diff --git a/PS1/doc.tex b/PS1/doc.tex index 2552432..b3bd03f 100644 --- a/PS1/doc.tex +++ b/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 diff --git a/PS1/q1b.m b/PS1/q1b.m index 68f8a3d..037ee6e 100644 --- a/PS1/q1b.m +++ b/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; diff --git a/PS1/q2_0V05.png b/PS1/q2_0V05.png index d63f8b7..8058bdc 100644 Binary files a/PS1/q2_0V05.png and b/PS1/q2_0V05.png differ diff --git a/PS1/q2_0V1.png b/PS1/q2_0V1.png index 15ac014..8b37d7e 100644 Binary files a/PS1/q2_0V1.png and b/PS1/q2_0V1.png differ diff --git a/PS1/q2_0V2.png b/PS1/q2_0V2.png index c302440..962fb26 100644 Binary files a/PS1/q2_0V2.png and b/PS1/q2_0V2.png differ diff --git a/PS1/q2_0V3.png b/PS1/q2_0V3.png index 6b5045e..cffd8ab 100644 Binary files a/PS1/q2_0V3.png and b/PS1/q2_0V3.png differ diff --git a/PS1/q3.m b/PS1/q3.m new file mode 100644 index 0000000..c084830 --- /dev/null +++ b/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