pcb_writeup.tex

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\title{Task 6\\ \vspace{2mm}\Large{16-681 MRSD Project Course }}
\author{
  Daniel Arnett \\
    Karthik (Venkata Rama) Paga\\
  Dicong (David) Qiu\\
  Matthew Swenson\\
  Abdul Zafar
}

\begin{document}
  \maketitle
    \newpage

\section{Regulator Efficiency} Linear voltage regulator efficiency is given by $1-\frac{V_{in}-V_{out}}{V_{in}}$.  So: 
$$\eta_{3.3V} = 1-\frac{24V-3.3V}{24V} = .1375$$
$$\eta_{5V} = 1-\frac{24V-5V}{24V} = .20833$$
$$\eta_{12V} = 1-\frac{24V-12V}{24V} = .5$$


\section{Input Power}

The input power for the motor at maximum rated output is 
$$
P_{\textit{IN},\textit{motor}} = \frac{\textit{\textit{motor}} \times I_{\textit{motor}, \textit{max}} }{\eta_{motor}} = \frac{24V \times 10A}{1.0} = 240W
$$

The input power for the CPU board at maximum rated output is 
$$
P_{\textit{IN},\textit{3.3V}} = \frac{\textit{\textit{3.3V}} \times I_{\textit{3.3V}} }{\eta_{3.3V}} = \frac{3.3V \times 1A}{0.1375} = 24W
$$

The input power for the Wifi and Encoder subsystem at maximum rated output is 
$$
P_{\textit{IN},\textit{5V}} = \frac{\textit{\textit{5V}} \times I_{\textit{5V}} }{\eta_{5V}} = \frac{5V \times 1A}{0.20833} = 24W
$$

The input power for the LIDAR at maximum rated output is 
$$
P_{\textit{IN},\textit{12V}} = \frac{\textit{\textit{12V}} \times I_{\textit{12V}} }{\eta_{12V}} = \frac{12V \times 2A}{0.5} = 48W
$$


\section{Total System Efficiency}

The total system efficiency at maximum rated output is 
$$
\begin{align*}
    \eta &= \frac{P_{\textit{motor}} + P_{3.3V} + P_{5V} + P_{12V}}{P_{\textit{IN}, \textit{motor}} + P_{\textit{IN}, 3.3V} + P_{\textit{IN}, 5V} + P_{\textit{IN}, 12V}} \\
    &= \frac{24V \timex 10A + 3.3V \times 1A + 5V \times 1A + 12V \times 2A}{240W + 24W + 24W + 48W} \\
    &= 0.8104
\end{align*}
$$


\end{document}