`, therefore, net no. of free electrons created,
`n_(e)^(1) =n_(e) -n_(h) = 5 xx 10^(22) - 5 xx 10^(20) = 4.95 xx 10^(22) m^(-3)`
Also net no. of holes created
`n_(h)^(1) =(n_(1)^(2))/(n_(e)^(1)) =((1.5 xx 10^(16)))/(4.95 xx10^(22)) = 4.55 xx10^(9) m^(-3)`
As `n_(e)^(1) gt n_(h)^(1)`, the resulting material is n-type semiconductor.