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Next: Fermi statistics, charge carrier Up: From Semi-conductivity to Micro-electronics Previous: Crystal lattices, periodic potentials,

Band structure, mobility, effective mass, holes

The travelling Bloch wave function for the electron is reflected from planes in the crystal when $k_x = \frac{n\pi}{a}$. A similar condition will hold for $k_y$ and $k_z$. It therefore becomes a standing wave for these values of $k$. Imagine for convenience a one-dimensional lattice. We would have standing wave functions for the n$^{th}$ order Bragg condition as follows :-
$\displaystyle \psi$ $\textstyle \propto$ $\displaystyle \cos\left(\frac{n\pi x}{a}\right)$ (37)
$\displaystyle \psi$ $\textstyle \propto$ $\displaystyle \sin\left(\frac{n\pi x}{a}\right) .$  

Both of these standing wave solutions are correct, but in each of them, the electron has a very different probability distribution. The standing wave solutions are displayed in figure 8 for the first order Bragg condition. The sine wave has its probability density between the cores, and the cosine wave has its probability density dominantly over the cores. The energy state of the sine distribution is therefore higher than that of the cosine distribution, as in this case, the negative electron charge density is further from the positive cores. We have a situation where the wave function $\psi(x) = u_k(x)e^{-kx}$ has two different energies for the same value of $x$.

Figure 8: The two standing wave solutions for Bloch wave satisfying the first order Bragg condition.

In the case of bound states for the non-periodic potentials we have treated up to now, (the particle in a box, the square well, the Coulomb potential of the hydrogen atom), we have seen the energy spectrum is discretised into a sequence of points. Now, with the case of bound states in a periodic potential, the energy spectrum has regions where it has allowed continuous bands, and forbidden band gaps. The band gaps arise at particular values of the particle momentum or wavenumber $(p=\hbar k = \hbar \frac{n\pi}{a})$. The significance of the zone in $k$-space where each energy band occurs is so great that these zones have a special name. They are called ``Brillouin zones''.

$\vert k\vert < \frac{\pi}{a}$ 1st Brillouin zone
$\frac{\pi}{a} < \vert k\vert < \frac{2\pi}{a}$ 2nd Brillouin zone
and so on. You may often see the energy band diagrams plotted so that all the energy states are mapped over the first Brillouin zone, as in figure 9.
Figure 9: The energy band diagram, with bands for successive Brillouin zones mapped over the first Brillouin zone. (Compare to figure 6.)
\includegraphics[width=0.5\textwidth]{bloch-band3.eps}

The issue of the density of states $g(\epsilon)d\epsilon$ will arise later, in discussions of the quantum statistics of electrons (fermions) in energy bands, just as the issue arose in connection with ``gases'' of fermions and bosons in the preceding section on statistical mechanics. The discussion now will refer to a one dimensional band. Later, the discussion will be broadened to cover a more realistic three dimensional band.

As before, imagine $a$ is the periodicity of the lattice, and $l$ is its length. The lattice contains $M$ atoms.

\begin{displaymath}
l = Ma
\end{displaymath} (38)

The Bloch wave function is
\begin{displaymath}
\psi(x) = u_k(x)e^{-ikx}
\end{displaymath} (39)

and it must satisfy the boundary condition
\begin{displaymath}
\psi(0) =\psi(Ma)
\end{displaymath} (40)

and the periodicity condition
\begin{displaymath}
u_k(x) = u_k(x+a).
\end{displaymath} (41)

It follows that $\exp(ikMa) = 1$

Exercise 6.2
Verify this.

Therefore $k=\frac{2\pi n}{Ma}$ and

$\displaystyle -\frac{\pi}{a} <$ $\textstyle k$ $\displaystyle < \frac{\pi}{a} \qquad \mbox{ 1st Brillouin zone}$ (42)
$\displaystyle \mbox{i.e.} \qquad -\frac{\pi}{a} <$ $\textstyle \frac{2\pi n}{ML}$ $\displaystyle < \frac{\pi}{a}$  
$\displaystyle \mbox{i.e.} \qquad -\frac{M}{2} <$ $\textstyle n$ $\displaystyle < \frac{M}{2}$  

Therefore there are $M$ states in the band spaced $\Delta k = \frac{2\pi}{ML}$ apart. Each primitive cell in one dimension therefore contributes exactly one independent value of $k$ to each band. Taking spin into account, the total number of electron states in each one-dimensional band is $2M$.

It is worthwhile considering the origin of band structure from a third point of view. Consider the formation of molecular orbitals when two hydrogen atoms approach each other to form a chemical bond. The two original single atom $\psi^{AO}_{1s}(r)$ wave functions evolve into the two molecular orbitals $\psi^{MO}_1(r)$ and $\psi^{MO}_2(r)$. Whereas the two single atom orbitals had the same energy when the atoms were separated, the two molecular orbitals have different energies. The bonding molecular orbital has a lower energy than the non-bonding molecular orbital. We extract the principle that the process of bonding generates new molecular orbitals, whose proliferation increases with the number of atoms participating in the bond. Moreover, the energies of these new molecular orbitals depends on the inter nuclear separation. The inter atomic distance in the molecule is that distance for which the molecular orbital energies are minimised. If we have a solid, we may imagine that we have a very large number $N_A$ of atoms participating in the bond. There will therefore be very many molecular orbitals, so many that they form a quasi-continuous band of available energy states for the electrons. The concept of band formation via many molecular orbitals is illustrated for silicon and diamond in figure 10.

Figure 10: Band formation via many molecular orbitals illustrated for silicon and diamond as a function of inter atomic spacing.

Note the formation of quasi-continuous bands separated by forbidden gaps. This method of viewing band formation is instructive, as it illuminates the mechanism of filling of the bands by electrons, according to the Pauli Exclusion Principle for fermions.

For example, the 6$M$ electrons of carbon must be allocated to the bands for a solid carbon lattice consisting of $M$ atoms. It turns out that the valence electrons are all allocated to the lower valence band, and no electrons are allocated to the upper band, which remains empty. The empty band is separated from the valence band by a forbidden gap.

The width of the band gap is 6eV in the case of diamond. It is not possible for the valence electrons to be transported, as there are no empty states in the valence band for the electrons to move into.

In the case of a metal, some of the energy bands overlap, without any gap formation. There are thus many unoccupied energy states. Metals are conductors, as these unoccupied energy states offer a pathway for transport of the electrons.

Silicon is a semi-conductor, as its energy gap is only 1 eV. At room temperature, some electrons acquire sufficient energy to become excited across the forbidden gap, into the empty band above. These electrons then become conduction electrons, and this empty band is known as the valence band. The distinction between insulators, semiconductors and metals discussed above is illustrated in figure 11

Figure 11: The distinction between insulators, semiconductors and metals.
(a) insulator (wide band gap, filled valence band, empty conduction band).
(b) semiconductor (narrow band gap, filled valence band, empty conduction band).
(c) metal (narrow band gap, filled valence band, partially filled conduction band).
(d) metal (overlapping bands, accessible states in conduction band).
The shading represents filled bands, or in the case (c), also partially filled bands.
\includegraphics[width=0.8\textwidth]{band-all.eps}

Electron velocity
The velocity of an electron in a band is the group velocity of the wave packet.

\begin{displaymath}
v_g = \frac{d\omega}{dk} = \frac{1}{\hbar} \frac{dE}{dk}
\end{displaymath} (43)

Exercise 6.3
Verify this. Indicate on an energy-band diagram the maximum and and minimum electron velocities.

Electron effective mass
A free electron has a rest mass of $m_0=511$keV. An electron in a material interacts with its surroundings. The electron responds to a given force with an acceleration dependent on its inertial mass as indicated in Newton's 2nd Law of motion :-

\begin{displaymath}
F = m^*\frac{dv}{dt}
\end{displaymath} (44)

The electrons inertia will have to take into account its interaction with the material, as any motion of the electron will have an effect on the material. Equation 44 recognises this fact by considering that the electron's effective mass $m^*$ in the medium will be different to its free mass at rest $m_0$. Accordingly we evaluate the electron's effective mass $m^*$ by separately writing down the expressions for the force on the electron, and its acceleration, and then solving for the electron's effective mass $m^*$ using equation 44.
\begin{displaymath}
F = \frac{dp}{dt} = \frac{\hbar k}{dt}
\end{displaymath} (45)

and
\begin{displaymath}
a = \frac{dv}{dt} = \frac{dv}{dk} \frac{dk}{dt}
\end{displaymath} (46)

using equation 43
\begin{displaymath}
a = \frac{d}{dk} \left( \frac{1}{\hbar} \frac{dE}{dk}\right) \frac{dk}{dt}
\end{displaymath} (47)

and so finally, substitution for $a$ and $F$ in equation 44
\begin{displaymath}
m^* = \frac{\hbar^2}{ \left(\frac{d^2E}{dk^2}\right)}.
\end{displaymath} (48)

Exercise 6.4
Verify this.

The quantity $\frac{d^2E}{dk^2}$ is the curvature of the band. We see the effective mass is inversely proportional to the band curvature.

Exercise 6.5
Indicate where the effective mass of the electron is greatest and least on the band diagram. Note that it becomes infinite at points of inflection, and that it can also be negative.

The frequency response of semi-conductors is understood in terms of the effective mass of the electron. Optimisation of the frequency response requires designing semiconductor materials with the appropriate band structure.

Equations 43 and 48 give the dependence of the electron wave function group velocity, $v_g$, and the electron effective mass, $m^*$, on the band energy $E(k)$. This is also depicted graphically in Figure 12

Figure 12: The dependence of the electron wave function group velocity $v_g$ and the electron effective mass $m^*$ on the band energy $E(k)$.
\includegraphics[width=0.45\textwidth]{band_m-star.eps}

Holes
In the case of a completely filled band, $\Sigma k =0$. This is because there is an energy state labelled by $-k$ for every energy state labelled by $k$. If an electron is excited from the valence band to the conduction band, then $\Sigma k$ is no longer zero. Because an electron of momentum $k$ is now missing from the summation, we have $\Sigma k =-k$ for the remaining electrons in the valence band. (See figure 13). The hole that is left behind in the valence band therefore has the following properties.

  1. The hole has a +ve charge
  2. The energy of the hole is equal and opposite to that of the missing electron
  3. The effective mass of the hole is equal and opposite to that of the missing electron. (The hole mass will be the curvature of the band near the top of the band.)
  4. The momentum of the hole is $-k$

Figure 13: The formation of a hole in the valence band.
\includegraphics[width=0.4\textwidth]{bloch-band4.eps}

Figure 14: Properties of various semiconductors at room temperature. The (I) and (D) denote the band gap as indirect and direct respectively.
\includegraphics[width=0.99\textwidth]{table.eps}


next up previous
Next: Fermi statistics, charge carrier Up: From Semi-conductivity to Micro-electronics Previous: Crystal lattices, periodic potentials,
Simon Connell 2004-10-04