<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[paper=A4]{article}
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\lhead{Car 313, 23.04, Rev. 1}
\chead{\large TS Discharge Circuit Schematic}

\begin{document}

\begin{figure}[H]
	\centering
	\includegraphics[width=0.98\textwidth]{./Documents/DC.pdf}
	\caption{Schematic of the Discharge Circuit PCB}
\end{figure}

As seen in the Schematic, for our discharge circuitry a PTC (PTCEL13R251NxE) is used. The total capacitance of the DC-link capacitor from the two inverters that we are using is \SI{200}{\micro\farad}, and the maximum voltage of the accumulator is 403.2V. Using the RC discharging circuit equation, we obtain the highest resistance that the PTC can have so that we are still within the 5s discharge limit.

\begin{align}
	V_C =            &amp; V_0 \cdot e^{-t/RC}                                                                 \\
	\SI{60}{\volt} = &amp; \SI{403.2}{\volt} \cdot e^{-\SI{5}{\second}/(R_{PTC} \cdot \SI{200}{\micro\farad})} \\
	R_{PTC} \approx  &amp; \SI{13123}{\ohm}
\end{align}

\begin{wrapfigure}{r}{0.4\textwidth}
	\includegraphics[width=1\linewidth]{./Pictures/PTC-R-T.png}
	\caption{Resistance vs. Temperature for PTCEL13 (typical)}
	\label{fig:PTC_T_R}
\end{wrapfigure}

To Calculate how much discharge attempts we can have before the discharge reaches 5s, we first see at what temperature the PTC the resistance of \SI{13123}{\ohm} have. We can get this from the datasheet of the PTC (fig. \ref{fig:PTC_T_R}).

As seen in the datasheet, the temperature is about \SI{165}{\celsius}. Assuming that the temperature rises instantly after the discharge, and the heat dissipation is negligible (since the thermal time constant $\tau_{th}$ is \SI{130}{\second}). As we are calculating the lest amount of precharge allowed, we assume the temperature of the environment to be \SI{45}{\celsius}. Since the thermal capacity $C_{th}$ is \SI{1.45}{\joule/\kelvin}, we can see that the total energy that we can generate from discharge is $E = \SI{120}{\kelvin} \cdot \SI{1.45}{\joule/\kelvin} = \SI{174}{\joule}$.

We then calculate how much energy in produced in one discharge:

\begin{align}
	E = &amp; \frac{1}{2} \cdot C \cdot V^2 \\
	=   &amp; \SI{16.26}{\joule}
\end{align}

Therefore, the total amount of discharge we can carry out before the time exceeds \SI{5}{\second} is $\SI{174}{\joule} / \SI{16.26}{\joule} = 10.7$ --- ten times.

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