In the circuit shown below, \(R\) is \(1\ \text{k}\Omega\) and \(R_f\) is \(10\ \text{k}\Omega\). If \(V_{in}\) is \(100\) mV and the op-amps have supply voltages of \(\pm15\) V, then \(V_{out}\) is ______ V.
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This is a two-op-amp instrumentation-amplifier stage: \(V_{out}=V_{in}\left(1+\dfrac{2R_f}{R}\right)\).
Step 1: Name the two inverting nodes.
Call the inverting-input node of the top op-amp $A$ and that of the bottom op-amp $B$. These two nodes are joined only by the resistor $R$, and $V_{in}$ sits directly between them, so
\[ V_A-V_B=V_{in} \]
Step 2: Find the current through $R$.
\[ I_R=\frac{V_A-V_B}{R}=\frac{V_{in}}{R} \]
This current has nowhere else to go except through the two feedback resistors $R_f$, since an ideal op-amp draws no input current and forces zero net current into its input node.
Step 3: Relate each output to its own inverting node.
At the top op-amp, the same current $I_R$ that leaves node $A$ through $R$ must be supplied through its own $R_f$ from the output, so
\[ V_{out,top}=V_A+I_R R_f \]
At the bottom op-amp, by the mirror-image argument,
\[ V_{out,bottom}=V_B-I_R R_f \]
Step 4: Subtract to get the overall output.
\[ V_{out}=V_{out,top}-V_{out,bottom}=(V_A-V_B)+2I_R R_f=V_{in}+2\left(\frac{V_{in}}{R}\right)R_f \]
\[ V_{out}=V_{in}\left(1+\frac{2R_f}{R}\right) \]
Step 5: Substitute the given values.
With $V_{in}=0.1$ V, $R=1$ k$\Omega$ and $R_f=10$ k$\Omega$,
\[ \frac{2R_f}{R}=\frac{2\times10}{1}=20 \]
\[ V_{out}=0.1\times(1+20)=0.1\times21=2.1\text{ V} \]
Since this sits comfortably inside the $\pm15$ V rails, there is no clipping.
\[ \boxed{V_{out}=2.1\text{ V}} \]
