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Add quest.tex for fa20 3 week quest
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‎ee16b/images/capacitor.png

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‎ee16b/images/opamp_model.png

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‎ee16b/images/span1.png

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‎ee16b/images/span2.png

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‎ee16b/images/tomography-problem.png

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‎ee16b/images/tomography.png

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‎ee16b/quest.tex

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\input{header}
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\title{EECS 16B Quest Review Session}
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\author{Presented by George Hutchinson, Naomi Sagan}
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\date{\today}
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\newcommand{\SlideAccessingLogistics}{@\#}
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\begin{document}
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\input{../standard-content}
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\input{topics/16a-import/circuit-cheat-sheet}
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\input{topics/16a-import/resistors}
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\input{topics/16a-import/kcl-kvl}
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\input{topics/16a-import/nodal-analysis}
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\input{topics/16a-import/divider-circuits}
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\input{topics/16a-import/superposition}
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\input{topics/16a-import/equivalent-circuits}
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\input{topics/16a-import/capacitors}
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%\input{topics/16a-import/charge-sharing}
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\input{topics/16a-import/power}
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%\input{topics/16a-import/op-amps}
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\input{topics/16a-import/loading}
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\input{topics/16a-import/op-amp-configs}
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\input{topics/cmos-logic}
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\input{topics/minimum-power}
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\end{document}

‎ee16b/topics/16a-import/capacitors.tex

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\section{Capacitors and Capacitance}
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\begin{frame}{Introduction to Capacitors}
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\textbf{Capacitor}: generally two surfaces that \textbf{store charge}, with non-conductive material between plates.\\[5pt]
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\begin{tabular}{m{0.4\textwidth} m{0.5\textwidth}}
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& \multirow{3}{*}{
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\includegraphics[width=0.5\textwidth]{images/capacitor.png}
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} \\
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$CV = Q$ \\
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$C = \epsilon A/d$ & \\[8pt]
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\end{tabular}
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\begin{itemize}
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\item $C$: capacitance
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\item $A$: area of capacitor (one plate)
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\item $d$: distance between plates
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\item $\epsilon$: “permittivity”, a constant depending on the material in the space between the two plates
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\end{itemize}
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\end{frame}
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\begin{frame}{Circuit model of a capacitor}
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\begin{tabular}{m{0.75\textwidth} m{0.15\textwidth}}
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Unit is the \textbf{Farad} (F) $\rightarrow$ Coulombs per volt ($C/V$) & \multirow{5}{*}{
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\begin{circuitikz}
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\draw (0, 2) to[C] (0, 0);
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\end{circuitikz}
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} \\[10pt]
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$C = Q/V$ & \\
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capacitance = charge/voltage ($F = C/V$) & \\[10pt]
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$E = \frac{1}{2} QV = \frac{1}{2} CV^2$ & \\
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\multicolumn{2}{c}{energy = 1/2 * capacitance * voltage squared ($J=C/V \cdot V^2 = CV$)}
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\end{tabular}
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\end{frame}
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\begin{frame}{Sanity Check: Parallel Plate Capacitor}
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\begin{align*}
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C = \epsilon \frac{A}{d}
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\end{align*}
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What is the \textbf{capacitance} of pair of parallel plates when
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\begin{itemize}
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\item $A \to 0$?
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\item $A \to \infty$?
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\item $d \to 0$?
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\item $d \to \infty$? \\[10pt]
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\end{itemize}
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\textit{Does this make sense intuitively?}
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\end{frame}
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\begin{frame}{Sanity Check: Parallel Plate Capacitor [Solution]}
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\begin{align*}
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C = \epsilon \frac{A}{d}
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\end{align*}
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What is the \textbf{capacitance} of pair of parallel plates when
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\begin{itemize}
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\item $A \to 0$? \textcolor{blue}{$C \to 0$}
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\item $A \to \infty$? \textcolor{blue}{$C \to \infty$}
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\item $d \to 0$? \textcolor{blue}{$C \to \infty$}
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\item $d \to \infty$? \textcolor{blue}{$C \to 0$}\\[10pt]
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\end{itemize}
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\textit{Does this make sense intuitively?} \\
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\textcolor{blue}{
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Yes, since as the area of the capacitor increases, the capacitor can hold more charge and vice versa. As the distance between the plates decreases, the charges escape to the other plate more easily.
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}
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\end{frame}
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\begin{frame}{Capacitors in Parallel}
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We know that the two capacitors must be at the \textbf{same voltage} but \textit{not necessarily have the same charge}. So: \\[10pt]
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\begin{tabular}{m{0.7\textwidth} m{0.2\textwidth}}
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$C_{eq} = Q/V$ & \multirow{4}{*}{
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\begin{circuitikz}[scale=0.7, transform shape]
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\draw (1, 3.75) to[short, -*] (1, 3)
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(0, 3) to[short] (2, 3)
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(0, 3) to[C=$C_1$] (0, 1)
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(2, 3) to[C=$C_2$] (2, 1)
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(0, 1) to[short] (2, 1)
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(1, 1) to[short, *-] (1, 0.25);
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\end{circuitikz}
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} \\
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$C_{eq} = (Q_1 + Q_2)/V$ & \\
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$C_{eq} = Q_1/V + Q_2/V$ & \\
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$C_{eq} = C_1 + C_2$ & \\
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\end{tabular}
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\end{frame}
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\begin{frame}{Capacitors in Parallel}
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\LARGE{
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TLDR: Just add them \\[5pt]
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$C_{eq} = \sum_n C_n$
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}
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\end{frame}
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\begin{frame}{Capacitors in Series}
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We know that both $C_1$ and $C_2$ have the \textbf{same charge} $Q$ stored in them since the \textit{current going through each of the capacitors must leave through the other}. On the other hand, the voltages \textbf{sum to the total voltage}. \\[10pt]
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\begin{tabular}{m{0.6\textwidth} m{0.3\textwidth}}
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Knowing this: & \multirow{4}{*}{
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\begin{circuitikz}[scale=0.7, transform shape]
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\draw (0, 2) to[V=$V$] (0, 0)
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(0, 2) to[C=$C_1$] (2, 2)
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(2, 2) to[C=$C_2$] (4, 2)
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(4, 2) to[short] (4, 0)
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(0, 0) to[short] (4, 0);
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\end{circuitikz}
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} \\
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$1/C_{eq} = (V_1 + V_2)/Q$ & \\
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$V_1/Q + V_2/Q$ & \\
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$1/C_{eq} = 1/C_1 + 1/C_2$ & \\
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\end{tabular}
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\end{frame}
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\begin{frame}{Capacitors in Series}
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\LARGE{
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TLDR: \\[5pt]
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$1/C_{eq} = \sum_n 1/C_n$
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}
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\end{frame}
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\begin{frame}{Equivalent Capacitance: Steps to Solve}
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\begin{itemize}
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\item \textbf{Decide} what \textbf{two nodes} you’re finding your capacitance over. \\
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\begin{itemize}
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\item Normally, it will be the capacitance between the \textit{terminals of a voltage or current source, or between two open terminals}.
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\end{itemize}
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\item \textbf{Break the problem down}: which capacitors are in \textbf{parallel}? Which capacitors are in \textbf{series}?
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\item \textbf{Use these equivalent capacitance equations} to simplify capacitances one “group” at a time until you are left with a single capacitance.
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\end{itemize}
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\textit{Note: Capacitor equations are exactly opposite of resistor equations!}
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\end{frame}
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\begin{frame}{Practice: Equivalent Capacitance}
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Find the \textbf{total capacitance} in this circuit.
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\begin{center}
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\begin{circuitikz}[scale=0.8, transform shape]
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\draw (5, 0) to[V=$V$] (0, 0)
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(0, 3) to[short] (0, 0)
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(0, 3) to[short] (0.5, 3)
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(0.5, 3.75) to[short] (0.5, 2.25)
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(0.5, 3.75) to[C=$C_1$] (2.5, 3.75)
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(0.5, 2.25) to[C=$C_1$] (2.5, 2.25)
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(2.5, 3.75) to[short] (2.5, 2.25)
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(2.5, 3) to[short] (3, 3)
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(3, 3) to[C=$C_3$] (5, 3)
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(5, 3) to[short] (5, 0);
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\end{circuitikz}
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\end{center}
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\end{frame}
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\begin{frame}{Practice: Equivalent Capacitance [Solution]}
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Find the \textbf{total capacitance} in this circuit. \\[5pt]
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\color{blue}
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\begin{tabular}{m{0.55\textwidth} m{0.35\textwidth}}
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The parallel portion becomes: &
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\multirow{5}{*}{
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\color{black}
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\begin{circuitikz}[scale=0.55, transform shape]
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\draw (5, 0) to[V=$V$] (0, 0)
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(0, 3) to[short] (0, 0)
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(0, 3) to[short] (0.5, 3)
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(0.5, 3.75) to[short] (0.5, 2.25)
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(0.5, 3.75) to[C=$C_1$] (2.5, 3.75)
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(0.5, 2.25) to[C=$C_1$] (2.5, 2.25)
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(2.5, 3.75) to[short] (2.5, 2.25)
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(2.5, 3) to[short] (3, 3)
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(3, 3) to[C=$C_3$] (5, 3)
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(5, 3) to[short] (5, 0);
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\end{circuitikz}
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} \\
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$C_{par} = C_1 + C_2$ & \\[5pt]
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Then we use our series equation: \\
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$1/C_{eq} = 1/C_{par} + 1/C_3 = \frac{C_{par} + C_3}{C_{par} C_3}$ & \\[5pt]
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$C_{eq} = \frac{C_{par}C_3}{C_{par} + C_3} = \frac{C_1 C_3 + C_2 + C_3}{C_1 + C_2 + C_3}$ & \\
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\end{tabular}
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\end{frame}
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\begin{frame}{Charging a Capacitor}
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\begin{itemize}
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\item When a capacitor is supplied with current, it \textbf{charges up}.
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\item $V = Q/C$, so the voltage \textit{increases with time}.
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\item When the capacitor \textbf{discharges}, it \textit{loses charge and (therefore) voltage}. \\[5pt]
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\end{itemize}
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\underline{Working with charges over time}:
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\begin{itemize}
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\item Charge on capacitor after $t$ seconds (constant current): $I \cdot t$.
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\item $Q_{final} = C(V_{final} - V_{init})$
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\item $I = C dV/dt = C \Delta V / \Delta t$
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\item $C = I \Delta t / \Delta V$
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\item $Q_{final} = C \Delta V = I \Delta t = I(T_{final} = t_{init})$
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\end{itemize}
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\textit{Note: we'll be working with discrete time in 16A.}
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\end{frame}

‎ee16b/topics/16a-import/change-of-basis.tex

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\section{Change of Basis and Diagonalization}
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\begin{frame}{Change of Basis}
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\end{frame}
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\begin{frame}{Practice: Change of Basis}
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\end{frame}
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\begin{frame}{Diagonalization}
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\end{frame}

‎ee16b/topics/16a-import/charge-sharing.tex

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\section{Charge Sharing}
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\begin{frame}{Charge Sharing}
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Capacitors can be first charged up, then \textit{reconfigured into a different circuit}, usually via switches. \\[5pt]
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\underline{States to Analyze}:
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\begin{enumerate}
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\item \textbf{Initial state}: after charging up
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\item \textbf{Final state}: after charges redistribute in new configuration
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\end{enumerate}
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\textbf{NOTE!} \textit{Charges are *always* conserved from phase 1 to 2.}
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\end{frame}
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\begin{frame}{Charge Sharing: Steps to Solve}
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\begin{itemize}
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\item\textbf{ Note how many states} you will have to find charges for, and \textbf{draw their equivalent circuits}. Generally, there are two: an \textit{initial state and a final state}.
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\begin{itemize}
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\item Sometimes, there will be intermediate stages--but you \textit{solve those much like you will a two-state problem}.
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\end{itemize}
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\item Find the \textbf{charges on all capacitors in your initial state} in terms of your knowns.
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\begin{itemize}
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\item Most often, you’ll be given an \textit{initial charge on one cap, or a voltage source, and the capacitances of all caps}.
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\item For 16A, this is generally taken once the charges stabilize.
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\end{itemize}
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\item Find the \textbf{charges on all capacitors in your second state}, knowing that $Q_{final, total} = Q_{init, total}$ in your system because \textbf{charge is conserved}, in terms of your knowns.
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\end{itemize}
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\end{frame}
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\begin{frame}{Practice: Charge Sharing}
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\begin{tabular}{m{0.55\textwidth} m{0.35\textwidth}}
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& \multirow{2}{*}{
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\begin{circuitikz}[scale=0.7, transform shape]
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\draw (0, 0) to[C=$C$, v_>=$\,\,V_i$] (0, 3)
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(0, 0) to[short] (3, 0)
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(0, 3) to[nos] (3, 3)
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(3, 0) to[C=$C$, v_>=$\,\,0\,V$] (3, 3);
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\end{circuitikz}
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} \\[5pt]
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Suppose the left capacitor has an initial voltage of $V_i$ & \\[30pt]
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\end{tabular}
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\textbf{What happens when we close the switch?}
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\end{frame}
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\begin{frame}{Practice: Charge Sharing [Solution]}
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\color{blue}
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\begin{tabular}{m{0.7\textwidth} m{0.2\textwidth}}
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Assuming ideal switch connection, \textbf{total charge must be conserved}. & \multirow{2}{*}{
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\color{black}
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\begin{circuitikz}[scale=0.6, transform shape]
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\draw (0, 0) to[C=$C$, v_>=$\,\,V_i$] (0, 3)
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(0, 0) to[short] (3, 0)
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(0, 3) to[nos] (3, 3)
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(3, 0) to[C=$C$, v_>=$\,\,0\,V$] (3, 3);
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\end{circuitikz}
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} \\[15pt]
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Phase 1 is when switch is open and Phase 2 is after we \textit{close the switch and reach steady state}. & \\[20pt]
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\end{tabular}
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$Q_{1, final} = Q_1 / 2$ because each capacitor has the same voltage and therefore the same charge in Phase 2. \\[10pt]
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However, energy at Phase 2 = $\frac{CV^2}{2} = \frac{CQ^2}{2C^2} = \frac{Q^2}{2C}$. This is half the energy in Phase 1, so energy was dissipated between Phase 1 nd Phase 2!
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\end{frame}
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\begin{frame}{Challenge Practice: Charge Sharing}
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Phase 1: all $\phi_1$ switches are \textbf{closed}, and all $\phi_2$ switches are \textbf{open}. \\[5pt]
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Phase 2: all $\phi_1$ switches are \textbf{open}, and all $\phi_2$ switches are \textbf{closed}. \\[5pt]
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What is $V_x$ during phase 2 if $C_1 = C$ and $C_2 = 9C$?
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\begin{center}
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\begin{circuitikz}[scale=0.75, transform shape]
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\draw (0, 3) to[V=$V_{source}$] (0, 0.5) node[ground] {}
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(0, 3) to[short] (2, 3)
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(2, 3) to[ospst, l=$\phi_1$] (4, 3)
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(2, 3) to[short] (2, 1)
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(2, 1) to[cspst, l=$\phi_2$] (4, 1)
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(4, 3) to[C=$C_1$] (4, 1)
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(4, 3) to[cspst, l=$\phi_2$] (6, 3)
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(4, 1) to[ospst, l=$\phi_1$] (6, 1)
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(6, 3) to[C=$C_2$] (6, 1) node[ground] {}
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(6, 3) to[short] (8, 3) node[label={[font=\footnotesize] above:$V_x$}] {}
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(8, 3) to[ospst, l=$\phi_1$] (8, 1) node[ground] {};
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\end{circuitikz}
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\end{center}
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\end{frame}
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\begin{frame}{Challenge Practice: Charge Sharing [Solution]}
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\color{blue}
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\underline{Phase 1}
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\begin{align*}
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Q_1 = C_1 V_1 = C_1(V_S - 0) = C_1 V_S \\
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Q_2 = C_2 V_2 = (9C)(0 - 0) = 0
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\end{align*}
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\begin{center}
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\color{black}
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\begin{circuitikz}[scale=0.75, transform shape]
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\draw (0, 3) to[V=$V_{source}$] (0, 0.5) node[ground] {}
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(0, 3) to[short] (2, 3)
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(2, 3) to[short] (4, 3)
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(4, 3) to[C=$C_1$, v=$\,$] (4, 1)
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(4, 1) to[short] (6, 1)
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(6, 3) to[C=$C_2$, v=$\,$] (6, 1) node[ground] {}
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(6, 3) to[short] (8, 3) node[label={[font=\footnotesize] above:$V_x$}] {}
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(8, 3) to[short] (8, 1) node[ground] {};
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\end{circuitikz}
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\end{center}
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\end{frame}
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\begin{frame}{Challenge Practice: Charge Sharing [Solution]}
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\color{blue}
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\underline{Phase 2}
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\begin{align*}
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V_1 = V_x - V_S \,\, \rightarrow \,\, Q_1 = C_1(V_x - V_S) = C(V_X - V_S) \\
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V_2 = V_X \,\, \rightarrow \,\, Q_2 = C_2 V_X = 9CV_X \qquad \qquad \qquad
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\end{align*}
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\begin{center}
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\color{black}
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\begin{circuitikz}[scale=0.75, transform shape]
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\draw (0, 3) to[V=$V_{source}$] (0, 0.5) node[ground] {}
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(0, 3) to[short] (2, 3)
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(2, 3) to[short] (2, 1)
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(2, 1) to[short] (4, 1)
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(4, 3) to[C=$C_1$, v=$\,$] (4, 1)
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(4, 3) to[short] (6, 3)
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(6, 3) to[C=$C_2$, v=$\,$] (6, 1) node[ground] {}
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(6, 3) to[short] (8, 3) node[label={[font=\footnotesize] above:$V_x$}] {};
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\end{circuitikz}
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\end{center}
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\end{frame}
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\begin{frame}{Challenge Practice: Charge Sharing [Solution]}
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\color{blue}
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Total charge in Phase 1 = Total charge in Phase 2
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\begin{align*}
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Q_{1, phase1} + Q_{2, phase1} = Q_{1, phase2} + Q_{2, phase2} \\
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CV_ + 0 = C(V_X - V_S) + 9CV_X \\
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2CV_S = 10CV_X \,\, \rightarrow \,\, V_X = V_S / 5
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\end{align*}
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\begin{center}
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\color{black}
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\begin{circuitikz}[scale=0.75, transform shape]
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\draw (0, 3) to[V=$V_{source}$] (0, 0.5) node[ground] {}
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(0, 3) to[short] (2, 3)
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(2, 3) to[short] (2, 1)
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(2, 1) to[short] (4, 1)
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(4, 3) to[C=$C_1$, v=$\,$] (4, 1)
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(4, 3) to[short] (6, 3)
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(6, 3) to[C=$C_2$, v=$\,$] (6, 1) node[ground] {}
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(6, 3) to[short] (8, 3) node[label={[font=\footnotesize] above:$V_x$}] {};
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\end{circuitikz}
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\end{center}
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\end{frame}

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