Wireless Communication

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 . Ho

An intrinsic or pure semiconductor (eg., pure Silicon) is an insulator at T=0K. This is because all electrons are bound to their atoms and no free electron is available for conduction. At higher temperatures (T>0K), for example, at room temperature(T=300K), some electrons gain enough thermal energy to break their bonds, cross the band gap energy and enter the conduction band. These electrons are free to conduct current. Free electrons and holes are always generated in pairs. So, for every free electron in the conduction band, there is a hole in the valance band. Let us denote the free electron concentration by n and free hole concentration by p . Since free electrons and holes are generated in pairs, we can say, n=p=n_i where n_i is called intrinsic carrier concentration.Also, np=n_i^2 Typically, number of free electrons in an intrinsic Si at room temperature is of the order of 10 10 /cm 3 . But the number of Si atoms per cm 3 is of the order of 10 22 . Comparing the