Suppose a 'n'- type wafer is created by doping Si crystal having `5xx10^(28) "atoms"//m^(3)` with 1 ppm concentration of As. On the surfabe 200 ppm Boron is added to create 'p' region in this wafer. Considering `n_(i)=1.5xx10^(16)m^(-3)`, (i) Calculate the densities of the charge carriers in the n & p regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.

Explanation

(i) When As is implanted in Si-crystal, n-type wafer is created. The no. of majority carried electrons due to doping of As is
`n_(e)=N_D=1/10^(6)xx5xx10^(28)=5xx10^(22)//m^(3)`
No. of minority carriers (holes) in n-type wafer is
`n_(h)=n_(i)^(2)/n_(e)=((1.5xx10^(16))^(2))/(5xx10^(22))=0.45xx10^(10)//m^(3)`
When B is implanted in Si-crystal, p-type wafer is created with no. of holes,
`n_(h)=N_A=200/10^(6)xx(5xx10^(28))=1xx10^(25)//m^(3)`
Minority carriers (electrons) created in p-type wafer is `n_(e)=n_(i)^(2)/n_(h)=((1.5xx10^(16))^(2))/(1xx10^(25))=2.25xx10^(7)//m^(3)`
(ii) When p-n junction is reverse boased, the minority carried holes of n-region wafer `(n_(h)=0.45xx10^(10)//m^(3))` would contribute more to the reverse saturation current than minority carrier electrons `(n_(e)=2.25xx10^(7)//m^(3))` of p-region wafer.