589 lines
23 KiB
TeX
589 lines
23 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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\renewcommand{\thesubsection}{\indent(\alph{subsection})}
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%\lstset{language=Matlab,%
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%}
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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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\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{stuff and things}
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\end{figure}
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\subsection*{(b)}
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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 measn 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}
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\subsection*{(a)}
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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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\subsection*{(b)}
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.5\linewidth}
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\centering
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\includegraphics[width=\textwidth]{q3_ne.png}
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\caption{Fermi Functions at $0.3V$.}
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\label{fig:q3_ne}
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\end{subfigure}%
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\begin{subfigure}{0.5\linewidth}
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\centering
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\includegraphics[width=\textwidth]{q3_current.png}
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\caption{Fermi Functions at $0.3V$.}
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\label{fig:q3_current}
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\end{subfigure}
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\end{figure}
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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\linewidth}
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\centering
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\includegraphics[width=\textwidth]{q3_0V05.png}
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\caption{Fermi Functions at $0.3V$.}
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\label{fig:q3_0V05}
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\end{subfigure}%
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\begin{subfigure}{0.5\linewidth}
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\centering
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\includegraphics[width=\textwidth]{q3_0V.png}
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\caption{Fermi Functions at $0.3V$.}
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\label{fig:q3_0V}
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\end{subfigure}
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\end{figure}
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\subsection*{(d)}
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The difference in temperature causes a difference in the sharpness of the contact Fermi functions.
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This in turn leads to the behaviour in $f_2 - f_1$ seen in Figures \ref{fig:q3_0V05} and \ref{fig:q3_0V}.
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As seen in the previous questions, the current is proportional to the area under the curve of the $D(E) * [f_2 - f_1]$.
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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.
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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.
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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
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$[f_2 - f_1]$, explaining why we get a reverse current flow at that channel level.
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The maximum current occurs at $\varepsilon = \pm0.4$ eV (relative to $\mu$).
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\section*{Question 4}
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\subsection*{(a)}
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\subsubsection*{(i)}
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\begin{equation*}
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I = 0, N = 0
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\end{equation*}
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This is because the only allowed energy level in the channel is at a higher energy
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level than exists in either of the contacts, thus there are no electrons that would
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flow into the channel from either contact, thus no current {\em and} no electrons in the channel.
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\subsubsection*{(ii)}
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\begin{equation*}
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I = 608nA, N = 0.5
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\end{equation*}
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Given that we are operating with a single energy level in the channel, we can use
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the equations 9 and 10 (provided in the assignment) directly.
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\begin{equation*}
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I = \frac{q}{\hbar} \cdot \frac{0.005eV \cdot 0.005eV}{0.005eV + 0.005eV}
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\cdot [1 - 0] = 608nA
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\end{equation*}
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|
\begin{equation*}
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N = \frac{0.005eV \cdot 1 + 0.005eV \cdot 0}{0.005eV + 0.005eV} = 0.5
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\end{equation*}
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|
\subsubsection*{(iii)}
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|
\begin{equation*}
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|
I = 0A, N = 1
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\end{equation*}
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|
|
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.
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|
|
|
\begin{equation*}
|
|
I = \frac{q}{\hbar} \cdot \frac{0.005eV \cdot 0.005eV}{0.005eV + 0.005eV}
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\cdot [1 - 1] = 0A
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|
\end{equation*}
|
|
|
|
\begin{equation*}
|
|
N = \frac{0.005eV \cdot 1 + 0.005eV \cdot 1}{0.005eV + 0.005eV} = 1
|
|
\end{equation*}
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|
|
|
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|
\subsection*{(b)}
|
|
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|
\subsubsection*{(i)}
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|
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|
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.
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|
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.
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|
|
|
\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 $f_2(E + U)$ 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.
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|
This assumes $U = 0$ eV.
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|
|
|
\subsection*{(c)}
|
|
|
|
\subsubsection*{(i)}
|
|
|
|
\begin{figure}
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q4c.jpg}
|
|
\caption{}
|
|
\label{fig:q4c}
|
|
\end{figure}
|
|
|
|
\subsubsection*{(ii)}
|
|
\subsubsection*{(iii)}
|
|
|
|
|
|
\subsection*{(d)}
|
|
|
|
Using equation (5) in the assignment and plugging in the given values we obtain:
|
|
|
|
\begin{equation*}
|
|
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*0 + 0.5*0\:eV + 0.5*0.6\:eV + 0.25\:eV*(0.325 - 0)\:\:[eV]
|
|
U = 0.38125 eV$ $
|
|
\end{equation*}
|
|
|
|
Then, starting with Equation \ref{eq:N_old} and plugging in $D(E-U) = \delta(E - U - \varepsilon)$, we obtain
|
|
|
|
|
|
\subsection*{(e)}
|
|
\subsection*{(f)}
|
|
\subsection*{(g)}
|
|
|
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|
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|
|
|
|
% TODO appendix for Part C code
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|
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\end{document} |