commit

PS2/doc.tex

@@ -581,8 +581,6 @@ axis([0 1e-9 0 0.04]);
                 \item %3biv
 
                 \( sinc \) is a purely real function, so we can ignore the normalizing portion of the integral.
-
-                % TODO just replace these in the formulas themselves
                 As well, to simplify the interim equations we will assign the constants \( A = \frac{1}{2} \sqrt{\frac{L}{\pi \hbar}} \)
                 and \( B = \frac{L}{2 \pi \hbar} \).
 
@@ -590,13 +588,14 @@ axis([0 1e-9 0 0.04]);
                     \int _{-\infty} ^{\infty} \Phi (p')^2 dp = \int _{-\infty} ^{\infty} A^2 \left[ sinc\left( B(p'+p_1) \right)^2 + 2 sinc \left( B(p'+p_1) \right) sinc \left( B(p'-p_1) \right) + sinc \left( B(p'-p_1) \right) \right] dp
                 \end{equation*}
 
-                Given property (26) of the \( sinc \) function, we can simplify the left and right elements. % TODO better word than elements
+                Given property (26) of the \( sinc \) function, we can simplify the left and right terms.
-                Using a change of variable \( p'' = p_1 - p' \), and properties (27) and (28), we can further evaluate the central element.
+                Using a change of variable \( p'' = p_1 - p' \), and properties (27) and (28), we can further evaluate the central term.
 
                 \begin{align}
                     \int _{-\infty} ^{\infty} \Phi (p')^2 dp = \int _{-\infty} ^{\infty} A^2 \left[2 sinc \left( B(2p_1 - p'') \right) sinc \left( B(-p'') \right) \right] dp + A^2 \frac{2}{B} \\
                     &= \int _{-\infty} ^{\infty} A^2 \left[ 2 sinc \left(B(2p_1 - p'') \right) sinc(B p'') \right] dp + A^2 \frac{2}{B} \\
                     &= \frac{A^2}{B} \left[ sinc(2p_1) + 2 \right] \\
+                    &= \frac{1}{2} \left[ sinc(2p_1) + 2 \right]
                 \end{align}
 
                 \item %3bv
@@ -679,36 +678,47 @@ axis([0 1e-9 0 0.04]);
         \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
            
             \item %4a
+            \begin{minipage}[t]{\linewidth}
 
             \begin{figure}[H]
                 \centering
-                \begin{subfigure}{0.5\textwidth}
+                \begin{subfigure}[b]{0.5\textwidth}
                     \centering
                     \includegraphics[width=\textwidth]{q4a_e1.png}
                     \caption{}
                 \end{subfigure}%
-                \begin{subfigure}{0.5\textwidth}
+                \begin{subfigure}[b]{0.5\textwidth}
                     \centering
-                    \includegraphics[width=\textwidth]{q.png}
+                    \includegraphics[width=\textwidth]{q4a_e2.png}
                     \caption{}
                 \end{subfigure}
-                \caption{(a),(b) Previous plots but with the classical momentum marked in red.}
+                \caption{Probability Density Plots for first two energy levels.}
             \end{figure}
+            \end{minipage}
 
-            % TODO I really don't like this explanation...
             An \( a \) value of \( 0.53 \angstrom \) was chosen in order to provide an adequately
             shaped graph without sacrificing too much computation time and ensure that the first two
             numerical energies correspond to the given experimental results. The experimental results are
             \(\SI{0.14395}{\electronvolt}\) and \(\SI{0.43185}{\electronvolt} \) for the first and second energy levels 
-            respectively, and the numerical results with our chosen \(a\) are \(\SI{0.0.14386}{\electronvolt}\) and \(\SI{0.43140}{\electronvolt} \)
+            respectively, and the numerical results with our chosen \(a\) are \(\SI{0.14386}{\electronvolt}\) and \(\SI{0.43140}{\electronvolt} \),
+            which are in agreement.
+
 
 
-            % maybe talk about normalization?
 
             \item %4b
             \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
                 \item %4bi
-                \lipsum[1]
+
+                The energies were used were \( 0.14395eV \) and \( 0.43185eV \) for the first and second energy levels,
+                respectively.
+
+                \begin{minipage}[H]{\linewidth}
+                    \centering
+                    \includegraphics[width=\textwidth]{q4bi_I-V.png}
+                    \captionof{figure}{Current vs. Voltage.}
+                \end{minipage}
+
                 \item %4bii
 
                 Between \( 0V \) and \( 0.25V \), only the first energy level is carrying any current.
@@ -716,14 +726,13 @@ axis([0 1e-9 0 0.04]);
                 energy level drops to 0, meaning no electrons can transfer.
 
                 Between \(0.4V\) and \(0.65V\), only the second energy level is carrying current.
-                This energy level stops dropping current because it's shifted energy drops below
+                This energy level stops conducting current because it's shifted energy drops below
                 the threshold where the contacts have any coupling with it.
 
                 \item %4biii
                 
                 Negative Drain Resistance (NDR) is present in this design from a \(V_D\) of
-                approximately \(0.27V\) to \(0.45V\), with a mostly linear region from
+                approximately \(0.27V\) to \(0.45V\), as well as from \(0.65V\) to \(0.8V\)
-                aroung \(0.28V\) to \(0.3V\).
 
             \end{enumerate}
 

PS2/q4b_main.m

@@ -19,7 +19,7 @@ U0 = 0.025;
 kBT = 0.025;
 mu = 0;
 cal_E1 = 0.2879 / 2;
-cal_E2 = 0.2879 * 3 / 2;
+cal_E2 = 0.2879 * 3 / 2 - 0.1;
 
 % Capacitance parameters