diff --git a/PS1/doc.tex b/PS1/doc.tex index b3bd03f..604b74b 100644 --- a/PS1/doc.tex +++ b/PS1/doc.tex @@ -27,7 +27,7 @@ % commentstyle=\color{mygreen},% % showstringspaces=false,%without this there will be a symbol in the places where there is a space % numbers=left,% -% numberstyle={\tiny \color{black}},% size of the numbers +% numbersty_sumle={\tiny \color{black}},% size of th_sume numbers % numbersep=9pt, % this defines how far the numbers are from the text % emph=[1]{for,end,break},emphstyle=[1]\color{red}, %some words to emphasise % %emph=[2]{word1,word2}, emphstyle=[2]{style}, @@ -432,16 +432,151 @@ ylabel('Current [A]'); \section*{Question 3} \subsection*{(a)} + + Below is our code. Note that some variable names are different from those in the example code. + \subsection*{(b)} + + \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} + \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 $f_2 - f_1$ 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 $D(E) * [f_2 - f_1]$. + 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. + 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. + 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 + $[f_2 - f_1]$, 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 = 608nA, 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] = 608nA + \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 = 0A, 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] = 0A + \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 $f_2(E + U)$ 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. + This assumes $U = 0$ eV. + \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)} diff --git a/PS1/q4c.jpg b/PS1/q4c.jpg new file mode 100644 index 0000000..1c8718b Binary files /dev/null and b/PS1/q4c.jpg differ