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@ -403,11 +403,14 @@ ylabel('Current [A]');
\subsection*{(b)}
We can see from these figures that the area under the $f_1(E + U) - f_2(E + U)$ curve
gradually moves down in energy, this measn that a smaller and smaller portion of it is
gradually moves down in energy, this means that a smaller and smaller portion of it is
above $E = 0eV$, which means there are fewer total energy levels through which current
can flow. From this we can see that with a high enough $V_D$ we will hit saturation, and
be unable to pass more current.
% TODO re-word this
The self-consistent potential will decrease, which is what causes this shift in levels.
\subsection*{(c)}
At $V_D = 0.3V$, the energy levels from approximately $0$ to $0.2$ $eV$ are being
@ -474,12 +477,12 @@ ylabel('Current [A]');
\subsection*{(d)}
The difference in temperature causes a difference in the sharpness of the contact Fermi functions.
This in turn leads to the behaviour in $f_2 - f_1$ seen in Figures \ref{fig:q3_0V05} and \ref{fig:q3_0V}.
As seen in the previous questions, the current is proportional to the area under the curve of the $D(E) * [f_2 - f_1]$.
When the channel level is $\varepsilon = -0.05$ eV, the positive region of $[f_2 - f_1]$ overlaps the non-zero part of $D(E)$, giving a positive current.
Alternatively, when $\varepsilon = 0$ eV, half of the area under $D(E)$ overlaps with the negative part of $[f_2 - f_1]$, while half overlaps the positive part.
This in turn leads to the behaviour in $f_1 - f_2$ seen in Figures \ref{fig:q3_0V05} and \ref{fig:q3_0V}.
As seen in the previous questions, the current is proportional to the area under the curve of the $D(E) * [f_1 - f_2]$.
When the channel level is $\varepsilon = -0.05$ eV, the positive region of $[f_1 - f_2]$ overlaps the non-zero part of $D(E)$, giving a positive current.
Alternatively, when $\varepsilon = 0$ eV, half of the area under $D(E)$ overlaps with the negative part of $[f_1 - f_2]$, while half overlaps the positive part.
Thus the areas cancel out completely, and the resultant current is 0. A plot of $\varepsilon = +0.05$ eV would show overlap of $D(E)$ with the negative part of
$[f_2 - f_1]$, explaining why we get a reverse current flow at that channel level.
$[f_1 - f_2]$, explaining why we get a reverse current flow at that channel level.
The maximum current occurs at $\varepsilon = \pm0.4$ eV (relative to $\mu$).
\section*{Question 4}
@ -572,11 +575,22 @@ ylabel('Current [A]');
I = 1.8mA
\end{equation*}
(Note that 1)
\subsubsection*{(iii)}
The graph for the fermi functions at the saturation potentials will look much the same,
except that they will be shifted up. This means that there is more area "under" the
difference between the fermi functions and thus the bounds of integration can shift and
become $[-0.1, 0.3]$.
\begin{equation*}
I = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot \int_{-0.1}^{0.3} [1 - 0] \cdot 10^4 dE \\
\end{equation*}
\begin{equation*}
I = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot [10^4 \cdot 0.3 - 10^4 \cdot -0.1]
\end{equation*}
\begin{equation*}
I = 2.4mA
\end{equation*}
\subsection*{(d)}
@ -614,9 +628,78 @@ ylabel('Current [A]');
\subsection*{(e)}
Assuming the density of states for each molecule can be modeled by $D(E) = \delta(E-\varepsilon)$,
Equation (12) in the assignment is valid here. Thus the current will be maximized when $[f_1(E) - f_2(E)]|_{E=\varepsilon}$ is maximized.
Solving $[f_1(E) - f_2(E)]|_{E=\varepsilon}$ for each molecule, we obtain:
Molecule A:
\begin{equation*}
[f_{1}(E) - f_{2}(E)]|_{E=\varepsilon_A} =
[\frac{1}{1 + e^{frac{\varepsilon_A}{k_BT_1}}} - [\frac{1}{1 + e^{frac{\varepsilon_A}{k_BT_2}}}]
= [\frac{1}{1 + e^{frac{0}{0.024}}} - [\frac{1}{1 + e^{frac{0}{0.027}}}] = 0
\end{equation*}
Molecule B:
\begin{equation*}
[f_{1}(E) - f_{2}(E)]|_{E=\varepsilon_B} =
[\frac{1}{1 + e^{frac{\varepsilon_B}{k_BT_1}}} - [\frac{1}{1 + e^{frac{\varepsilon_B}{k_BT_2}}}]
= [\frac{1}{1 + e^{frac{-0.05}{0.024}}} - [\frac{1}{1 + e^{frac{-0.05}{0.027}}}] = 0.02493
\end{equation*}
Molecule C:
\begin{equation*}
[f_{1}(E) - f_{2}(E)]|_{E=\varepsilon_C} =
[\frac{1}{1 + e^{frac{\varepsilon_C}{k_BT_1}}} - [\frac{1}{1 + e^{frac{\varepsilon_C}{k_BT_2}}}]
= [\frac{1}{1 + e^{frac{-0.1}{0.024}}} - [\frac{1}{1 + e^{frac{-0.1}{0.027}}}] = 0.00877
\end{equation*}
Molecule B should be chosen.
\subsection*{(f)}
Referring again to equation 2 in the assignment, we know that $D(E)$ is valid for all $E$, however
since $\gamma_1$ is dependant on the energy level, we can use it to reduce our limits. For material
A, this means we only care about $E$ above $0eV$, and for B, $E$ below $0eV$.
Since we know the voltages on the contacts, we can calculate the effective fermi level at each.
\begin{equation*}
\mu_1 = \mu_0 - V_S = 0 - (-2V) = 2eV
\end{equation*}
\begin{equation*}
\mu_2 = \mu_0 - V_D = 0 - 1V = -1eV
\end{equation*}
With the fermi levels of each contact, we can adjust the limits of integration further for each
material. Material A can be evaluated on $[0eV, 2eV]$, and material B can be evaluated on $[-1eV, 0eV]$.
\begin{equation*}
I_A = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot \int_{0}^{2} [1 - 0] \cdot 10^4 dE
\end{equation*}
\begin{equation*}
I_A = 12.2mA
\end{equation*}
\begin{equation*}
I_B = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot \int_{0}^{2} [1 - 0] \cdot 10^4 dE
\end{equation*}
\begin{equation*}
I_A = 6.1mA
\end{equation*}
Material A should be chosen.
\subsection*{(g)}
Using Ohm's law we find the corresponding conductance:
\begin{equation*}
\sigma{max} = \frac{I_{max}}{V} = \frac{500\:nA}{4.3\:mV} = 116\:\mu S.
\end{equation*}
We expect this result to be an integer multiple of $G_0 = 38.76$ $\mu S$, the quantum of conductance. We find
\begin{equation*}
\frac{\sigma{max}}{G_0} = 3.
\end{equation*}
We conclude there are 3 levels.