saving state

PS1/doc.tex

@@ -6,7 +6,7 @@
 \usepackage{color}
 \usepackage{amsmath}
 \usepackage{float}
-\usepackage{caption}
+\usepackage[justification=centering]{caption}
 \usepackage{subcaption}
 \usepackage[margin=0.75in]{geometry}
 
@@ -89,7 +89,7 @@
 
         \begin{equation}
             \label{eq:I_new}
-            I = \frac{q}{\hbar}\frac{\gamma_1 \gamma_2}{\gamma_1 + \gamma_2} \int_{-\infty}^{\infty}[f_1(E + U) - f_2(E + U)] DE dE,
+            I = \frac{q}{\hbar}\frac{\gamma_1 \gamma_2}{\gamma_1 + \gamma_2} \int_{-\infty}^{\infty}[f_1(E + U) - f_2(E + U)] DE dE.
         \end{equation}
 
         \subsection*{(b)}
@@ -230,6 +230,7 @@ ylabel('Current [A]');
 
         \subsection*{(c)}
 
+        \subsubsection*{(i)}
         \begin{figure}[H]
             \centering
             \begin{subfigure}{0.5\textwidth}
@@ -245,7 +246,7 @@ ylabel('Current [A]');
                 \centering
                 \includegraphics[width=\textwidth]{q1c_2.png}
                 \caption{Plot of channel electrons vs. drain voltage.}
-                \label{fig:q1c_1}
+                \label{fig:q1c_2}
                 % Note that the comment after \end{subfigure} is required for side by side figures
             \end{subfigure}
             \begin{subfigure}{0.5\textwidth}
@@ -253,7 +254,7 @@ ylabel('Current [A]');
                 \centering
                 \includegraphics[width=\textwidth]{q1c_3.png}
                 \caption{Plot of channel electrons vs. drain voltage.}
-                \label{fig:q1c_1}
+                \label{fig:q1c_3}
                 % Note that the comment after \end{subfigure} is required for side by side figures
             \end{subfigure}%
             \begin{subfigure}{0.5\textwidth}
@@ -261,7 +262,7 @@ ylabel('Current [A]');
                 \centering
                 \includegraphics[width=\textwidth]{q1c_4.png}
                 \caption{Plot of channel electrons vs. drain voltage.}
-                \label{fig:q1c_1}
+                \label{fig:q1c_4}
                 % Note that the comment after \end{subfigure} is required for side by side figures
             \end{subfigure}
             \begin{subfigure}{0.5\textwidth}
@@ -269,13 +270,183 @@ ylabel('Current [A]');
                 \centering
                 \includegraphics[width=\textwidth]{q1c_5.png}
                 \caption{Plot of channel electrons vs. drain voltage.}
-                \label{fig:q1c_1}
+                \label{fig:q1c_5}
                 % Note that the comment after \end{subfigure} is required for side by side figures
             \end{subfigure}%
 
-            \caption{Visual representation of the fermi functions of the contacts and channel.}
+            \caption{Visual representation of the Fermi functions of the contacts and channel.}
         \end{figure}
         
+        Table \ref{table:q1ci} shows the variation in the difference between $f_1$ and $f_2$ at $E = \varepsilon$, and $I$ at different drain voltages. 
+        We compare the differences between values at $V_D = 1.0\:V$ and $V_D = 0.8\:V$:
+
+        \begin{equation*}
+            \frac{([f_1(E + U) - f_2(E + U)]|_{E = \varepsilon})|_{V_D = 1.0\:V}}{([f_1(E + U) - f_2(E + U)]|_{E = \varepsilon})|_{V_D = 0.8\:V}} = 1.0363,
+        \end{equation*}
+        \begin{equation*}
+        \frac{I|_{V_D = 1.0\:V}}{I|_{V_D = 0.8\:V}} = 1.0504,
+        \end{equation*}
+        %
+        and between values at $V_D = 0.8\:V$ and $V_D = 0.5\:V$:
+        %
+        \begin{equation*}
+            \frac{([f_1(E + U) - f_2(E + U)]|_{E = \varepsilon})|_{V_D = 0.8\:V}}{([f_1(E + U) - f_2(E + U)]|_{E = \varepsilon})|_{V_D = 0.5\:V}} = 2.1909,
+        \end{equation*}
+        \begin{equation*}
+        \frac{I|_{V_D = 0.8\:V}}{I|_{V_D = 0.5\:V}} = 2.1218.
+        \end{equation*}
+        %
+        In both comparisons we see that $I$ changes in proportion to $[f_1(E + U) - f_2(E + U)]|_{E = \varepsilon}$, as predicted by Equation (9).
+        
+        \begin{center}
+        \begin{tabular}{l | c | c}
+                
+                $V_D$ [V]       & $[f_1(E + U) - f_2(E + U)]|_{E = \varepsilon}$    & I [nA]          \\
+                \hline
+                0.0         & 0.000                                                   & 0.0           \\
+                0.2         & 0.015                                                   & 17.0          \\
+                0.5         & 0.440                                                   & 271           \\
+                0.8         & 0.964                                                   & 575           \\
+                1.0         & 0.999                                                   & 604           \\
+            
+            \end{tabular}
+            \captionof{table}{Differences in contact Fermi functions evaluated at $E=\varepsilon$ and current $I$ at different drain voltages $V_D$. 
+            Values are taken from Figures \ref{fig:q1b_current} and \ref{fig:q1c_1} to \ref{fig:q1c_5}.}
+            \label{table:q1ci}
+        \end{center}
+
+
+        \subsubsection*{(ii)}
+            Figure \ref{fig:q1c_ii} has the Fermi functions marked at their "step points", or when they are equal to $0.5$.
+            This can be used to find the self-consistant potential $U$, via the equation for the source Fermi function:
+
+            \begin{equation*}
+                f(E + U) = {\frac{1}{1 + e^{(E + U - \mu_1) / k_B T}}}
+            \end{equation*}
+            We could also use the drain Fermi function but since $V_S = 0$, it is simpler to use the source.
+            %
+            The function will "step" when $E = \mu_1 - U$, which from Figure \ref{fig:q1c_ii} occurs at $E = \mu_1 - U = 0.4$ for the source contact.
+            Since $\mu_1 = \mu - q V_S = 0$ We can simply rearrange to find $U = - E = -0.4\:eV$.
+
+            \begin{figure}[H]
+                % Mess around with widths later
+                \centering
+                \includegraphics[width=0.7\textwidth]{q1c_ii.png}
+                \caption{Marked step points of contact Fermi functions.}
+                \label{fig:q1c_ii}
+                % Note that the comment after \end{subfigure} is required for side by side figures
+            \end{figure}%
+
+        \subsubsection*{(iii)}
+
+            At $V_D = 1V$, the source is trying to {\em fill} the channel level while the drain is trying to {\em empty} it.
+            This is because the source has electrons at the channel level, and is filling these in while the drain does not have
+            any and is attempting to bring the channel level back down to where it is.
+
+        \subsubsection*{(iv)}
+
+            Referring to figure \ref{fig:q1c_ii} again, we can see the areas where the difference between the Fermi functions of each contact
+            are 1. Roughly, this means that the channel current $I$ would remain the same if the channel energy level was anywhere between
+            $0.3eV$ and $-0.5eV$.
+
+        \subsubsection*{(v)}
+
+            There is no current when $V_D = 0$ because the source does not want to fill the channel. It has no electrons
+            at the channel energy level, and thus there is no impetus to fill the channel. A similar story occurs with the
+            drain, in that it has no electrons at the appropriate energy level, and there are none in the channel for it to pull out.
+
+    \section*{Question 2}
+
+        \subsection*{(a)}
+
+        \begin{figure}[H]
+            \centering
+            \begin{subfigure}{0.7\textwidth}
+                \centering
+                \includegraphics[width=0.7\textwidth]{q2_I-V.png}
+                \caption{I-V curves.}
+                \label{fig:q2_iv}
+            \end{subfigure}
+            \begin{subfigure}{0.33\textwidth}
+                \centering
+                \includegraphics[width=\textwidth]{q2_0V.png}
+                \caption{Fermi Functions at $0V$.}
+                \label{fig:q2_0v}
+            \end{subfigure}%
+            \begin{subfigure}{0.33\textwidth}
+                \centering
+                \includegraphics[width=\textwidth]{q2_0V05.png}
+                \caption{Fermi Functions at $0.05V$.}
+                \label{fig:q2_0v}
+            \end{subfigure}%
+            \begin{subfigure}{0.33\textwidth}
+                \centering
+                \includegraphics[width=\textwidth]{q2_0V1.png}
+                \caption{Fermi Functions at $0.1V$.}
+                \label{fig:q2_0v}
+            \end{subfigure}
+            \begin{subfigure}{0.33\textwidth}
+                \centering
+                \includegraphics[width=\textwidth]{q2_0V2.png}
+                \caption{Fermi Functions at $0.2V$.}
+                \label{fig:q2_0v}
+            \end{subfigure}%
+            \begin{subfigure}{0.33\textwidth}
+                \centering
+                \includegraphics[width=\textwidth]{q2_0V3.png}
+                \caption{Fermi Functions at $0.3V$.}
+                \label{fig:q2_0v}
+            \end{subfigure}
+            \caption{stuff and things}
+        \end{figure}
+
+        \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
+        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.
+
+        \subsection*{(c)}
+
+        At $V_D = 0.3V$, the energy levels from approximately $0$ to $0.2$ $eV$ are being
+        used for electron transport. This is the range where the difference in the contact Fermi
+        functions is more than 0 and the energy is more than 0.
+
+        The difference between the two contacts is the greatest at $0eV$, because this is the
+        point at which their Fermi functions have the greatest difference, and thus will be making
+        the most "effort" to equalize the channel potential.
+
+        \subsection*{(d)}
+
+        Based on the supposed material changes, we would choose material A to maximize the
+        drain current. The energy levels vanishing at $-0.4eV$ would mean that there is a
+        greater number of energy levels that would be able to be used for conduction. There
+        would be a larger {\em area} where there is a non-zero difference in the two Fermi
+        functions at the contacts and there are energy levels available in the channel.
+
+        This can be contrasted with material B, which would have {\em no} energy levels in
+        this "conduction" zone and thus no current would be able to flow at all.
+
+    \section*{Question 3}
+    
+        \subsection*{(a)}
+        \subsection*{(b)}
+        \subsection*{(c)}
+        \subsection*{(d)}
+
+    \section*{Question 4}
+
+        \subsection*{(a)}
+        \subsection*{(b)}
+        \subsection*{(c)}
+        \subsection*{(d)}
+        \subsection*{(e)}
+        \subsection*{(f)}
+        \subsection*{(g)}
+
+    
 
 
 % TODO appendix for Part C code

PS1/q1b.m

@@ -4,6 +4,7 @@ clear all;
 
 % Physical constants
 hbar = 1.052e-34;
+q = 1.602e-19;
 
 % Single-charge coupling energy (eV)
 U_0 = 0.25;

PS1/q3.m

@@ -0,0 +1,32 @@
+% Thermo-electric current
+
+% Physical constants
+hbar = 1.054e-34;
+q = 1.602e-19;
+
+% Parameters (eV)
+kBT_1 = 0.025;
+kBT_2 = 0.026;
+mu = 0;
+gamma_1 = 0.005;
+gamma_2 = gamma_1;
+gamma_sum = gamma_1 + gamma_2;
+
+% Channel energy levels, varying between -0.25eV and 0.25eV
+epsilon = linspace(-0.25, 0.25, 101);
+
+% Energy grid
+E = linspace(-1, 1, 501);
+
+% Contact fermi functions
+f_1 = 1 ./ (1 + exp((E - mu)./kBT1));
+f_2 = 1 ./ (1 + exp((E - mu)./kBT2));
+
+% Iterate through channel energy levels
+for n = 1:length(epsilon)
+    % Compute energy level density functions - integral normalized to unity
+    D = (gamma./(2*pi))./((E-cal_E(count)).ˆ2+((gamma./2).ˆ2));
+    D = D./(dE*sum(D));
+
+
+end