Questions: The donor concentration in a sample of n-type silicon is increased by a factor of 100. The shift in the position of the Fermi level at 300 K, assuming the sample to non degenerate is meV.

Solution Steps

We need to determine the shift in the position of the Fermi level in an n-type silicon sample when the donor concentration is increased by a factor of 100 at 300 K. The sample is assumed to be non-degenerate.

For a non-degenerate n-type semiconductor, the Fermi level \( E_F \) is given by: \[ E_F = E_C - kT \ln\left(\frac{N_C}{N_D}\right) \] where:

• \( E_C \) is the conduction band edge,
• \( k \) is the Boltzmann constant,
• \( T \) is the temperature in Kelvin,
• \( N_C \) is the effective density of states in the conduction band,
• \( N_D \) is the donor concentration.

Let \( N_{D1} \) be the initial donor concentration and \( N_{D2} = 100 N_{D1} \) be the final donor concentration. The initial Fermi level \( E_{F1} \) and the final Fermi level \( E_{F2} \) are: \[ E_{F1} = E_C - kT \ln\left(\frac{N_C}{N_{D1}}\right) \] \[ E_{F2} = E_C - kT \ln\left(\frac{N_C}{N_{D2}}\right) = E_C - kT \ln\left(\frac{N_C}{100 N_{D1}}\right) \]

Simplify the expression for \( E_{F2} \): \[ E_{F2} = E_C - kT \ln\left(\frac{N_C}{100 N_{D1}}\right) = E_C - kT \left(\ln\left(\frac{N_C}{N_{D1}}\right) - \ln(100)\right) \] \[ E_{F2} = E_C - kT \ln\left(\frac{N_C}{N_{D1}}\right) + kT \ln(100) \]

The shift in the Fermi level \( \Delta E_F \) is: \[ \Delta E_F = E_{F2} - E_{F1} = \left(E_C - kT \ln\left(\frac{N_C}{N_{D1}}\right) + kT \ln(100)\right) - \left(E_C - kT \ln\left(\frac{N_C}{N_{D1}}\right)\right) \] \[ \Delta E_F = kT \ln(100) \]

At \( T = 300 \) K, \( k = 8.6173 \times 10^{-5} \) eV/K: \[ \Delta E_F = (8.6173 \times 10^{-5} \, \text{eV/K}) \times 300 \, \text{K} \times \ln(100) \] \[ \Delta E_F = 0.02585 \, \text{eV} \times \ln(100) \] \[ \Delta E_F = 0.02585 \, \text{eV} \times 4.6052 \] \[ \Delta E_F \approx 0.1190 \, \text{eV} = 119.0 \, \text{meV} \]

Final Answer