Ncert Exercises & Solutions

Suppose an n-type wafer is created by doping Si crystal having $5 \times 10^{28} atoms / m^{3}$ with 1 ppm concentration of As. On the surface, 200 ppm boron is added to create 'P' region in this wafer. Considering $n_{i} = 1.5 \times 10^{16} m^{- 3}$ ,

(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 the diode is reverse biased.

Hint: Use

n_{e} n_{h} = {n_{i}}^{2}

Step 1: Find the no. of majority and minority charge carriers in the n-type wafer.

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

n_{e} = N_{D} = \frac{1}{10^{6}} \times 5 \times 10^{28}

= 5 \times 10^{22} / m^{3}

The number of minority carriers (holes) in the n-type wafer is:

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

Step 2: Find the no. of majority and minority charge carriers in the p-type wafer.

When B is implanted in Si crystal, the p-type wafer is created with the number of holes,

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

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

n_{e} = \frac{{n_{i}}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2}}{1 \times 10^{25}} = 2.25 \times 10^{27} / m^{3}

Step 3: Find the type of charge carriers contributing more to the reverse saturation current.

When the p-n junction is reverse biased, the minority carrier holes of the n-region wafer

(n_{h} = 0.45 \times 10^{10} / m^{3})

would contribute more to the reverse saturation current than minority carrier electrons

(n_{e} = 2.25 \times 10^{7} / m^{3})

of the p-region wafer.

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