Q4 (c) graph

PS1/doc.tex

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@@ -432,16 +432,151 @@ ylabel('Current [A]');
     \section*{Question 3}
     
         \subsection*{(a)}
+
+        Below is our code. Note that some variable names are different from those in the example code.
+
         \subsection*{(b)}
+
+        \begin{figure}[H]
+            \centering
+            \begin{subfigure}{0.5\linewidth}
+                \centering
+                \includegraphics[width=\textwidth]{q3_ne.png}
+                \caption{Fermi Functions at $0.3V$.}
+                \label{fig:q3_ne}
+            \end{subfigure}%
+            \begin{subfigure}{0.5\linewidth}
+                \centering
+                \includegraphics[width=\textwidth]{q3_current.png}
+                \caption{Fermi Functions at $0.3V$.}
+                \label{fig:q3_current}
+            \end{subfigure}
+        \end{figure}
+
         \subsection*{(c)}
+
+        \begin{figure}[H]
+            \centering
+            \begin{subfigure}{0.5\linewidth}
+                \centering
+                \includegraphics[width=\textwidth]{q3_0V05.png}
+                \caption{Fermi Functions at $0.3V$.}
+                \label{fig:q3_0V05}
+            \end{subfigure}%
+            \begin{subfigure}{0.5\linewidth}
+                \centering
+                \includegraphics[width=\textwidth]{q3_0V.png}
+                \caption{Fermi Functions at $0.3V$.}
+                \label{fig:q3_0V}
+            \end{subfigure}
+        \end{figure}
+
         \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.
+        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.
+        The maximum current occurs at $\varepsilon = \pm0.4$ eV (relative to $\mu$).
 
     \section*{Question 4}
 
         \subsection*{(a)}
+
+            \subsubsection*{(i)}
+
+                \begin{equation*}
+                    I = 0, N = 0
+                \end{equation*}
+
+                This is because the only allowed energy level in the channel is at a higher energy
+                level than exists in either of the contacts, thus there are no electrons that would
+                flow into the channel from either contact, thus no current {\em and} no electrons in the channel.
+
+            \subsubsection*{(ii)}
+
+                \begin{equation*}
+                    I = 608nA, N = 0.5
+                \end{equation*}
+
+                Given that we are operating with a single energy level in the channel, we can use
+                the equations 9 and 10 (provided in the assignment) directly.
+
+                \begin{equation*}
+                    I = \frac{q}{\hbar} \cdot \frac{0.005eV \cdot 0.005eV}{0.005eV + 0.005eV}
+                        \cdot [1 - 0] = 608nA
+                \end{equation*}
+
+                \begin{equation*}
+                    N = \frac{0.005eV \cdot 1 + 0.005eV \cdot 0}{0.005eV + 0.005eV} = 0.5
+                \end{equation*}
+
+            \subsubsection*{(iii)}
+
+                \begin{equation*}
+                    I = 0A, N = 1
+                \end{equation*}
+
+                Given that we are operating with a single energy level in the channel, we can use
+                the equations 9 and 10 (provided in the assignment) directly.
+
+                \begin{equation*}
+                    I = \frac{q}{\hbar} \cdot \frac{0.005eV \cdot 0.005eV}{0.005eV + 0.005eV}
+                        \cdot [1 - 1] = 0A 
+                \end{equation*}
+
+                \begin{equation*}
+                    N = \frac{0.005eV \cdot 1 + 0.005eV \cdot 1}{0.005eV + 0.005eV} = 1
+                \end{equation*}
+
+
         \subsection*{(b)}
+
+            \subsubsection*{(i)}
+
+            For $f_1(E + U)$ the step point occurs at $E = \mu_1 - U = 0.25$ eV. Since $U = -0.25$ eV, it follows that 
+            $\mu_1 = 0$ eV. Since $\mu_1 = \mu - qV_S$ and $\mu = 0$ eV, $V_S = 0$ V.
+            For $f_2(E + U)$ the step point occurs at $E = \mu_2 - U = -0.25$ eV. Since $U = -0.25$ eV, it follows that 
+            $\mu_2 = -0.5$ eV. Since $\mu_1 = \mu - qV_D$ and $\mu = 0$ eV, $V_D = 0.5$ V.
+            
+            \subsubsection*{(ii)}
+
+            For $f_1(E)$ the step point occurs at $E = \mu_1 = 0.25$ eV. Since $\mu_1 = \mu - qV_S$ and $\mu = 0$ eV, $V_S = -0.25$ V.
+            For $f_2(E + U)$ the step point occurs at $E = \mu_2 = -0.25$ eV. Since $\mu_1 = \mu - qV_D$ and $\mu = 0$ eV, $V_D = 0.25$ V.
+            This assumes $U = 0$ eV.
+
         \subsection*{(c)}
+        
+            \subsubsection*{(i)}
+
+            \begin{figure}
+                \centering
+                \includegraphics[width=\textwidth]{q4c.jpg}
+                \caption{}
+                \label{fig:q4c}
+            \end{figure}
+
+            \subsubsection*{(ii)}
+            \subsubsection*{(iii)}
+
+        
         \subsection*{(d)}
+
+        Using equation (5) in the assignment and plugging in the given values we obtain:
+
+        \begin{equation*}
+            U = -q[\frac{C_SV_S+C_GV_G+C_DV_D}{C_E}] + \frac{q^2(N-N_0)}{C_E}
+            U = \alpha_SV_S + \alpha_GV_G + \alpha_DV_D + U_0(N - N_0)\:\:[eV]
+            U = 0*0 + 0.5*0\:eV + 0.5*0.6\:eV + 0.25\:eV*(0.325 - 0)\:\:[eV]
+            U = 0.38125 eV$ $
+        \end{equation*}
+
+        Then, starting with Equation \ref{eq:N_old} and plugging in $D(E-U) = \delta(E - U - \varepsilon)$, we obtain
+        
+
         \subsection*{(e)}
         \subsection*{(f)}
         \subsection*{(g)}