Question

Suppose a ‘n’-type wafer is created by doping Si crystal having 5 × 10²⁸ atoms/m3 with 1 ppm concentration of As. On the surface 200 ppm Boron is added to create ‘P’ region in this wafer. Considering n i = 1.5 × 10¹⁶ m^–3, (i) Calculate the densities of the charge carriers in the n and p regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.

Answer 1

(i) n-type wafer is created when As is implanted in Si crystal. The number of majority carriers electrons due to doping of As is

`n_e = N_D`

= `10^-6 xx 5 xx 10^28` atoms/m³

= `5 xx 10^22/m^3`

The number of minority carriers (holes) in an 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`

The p-type wafer is created with the number of holes, when Boron is implanted in Si crystal,

`n_h = N_A`

= `200 xx 10^-6 xx (5 xx 10^28)`

= `1 xx 10^25/m^3`

Minority carriers (electrons) created in the p-type wafer is

`n_e = n_i^2/n_h`

= `(1.5 xx 10^16)^2/(1 xx 10^25)`

= `2.25 xx 10^7/m^3`

(ii) The minority carrier holes of the n-region wafer (n_h = 0.45 × 10¹⁰/m³) would contribute more to the reverse saturation current than minority carrier electrons (n_e = 2.25 × 10⁷/m³) of the p-region wafer when the p-n junction is reverse biased.