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

Solutions for lecture 13 exercises¶

Warm-up exercises*¶

In this set of warm-up exercises we consider a two-dimensional semiconductor.

Question 1.¶

The dimensionality of the system enters the integral implicitly with the density of states, which in 2D and expressed per unit volume is a constant:

\[ g_{2 D}(E)=\frac{2_s}{2 \pi} \frac{m}{\hbar^2} \]

\[ n_e = \int_{E_c}^\infty f(E)g_e(E)dE = \int_{E_c}^\infty\frac{m_e}{\pi\hbar^2}\frac{1}{e^{(E-E_F)/k_BT}+1}dE. \]

Question 2.¶

Following the derivation in the lecture notes, whenever calculating the hole-dependent quantities, replace all relevant physical quantities with their hole equivalents.

For example, state the integral in terms of the electron energy scale using $E \equiv E_e $ and $E_F \equiv E_{F,e}$ with $f_n(E) = 1 -f(E)$: $$ n_h = \int_{-\infty}^{E_v} f_h(E)g_h(E)dE = \int_{-\infty}^{E_v}\frac{m_h}{\pi\hbar^2}\left(\frac{1}{e^{-(E-E_F)/k_BT}+1}\right)dE $$
Next, using $E_h = -E$, change the integration element and the corresponding integration limits, and the energy variable: $$ n_h = \int_{-E_v}^\infty\frac{m_h}{\pi\hbar^2}\frac{1}{e^{(E_h+E_F)/k_BT}+1}dE_h, $$

remembering that $E_{F,h} = - E_F$.

Question 3.¶

The sketch of the density of states (=band structure) for a 2D semiconductor:

from matplotlib import pyplot

import numpy as np
from scipy.optimize import curve_fit
from scipy.integrate import quad

from common import draw_classic_axes, configure_plotting

configure_plotting()

# Band structure parameters.
E_V, E_C, E_F = -1.2, 1.8, .4
E_D, E_A = E_C - .7, E_V + .5
m_h, m_e = 1, .5

default_colors = pyplot.rcParams['axes.prop_cycle'].by_key()['color']
blue, red = default_colors[0], default_colors[3]
E = np.linspace(-3, 3, 1000)
fig, ax = pyplot.subplots()

n_F = 1/(np.exp(2*(E - E_F)) + 1)
# define g_e and g_h as step functions at E_C and E_V, respectively:
g_e = m_e * np.heaviside(E - E_C, 0)
g_h = m_h * np.heaviside(E_V - E, 0)
ax.plot(E, g_e, label="$g_e$", c=blue)
ax.plot(E, g_h, label="$g_h$", c=red)
ax.fill_between(E, 10 * g_e * n_F, 0, alpha=.7, label="$n_e$", color=blue)
ax.fill_between(E, 10 * g_h * (1-n_F), 0, alpha=.7, label="$n_h$", color=red)
ax.plot(E, n_F, label="$n_F$", linestyle='dashed', c='k')
ax.axvline(E_F, color='gray', linestyle='dotted', linewidth=1)
ax.set_ylim(bottom=0,top=1.4)

ax.set_xlabel('$E$')
ax.set_ylabel('$g$')
ax.set_xticks([E_V, E_C, E_F])
ax.set_xticklabels(['$E_V$', '$E_C$', '$E_F$'])
ax.set_yticks([])
ax.legend()
draw_classic_axes(ax, xlabeloffset=.1)

svg

Question 4.¶

Assuming that the Fermi energy is in the gap $(E_v < E_F < E_c)$ and far away from the band edges compared to $k_BT$;

\[ E_F-E_v \gg k_BT \text{ and } E_c - E_F \gg k_BT, \]

the Fermi-Dirac distribution can be approximated as Boltzmann distribution:

\[ f(E)_{e/h} \approx \exp\left[-(E_{e/h}\pm E_F)/k_BT\right]. \]

and the integrals for $n_h$ and $n_e$ simplify to:

\[ n_h \approx \int_{-E_v}^\infty\frac{m_h}{\pi\hbar^2}e^{-(E_h+E_F)/k_BT}dE_h = N_V e^{(E_v-E_F)/k_BT} \]

\[ n_e \approx \int_{E_c}^\infty\frac{m_e}{\pi\hbar^2}e^{-(E-E_F)/k_BT}dE = N_C e^{-(E_c+E_F)/k_BT} \]

where

\[ N_{V,C}= \frac{m_{h,e} k_B T}{\pi\hbar^2} \]

Question 5.¶

Under the assumption of the charge balance ($n_e = n_h$), the Fermi energy can be found by dividing the results of the previous question: $$ 1 = \frac{n_h}{n_e} = \left(\frac{N_V}{N_C}\right) e^{(E_v+E_c-2E_F)/k_BT} = \left(\frac{m_h}{m_e}\right) e^{(E_v+E_c-2E_F)/k_BT}, $$ and taking the logarithm of both sides of the equation: $$ E_F = \frac{E_V+E_C}{2} + \frac{k_B T}{2}\ln{\left(\frac{m_h}{m_e}\right)}. $$

Exercise 1: Energy, mass, velocity and cyclotron motion of electrons and holes¶

Question 1.¶

Electrons near the top of the valence band have a negative effective mass, their energy decreases as $k$ increases from 0, and they have a negative group velocity for $k>0$.

Question 2.¶

Holes near the top of the valence band have a positive effective mass, their energy increases as $k$ increases from 0, and they have a negative group velocity for $k>0$.

Question 3.¶

The equation of motion for an electron near the bottom of the conduction band is:

\[ m^* \frac{d\mathbf{v}}{dt} = -e\mathbf{v} \times \mathbf{B} \]

and when replacing we get two coupled equations:

\[\begin{align*} \dot{k_x} &= -\frac{e}{m^*}B k_y\\ \dot{k_y} &= +\frac{e}{m^*}B k_x \end{align*}\]

The solution to this equation is circular motion of cyclotron frequency of $\omega_c = \frac{eB}{m^*}$, where the Lorentz force is perpendicular to $\nabla_\mathbf{k} E$.

Question 4.¶

A hole near the bottom of the conduction band will have the same chirality as an electron. The chirality would be just the opposite if we would consider the valence band (for both electrons and holes).

Exercise 2*: a 1D semiconductor¶

def dispersion(EG, tcb, tvb, N=100, kmax=np.pi/2):
    a = 1
    kx = np.linspace(-kmax, kmax, N)
    Ecb = EG - 2*tcb*(np.cos(kx*a)-1)
    Evb = 2*tvb*(np.cos(kx*a)-1)

    # Plot dispersion
    plt.figure(figsize=(6,5))
    cb, = plt.plot(kx, Ecb, label="Conduction B.")
    vb, = plt.plot(kx, Evb, label="Valence B.")

    plt.xlabel('$k_x$', fontsize=20)
    plt.ylabel('$E$', fontsize=20)
    plt.title('E(k) for tcb:'+str(tcb)+' tvb:'+str(tvb))
    plt.legend(handles=[cb, vb])

dispersion(10, 2, 8)

svg

Question 1.¶

For the electrons in the conduction band we find

\[ v_{cb,e} = \frac{2at_{cb}}{\hbar}\sin (ka) \]

\[ m_{cb,e} = \frac{\hbar^2}{2a^2t_{cb}\cos (ka)}. \]

For the holes in the valence band we obtain

\[ v_{vb,h} = -\frac{2at_{vb}}{\hbar}\sin (ka) \]

\[ m_{vb,h} = -\frac{\hbar^2}{2a^2t_{vb}\cos (ka)}. \]

Question 2.¶

This approximation indicates the chemical potential is "well below" the conduction band and "well above" the valence band. This way, only a few electrons occupy the the states near the bottom of the conduction band and only a few holes occupy the states near the top of the valence band. This allows us the approximate the bottom of the conduction band and the top of the valence band as parabolic dispersions. In addition, the Fermi-Dirac distribution can be approximated as the Boltzman distribution.

The dispersions of the valence and conduction band are approximately equal to

\[ E_{cb} \approx E_G + t_{cb}(ka)^2 \]

\[ E_{vb} \approx -t_{vb}(ka)^2 \]

Question 3.¶

For the density of states for both the electrons and holes we find

\[ g_e(E) = \frac{1}{\pi a \sqrt{t_{cb}(E-E_G)}}, \:\: \text{if} \:\: E > E_G, \]

and

\[ g_h(E_h) = \frac{1}{\pi a \sqrt{t_{vb}E_h}}, \:\: \text{if} \:\: E_h > 0, \]

where $E_h = -E$.

Question 4.¶

The electron density in the conduction band is given by

\[\begin{align*} n_e &= \int_{E_G}^{\infty} f(E)g_c(E)dE\\ &\approx \int_{E_G}^{\infty} e^{-\beta (E-\mu)} g_c(E) dE\\ &=\sqrt{\frac{k_B T}{\pi t_{cb}}}\frac{e^{\beta (\mu - E_G)}}{a}. \end{align*}\]

Analoguous, we find for the hole density

\[\begin{align*} n_h &= \int_{0}^{\infty} f(E_h) g_h(E_h) dE_h\\ &\approx \sqrt{\frac{k_B T}{\pi t_{vb}}}\frac{e^{-\beta \mu}}{a}. \end{align*}\]

Question 5.¶

In the intrinsic regime $n_e = n_h$. Solving this for $\mu$ results in

\[ \mu = \frac{1}{2}E_G + \frac{k_B T}{4}ln(\frac{t_{cb}}{t_{vb}}) \]

Exercise 3*. Holes and electrons in a 1D tight-binding energy band¶

Question 1.¶

For electrons we find

\[ m_e = -\frac{\hbar^2}{2ta^2\cos(ka)} \]

\[ v_e = -\frac{2ta\sin(ka)}{\hbar}, \]

and for holes we obtained

\[ m_h = -m_e \]

\[ v_h = v_e. \]

Thus the effective masses of electrons an hole will be of opposite sign, while the group velocities will be the same!

Question 2.¶

The number of holes is defined as

\[ N_h = \int_{-\varepsilon-2t}^{-\varepsilon+2t} f(E_h, E_{F,h}) g_h(E_h)d E_h, \]

where $g_h(E_h)$ is the density of states of the holes in terms of hole energy $E_h$.

Question 3.¶

Small hint

It is convenient to write the hole integral in terms of the electron energy.

Exercise 4*: The Hall coefficient when both electrons and holes are present in the system¶

Question 1.¶

\[ R_{H,h} = \frac{1}{ne} \]

Question 2.¶

Filling in $n_e = n_h$ and $\mu_e=\mu_h$ results in $R_H = 0$.