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 surface 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

When As is implanted in Si crystal, `n -`type wafer is created. The number of majority carriers electrons due to doping of As is
`n_(e) = N_(D) = (1)/(10^(6)) xx 5 xx 10^(28)`
`= 5 xx 10^(22) //m^(3)`
Number of minority carriers (holes) in `n -` type wafer is
`n_(h) = (n_(i)^(2))/(n_(e)) = ((1.5 xx 10^(16))^(2))/(5 xx 10^(22))`
`= 0.45 xx 10^(10) //m^(3)`
When B is implanted in Si crystal, `p-`type wafer is created with number of holes,
`n_(h) = N_(A) = (200)/(10^(6)) xx (5 xx 10^(28)) = 1 xx 10^(25) //m^(3)`
Minority carriers (electrons) created in `p -` type wafer is
`n_(e) = (n_(i)^(2))/(n_(h)) = ((1.5 xx 10^(16))^(2))/(1 xx 10^(25))`
`= 2.25 xx 10^(27)//m^(3)`
When `p - n` junction is reverse biased, the minority carrier holes of `n -`region wafer `(n_(h) = 0.45 xx 10^(10) //m^(3))` would contribute more to the reverse saturation current than minority carrier electrons `(n_(e) = 2.25 xx 10^(7) //m^(3))` of `p` region wafer.