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%TCIDATA{Created=Sat Feb 20 21:25:39 1999}
%TCIDATA{LastRevised=Sat Feb 20 22:00:18 1999}

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\SetTitle{Physics 312 -- Assignment 6} \SetAuthor{} \Setdate{}
\TitlePage{}

\vspace{0.5cm}

\textbf{1}. Explain why a transient current flows when you touch a
piece of $n$-type semiconductor to a piece of $p$-type
semiconductor. What is the direction of current flow? What stops
the current after a while? Similar questions are in
\textit{Bloomfield}, p. 439.

\vspace{0.5cm}

If we bring an $n$-type semiconductor (for instance,
$As$\thinspace -\thinspace doped $Ge$) into contact with a
$p$-type semiconductor (for instance, $Ga\,$-\thinspace doped
$Ge$), we have electrons in the conduction band on the $n$-type
side and holes in the valence band on the $p$-type side. We can
lower the energy by moving some of the former into the latter.
Since the electrons are negatively charged, the current will be from the $p$%
-type side to the $n$-type side. There is also a concentration
gradient, i.e. more electrons on one side, more holes on the
other, which leads to a diffusion that from a region of high
concentration to one of lower concentration. This effect pushes
electrons in the same direction as the energy considerations. In
fact, the two effects are related, because the
carrier concentrations depend on the concentration of  impurities, such as $%
As$ and $Ga,$and their ``donor'' and ``acceptor'' levels (see
below).  The real quantity driving the migration of electrons is
the ``local chemical potential'', which depends both on the
''impurity levels'' available and on the carrier concentrations.

Before this electron migration started both sides were
electrically neutral, but now negative charges are building up on
the $p$ side and positive on the $n$ side. It costs energy to
separate charge. So when the gain in energy from moving electrons
from conduction to valence bands balances the loss in energy from
separating the charges, the current stops. (The current stops when
the chemical potential is the same on both sides.)

This process creates a ``depletion region'' or ``depletion zone,''
and it is important in the operation of diodes.\smallskip

\textit{About n-type and p-type materials\smallskip }

A somewhat naive picture is as follows. Start with silicon or germanium ($Ge$%
) atoms which have four valence electrons. Take the $Ge$ ions (the
nuclei plus the bound electrons) and construct a lattice. Then
find the states associated with this structure and fill them up
(obeying the Pauli exclusion principle and starting with the
lowest energy). One finds that the energies of these states fall
into ``bands'' separated by ``gaps.'' Furthermore, the last filled
state completes a band (the valence band) and the next available
state is $E_{g}$ higher in energy.

Now let us consider substituting a small percentage of $As$ (or $P$, or $Sb$%
) atoms for some of the $Ge$. Since $As$ has five valence
electrons, we have some extra electrons which could go in the
higher (conduction) band. This isn't quite right because the $As$
ions are different from the $Ge$ ions (they have more charge);
however, it is true that the $Ge$ doped with $As$ has very similar
energy levels as pure $Ge$ and that there are additional states,
localized near the $As$ impurities, that are close in energy to
the conduction band of pure $Ge$. These are called ''donor''
states. At ordinary temperatures, thermal energy is sufficient to
promote electrons from the donor states to the conduction band. We
end up with extra electrons in the conduction band (''carriers''),
although their concentration is not the same as that of the $As$
impurities, in general. This is called an $n$-type semiconductor.

If we substituted $Ga$ atoms instead (or another atom with three
valence electrons, such as $B,$ $Al$, or $In$), the energy levels
would again be quite similar to those in pure $Ge$, but this time
there would be ``acceptor'' states associated with the $Ga$
impurities, as well extra unfilled levels (holes) in the valence
band at finite temperature. This is called a $p$-type
semiconductor.

\vspace{0.5cm} \hrule
\vspace{0.5cm}

\textbf{2}. \textit{Melissinos}, exercise 1.1.

(a) Look up the atomic mass number $A$, and density $\rho$ of $Si$
and $Ge$ and find the number of atoms per $\mathrm{cm}^3$.

(b) Assuming that the atoms are in a diamond structure ($8$
atoms/unit cell) find the lattice spacing.

(c) Find the resistivity of $Ge$ at room temperature if it is doped with $%
10^{15}~\mathrm{atoms/cm^3}$ of $Sb$. Assume a mobility of the donor's
electrons of $\mu_e=1200~\mathrm{Cm^2/ V\cdot s}$.

\vspace{0.5cm}

(a) From the CRC Handbook we find the mass numbers $A$ and
densities $\rho $ for $Si$ and $Ge$
\begin{eqnarray}
A_{Si} &=&28.09\ \ \ \ \ \ \ \ \rho _{Si}=2.33~\mathrm{g/cm^{3}}  \nonumber
\\
A_{Ge} &=&72.59\ \ \ \ \ \ \ \ \rho _{Ge}=5.32~\mathrm{g/cm^{3}}
\end{eqnarray}
Noting that there are $N_{A}$ atoms in a mole where $N_{A}$ is Avogadro's
number, we can calculate the number density from the mass density and the
atomic mass, as follows
\begin{eqnarray}
Si:\ \ \left( \frac{2.33~\mathrm{g}}{\mathrm{cm^{3}}}\right) \left( \frac{%
6.022\times 10^{23}~\mathrm{atoms}}{28.09~\mathrm{g}}\right)  &=&5.00\times
10^{22}~\mathrm{atoms/cm^{3}};  \nonumber \\
Ge:\ \ \left( \frac{5.32~\mathrm{g}}{\mathrm{cm^{3}}}\right) \left( \frac{%
6.022\times 10^{23}~\mathrm{atoms}}{72.59~\mathrm{g}}\right)  &=&4.41\times
10^{22}~\mathrm{atoms/cm^{3}}.
\end{eqnarray}

(b) Next, from the number density we can calculate the volume of a unit cell
which contains $8$ atoms in this instance
\begin{eqnarray}
V_{cell-Si} &=&8~\mathrm{atoms}\left( \frac{\mathrm{cm^{3}}}{5.00\times
10^{22}~\mathrm{atoms}}\right) =1.60\times 10^{-22}\mathrm{cm^{3}};
\nonumber \\
V_{cell-Ge} &=&8~\mathrm{atoms}\left( \frac{\mathrm{cm^{3}}}{4.41\times
10^{22}~\mathrm{atoms}}\right) =1.81\times 10^{-22}\mathrm{cm^{3}}.
\end{eqnarray}
Then from the volume of the unit cell we can calculate the lattice spacing
\begin{eqnarray}
a_{Si} &=&\left( 1.60\times 10^{-22}~\mathrm{cm^{3}}\right)
^{1/3}=5.43\times 10^{-8}~\mathrm{cm}=5.43~\mathrm{angstroms};  \nonumber \\
a_{Ge} &=&\left( 1.81\times 10^{-22}~\mathrm{cm^{3}}\right)
^{1/3}=5.66\times 10^{-8}~\mathrm{cm}=5.66~\mathrm{angstroms}.
\end{eqnarray}

(c) First of all we can see by the position of Antimony ($Sb$) on the
periodic table that it is a ``donor'' and that we are talking about an ``$n$%
-type'' (doped) semiconductor. The intrinsic carrier density $n_i$ of $Ge$
calculated at the end of Section 1.1 in \textit{Melissinos} is $10^{13}/~%
\mathrm{cm^3}$; it is much less than donor density $N_D$ given in the
problem as $10^{15}/~\mathrm{cm^3}$. Therefore, the carrier density of
negative charges $n$ is essentially equal to $N_D$. The carrier density of
positive charges is approximately $n_i^2/N_D$ (see \textit{Melissinos}, p.
11) which is very much smaller, and so we will neglect any contribution from
positive carriers hereafter.

When only one carrier contributes the conductivity $\sigma$ is given by
\begin{equation}
\sigma= q n \mu ,
\end{equation}
where $q$ is the charge, $n$ the carrier density, and $\mu$ the mobility
(see \textit{Melissinos}, p. 13). Thus
\begin{equation}
\sigma= \left(1.60 \times 10^{-19}~\mathrm{C} \right) \left(10^{15}~/\mathrm{%
cm^3} \right) \left( 1200~\mathrm{cm^2/V \cdot s} \right) = \frac{0.192~%
\mathrm{C}} {\mathrm{V \cdot s\cdot cm}}.
\end{equation}
Converting the $\mathrm{cm}$ to $\mathrm{m}$ and taking the reciprocal gives
the resistivity $\rho = 5.21 \times 10^{-2}~\Omega \cdot \mathrm{m}$.

\vspace{0.5cm} \hrule
\vspace{0.5cm}

\textbf{3}. \textit{Melissinos}, exercise 1.3.

Consider germanium doped with $10^{14}/\mathrm{cm^3}$ atoms of arsenic.

(a) Find the conductivity assuming a reasonable value of the mobility of the
impurities.

(b) The energy gap of germanium is $E_g=0.67~\mathrm{eV}$ and the density of
states at the edge of the conduction band can be taken as $N_c=10^{19}/%
\mathrm{cm^3}$. Estimate the intrinsic carrier density for germanium at room
temperature.

(c) Use the result of (b) to find the density of holes in the
doped sample.

\textit{(Note that part (a) is closely related to part (c) of exercise 1.1.)}

\vspace{0.5cm}

(b) I think it's more sensible to answer part (b) first. The
intrinsic carrier density $n_i$ is related to the density of
states at the edge of the conduction band $N_C$ through
\begin{equation}
n_i = N_C \exp \left( -\frac{E_g}{2kT}\right).
\end{equation}
(See \textit{Melissinos}, p. 8.) So
\begin{equation}
n_i = 10^{19} /\mathrm{cm^3} \times \exp \left( -\frac{0.67~\mathrm{eV}}{2 (%
\mathrm{eV}/40)}\right) \approx 1.5 \times 10^{13} /\mathrm{cm^3}.
\end{equation}

(a) Arsenic ($As$) is a donor. The density of donors $N_D=10^{14}/ \mathrm{%
cm^3}$ is many times larger than the intrinsic carrier density, so the
carrier density we need in the calculation of the conductivity ($\sigma=q n
\mu $) is approximately the donor density ($n=N_D$). Next we must assume a
reasonable value for the mobility of donor electrons in $Ge$. Let us take as
an order-of-magnitude approximation $10^{3}~\mathrm{cm^2/V \cdot s}$. Then
\begin{equation}
\sigma= \left(1.60 \times 10^{-19}~\mathrm{C} \right) \left(10^{14}~/\mathrm{%
cm^3} \right) \left( 10^{3}~\mathrm{cm^2/V \cdot s} \right) = \frac{0.016~%
\mathrm{C}} {\mathrm{V \cdot s\cdot cm}}= 1.6/\Omega \cdot \mathrm{m}.
\end{equation}

Using a relationship between the hole and electron densities and the
intrinsic carrier density ($np=n_i^2$) and the approximation $n\approx N_D$
led to eq. (1.12) in \textit{Melissinos}, which yields in this particular
case
\begin{equation}
p_n \approx \frac{n_i^2}{N_D} \approx 2.3 \times 10^{12}/\mathrm{cm^3}.
\end{equation}

\vspace{0.5cm} \hrule
\vspace{0.5cm}

\textbf{4}. \textit{Melissinos} exercise 1.4.

Make a plot of the Fermi-Dirac distribution at $T=-78^{\circ}\mathrm{C}$,
room temperature, and a $T=500^{\circ}\mathrm{C}$ when $E_F=1~\mathrm{eV}$.

\textit{(Accurate plots are expected. First plot for $0 < E < 2~\mathrm{eV}$
and then for $0.8~\mathrm{eV}<E<1.2~\mathrm{eV}$.)}

\vspace{0.5cm}

The Fermi-Dirac distribution is given by
\begin{equation}
f(E)=\frac{1}{\mathrm{exp}[(E-E_{F})/kT]+1},
\end{equation}
where $E_{F}$ is the Fermi energy. Let us express $kT$ for the various
temperatures in $eV$'s:

\begin{tabular}{cc}
$T$ (in $\mathrm{C}$) & $kT$ in $\mathrm{eV}$'s \\
$-78^{\circ }~\mathrm{C}$ & $0.017~\mathrm{eV}$ \\
$17^{\circ }~\mathrm{C}$ & $0.025~\mathrm{eV}$ \\
$500^{\circ }~\mathrm{C}$ & $0.067~\mathrm{eV}$%
\end{tabular}

Now let us plot the Fermi distribution for the range $0<E<2$

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and then in a smaller region surrounding the Fermi energy ($0.8<E<1.2$)

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This function gives the average number of fermions filling a level
with energy $E$. Because we are talking about fermions, $f(E)$
varies between
zero and one. At $T=0$, $f(E)$ is a step function --- equal to one for $%
E<E_{F\text{ }}$ and equal to zero for $E>E_{F}$.

\end{document}
