#### Suppose a ‘n’-type wafer is created by doping Si crystal having $5 \times 10^{28}$ atoms/m3 with 1ppm concentration of As. On the surface, 200 ppm of Boron is added to create the ‘P’ region in this wafer. Considering $n_{i} = 1.5 \times 10^{16} m^{-3}$, (i) Calculate the densities of the charge carriers in the n & p regions. (ii) Comment on which charge carriers would contribute largely to the reverse saturation current when the diode is reverse biassed.

$n_{e}=N_{D}=10^{-6}\times 5 \times 10^{28}atoms/m^{3}\\ =5 times 10^{22}/m^{3}$

number of minimum carriers (holes) in n-type wafer is

$n_{h}=\frac{n_{i}^{2}}{n_{e}}=\frac{\left ( 1.5 \times 10^{16} \right )^{2}}{5 \times 10^{22}}=0.45 \times 10^{10}/m^{3}$

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

$n_{h}=N_{A}=200 \times 10^{-6} \times \left ( 5 \times 10^{28} \right )=1 \times 10^{25}/m^{3}$

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

$n_{h}=\frac{n_{i}^{2}}{n_{e}}=\frac{\left ( 1.5 \times 10^{16} \right )^{2}}{1 \times 10^{25}}=2.25 \times 10^{7}/m^{3}$

(ii) The minority carrier holes of n-region wafer $\left (n_{h}=0.45 \times 10^{10}/m^{3} \right )$ would contribute more to the reverse saturation current than minority carrier electrons $\left (n_{e}=2.25 \times 10^{7}/m^{3} \right )$ of p-region wafer when p-n junction is reverse biassed