A silicon sample is doped with Boron to achieve impurity concentration of 1.5 × 10¹⁴ cm^-3. Assuming that complete ionization takes place and that intrinsic carrier concentration of Silicon is 1.5 × 10¹⁰ cm^-3, what is the concentration of electrons?

The relationship between carrier concentrations in a semiconductor is given by the mass action law, which states:

n × p = n_i²

Where:

• n = concentration of electrons (minority carriers in p-type material, in cm^-3).
• p = concentration of holes (majority carriers in p-type material, in cm^-3).
• n_i = intrinsic carrier concentration of silicon (in cm^-3).

Given data:

• Intrinsic carrier concentration, n_i = 1.5 × 10¹⁰ cm^-3.
• Acceptor concentration (boron doping), N_A = 1.5 × 10¹⁴ cm^-3.
• In a p-type semiconductor, the majority carrier concentration (holes) is approximately equal to the acceptor concentration: p ≈ N_A.

Step 1: Apply the mass action law

Using the mass action law:

n × p = n_i²

Substitute p ≈ N_A:

n × N_A = n_i²

Solve for the electron concentration, n:

n = n_i² / N_A

Step 2: Substitute the given values

Intrinsic carrier concentration: n_i = 1.5 × 10¹⁰ cm^-3

Acceptor concentration: N_A = 1.5 × 10¹⁴ cm^-3

Substitute these values into the equation:

n = (1.5 × 10¹⁰)² / (1.5 × 10¹⁴)

First, calculate n_i²:

n_i² = (1.5 × 10¹⁰) × (1.5 × 10¹⁰) = 2.25 × 10²⁰

Now, divide by N_A:

n = (2.25 × 10²⁰) / (1.5 × 10¹⁴)

n = 1.5 × 10⁶ cm^-3

The concentration of electrons in the silicon sample is 1.5 × 10⁶ cm^-3.