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 \(5 \times 10^{28}\) atoms m\(^{-3}\), find the number of holes created per cubic centimetre in the specimen due to doping. Also give one example of such dopants.

Hint:

Doping with trivalent impurities like Boron introduces holes in the semiconductor, making it p-type.

Answer 1

Hole Generation in P-type Silicon via Doping

Provided Data:

• Average dopant atoms per \( 5 \times 10^7 \) silicon atoms
• Silicon atom density: \( 5 \times 10^{28} \) atoms/m\(^3\)
• Objective: Calculate hole density per cubic centimetre from doping.

Calculation Process:

Step 1: Dopant-to-Silicon Atom Ratio

The provided ratio is: \[ \frac{\text{Dopant Atoms}}{\text{Silicon Atoms}} = \frac{1}{5 \times 10^7} \]

Step 2: Dopant Atom Density per Cubic Metre

Using the silicon atom density \( 5 \times 10^{28} \) atoms/m\(^3\), the dopant concentration per cubic metre is: \[ \text{Dopant Atoms/m}^3 = \left( \frac{1}{5 \times 10^7} \right) \times (5 \times 10^{28} \, \text{atoms/m}^3) = 10^{21} \, \text{dopant atoms/m}^3 \]

Step 3: Hole Density per Cubic Metre

Each dopant atom introduces one hole. Therefore, the hole density per cubic metre equals the dopant atom density: \[ \text{Holes/m}^3 = 10^{21} \, \text{holes/m}^3 \]

Step 4: Conversion to Hole Density per Cubic Centimetre

With \( 1 \, \text{m}^3 = 10^6 \, \text{cm}^3 \), the hole density per cubic centimetre is: \[ \text{Holes/cm}^3 = \frac{10^{21} \, \text{holes/m}^3}{10^6 \, \text{cm}^3/\text{m}^3} = 10^{15} \, \text{holes/cm}^3 \]

Dopant Example for P-type Silicon:

A common dopant for creating p-type silicon is boron (B). Boron possesses one fewer valence electron than silicon. When a boron atom replaces a silicon atom in the crystal lattice, it creates a deficiency in electrons, manifesting as a hole.

Conclusion:

The calculated hole concentration per cubic centimetre due to doping in the silicon specimen is \( \boxed{10^{15}} \, \text{holes/cm}^3 \).