Question

A p-type Si semiconductor is made by doping an average of one dopant atom per \( 5 \times 10^7 \) silicon atoms. If the number density of silicon atoms in the specimen is \( 5 \times 10^{28} \, \text{atoms m}^{-3} \), find the number of holes created per cubic centimeter in the specimen due to doping. Also give one example of such dopants.

Hint:

In a p-type semiconductor, the dopant atoms create holes by accepting electrons. The dopant concentration determines the number of holes.

Answer 1

The doping concentration is \( \frac{1}{5 \times 10^7} \) dopant atoms per silicon atom. The number of holes per cubic meter is calculated by multiplying this concentration by the number of silicon atoms per cubic meter: \[ n_{\text{holes}} = \left( \frac{1}{5 \times 10^7} \right) \times (5 \times 10^{28}) = 10^{21} \, \text{holes m}^{-3} \] To express this in holes per cubic centimeter, we convert units: Since \( 1 \, \text{m}^3 = 10^6 \, \text{cm}^3 \): \[ n_{\text{holes}} = \frac{10^{21}}{10^6} = 10^{15} \, \text{holes cm}^{-3} \] Consequently, the doping results in \( 10^{15} \, \text{holes cm}^{-3} \). A common dopant for p-type semiconductors is Boron.