Solutions for lecture 14 exercises¶

    from matplotlib import pyplot as plt 
import numpy as np
from math import pi

Exercise 1: Crossover between extrinsic and intrinsic regimes¶

Question 1.¶

Law of mass action:

\[ n_e n_h = n_i^2 = N_C N_V e^{-E_G/(k_B T)} \]

Charge balance condition:

\[ n_e - n_h + n_D - n_A = N_D - N_A \]

Question 2.¶

Since \(E_G \gg k_B T\), we can use the law of mass action together with charge balance. The question gives us another useful piece of information: we are around \(|N_D-N_A| \approx n_i\). That means we are near the transition between extrinsic and intrinsic regimes. In this regime we can neglect \(n_D\) and \(n_A\), just like in the lecture. Writing \(n_e n_h = n_i^2\) and \(n_e - n_h = N_D - N_A\) and solving these together, we obtain

\[\begin{align*} n_{e} = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D),\\ n_{h} = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D) \end{align*}\]

where \(D = N_D - N_A\).

For \(n_i \gg |N_D - N_A|\) we recover the intrinsic regime, while the opposite limit gives the extrinsic expressions.

Question 3.¶

If \(|D| \ll n_i\), then doping is not important and the intrinsic result is reproduced. Conversely, if \(|D| \gg n_i\), it is mostly the doping that determines \(n_e\) and \(n_h\). The thermal factor becomes unimportant. You can check both limits by Taylor expanding the exact solution.

Exercise 2: Donor ionization¶

Question 1.¶

If all the donors are ionized (\(n_D \approx 0\)), the Fermi level is given by the extrinsic-regime expression from the lecture.

Question 2.¶

The donor-bound electron concentration is

\[ n_D = \frac{N_D}{\exp\!\left[(E_D-E_F)/(k_B T)\right]+1} \]

Question 3.¶

Using \(n_D \sim N_D\) we get

\[ \frac{E_C - E_D}{k_B T} \approx \log\!\left(\frac{N_C}{N_D - N_A}\right). \]

For germanium at room temperature, \((E_C - E_D)/(k_B T) \approx 1/3\) (as derived in the lecture), so assuming that all donors are ionized is safe as long as \(N_D - N_A \ll N_C\).

Exercise 3: Performance of a diode¶

Question 1.¶

Intrinsic semiconductors have no impurities. Adding dopant atoms creates extra unbound electrons or holes, depending on whether the dopant is a donor or an acceptor. Impurity eigenstates appear and the \(E_F\) level shifts up or down.

A diode requires a p-n junction, so intrinsic semiconductors alone are not enough. Drawing a diagram is very helpful.

Question 2.¶

Under reverse bias, only two processes carry current: electrons that are thermally excited into the conduction band on the p-doped side and holes that are thermally excited into the valence band on the n-doped side.

Question 3.¶

\[ I_s(T) \propto e^{-E_G/(k_B T)}\]

Exercise 4: Quantum well heterojunction in detail¶

Question 1.¶

Include the energy bands here. You can find them at the book's section 18.2

Question 2.¶

This is a "particle in a box" problem.

\[\begin{align*} -\frac{\hbar^2}{2m_e^{\ast}} \nabla^2 \Psi_e &= (E_e-E_C)\Psi_e\\ -\frac{\hbar^2}{2m_h^{\ast}} \nabla^2 \Psi_h &= (E_h+E_V)\Psi_h \end{align*}\]

Here and below \(E_h\) is the energy of a hole in the valence band.

Question 3.¶

\[\begin{align*} E_e = E_C + \frac{\hbar^2}{2m_e^{\ast}} \left[\left(\frac{\pi n}{L}\right)^2+k_x^2+k_y^2\right],\\ E_h = -E_V + \frac{\hbar^2}{2m_h^{\ast}} \left[\left(\frac{\pi n}{L}\right)^2+k_x^2+k_y^2\right] \end{align*}\]

Question 4.¶

This is a 2D electron/hole gas, therefore the DOS per unit area expression is the same as in 2D parabolic dispersion:

\[\begin{align*} g_e = \frac{m_e^{\ast}}{\pi\hbar^2},\\ g_h = \frac{m_h^{\ast}}{\pi\hbar^2} \end{align*}\]

Question 5.¶

\(L\) can be found here using the previous questions, by setting:

\[ E_e + E_h - E_C + E_V = 0.1 \,\text{eV} = \frac{\hbar^2}{2}\left[\left(\frac{\pi n}{L}\right)^2+k_x^2+k_y^2\right] \left(\frac{1}{m_e^{\ast}}+\frac{1}{m_h^{\ast}}\right) \]

By choosing the lowest subband with \(n=1\) and \(k_x = k_y = 0\), \(L\) can be found as \(\approx 6.85\) nm.

Question 6.¶

For a laser, one wants to fix the emission wavelength to a certain value. With this setup, the band gap is easy to design because it is set by \(L\), which is fixed.

Question 7.¶

If donor impurities are put outside the well (on both sides, for example), the donated electrons can lower their energy by falling into the well, but the ionized dopants remain behind. This gives an advantage because very high electron mobility can be obtained within the quantum well (there are no ionized dopants in the well to scatter off). This is called modulation doping.