Finding the fermi energy in doped quantum well

[nkk2008]

Homework Statement

Regular Doping: Let's say that we want to introduce some electrons into a quantum well, by
adding dopant atoms (donors) to the semiconductor. The "regular" way to do this would be to
deposit our semiconductor films, first a "barrier" layer with large band gap, then the quantum
well with small band gap, and then another barrier layer, and during the deposition of the
quantum well we simultaneously deposit some low concentration of dopant atoms. The situation
for the conduction band edge as a function of distance through the structure would then look like

where E1 is the energy of the bottom of the lowest quantum well band and EC is the energy
difference between the conduction band minima in the barrier and the quantum well.

Consider a doping density of 1019 cm-3 in the well. For simplicity we'll assume zero binding
energy for the electrons on the donors, so that that all of these electrons are introduced into
propagating states of the quantum well. Compute the position of the Fermi-energy (at zero
temperature) relative to E1. (Hint: use the density of states in 2D).

Homework Equations

Relevant equations are given below in the first few lines of the solution.

The Attempt at a Solution

\begin{align*}<br /> &amp; D_{2D}(E) = \frac{mL^2}{\hbar^2 \pi}\\<br /> &amp; N_h + N_D^+ = N_e + N_A^- \to N_D^+ = N_e\\<br /> &amp; N_D = 10^{19} \; cm^{-3}\\<br /> &amp; N_D^+ = \frac{N_D}{1+2e^{(-E_D + E_F)/k_b T}}\\<br /> &amp; N_e = \int_{0}^\infty D_{2D}(E)f(E)\;\mathrm{d}E =\int_{0}^\infty \frac{mL^2}{\hbar^2 \pi}\frac{1}{1+e^{(E-E_F)/k_bT}}\;\mathrm{d}E\\<br /> &amp; \text{very low T}\to \frac{1}{1+e^{(E-E_F)/k_bT}} \approx e^{-(E-E_F)/k_bT}\\<br /> &amp; \frac{N_D}{1+2e^{(-E_D + E_F)/k_b T}} =\int_{0}^\infty \frac{mL^2}{\hbar^2 \pi}e^{-(E-E_F)/k_bT}\;\mathrm{d}E\\<br /> &amp; \frac{N_D}{1+2e^{(-E_D + E_F)/k_b T}} = \frac{mL^2}{\hbar^2 \pi} \int_{0}^\infty e^{-E/k_bT}e^{E_F/k_bT} \mathrm{d}E = \frac{mL^2}{\hbar^2 \pi} e^{E_F/k_bT} \int_{0}^\infty e^{-E/k_bT}\mathrm{d}E\\<br /> &amp; \text{letting $u=E/k_bT$,} \int_{0}^\infty e^{-E/k_bT}\mathrm{d}E \to k_bT \int_{0}^\infty e^{-u}\mathrm{d}u = k_bT (1) = k_bT\\<br /> &amp; \text{Therefore, we have } \frac{N_D}{1+2e^{(-E_D + E_F)/k_b T}} = \frac{mL^2}{\hbar^2 \pi} e^{E_F/k_bT} k_bT\\<br /> &amp; \text{Rearranging, we get: } \frac{N_D \hbar^2 \pi}{mL^2k_bT} = e^{2E_F/kt} + e^{E_F/kt}\\<br /> &amp; \text{Solving the fraction term, we get:} \frac{N_D \hbar^2 \pi}{mL^2k_bT} = <br /> \end{align*}

So I have no idea where to go from here. The question says zero temperature, and I cannot reconcile that with the fact that if I have 0K as T, my equations make no sense.

Thanks
Nkk

 

[mfb]

$$\text{very low T}\to \frac{1}{1+e^{(E-E_F)/k_bT}} \approx e^{-(E-E_F)/k_bT}\\ $$
Why don't you use the limit T=0 here?
$$\frac{1}{1+e^{(E-E_F)/k_bT}} \to
\begin{cases}
\dots &\mbox{if E>E_F} \\
\dots &\mbox{if E=E_F} \\
\dots &\mbox{if E<E_F}
\end{cases} $$

I'm not sure what the result will be, but it will certainly make calculations easier.