\ifthenelse {\boolean{paper}} { \abstract{ This paper analyses a non-inverting op-amp configuration, with the opamp and gain resistors using the FMMD methodology. It has three base components, two resistors and one op-amp. The two resistors are used as a potential divider to program the gain of the amplifier. We consider the two resistors as a functional group where the function is provides is to operate as a potential divider. The base component error modes of the resistors are used to model the potential divider from a failure mode perspective. We determine the failure symptoms of the potential divider and consider these as failure modes of a derived component. We can now create a functional group representing the amplifier, by bringing the failure modes from the potential divider and the op-amp into a functional group. This can now be analysed and a derived component to represent the non inverting amplifier determined. } } { } \section{Introduction} Standard non inv op amp from ``art of electronics'' ~\cite{aoe}[pp.234] shown in figure \ref{fig:noninvamp}. \begin{figure}[h] \centering \includegraphics[width=200pt,keepaspectratio=true]{./noninvopamp/noninv.jpg} % noninv.jpg: 341x186 pixel, 72dpi, 12.03x6.56 cm, bb=0 0 341 186 \caption{Standard non inverting amplifier configuration} \label{fig:noninvamp} \end{figure} The functional of the resistors in this amplifier is to set the gain. They operate as a potential divider and program the minus input on the op-amp to balance them against the positive input, giving the voltage gain ($G_v$) defined by $$ G_v = 1 + \frac{R2}{R1} $$ at the output. As the resistors work to provide a specific function, that of a potential divider, we can treat them as a functional group. Using the EN298 specification for resistor failure ~\cite{en298}[App.A] we can assign failure modes of $OPEN$ and $SHORT$ to the resistors. Thus $R1$ has failure modes $\{R1\_OPEN, R1\_SHORT\}$ and $R2$ has failure modes $\{R2\_OPEN, R2\_SHORT\}$. Modelling this as a functional group, we can draw a circle to represent each failure mode in the potential divider, shown in figure \ref{fig:fg1}. \begin{figure}[h] \centering \includegraphics[width=200pt,keepaspectratio=true]{./noninvopamp/fg1.jpg} % fg1.jpg: 430x271 pixel, 72dpi, 15.17x9.56 cm, bb=0 0 430 271 \caption{potential divider `functional group' failure modes} \label{fig:fg1} \end{figure} We can now look at each of these base component failure modes, and determine how they will affect the operation of the potential divider. Each failure mode scenario we look at will be given a teat case number, which is represented on the diagram, with an asterisk marking which failure modes is is modelling (see figure \ref{fig:fg1a}). \begin{figure}[h] \centering \includegraphics[width=200pt,keepaspectratio=true]{./noninvopamp/fg1a.jpg} % fg1a.jpg: 430x271 pixel, 72dpi, 15.17x9.56 cm, bb=0 0 430 271 \caption{potential divider with test cases} \label{fig:fg1a} \end{figure} \begin{table}[ht] \caption{Potential Divider: Failure Mode Effects Analysis: Single Faults} % title of Table \centering % used for centering table \begin{tabular}{||l|c|c|l|l||} \hline \hline \textbf{Test} & \textbf{Pot.Div} & \textbf{ } & \textbf{General} \\ \textbf{Case} & \textbf{Effect} & \textbf{ } & \textbf{Symtom Description} \\ % R & wire & res + & res - & description \hline \hline TC1: $R_1$ SHORT & LOW & & Low PD \\ TC2: $R_1$ OPEN & HIGH & & High PD \\ \hline TC3: $R_2$ SHORT & HIGH & & High PD \\ TC4: $R_2$ OPEN & LOW & & Low PD \\ \hline \hline \end{tabular} \label{pdfmea} \end{table} We can now collect the symptoms of failure. From the four base component failure modes, we now have two symptoms, $LOW\;PD, HIGH\;PD$. We can represent the collection of these symptoms by drawing connecting lines between the test cases and naming them (see figure \ref{fig:fg1b}). \begin{figure}[h] \centering \includegraphics[width=200pt,keepaspectratio=true]{./noninvopamp/fg1b.jpg} % fg1b.jpg: 430x271 pixel, 72dpi, 15.17x9.56 cm, bb=0 0 430 271 \caption{Collection of potential divider failure mode symptoms} \label{fig:fg1b} \end{figure} \vspace{60pt} $$ \int_{0\-}^{\infty} f(t).e^{-s.t}.dt \; | \; s \in \mathcal{C}$$ \today $$\frac{-b\pm\sqrt{ {b^2-4ac}}}{2a}$$ \today