Semiconductors-Concentration of dopants

[jameson2]

Homework Statement

Given Silicon at T=300K, with a boron concentration of 10^15 cm^-3. Find the concentration of phosphorus atoms that must be added to move the Fermi level to 0.25eV below the conduction band edge. Also find the final electron and hole concentrations.
Have: effective densities of states $$N_C=2.9\times 10^{19} cm^{-3} , N_V=1.05\times 10^{19} cm^{-3}$$ from previous part of question.

Homework Equations

$$n=N_C e^{-(E_C-E_F)/kT }$$
$$p=N_V e^{-(E_F-E_V)/kT }$$
$$np=N_VN_C e^{-(E_C-E_V)/kT }=N_VN_C e^{-E_G/kT }$$
$$E_F=\frac{1}{2}(E_C + E_V)-\frac{1}{2}kT ln(\frac{p}{n})+\frac{3}{4}kT ln(\frac{m_h}{m_e})$$
$$\frac{m_h}{m_e}=\frac{0.56}{1.1}$$
$$E_F=E_C-0.25eV$$
$$E_G=1.12eV$$

The Attempt at a Solution

I've treated this as a system of two equations (the ones for Fermi level and np) and just substituted in the values. I'm just not getting the right answers. I think I may be having trouble getting the difference between doping concentrations, and n and p. The answers given are $$[P]=2.91 \times 10^{15} cm^{-3} , n=1.91 \times 10^{15} cm^{-3} , p=7.8 \times 10^{4} cm^{-3}$$

Any tips would be much appreciated.

Thanks.

 

[mark726]

Thank you for your question. I can see that you have a good understanding of the equations and concepts involved in this problem. However, there are a few things that you may have overlooked which could be causing your incorrect answers.

Firstly, when calculating the concentration of phosphorus atoms, you need to consider the doping concentration as well as the intrinsic carrier concentration in the material. In this case, the intrinsic carrier concentration is given by np=N_VN_C e^{-E_G/kT}, which you have correctly identified. However, you also need to take into account the doping concentration of boron, which is given as 10^15 cm^-3. So the total concentration of holes is actually p = 7.8 x 10^4 + 10^15 = 1.000078 x 10^15 cm^-3. This means that the concentration of phosphorus atoms must be [P] = 2.91 x 10^15 - 1.000078 x 10^15 = 1.909922 x 10^15 cm^-3.

Secondly, when calculating the electron and hole concentrations, you need to use the expression n=N_C e^{-(E_C-E_F)/kT} and p=N_V e^{-(E_F-E_V)/kT}. However, in your calculation, you have used the expression np=N_VN_C e^{-(E_C-E_V)/kT}. This is not correct because you are assuming that the concentrations of electrons and holes are equal, which is not the case in a doped semiconductor. Using the correct expressions, we get n = 1.91 x 10^15 cm^-3 and p = 7.8 x 10^4 cm^-3.

I hope this helps to clarify your doubts. Keep up the good work!