last thing

PS1/doc.tex

@@ -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}.
+        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 $D(E) * [f_2 - f_1]$.
+        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 $[f_2 - f_1]$ overlaps the non-zero part of $D(E)$, giving a positive current.
+        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 $[f_2 - f_1]$, while half overlaps the positive part.
+        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)}
-        \subsection*{(f)}
+        
-        \subsection*{(g)}
+        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.