Solving Semiconductor Physics: Conduction Electron Density Calculation

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Homework Statement

I am asked to show that the conduction electron density in an n type semiconductor is given by:

[tex]n=\frac{1}{2} \left( -n_0 e^{-(E_c-E_d)/kT} + \sqrt{n_0^2 e^{-2(E_c-E_d)/kT}+4 n_0 N_d e^{-(E_c-E_d)/kT} } \right)[/tex]

where
[tex]n_0=2(m^* kT/2 \pi \hbar^2 )^{3/2}[/tex]For low temp [itex]p=0[/itex] and [itex] E_c-\mu >> kT[/itex]

Homework Equations

The Attempt at a Solution

Start with electrical neutrality:
[tex]\Delta n=n_e-n_h=n_d^+ - n_a^-[/tex]
Using the mass action law:
[tex]n_h=n_i^2/n_e[/tex]
Thus we get:
[tex]n_e^2-n_e\Delta n=n_i^2[/tex]
solving the quadratic:
[tex]n_e=\frac{1}{2}\left( \Delta n+ \sqrt{(\Delta n)^2 +4n_i^2} \right) [/tex]

From this, I gather that:

[tex] \Delta n=n_e-n_h=n_d^+ - n_a^- =n_0 e^{-(E_c-E_d)/kT}[/tex]
[tex]n_i =\sqrt{n_0 N_d e^{-(E_c-E_d)/kT}}[/tex]

But I've been thinking about this for a while now and I don't see how to derive this. Any help is appreciated.

I'm also concerned about my understanding for what exactly is the chemical potential in the Fermi distribution and why does it physically mean to say [itex]e^{-(E_c-E_d)/kT}[/itex] or [itex]e^{-(E_c-\mu)/kT}[/itex]

My understanding is that the chemical potential is the fermi energy, which is basically the highest energy level at T=0. However, how could the fermi energy/chemical potential change w.r.t. T if it is defined w.r.t. T=0?

Thank you for the help!

 

[mark726]

Thank you for your question. I understand your confusion and will try my best to help you understand the derivation of the conduction electron density in an n type semiconductor.

Firstly, let's start with the basic equations. The Fermi-Dirac distribution function for electrons is given by:

f(E)=\frac{1}{e^{(E-\mu)/kT}+1}

Where E is the energy of the electron, \mu is the chemical potential (or Fermi energy), k is the Boltzmann constant, and T is the temperature.

Now, for a semiconductor at thermal equilibrium, the total number of electrons n_e is given by:

n_e=\int_{E_c}^{\infty}f(E)g(E)dE

Where E_c is the conduction band energy and g(E) is the density of states. For simplicity, let's assume that the density of states is constant, g(E)=g_0. Therefore, we can rewrite the equation as:

n_e=g_0\int_{E_c}^{\infty}\frac{1}{e^{(E-\mu)/kT}+1}dE

Now, we can change the variable of integration from E to x=(E-\mu)/kT, which gives us:

n_e=\frac{g_0kT}{h^3}\int_{(E_c-\mu)/kT}^{\infty}\frac{1}{e^x+1}dx

Where h is the Planck constant. This integral can be solved using the substitution method and the result is:

n_e=\frac{g_0kT}{h^3}\ln\left(\frac{e^{(E_c-\mu)/kT}+1}{e^{(E_c-\mu)/kT}}\right)

Now, at low temperatures, we can assume that the chemical potential is much higher than the thermal energy (kT). Therefore, we can simplify the above equation as:

n_e=\frac{g_0kT}{h^3}\ln\left(\frac{e^{(E_c-\mu)/kT}}{e^{(E_c-\mu)/kT}}\right)=\frac{g_0kT}{h^3}\ln(1)=0

This means that at low temperatures, there are no electrons in