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quad!!!!

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acereca 2018-08-15 08:53:45 +02:00
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parts/theory.tex
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@ -50,7 +50,7 @@ In case of the current measurement circuit we require the following:
Here we use the same Amplifiers and so we can use the following equation for our input current: Here we use the same Amplifiers and so we can use the following equation for our input current:
\begin{align} \begin{align}
I_\text{48V IN}\cdot R_\text{shunt} \cdot r_\text{diffOpAmp} \cdot r_\text{OpAmp} = V_\text{48I pin}\nonumber\\ I_\text{48V IN}\cdot R_\text{shunt} \cdot r_\text{diffOpAmp} \cdot r_\text{OpAmp} = V_\text{48I pin}\nonumber\\
\Leftrightarrow\qud \frac{V_\text{48I pin}}{R_\text{shunt}} \cdot \frac{1}{r_\text{diffOpAmp}\cdot r_\text{OpAmp}} = I_\text{48V IN} \Leftrightarrow\quad \frac{V_\text{48I pin}}{R_\text{shunt}} \cdot \frac{1}{r_\text{diffOpAmp}\cdot r_\text{OpAmp}} = I_\text{48V IN}
\end{align} \end{align}
\subsection{9.6V Output Voltage} \subsection{9.6V Output Voltage}