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  • 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 Boron is added to create ‘P’ region in this wafer. Considering n_{i} = 1.5 times 10^{16} m^{-3}

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.

Answers (1)

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

 

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