From e1041ef4f4ab42593985018cb0b31e9cb67bd882 Mon Sep 17 00:00:00 2001 From: David Lenfesty Date: Sat, 6 Feb 2021 22:20:53 -0700 Subject: [PATCH] last thing --- PS1/doc.tex | 105 ++++++++++++++++++++++++++++++++++++++++++++++------ 1 file changed, 94 insertions(+), 11 deletions(-) diff --git a/PS1/doc.tex b/PS1/doc.tex index 52e4a50..3a1a634 100644 --- a/PS1/doc.tex +++ b/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}. - 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)} - \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.