283 lines
8.9 KiB
TeX
283 lines
8.9 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{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{gray}{rgb}{0.5,0.5,0.5}
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\definecolor{mauve}{rgb}{0.58,0,0.82}
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%\lstset{language=Matlab,%
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% %basicstyle=\color{red},
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% breaklines=true,%
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% morekeywords={matlab2tikz},
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% keywordstyle=\color{blue},%
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% morekeywords=[2]{1}, keywordstyle=[2]{\color{black}},
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% identifierstyle=\color{black},%
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% stringstyle=\color{mylilas},
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% commentstyle=\color{mygreen},%
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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 the 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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%}
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\lstset{basicstyle=\small,
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keywordstyle=\color{mauve},
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identifierstyle=\color{dkgreen},
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stringstyle=\color{gray},
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numbers=left,
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xleftmargin=5em
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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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\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_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_3.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_4.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_5.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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\caption{Visual representation of the fermi functions of the contacts and channel.}
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\end{figure}
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% TODO appendix for Part C code
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\end{document} |