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Copyright, 1995-2017, all rights reserved, Mark Jarrell (Dept.of
Physics and Atronomy, Louisiana State University, LA 70803). This
material may not be reproduced for profit, modified or published in any
form (this includes electronic redistribution) without the prior written
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}
\title{Chapter 12: Semiconductors }
\author{Bardeen \& \ Shottky}
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\tableofcontents
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\Large
Semiconductors are of obvious technological importance - so much so, that a
whole chapter will be dedicated to them.
Semiconductors are distinguished from metals in that they have a gap at
the Fermi surface, and are distinguished from insulators in that the gap is
small $\alt \ 1eV$.
\begin{figure}[htb]
\centerline{\includegraphics[height=2.0in,keepaspectratio,clip=true]{gap.pdf}}
\caption[]{\em{There is no band gap at the Fermi energy in a metal, while there is a band
gap in an insulator. Semiconductors on the other hand have a
band gap, but it is much smaller than those found in
insulators.}\label{fig:gap}}
\end{figure}
Most condensed matter physicists make the distinction on the basis of the
conductivity and its temperature dependence. In the Drude model (parabolic
band)
\beq
\sigma \ = \frac{ne^2\tau }{m^*}, \qquad \mu \ = \frac{e\tau }{m^*},
\qquad \sigma \ = ne\tau
\eeq
Almost always $\frac 1{\tau}$ increases as $T$ increases, ie the thermal
excitations increase the scattering rate and decrease the lifetime of the quasiparticle.
For example, we have seen that $\frac 1{\tau } \sim \ T$ at high temperatures
due to electron-phonon interactions. In metals, $n$ is about constant, so
the temperature dependence of metals is dictated by $\tau $.
\beq
\mbox{metals } \qquad \sigma \ \sim \ \tau \sim \frac{1}{T}, \qquad \sigma \downarrow \
\mbox{as } T \uparrow
\eeq
However, in semiconductors, the population of free carriers $n$ is temperature
dependent.
\begin{figure}[htb]
\centerline{\includegraphics[height=1.5in,keepaspectratio,clip=true]{semicond.pdf}}
\caption[]{\em{This shows the temperature dependence for the excitation
of electrons, thus allowing the number of free carriers to vary
with changes in temperature.}\label{fig:semicond}}
\end{figure}
The exponential always will dominate the power law dependence of $\tau $.
\beq
\sigma \ \sim \ \tau n \sim \ \frac 1Te^{-E_g/kT}
\eeq
\beq
\sigma \ \uparrow \quad \mbox{as} \quad T \uparrow \nonumber
\eeq
The same is true for insulators, of course, except here $n$ is so small that
for all realistic purposes $\sigma \ \sim \ 0$.
\section{ Band Structure }
Clearly the band structure of the semiconductors is crucial then for their
device applications. Semiconductors fall into several categories, depending
upon their composition, the simplest, {\it type IV} include silicon and
germanium. The type refers to their valence.
\begin{figure}[htb]
\centerline{\includegraphics[height=2.25in,keepaspectratio,clip=true]{split.pdf}}
\caption[]{\em{Sketch of the sp$^3$ bands in Si vs.\ Si-Si separation.}\label{fig:split}}
\end{figure}
Recall that Si and Ge have a $s^2p^2$ atomic shell, which forms highly
directional $sp^3$ hybrid bonds in the solid state (with tetragonal symmetry).
It is the covalent bonding, or rather the splitting between the bonding and
antibonding bands, that forms the gap. The band structure is also quite rich
\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{structure.pdf}}
\caption[]{\em{Sketch of quasiparticle bands in Si (right) along the
high symmetry directions (left).}\label{fig:structure}}
\end{figure}
\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{structure2.pdf}}
\caption[]{\em{Sketch of quasiparticle bands in Ge along the high symmetry
directions. Note the indirect, roughly $\Gamma\to L$, minimum gap energy.}\label{fig:structure2}}
\end{figure}
\beq
\frac 1{m_{ij}^*} = \frac 1{\hbar ^2} \frac{\partial ^2E(\k )}{\partial
k_i\partial k_j}
\eeq
%\beq
% E(\k ) = \hbar ^2\left( \frac{k_x^2 + k_y^2}{2m_t^*} +
% \frac{k_z^2}{2m_l^*}\right)
%\eeq
%\begin{table}[htb]
% \centering
% \begin{tabular}{|ccc|} \hline
% Z: & $\Gamma \ \rightarrow \ $ hole minimum & Element
% \\ \hline
% Z: & 111 & Ge \\
% Z: & 100 & Si \\ \hline
% \end{tabular}
% \caption[]{{\em The $\k $-vectors which give the minimum gap
% between the conduction and valence bands of silicon and
% germanium.}}
%\end{table}
%\begin{table}[htb]
% \centering
% \begin{tabular}{|l|cr|} \hline
% & $\frac{m_t^*}m $ & $\frac{m_l^*}m $ \\ \hline
% Si & 0.19 & 0.98 \\
% Ge & 0.082 & 1.57 \\ \hline
% \end{tabular}
% \caption{{\em The ratios between the effective masses ($m_t^*$
% and $m_l^*$) and the free electron mass in the
% corresponding elements.}}
\begin{figure}[htb]
\centerline{\includegraphics[height=2.75in,keepaspectratio,clip=true]{structure3.pdf}}
\caption[]{\em{Sketch of quasiparticle bands in GaAs along the
high symmetry directions of the Brillouin zone. Note the direct,
$\Gamma\to \Gamma$, minimum gap energy. The nature of the gap can
be tuned with Al doping.}\label{fig:structure3}}
\end{figure}
The situation in III-V semiconductors such as GaAs is similar, in that
covalent $sp^3$ bands still form. However, the gap is direct.
For this reason GaAs makes more efficient optical devices than does either
Si or Ge. A particle-hole excitation across the gap can readily recombine,
emit a photon (which has essentially no momentum) and conserve momentum in
GaAs; whereas, in an indirect gap semiconductor, this recombination requires
the addition creation or absorption of a phonon or some other lattice
excitation to conserve momentum.
\begin{figure}[htb]
\centerline{\includegraphics[height=1.25in,keepaspectratio,clip=true]{optical.pdf}}
\caption[]{\em{A particle-hole excitation across the gap can readily
recombine, emit a photon (which has essentially no momentum) and conserve
momentum in a direct gap semiconductor (left) such as GaAs. Whereas, in
an indirect gap semiconductor (right), this recombination requires
the addition creation or absorption of a phonon or some other lattice
excitation to conserve momentum.}\label{fig:optical}}
\end{figure}
For the same reason, excitons live much longer in Si and especially Ge
than they do in GaAs.
\begin{table}[htb]
\centering
\begin{tabular}{|c|c|} \hline
material & $\tau _{\mbox{exciton}}$ \\ \hline
GaAs & $1 ns (10^{-9}s)$ \\
Si & $19 \mu s (10^{-5}s)$ \\
Ge & $1 ms (10^{-3}s)$ \\ \hline
\end{tabular}
\caption[]{}
\end{table}
\section{ Charge Carrier Density in Intrinsic Semiconductors.}
Both electrons and holes contribute to the conductivity with the same sign.
\begin{figure}[htb]
\centerline{\includegraphics[height=1.0in,keepaspectratio,clip=true]{contribute.pdf}}
\caption[]{{\em The contribution of the electrons and holes to the
conductivity.}\label{fig:contribute}}
\end{figure}
Here the mobilities are assumed to be constant. This is valid since for
semiconductors all of the conducting carriers full near the top or bottom of
bands, where $E_k \sim \frac{\hbar ^2\k ^2}{2m_p^*}$ and the effective mass
approximation is valid. Here, we found that $\mu\sim e\tau/m*$
\begin{figure}[htb]
\centerline{\includegraphics[height=1.75in,keepaspectratio,clip=true]{energy.pdf}}
\caption[]{}
\end{figure}
However, as mentioned before, the carrier concentrations are highly
$T$-dependent since all of the carriers in an intrinsic (undoped) semiconductor
are thermally induced (i.e. $n = p = 0$ at $T = 0$).
\beq
n = \int _{E_c}^{E_{top}} D_C(E)f(E,T) dE \rightarrow \ \int
_{E_c}^{\infty} D_C(E)f(E,T) dE
\eeq
\beq
p = \int _{E_{bottom}}^{E_{v}} D_V(E)\left\{ 1 - f(E,T)\right\} dE
\rightarrow \ \int _{-\infty}^{E_{v}} D_V(E)\left\{ 1 -
f(E,T)\right\} dE
\eeq
To proceed further we need forms for $D_C$ and $D_V$. Recall that in the
parabolic approximation $E_k \simeq \ \frac{\hbar ^2 \k ^2}{2m^*}$ we found
that $D(E) = \frac{\left( 2m^*\right) ^{\frac 32}}{2\pi ^2\hbar ^3}\sqrt{E}$.
Thus,
\beq
D_C(E) = \frac{\left( 2m_n^*\right) ^{\frac 32}}{2\pi ^2\hbar
^3}\sqrt{E - E_C}
\eeq
\beq
D_V(E) = \frac{\left( 2m_p^*\right) ^{\frac 32}}{2\pi ^2\hbar
^3}\sqrt{E_V - E}
\eeq
for $E > E_C$ and $E < E_V$ respectively, and zero otherwise $E_V < E < E_C$.
In an intrinsic (undoped) semiconductor $n = p$, and so $E_F$ must lie in the band
gap. However, if $m_n^* \neq \ m_p^*$ (ie. $D_C \neq \ D_V$), then the chemical
potential, $E_F$, must be adjusted up or down from the center of the gap
so that $n = p$.
Furthermore, the carriers which are induced across the gap are relatively
high in energy, compared to $k_BT$, since typically
$E_g = E_C - E_V \gg \ k_BT$.
\begin{table}[htb]
\centering
\begin{tabular}{|ccc|} \hline
& $E_g (eV)$ & $n_i (cm^{-3})(300 {}^{\circ}K)$ \\ \hline
Ge & $0.67$ & $2.4 \times \ 10^{13}$ \\
Si & $1.1$ & $1.5 \times \ 10^{10}$ \\
GaAs & $1.43$ & $5 \times \ 10^7$ \\ \hline
\end{tabular}
\caption[]{}
\end{table}
\beq
\frac{1 eV}{k_B} \sim \ 10000{}^{\circ}K \gg \ 300{}^{\circ}K \sim \ T
\eeq
Thus, assuming that $E - E_F \agt \ \frac{E_g}2 \gg \ k_BT$
\beq
\frac 1{e^{(E - E_F)/k_BT} + 1} \simeq \ \frac 1{e^{(E - E_F)/k_BT}} =
e^{-(E - E_F)/k_BT}
\eeq
ie., Boltzmann statistics. A similar relationship holds for holes where $-(E -
E_F) \agt \ \frac{E_g}2 \gg \ k_BT$
\beq
1 - \frac 1{e^{(E - E_F)/k_BT} + 1} \simeq \ 1 - \left\{ 1 - e^{-(E -
E_F)/k_BT}\right\} = e^{-(E - E_F)/k_BT}
\eeq
since $(1 - f(E)) = f(-E)$ and $e^{(E - E_F)/k_BT}$ is small. Thus, the
concentration of electrons $n$
\beqa
n & \simeq \ & \frac{\left( 2m_n^*\right) ^{\frac 32}}{2\pi ^2\hbar ^3}
e^{E_F/k_BT} \int _{E_C}^{\infty} \sqrt{E - E_C}e^{-E/k_BT} dE
\\
& = & \frac{\left( 2m_n^*\right) ^{\frac 32}}{2\pi ^2\hbar ^3}\left(
k_BT\right) ^{\frac 32} e^{-\beta (E_C - E_F)} \int _0^{\infty}
x^{\frac 12}e^{-x} dx \\
& = & 2\left( \frac{2\pi m_n^*k_BT}{h^2}\right) ^{\frac 32}e^{-\beta
(E_C - E_F)} = N_{eff}^C e^{-\beta (E_C - E_F)}
\eeqa
Similarly
\beq
p = 2\left( \frac{2\pi m_p^*k_BT}{h^2}\right) ^{\frac 32}e^{-\beta
(E_V - E_F)} = N_{eff}^V e^{-\beta (E_V - E_F)}
\eeq
where $N_{eff}^C$ and $N_{eff}^V$ are the partition functions for a classical
gas in 3-d and can be regarded as "effective densities of states" which are
temperature-dependent. Within this interpretation, we can regard the holes and
electrons statistics as classical. This holds so long as $n$ and $p$ are
small, so that the Pauli principle may be ignored - the so called {\it
nondegenerate limit}.
In general, in the nondegenerate limit,
\beq
np = 4\left( \frac{k_BT}{2\pi \hbar ^2}\right) ^3 \left(
m_n^*m_p^*\right) ^{\frac 32} e^{-\beta \E_g}
\eeq
this, the {\it law of mass action}, holds for both doped and intrinsic semiconductor so
long as we remain in the nondegenerate limit. However, {\it for an intrinsic
semiconductor}, where $n = p$, it gives us further information.
\beq
n_i = p_i = 2\left( \frac{k_BT}{2\pi \hbar ^2}\right) ^{\frac 32}
\left( m_n^*m_p^*\right) ^{\frac 34} e^{-\beta \E_g/2}
\eeq
%(Complete table on pg 10!)!
However, we already have relationships for $n$ and $p$ involving $E_C$ and $E_V$
\beq
n = p = N_{eff}^C e^{-\beta (E_C - E_F)} = N_{eff}^V e^{\beta (E_V -
E_F)}
\eeq
\beq
e^{2\beta E_F} = \frac{N_{eff}^V}{N_{eff}^C} e^{\beta (E_V + E_C)}
\eeq
or
\beq
E_F = \frac 12(E_V + E_C) + \frac 12k_BT \ln \left(
\frac{N_{eff}^V}{N_{eff}^C}\right)
\eeq
\beq
E_F = \frac 12(E_V + E_C) + \frac 34k_BT \ln \left(
\frac{m_p^*}{m_n^*}\right)
\eeq
Thus if $m_p^* \neq m_n^*$, the chemical potential $E_F$ in a semiconductor is
temperature dependent. Recall that this $T$-dependence was important for the
transport of a semiconductor in the presence of a thermal gradient $\grad T$.
\section{ Doping of Semiconductors}
$\sigma \ = ne\mu $, so the conductivity depends linearly upon the doping (it
may also effect $\mu $ in some materials, leading to a non-linear doping
dependence). A typical metal has
\beq
n_{metal} \sim \ 10^{23}/(cm)^3
\eeq
whereas we have seen that a typical semiconductor has
\beq
n_{i_{SeC}} \sim \frac{10^{10}}{cm ^3} \qquad \mbox{at} T \simeq
300{}^{\circ}K
\eeq
Thus the conductivity of an intrinsic semiconductor is quite small!
To increase $n$ (or $p$) to $\sim \ 10^{18}$ or more, dopants are used.
For example, in Si the elements used as dopants have either a $s^2p^1$ or
$s^2p^3$ atomic valence. Thus, in the tetrahedral bonding of Si there is
either an extra electron (half bond) or an unsatisfied bond or a hole.
\begin{figure}[htb]
\centerline{\includegraphics[height=1.75in,keepaspectratio,clip=true]{lattice.pdf}}
\caption[]{}
\end{figure}
Thus P or B will either donate or absorb additional electron (with the latter
called the creation of a hole). As in an exciton, these additional charges
will be localized around the donor or acceptor ion. The difference is
that here the donor/acceptor is fixed and may be treated as having infinite
mass, thus the binding energy is given by
\beq
E = \frac{m^*e^4}{2\ep ^2\hbar ^2n^2}
\eeq
\beq
m^* = \left\{ \begin{array}{ll}
\mbox{hole mass acceptor} (B) \\
\mbox{electron mass donor} (P)
\end{array} \right.
\eeq
Again, since $\frac{m^*}m < 1$ and $\ep \ \sim \ 10$ these energies are often
much less than $13.6 eV$ c.f. in Si $E \sim \ 30 MeV \sim \ 300{}^{\circ}K$ or
in Ge $E \sim \ GMeV \sim \ 60{}^{\circ}K$. Thus thermal excitations will
often ionize these dopant sites. In terms of energy levels
\begin{figure}[htb]
\centerline{\includegraphics[height=1.75in,keepaspectratio,clip=true]{levels.pdf}}
\caption[]{}
\end{figure}
\section{ Carrier Densities in Doped semiconductor }
The law of mass action is valid so long as the use of Boltzmann statistics
is valid i.e., if the degeneracy is small.
Thus, even for doped semiconductor
\beq
np = N_{eff}^CN_{eff}^Ve^{-\beta E_g} = n_i^2 = p_i^2
\eeq
Now imagine the temperature is finite so that some of the donors or acceptors
are ionized.
\begin{figure}[htb]
\centerline{\includegraphics[height=1.75in,keepaspectratio,clip=true]{ionized.pdf}}
\caption[]{}
\end{figure}
Furthermore, in equilibrium, the semiconductor is charge neutral so that
\beq
n + N_A^- = p + N_D^+
\eeq
The probability that a donor/acceptor is occupied by an electron is determined
by Fermi statistics
\beqa
n_D & = N_D^0 = & N_D \frac 1{1 + e^{\beta (E_D - E_F)}} \\
p_A & = N_A^0 = & N_A (1 - f(E_A)) = N_A \frac 1{1 + e^{\beta (E_F -
E_A)}}
\eeqa
To provide a solvable example, imagine that we have an $n$-type semiconductor
(no $p$-type dopants) so that $N_A = N_A^0 = N_A^+ = 0$, then
\beqa
n & = & N_{eff}^C e^{-\beta (E_C - E_F)} \\
N_D & = & N_D^0 + N_D^+ \\
N_D^0 & = & N_D \frac 1{e^{\beta (E_D - E_F)} + 1}
\eeqa
Furthermore, charge neutrality requires that
\beq
n = p + N_D^+
\eeq
An excellent approximation is to assume that for a (commercially) doped semiconductor
\beq
N_D^+ \gg \ n_i
\eeq
ie., many more carriers are provided by dopants than are thermally excited over
the entire gap, then as $np = n_i^2$, it must be that $N_D^+ \gg \ p$ so
that
\beqa
n & \approx & N_D^+ = N_D - N_D^0 \\
n & \approx & N_D \left( 1 - \frac 1{e^{\beta (E_D - E_F)} + 1}\right)
\eeqa
If we recall that thermally induced carriers satisfy the Boltzmann equation,
\beq
n = N_{eff}^C e^{\beta (E_F - E_C)}
\eeq
we can eliminate $E_F$ in $n$ (where $E_d=E_c-E_D$)
\beq
n = \frac{N_D}{1 + e^{\beta E_d}n/(N_{eff}^C)}
\eeq
This quadratic equation has only one meaningful solution
\beq
n = \frac{2N_D}{1 + \sqrt{1 + 4\left( N_D/N_{eff}^C\right) e^{\beta
E_d}}}
\eeq
At low $T \ll \ \frac{E_d}{k_B}$
\beq
n \simeq \ \sqrt{N_DN_{eff}^C}e^{-\beta E_d}
\eeq
at higher $T \gg \ \frac{E_d}{k_B}$
\beq
n = N_D
\eeq
At still higher $T$ our approximation breaks down that $N_D^+ \gg \ n$ since
thermally excited carriers will dominate.
\edo