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\title{ECE 456 - Problem Set 1}
\date{2021-02-06}
\author{David Lenfesty \\ lenfesty@ualberta.ca
    \and Phillip Kirwin \\ pkirwin@ualberta.ca}

\setcounter{tocdepth}{2} % Show subsections

\begin{document}

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    \section*{Question 1}
        \subsection*{(a)}
        Beginning with the following two equations:
        \begin{equation}
            \label{eq:N_old}
            N = \int_{-\infty}^{\infty}\frac{\gamma_1 f_1(E) + \gamma_2 f_2(E)}{\gamma_1 + \gamma_2} D(E-U) dE,
        \end{equation}

        \begin{equation}
            \label{eq:I_old}
            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,
        \end{equation}
        
        and changing the variable of integration to $E' = E - U$:
        
        \begin{equation*}
            N = \int_{-\infty}^{\infty}\frac{\gamma_1 f_1(E' + U) + \gamma_2 f_2(E' + U)}{\gamma_1 + \gamma_2} D(E') dE',
        \end{equation*}

        \begin{equation*}
            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'.
        \end{equation*}
        
        Replacing $E' \rightarrow E$, we obtain equations \ref{eq:N_new} and \ref{eq:I_new}.

        \begin{equation}
            \label{eq:N_new}
            N = \int_{-\infty}^{\infty}\frac{\gamma_1 f_1(E + U) + \gamma_2 f_2(E + U)}{\gamma_1 + \gamma_2} DE dE,
        \end{equation}

        \begin{equation}
            \label{eq:I_new}
            I = \frac{q}{\hbar}\frac{\gamma_1 \gamma_2}{\gamma_1 + \gamma_2} \int_{-\infty}^{\infty}[f_1(E + U) - f_2(E + U)] DE dE.
        \end{equation}

        \subsection*{(b)}

        For the provided constants, the plots of the number of channel electrons and the channel current follow:


        \begin{figure}[H]
            \centering
            \begin{subfigure}{0.5\textwidth}
                % Mess around with widths later
                \centering
                \includegraphics[width=\textwidth]{q1b_electrons.png}
                \caption{Plot of channel electrons vs. drain voltage.}
                \label{fig:q1b_electrons}
                % Note that the comment after \end{subfigure} is required for side by side figures
            \end{subfigure}%
            \begin{subfigure}{0.5\textwidth}
                % Mess around with widths later
                \centering
                \includegraphics[width=\textwidth]{q1b_current.png}
                \caption{Plot of channel current vs. drain voltage.}
                \label{fig:q1b_current}
            \end{subfigure}

            \caption{Number of electrons and current versus drain voltage.}
        \end{figure}

            Below is our code. Note that some variable names are different from those in the example code.
            \begin{lstlisting}[language=Matlab]
            
clear all;

%% Constants

% Physical constants
hbar = 1.052e-34;

% 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);

% 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);

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}

        \subsection*{(c)}

        \subsubsection*{(i)}
        \begin{figure}[H]
            \centering
            \begin{subfigure}{0.5\textwidth}
                % Mess around with widths later
                \centering
                \includegraphics[width=\textwidth]{q1c_1.png}
                \caption{Plot of channel electrons vs. drain voltage.}
                \label{fig:q1c_1}
                % Note that the comment after \end{subfigure} is required for side by side figures
            \end{subfigure}%
            \begin{subfigure}{0.5\textwidth}
                % Mess around with widths later
                \centering
                \includegraphics[width=\textwidth]{q1c_2.png}
                \caption{Plot of channel electrons vs. drain voltage.}
                \label{fig:q1c_2}
                % Note that the comment after \end{subfigure} is required for side by side figures
            \end{subfigure}
            \begin{subfigure}{0.5\textwidth}
                % Mess around with widths later
                \centering
                \includegraphics[width=\textwidth]{q1c_3.png}
                \caption{Plot of channel electrons vs. drain voltage.}
                \label{fig:q1c_3}
                % Note that the comment after \end{subfigure} is required for side by side figures
            \end{subfigure}%
            \begin{subfigure}{0.5\textwidth}
                % Mess around with widths later
                \centering
                \includegraphics[width=\textwidth]{q1c_4.png}
                \caption{Plot of channel electrons vs. drain voltage.}
                \label{fig:q1c_4}
                % Note that the comment after \end{subfigure} is required for side by side figures
            \end{subfigure}
            \begin{subfigure}{0.5\textwidth}
                % Mess around with widths later
                \centering
                \includegraphics[width=\textwidth]{q1c_5.png}
                \caption{Plot of channel electrons vs. drain voltage.}
                \label{fig:q1c_5}
                % Note that the comment after \end{subfigure} is required for side by side figures
            \end{subfigure}%

            \caption{Visual representation of the Fermi functions of the contacts and channel.}
        \end{figure}
        
        Table \ref{table:q1ci} shows the variation in the difference between $f_1$ and $f_2$ at $E = \varepsilon$, and $I$ at different drain voltages. 
        We compare the differences between values at $V_D = 1.0\:V$ and $V_D = 0.8\:V$:

        \begin{equation*}
            \frac{([f_1(E + U) - f_2(E + U)]|_{E = \varepsilon})|_{V_D = 1.0\:V}}{([f_1(E + U) - f_2(E + U)]|_{E = \varepsilon})|_{V_D = 0.8\:V}} = 1.0363,
        \end{equation*}
        \begin{equation*}
        \frac{I|_{V_D = 1.0\:V}}{I|_{V_D = 0.8\:V}} = 1.0504,
        \end{equation*}
        %
        and between values at $V_D = 0.8\:V$ and $V_D = 0.5\:V$:
        %
        \begin{equation*}
            \frac{([f_1(E + U) - f_2(E + U)]|_{E = \varepsilon})|_{V_D = 0.8\:V}}{([f_1(E + U) - f_2(E + U)]|_{E = \varepsilon})|_{V_D = 0.5\:V}} = 2.1909,
        \end{equation*}
        \begin{equation*}
        \frac{I|_{V_D = 0.8\:V}}{I|_{V_D = 0.5\:V}} = 2.1218.
        \end{equation*}
        %
        In both comparisons we see that $I$ changes in proportion to $[f_1(E + U) - f_2(E + U)]|_{E = \varepsilon}$, as predicted by Equation (9).
        
        \begin{center}
        \begin{tabular}{l | c | c}
                
                $V_D$ [V]       & $[f_1(E + U) - f_2(E + U)]|_{E = \varepsilon}$    & I [nA]          \\
                \hline
                0.0         & 0.000                                                   & 0.0           \\
                0.2         & 0.015                                                   & 17.0          \\
                0.5         & 0.440                                                   & 271           \\
                0.8         & 0.964                                                   & 575           \\
                1.0         & 0.999                                                   & 604           \\
            
            \end{tabular}
            \captionof{table}{Differences in contact Fermi functions evaluated at $E=\varepsilon$ and current $I$ at different drain voltages $V_D$. 
            Values are taken from Figures \ref{fig:q1b_current} and \ref{fig:q1c_1} to \ref{fig:q1c_5}.}
            \label{table:q1ci}
        \end{center}


        \subsubsection*{(ii)}
            Figure \ref{fig:q1c_ii} has the Fermi functions marked at their "step points", or when they are equal to $0.5$.
            This can be used to find the self-consistant potential $U$, via the equation for the source Fermi function:

            \begin{equation*}
                f(E + U) = {\frac{1}{1 + e^{(E + U - \mu_1) / k_B T}}}
            \end{equation*}
            We could also use the drain Fermi function but since $V_S = 0$, it is simpler to use the source.
            %
            The function will "step" when $E = \mu_1 - U$, which from Figure \ref{fig:q1c_ii} occurs at $E = \mu_1 - U = 0.4$ for the source contact.
            Since $\mu_1 = \mu - q V_S = 0$ We can simply rearrange to find $U = - E = -0.4\:eV$.

            \begin{figure}[H]
                % Mess around with widths later
                \centering
                \includegraphics[width=0.7\textwidth]{q1c_ii.png}
                \caption{Marked step points of contact Fermi functions.}
                \label{fig:q1c_ii}
                % Note that the comment after \end{subfigure} is required for side by side figures
            \end{figure}%

        \subsubsection*{(iii)}

            At $V_D = 1V$, the source is trying to {\em fill} the channel level while the drain is trying to {\em empty} it.
            This is because the source has electrons at the channel level, and is filling these in while the drain does not have
            any and is attempting to bring the channel level back down to where it is.

        \subsubsection*{(iv)}

            Referring to figure \ref{fig:q1c_ii} again, we can see the areas where the difference between the Fermi functions of each contact
            are 1. Roughly, this means that the channel current $I$ would remain the same if the channel energy level was anywhere between
            $0.3eV$ and $-0.5eV$.

        \subsubsection*{(v)}

            There is no current when $V_D = 0$ because the source does not want to fill the channel. It has no electrons
            at the channel energy level, and thus there is no impetus to fill the channel. A similar story occurs with the
            drain, in that it has no electrons at the appropriate energy level, and there are none in the channel for it to pull out.

    \section*{Question 2}

        \subsection*{(a)}

        \begin{figure}[H]
            \centering
            \begin{subfigure}{0.7\textwidth}
                \centering
                \includegraphics[width=0.7\textwidth]{q2_I-V.png}
                \caption{I-V curves.}
                \label{fig:q2_iv}
            \end{subfigure}
            \begin{subfigure}{0.33\textwidth}
                \centering
                \includegraphics[width=\textwidth]{q2_0V.png}
                \caption{Fermi Functions at $0V$.}
                \label{fig:q2_0v}
            \end{subfigure}%
            \begin{subfigure}{0.33\textwidth}
                \centering
                \includegraphics[width=\textwidth]{q2_0V05.png}
                \caption{Fermi Functions at $0.05V$.}
                \label{fig:q2_0v}
            \end{subfigure}%
            \begin{subfigure}{0.33\textwidth}
                \centering
                \includegraphics[width=\textwidth]{q2_0V1.png}
                \caption{Fermi Functions at $0.1V$.}
                \label{fig:q2_0v}
            \end{subfigure}
            \begin{subfigure}{0.33\textwidth}
                \centering
                \includegraphics[width=\textwidth]{q2_0V2.png}
                \caption{Fermi Functions at $0.2V$.}
                \label{fig:q2_0v}
            \end{subfigure}%
            \begin{subfigure}{0.33\textwidth}
                \centering
                \includegraphics[width=\textwidth]{q2_0V3.png}
                \caption{Fermi Functions at $0.3V$.}
                \label{fig:q2_0v}
            \end{subfigure}
            \caption{stuff and things}
        \end{figure}

        \subsection*{(b)}

        We can see from these figures that the area under the $f_1(E + U) - f_2(E + U)$ curve
        gradually moves down in energy, this measn that a smaller and smaller portion of it is
        above $E = 0eV$, which means there are fewer total energy levels through which current
        can flow. From this we can see that with a high enough $V_D$ we will hit saturation, and
        be unable to pass more current.

        \subsection*{(c)}

        At $V_D = 0.3V$, the energy levels from approximately $0$ to $0.2$ $eV$ are being
        used for electron transport. This is the range where the difference in the contact Fermi
        functions is more than 0 and the energy is more than 0.

        The difference between the two contacts is the greatest at $0eV$, because this is the
        point at which their Fermi functions have the greatest difference, and thus will be making
        the most "effort" to equalize the channel potential.

        \subsection*{(d)}

        Based on the supposed material changes, we would choose material A to maximize the
        drain current. The energy levels vanishing at $-0.4eV$ would mean that there is a
        greater number of energy levels that would be able to be used for conduction. There
        would be a larger {\em area} where there is a non-zero difference in the two Fermi
        functions at the contacts and there are energy levels available in the channel.

        This can be contrasted with material B, which would have {\em no} energy levels in
        this "conduction" zone and thus no current would be able to flow at all.

    \section*{Question 3}
    
        \subsection*{(a)}
        \subsection*{(b)}
        \subsection*{(c)}
        \subsection*{(d)}

    \section*{Question 4}

        \subsection*{(a)}
        \subsection*{(b)}
        \subsection*{(c)}
        \subsection*{(d)}
        \subsection*{(e)}
        \subsection*{(f)}
        \subsection*{(g)}

    


% TODO appendix for Part C code

\end{document}