Question

Suppose a pure Si crystal has \(5 \times 10^{28}\) atoms m\(^{-3}\). It is doped by 1 ppm concentration of boron. Calculate the concentration of holes and electrons, given that \(n_i = 1.5 \times 10^{16}\) m\(^{-3}\). Is the doped crystal n-type or p-type?

Hint:

Remember, doping a semiconductor with trivalent impurities (like Boron) increases the hole concentration, making it p-type. Doping with pentavalent impurities (like Phosphorus) would increase electron concentration, making it n-type.

Answer 1

First, calculate the concentration of boron dopants: \[ Boron concentration} = \frac{1 \, ppm}}{10^6} \times 5 \times 10^{28} = 5 \times 10^{22} \, m}^{-3} \] Since boron adds holes, the concentration of holes \( p \) will be approximately equal to the concentration of boron dopants. Assuming intrinsic carrier concentration \( n_i \) remains much smaller compared to the hole concentration due to doping: \[ p \approx 5 \times 10^{22} \, m}^{-3} \] Using the mass action law \( n \times p = n_i^2 \), calculate the concentration of electrons \( n \): \[ n = \frac{n_i^2}{p} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} = 4.5 \times 10^{9} \, m}^{-3} \] Thus, the concentration of holes \( p \) is \(5 \times 10^{22} \, m}^{-3}\) and electrons \( n \) is \(4.5 \times 10^{9} \, m}^{-3}\), indicating a p-type semiconductor.