956 lines
37 KiB
TeX
956 lines
37 KiB
TeX
\documentclass{article}
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\usepackage{graphicx}
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\usepackage{setspace}
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\usepackage{listings}
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\usepackage{color}
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\usepackage{amsmath}
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\usepackage{float}
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\usepackage[justification=centering]{caption}
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\usepackage{subcaption}
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\usepackage[margin=0.75in]{geometry}
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\renewcommand{\thesubsection}{\indent(\alph{subsection})}
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\definecolor{mygreen}{RGB}{28,172,0} % color values Red, Green, Blue
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\definecolor{mylilas}{RGB}{170,55,241}
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\lstset{language=Matlab,%
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breaklines=true,%
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keywordstyle=\color{blue},%
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identifierstyle=\color{black},%
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showstringspaces=false,%without this there will be a symbol in the places where there is a space
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numbers=left,%
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numberstyle={\tiny \color{black}},% size of th_sume numbers
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numbersep=9pt, % this defines how far the numbers are from the text
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emph=[1]{for,end,break},emphstyle=[1]\color{red}, %some words to emphasise
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%emph=[2]{word1,word2}, emphstyle=[2]{style},
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xleftmargin=5em
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}
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%\lstset{basicstyle=\small,
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% }
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\title{ECE 456 - Problem Set 1}
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\date{2021-02-06}
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\author{David Lenfesty \\ lenfesty@ualberta.ca
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\and Phillip Kirwin \\ pkirwin@ualberta.ca}
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\setcounter{tocdepth}{2} % Show subsections
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\begin{document}
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\doublespacing
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\pagenumbering{gobble}
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\maketitle
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\newpage
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\singlespacing
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\pagenumbering{arabic}
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\section*{Question 1}
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\subsection*{(a)}
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Beginning with the following two equations:
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\begin{equation}
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\label{eq:N_old}
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N = \int_{-\infty}^{\infty}\frac{\gamma_1 f_1(E) + \gamma_2 f_2(E)}{\gamma_1 + \gamma_2} D(E-U) dE,
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\end{equation}
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\begin{equation}
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\label{eq:I_old}
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I = \frac{q}{\hbar}\frac{\gamma_1 \gamma_2}{\gamma_1 + \gamma_2} \int_{-\infty}^{\infty}[f_1(E) - f_2(E)] D(E-U) dE,
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\end{equation}
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and changing the variable of integration to $E' = E - U$:
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\begin{equation*}
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N = \int_{-\infty}^{\infty}\frac{\gamma_1 f_1(E' + U) + \gamma_2 f_2(E' + U)}{\gamma_1 + \gamma_2} D(E') dE',
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\end{equation*}
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\begin{equation*}
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I = \frac{q}{\hbar}\frac{\gamma_1 \gamma_2}{\gamma_1 + \gamma_2} \int_{-\infty}^{\infty}[f_1(E' + U) - f_2(E' + U)] D(E') dE'.
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\end{equation*}
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Replacing $E' \rightarrow E$, we obtain equations \ref{eq:N_new} and \ref{eq:I_new}.
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\begin{equation}
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\label{eq:N_new}
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N = \int_{-\infty}^{\infty}\frac{\gamma_1 f_1(E + U) + \gamma_2 f_2(E + U)}{\gamma_1 + \gamma_2} DE dE,
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\end{equation}
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\begin{equation}
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\label{eq:I_new}
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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.
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\end{equation}
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\subsection*{(b)}
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For the provided constants, the plots of the number of channel electrons and the channel current follow:
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.5\textwidth}
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% Mess around with widths later
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\centering
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\includegraphics[width=\textwidth]{q1b_electrons.png}
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\caption{Plot of channel electrons vs. drain voltage.}
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\label{fig:q1b_electrons}
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% Note that the comment after \end{subfigure} is required for side by side figures
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\end{subfigure}%
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\begin{subfigure}{0.5\textwidth}
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% Mess around with widths later
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\centering
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\includegraphics[width=\textwidth]{q1b_current.png}
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\caption{Plot of channel current vs. drain voltage.}
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\label{fig:q1b_current}
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\end{subfigure}
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\caption{Number of electrons and current versus drain voltage.}
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\end{figure}
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Below is our code. Note that some variable names are different from those in the example code.
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\begin{lstlisting}[language=Matlab]
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clear all;
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%% Constants
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% Physical constants
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hbar = 1.052e-34;
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% Single-charge coupling energy (eV)
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U_0 = 0.25;
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% (eV)
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kBT = 0.025;
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% Contact coupling coefficients (eV)
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gamma_1 = 0.005;
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gamma_2 = gamma_1;
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gamma_sum = gamma_1 + gamma_2;
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% Capacitive gate coefficient
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a_G = 0.5;
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% Capacitive drain coefficient
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a_D = 0.5;
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a_S = 1 - a_G - a_D;
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% Central energy level
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mu = 0;
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% Energy grid, from -1eV to 1eV
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NE = 501;
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E = linspace(-1, 1, NE);
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dE = E(2) - E(1);
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% TODO name this better
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cal_E = 0.2;
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% Lorentzian density of states, normalized so the integral is 1
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D = (gamma_sum / (2*pi)) ./ ( (E-cal_E).^2 + (gamma_sum/2).^2 );
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D = D ./ (dE*sum(D));
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% Reference no. of electrons in channel
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N_0 = 0;
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voltages = linspace(0, 1, 101);
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% Terminal Voltages
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V_G = 0;
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V_S = 0;
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for n = 1:length(voltages)
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% Set varying drain voltage
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V_D = voltages(n);
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% Shifted energy levels of the contacts
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mu_1 = mu - V_S;
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mu_2 = mu - V_D;
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% Laplace potential, does not change as solution is found (eV)
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% q is factored out here, we are working in eV
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U_L = - (a_G*V_G) - (a_D*V_D) - (a_S*V_S);
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% Poisson potential must change, assume 0 initially (eV)
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U_P = 0;
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% Assume large rate of change
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dU_P = 1;
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% Run until we get close enough to the answer
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while dU_P > 1e-6
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% source Fermi function
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f_1 = 1 ./ (1 + exp((E + U_L + U_P - mu_1) ./ kBT));
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% drain Fermi function
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f_2 = 1 ./ (1 + exp((E + U_L + U_P - mu_2) ./ kBT));
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% Update channel electrons against potential
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N(n) = dE * sum( ((gamma_1/gamma_sum) .* f_1 + (gamma_2/gamma_sum) .* f_2) .* D);
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% Re-update Poisson portion of potential
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tmpU_P = U_0 * ( N(n) - N_0);
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dU_P = abs(U_P - tmpU_P);
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% Unsure why U_P is updated incrementally, perhaps to avoid oscillations?
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%U_P = tmpU_P;
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%U_P = U_P + 0.1 * (tmpU_P - U_P);
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end
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% Calculate current based on solved potential.
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% Note: f1 is dependent on changes in U but has been updated prior in the loop
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I(n) = q * (q/hbar) * (gamma_1 * gamma_1 / gamma_sum) * dE * sum((f_1-f_2).*D);
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end
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%%Plotting commands
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figure(1);
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h = plot(voltages, N,'k');
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grid on;
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set(h,'linewidth',[2.0]);
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set(gca,'Fontsize',[18]);
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xlabel('Drain voltage [V]');
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ylabel('Number of electrons');
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figure(2);
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h = plot(voltages, I,'k');
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grid on;
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set(h,'linewidth',[2.0]);
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set(gca,'Fontsize',[18]);
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xlabel('Drain voltage [V]');
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ylabel('Current [A]');
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\end{lstlisting}
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\subsection*{(c)}
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Code for this section can be found in appendix A.
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\subsubsection*{(i)}
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.5\textwidth}
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% Mess around with widths later
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\centering
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\includegraphics[width=\textwidth]{q1c_1.png}
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\caption{Plot of channel electrons vs. drain voltage.}
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\label{fig:q1c_1}
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% Note that the comment after \end{subfigure} is required for side by side figures
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\end{subfigure}%
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\begin{subfigure}{0.5\textwidth}
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% Mess around with widths later
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\centering
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\includegraphics[width=\textwidth]{q1c_2.png}
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\caption{Plot of channel electrons vs. drain voltage.}
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\label{fig:q1c_2}
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% Note that the comment after \end{subfigure} is required for side by side figures
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\end{subfigure}
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\begin{subfigure}{0.5\textwidth}
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% Mess around with widths later
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\centering
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\includegraphics[width=\textwidth]{q1c_3.png}
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\caption{Plot of channel electrons vs. drain voltage.}
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\label{fig:q1c_3}
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% Note that the comment after \end{subfigure} is required for side by side figures
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\end{subfigure}%
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\begin{subfigure}{0.5\textwidth}
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% Mess around with widths later
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\centering
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\includegraphics[width=\textwidth]{q1c_4.png}
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\caption{Plot of channel electrons vs. drain voltage.}
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\label{fig:q1c_4}
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% Note that the comment after \end{subfigure} is required for side by side figures
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\end{subfigure}
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\begin{subfigure}{0.5\textwidth}
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% Mess around with widths later
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\centering
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\includegraphics[width=\textwidth]{q1c_5.png}
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\caption{Plot of channel electrons vs. drain voltage.}
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\label{fig:q1c_5}
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% Note that the comment after \end{subfigure} is required for side by side figures
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\end{subfigure}%
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\caption{Visual representation of the Fermi functions of the contacts and channel.}
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\end{figure}
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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.
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We compare the differences between values at $V_D = 1.0\:V$ and $V_D = 0.8\:V$:
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\begin{equation*}
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\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,
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\end{equation*}
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\begin{equation*}
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\frac{I|_{V_D = 1.0\:V}}{I|_{V_D = 0.8\:V}} = 1.0504,
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\end{equation*}
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%
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and between values at $V_D = 0.8\:V$ and $V_D = 0.5\:V$:
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%
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\begin{equation*}
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\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,
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\end{equation*}
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\begin{equation*}
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\frac{I|_{V_D = 0.8\:V}}{I|_{V_D = 0.5\:V}} = 2.1218.
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\end{equation*}
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%
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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).
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\begin{center}
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\begin{tabular}{l | c | c}
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$V_D$ [V] & $[f_1(E + U) - f_2(E + U)]|_{E = \varepsilon}$ & I [nA] \\
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\hline
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0.0 & 0.000 & 0.0 \\
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0.2 & 0.015 & 17.0 \\
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0.5 & 0.440 & 271 \\
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0.8 & 0.964 & 575 \\
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1.0 & 0.999 & 604 \\
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\end{tabular}
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\captionof{table}{Differences in contact Fermi functions evaluated at $E=\varepsilon$ and current $I$ at different drain voltages $V_D$.
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Values are taken from Figures \ref{fig:q1b_current} and \ref{fig:q1c_1} to \ref{fig:q1c_5}.}
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\label{table:q1ci}
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\end{center}
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\subsubsection*{(ii)}
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Figure \ref{fig:q1c_ii} has the Fermi functions marked at their "step points", or when they are equal to $0.5$.
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This can be used to find the self-consistant potential $U$, via the equation for the source Fermi function:
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\begin{equation*}
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f(E + U) = {\frac{1}{1 + e^{(E + U - \mu_1) / k_B T}}}
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\end{equation*}
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We could also use the drain Fermi function but since $V_S = 0$, it is simpler to use the source.
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%
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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.
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Since $\mu_1 = \mu - q V_S = 0$ We can simply rearrange to find $U = - E = -0.4\:eV$.
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\begin{figure}[H]
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% Mess around with widths later
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\centering
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\includegraphics[width=0.7\textwidth]{q1c_ii.png}
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\caption{Marked step points of contact Fermi functions.}
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\label{fig:q1c_ii}
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% Note that the comment after \end{subfigure} is required for side by side figures
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\end{figure}%
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\subsubsection*{(iii)}
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At $V_D = 1V$, the source is trying to {\em fill} the channel level while the drain is trying to {\em empty} it.
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This is because the source has electrons at the channel level, and is filling these in while the drain does not have
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any and is attempting to bring the channel level back down to where it is.
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\subsubsection*{(iv)}
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Referring to figure \ref{fig:q1c_ii} again, we can see the areas where the difference between the Fermi functions of each contact
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are 1. Roughly, this means that the channel current $I$ would remain the same if the channel energy level was anywhere between
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$0.3eV$ and $-0.5eV$.
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\subsubsection*{(v)}
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There is no current when $V_D = 0$ because the source does not want to fill the channel. It has no electrons
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at the channel energy level, and thus there is no impetus to fill the channel. A similar story occurs with the
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drain, in that it has no electrons at the appropriate energy level, and there are none in the channel for it to pull out.
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\section*{Question 2}
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\subsection*{(a)}
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.7\textwidth}
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\centering
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\includegraphics[width=0.7\textwidth]{q2_I-V.png}
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\caption{I-V curves.}
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\label{fig:q2_iv}
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\end{subfigure}
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\begin{subfigure}{0.33\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q2_0V.png}
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\caption{Fermi functions at $0V$.}
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\label{fig:q2_0v}
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\end{subfigure}%
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\begin{subfigure}{0.33\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q2_0V05.png}
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\caption{Fermi functions at $0.05V$.}
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\label{fig:q2_0v}
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\end{subfigure}%
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\begin{subfigure}{0.33\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q2_0V1.png}
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\caption{Fermi functions at $0.1V$.}
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\label{fig:q2_0v}
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\end{subfigure}
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\begin{subfigure}{0.33\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q2_0V2.png}
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\caption{Fermi functions at $0.2V$.}
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\label{fig:q2_0v}
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\end{subfigure}%
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\begin{subfigure}{0.33\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q2_0V3.png}
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\caption{Fermi functions at $0.3V$.}
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\label{fig:q2_0v}
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\end{subfigure}
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\caption{(a) depicts I-V curves at $V_G = 0.5$ V (higher curve) and $V_G = 0.25$ V.
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(b) through (f) show Fermi functions and $D(E)$ at various drain voltages.}
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\end{figure}
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\subsection*{(b)}
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As the drain voltage increases, we see that the overlapping area
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under the "curve" of $f_1(E + U) - f_2(E+U)$ and $D(E)$ increases, up
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until the point where the drain only has electrons below the energy
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levels available in the channel. The difference function also reaches its maximum height and it widens downward rather than upward.
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At this point, the overlapping area no longer
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increases, and therefore the current does not increase either,
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hitting {\em saturation current}.
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%% WRONG EXPLANATION - kept for posterity :)
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% We can see from these figures that the area under the $f_1(E + U) - f_2(E + U)$ curve
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% gradually moves down in energy. This means that a smaller and smaller portion of it is
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% above $E = 0eV$, which means there are fewer total energy levels through which current
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% can flow. From this we can see that with a high enough $V_D$ we will hit saturation, and
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% be unable to pass more current. \\
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\subsection*{(c)}
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At $V_D = 0.3V$, the energy levels from approximately $0$ to $0.2$ $eV$ are being
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used for electron transport. This is the range where the difference in the contact Fermi
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functions is more than 0 and the energy is more than 0. \\
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The difference between the two contacts is the greatest at $0eV$, because this is the
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point at which their Fermi functions have the greatest difference, and thus will be making
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the most "effort" to equalize the channel potential.
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\subsection*{(d)}
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Based on the supposed material changes, we would choose material A to maximize the
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drain current. The energy levels vanishing at $-0.4eV$ would mean that there is a
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greater number of energy levels that would be able to be used for conduction. There
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would be a larger {\em area} where there is a non-zero difference in the two Fermi
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functions at the contacts and there are energy levels available in the channel. \\
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This can be contrasted with material B, which would have {\em no} energy levels in
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this "conduction" zone and thus no current would be able to flow at all.
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\section*{Question 3}
|
|
|
|
\subsection*{(a)}
|
|
|
|
Below is our code. Note that some variable names are different from those in the example code.
|
|
|
|
|
|
\begin{lstlisting}[language=Matlab]
|
|
% Thermo-electric current
|
|
|
|
% Physical constants
|
|
hbar = 1.054e-34;
|
|
q = 1.602e-19;
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|
|
|
% Parameters (eV)
|
|
kBT_1 = 0.025;
|
|
kBT_2 = 0.026;
|
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mu = 0;
|
|
gamma_1 = 0.005;
|
|
gamma_2 = gamma_1;
|
|
gamma_sum = gamma_1 + gamma_2;
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|
|
|
% Channel energy levels, varying between -0.25eV and 0.25eV
|
|
epsilon = linspace(-0.25, 0.25, 101);
|
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depsilon = epsilon(2) - epsilon(1);
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|
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% Energy grid
|
|
E = linspace(-1, 1, 501);
|
|
dE = E(2) - E(1);
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|
|
|
% Contact fermi functions
|
|
f_1 = 1 ./ (1 + exp((E - mu)./kBT_1));
|
|
f_2 = 1 ./ (1 + exp((E - mu)./kBT_2));
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|
|
% Iterate through channel energy levels
|
|
for n = 1:length(epsilon)
|
|
% Compute energy level density functions - integral normalized to unity
|
|
D = (gamma_sum./(2*pi))./((E-epsilon(n)).^2+((gamma_sum./2).^2));
|
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D = D./(dE*sum(D));
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|
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% Compute number of channel electrons
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|
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N(n) = dE*sum( ((gamma_1./gamma_sum).*f_1 + (gamma_2./gamma_sum).*f_2).*D );
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|
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% Compute the current in Amps; factor of q to resolve units
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|
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I(n) = q*(q/hbar)*dE*sum((f_1 - f_2).*D.*gamma_1.*gamma_2./gamma_sum);
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|
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%plot f_1 - f_2 and D/2
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if (abs(epsilon(n) + 0.05) <= depsilon / 2) & (epsilon(n) <= 0)
|
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figure(3);
|
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h = plot(f_1-f_2, E, 'x', D/2500, E, 'k-');
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set(gca, 'Fontsize', [18]);
|
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axis([-0.01 0.02 -1 1]);
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xlabel('f1(E) - f2(E), D(E)/2500');
|
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ylabel('ENERGY [eV]');
|
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legend('f1-f2', 'D(E)/2500');
|
|
title('CHANNEL LEVEL = -0.05 eV');
|
|
elseif (abs(epsilon(n)) <= depsilon / 2) & (epsilon(n) <= 0)
|
|
figure(4);
|
|
h = plot(f_1-f_2, E, 'x', D/2500, E, 'k-');
|
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set(gca, 'Fontsize', [18]);
|
|
axis([-0.01 0.02 -1 1]);
|
|
xlabel('f1(E) - f2(E), D(E)/2500');
|
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ylabel('ENERGY [eV]');
|
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legend('f1-f2', 'D(E)/2500');
|
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title('CHANNEL LEVEL = 0 eV');
|
|
end
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|
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end
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|
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% Final plots
|
|
figure(1);
|
|
\end{lstlisting}
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|
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\subsection*{(b)}
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|
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\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}
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|
|
|
\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 $f_1 - f_2$ 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
|
|
$[f_1 - f_2]$, 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}
|
|
|
|
\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 = 608\:nA, 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] = 608\:nA
|
|
\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 = 0\:A, 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] = 0\:A
|
|
\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 $f_2(E)$ 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}[H]
|
|
\centering
|
|
\includegraphics[width=0.7\textwidth, angle=90, origin=c]{q4c.jpg}
|
|
\caption{Visualisation of energy levels.}
|
|
\label{fig:q4c}
|
|
\end{figure}
|
|
|
|
\subsubsection*{(ii)}
|
|
Starting with equation 2 in the assignment we get:
|
|
|
|
\begin{equation*}
|
|
I = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot \int_{-0.1}^{0.2} [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 0.2 - 10^4 \cdot -0.1]
|
|
\end{equation*}
|
|
\begin{equation*}
|
|
I = 1.8\:mA
|
|
\end{equation*}
|
|
|
|
\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)}
|
|
|
|
Using equation (5) in the assignment and plugging in the given values we obtain:
|
|
|
|
|
|
\begin{align*}
|
|
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 \cdot 0 - 0.5 \cdot 0\:eV - 0.5 \cdot 0.6\:eV + 0.25\:eV\cdot(0.325 - 0)\:\:eV \\
|
|
U &= -0.21875\:eV
|
|
\end{align*}
|
|
%
|
|
Then, starting with Equation \ref{eq:N_new} and plugging in $D(E-U) = \delta(E - \varepsilon)$, we obtain
|
|
|
|
\begin{equation*}
|
|
N = \frac{\gamma_1f_1(\varepsilon + U) + \gamma_2f_2(\varepsilon + U)}{\gamma_1 + \gamma_2}
|
|
\end{equation*}
|
|
Recalling that the expressions for the contact Fermi functions are (with $\mu = 0$ eV):
|
|
|
|
\begin{align*}
|
|
f_1(\varepsilon + U) = \frac{1}{1 + e^{(\varepsilon + U + qV_S)/k_BT}} = \frac{1}{1 + e^{(0.2 - 0.21875 + 0)/0.025}} = 0.679 \\
|
|
f_2(\varepsilon + U) = \frac{1}{1 + e^{()\varepsilon + U + qV_D)/k_BT}} = \frac{1}{1 + e^{(0.2 - 0.21875 + 0.6)/0.025}} \simeq 0
|
|
\end{align*}
|
|
Then, solving for $\gamma_2$:
|
|
|
|
\begin{align*}
|
|
\gamma_2 = \gamma_2\frac{N - f_1(\varepsilon + U)}{f_2(\varepsilon + U) - N} = 5.45 \times 10^{-3} \: eV
|
|
\end{align*}
|
|
|
|
|
|
\subsection*{(e)}
|
|
|
|
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) = 2\:eV
|
|
\end{equation*}
|
|
\begin{equation*}
|
|
\mu_2 = \mu_0 - V_D = 0 - 1V = -1\:eV
|
|
\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 = 12.2\:mA
|
|
\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 = 6.1\:mA
|
|
\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.
|
|
|
|
|
|
|
|
\appendix
|
|
|
|
\newpage
|
|
\section{Question 1b Code}
|
|
|
|
\begin{lstlisting}[language=Matlab]
|
|
clear all;
|
|
|
|
%% Constants
|
|
|
|
% Physical constants
|
|
hbar = 1.052e-34;
|
|
q = 1.602e-19;
|
|
%epsilon_0 = 8.854e-12;
|
|
%epsilon_r = 4;
|
|
%mstar = 0.25 * 9.11e-31;
|
|
|
|
% Single-charge coupling energy (eV)
|
|
U_0 = 0.25;
|
|
% (eV)
|
|
kBT = 0.025;
|
|
% Contact coupling coefficients (eV)
|
|
gamma_1 = 0.005;
|
|
gamma_2 = gamma_1;
|
|
gamma_sum = gamma_1 + gamma_2;
|
|
% Capacitive gate coefficient
|
|
a_G = 0.5;
|
|
% Capacitive drain coefficient
|
|
a_D = 0.5;
|
|
a_S = 1 - a_G - a_D;
|
|
|
|
% Central energy level
|
|
mu = 0;
|
|
|
|
% Energy grid, from -1eV to 1eV
|
|
NE = 501;
|
|
E = linspace(-1, 1, NE);
|
|
dE = E(2) - E(1);
|
|
% TODO name this better
|
|
cal_E = 0.2;
|
|
|
|
% Lorentzian density of states, normalized so the integral is 1
|
|
D = (gamma_sum / (2*pi)) ./ ( (E-cal_E).^2 + (gamma_sum/2).^2 );
|
|
D = D ./ (dE*sum(D));
|
|
|
|
% Reference no. of electrons in channel
|
|
N_0 = 0;
|
|
|
|
voltages = linspace(0, 1, 101);
|
|
dV = voltages(2) - voltages(1);
|
|
|
|
% Terminal Voltages
|
|
V_G = 0;
|
|
V_S = 0;
|
|
|
|
for n = 1:length(voltages)
|
|
% Set varying drain voltage
|
|
V_D = voltages(n);
|
|
|
|
% Shifted energy levels of the contacts
|
|
mu_1 = mu - V_S;
|
|
mu_2 = mu - V_D;
|
|
|
|
% Laplace potential, does not change as solution is found (eV)
|
|
% q is factored out here, we are working in eV
|
|
U_L = - (a_G*V_G) - (a_D*V_D) - (a_S*V_S);
|
|
|
|
% Poisson potential must change, assume 0 initially (eV)
|
|
U_P = 0;
|
|
|
|
% Assume large rate of change
|
|
dU_P = 1;
|
|
|
|
% Run until we get close enough to the answer
|
|
while dU_P > 1e-6
|
|
% source Fermi function
|
|
f_1 = 1 ./ (1 + exp((E + U_L + U_P - mu_1) ./ kBT));
|
|
% drain Fermi function
|
|
f_2 = 1 ./ (1 + exp((E + U_L + U_P - mu_2) ./ kBT));
|
|
|
|
% Update channel electrons against potential
|
|
N(n) = dE * sum( ((gamma_1/gamma_sum) .* f_1 + (gamma_2/gamma_sum) .* f_2) .* D);
|
|
|
|
% Re-update Poisson portion of potential
|
|
tmpU_P = U_0 * ( N(n) - N_0);
|
|
dU_P = abs(U_P - tmpU_P);
|
|
|
|
% Unsure why U_P is updated incrementally, perhaps to avoid oscillations?
|
|
%U_P = tmpU_P;
|
|
U_P = U_P + 0.1 * (tmpU_P - U_P);
|
|
end
|
|
|
|
% Calculate current based on solved potential.
|
|
% Note: f1 is dependent on changes in U but has been updated prior in the loop
|
|
I(n) = q * (q/hbar) * (gamma_1 * gamma_1 / gamma_sum) * dE * sum((f_1-f_2).*D);
|
|
|
|
if (abs(V_D-0.0) <= dV/2)
|
|
figure(3); title('VD = 0.0 V');
|
|
subplot(2,3,1); plot(f_1,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f1(E+U)'); ylabel('ENERGY [eV]'); title('VD = 0.0 V');
|
|
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
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xlabel('D(E)/100'); ylabel('ENERGY [eV]'); title('VD = 0.0 V');
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subplot(2,3,3); plot(f_2,E,'k-'); axis([-0.1 1.1 -1 1]);
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xlabel('f2(E+U)'); ylabel('ENERGY [eV]'); title('VD = 0.0 V');
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subplot(2,3,5); plot(f_1-f_2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
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xlabel('f1(E+U)-f2(E+U), D(E)/100'); ylabel('ENERGY [eV]'); title('VD = 0.0 V');
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elseif (abs(V_D-0.2) <= dV/2)
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figure(4); title('VD = 0.2 V');
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subplot(2,3,1); plot(f_1,E,'k-'); axis([-0.1 1.1 -1 1]);
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xlabel('f1(E+U)'); ylabel('ENERGY [eV]'); title('VD = 0.2 V');
|
|
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('D(E)/100'); ylabel('ENERGY [eV]'); title('VD = 0.2 V');
|
|
subplot(2,3,3); plot(f_2,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f2(E+U)'); ylabel('ENERGY [eV]'); title('VD = 0.2 V');
|
|
subplot(2,3,5); plot(f_1-f_2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f1(E+U)-f2(E+U), D(E)/100'); ylabel('ENERGY [eV]'); title('VD = 0.2 V');
|
|
elseif (abs(V_D-0.5) <= dV/2)
|
|
figure(5); title('VD = 0.5 V');
|
|
subplot(2,3,1); plot(f_1,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f1(E+U)'); ylabel('ENERGY [eV]'); title('VD = 0.5 V');
|
|
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('D(E)/100'); ylabel('ENERGY [eV]'); title('VD = 0.5 V');
|
|
subplot(2,3,3); plot(f_2,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f2(E+U)'); ylabel('ENERGY [eV]'); title('VD = 0.5 V');
|
|
subplot(2,3,5); plot(f_1-f_2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f1(E+U)-f2(E+U), D(E)/100'); ylabel('ENERGY [eV]'); title('VD = 0.5 V');
|
|
elseif (abs(V_D-0.8) <= dV/2)
|
|
figure(6); title('VD = 0.8 V');
|
|
subplot(2,3,1); plot(f_1,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f1(E+U)'); ylabel('ENERGY [eV]'); title('VD = 0.8 V');
|
|
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('D(E)/100'); ylabel('ENERGY [eV]'); title('VD = 0.8 V');
|
|
subplot(2,3,3); plot(f_2,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f2(E+U)'); ylabel('ENERGY [eV]'); title('VD = 0.8 V');
|
|
subplot(2,3,5); plot(f_1-f_2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f1(E+U)-f2(E+U), D(E)/100'); ylabel('ENERGY [eV]'); title('VD = 0.8 V');
|
|
elseif (abs(V_D-1.0) <= dV/2)
|
|
figure(7); title('VD = 1.0 V');
|
|
subplot(2,3,1); plot(f_1,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f1(E+U)'); ylabel('ENERGY [eV]'); title('VD = 1.0 V');
|
|
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('D(E)/100'); ylabel('ENERGY [eV]'); title('VD = 1.0 V');
|
|
subplot(2,3,3); plot(f_2,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f2(E+U)'); ylabel('ENERGY [eV]'); title('VD = 1.0 V');
|
|
subplot(2,3,5); plot(f_1-f_2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
|
|
xlabel('f1(E+U)-f2(E+U), D(E)/100'); ylabel('ENERGY [eV]'); title('VD = 1.0 V');
|
|
end
|
|
|
|
end
|
|
|
|
|
|
%%Plotting commands
|
|
|
|
figure(1);
|
|
h = plot(voltages, N,'k');
|
|
grid on;
|
|
set(h,'linewidth',[2.0]);
|
|
set(gca,'Fontsize',[18]);
|
|
xlabel('Drain voltage [V]');
|
|
ylabel('Number of electrons');
|
|
|
|
figure(2);
|
|
h = plot(voltages, I,'k');
|
|
grid on;
|
|
set(h,'linewidth',[2.0]);
|
|
set(gca,'Fontsize',[18]);
|
|
xlabel('Drain voltage [V]');
|
|
ylabel('Current [A]');
|
|
\end{lstlisting}
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\end{document} |