PN junction - doping and carrier concentration - solved problem 1

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We have laid the foundation for calculation of carrier concentration in our previous post. In this section, let us solve a simple problem.
#1. A silicon pn junction employs the following doping levels. N_A = 10^{17}/cm^3 and N_D = 4\times10^{16}/cm^3. Determine the hole and electron concentration on the two sides.

Solution

To solve this problem all you need to know is tabulated below.

Type of semiconductor Majority carrier conc. Minority carrier conc.
p-type p_p=N_A n_p=\dfrac{n_i^2}{N_A}
n-type n_n=N_D p_n=\dfrac{n_i^2}{N_D}
where, n_i=1.5\times10^{10}/cm^3

Given data,
N_A = 10^{17}/cm^3
N_D = 4\times10^{16}/cm^3

For p-side,
Hole (Majority carrier) concentration = p_p=N_A = 10^{17}/cm^3.
Free electron (Minority carrier) concentration = n_p=\dfrac{n_i^2}{N_A} = \dfrac{(1.5\times10^{10})^2}{10^{17}}=2250/cm^3.

For n-side,
Free electron (Majority carrier) concentration = n_n=N_D = 4\times10^{16}/cm^3.
Hole (Minority carrier) concentration = p_n=\dfrac{n_i^2}{N_D} = \dfrac{(1.5\times10^{10})^2}{4\times10^{16}}=5625/cm^3.

We will solve more problems in the following posts.

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