Question

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.

• A. \(-15\)
• B. \(15\)
• C. \(2.1\)
• D. \(1.1\)

Hint:

This is a two-op-amp instrumentation-amplifier stage: \(V_{out}=V_{in}\left(1+\dfrac{2R_f}{R}\right)\).

Answer 1

The Correct Option is C

Step 1: Identify the circuit.
This is the classic two-op-amp instrumentation-amplifier front end. Each op-amp has its non-inverting input tied to a reference node, its inverting input joined to the other op-amp's inverting input through the single gain-setting resistor \(R\), and a feedback resistor \(R_f\) from each output back to its own inverting-input node. \(V_{in}\) is impressed directly across \(R\), between the two inverting-input nodes.

Step 2: Apply the virtual short at each op-amp.
Since no current flows into either op-amp's input pins, the whole of \(V_{in}\) appears across \(R\) itself, so the current through \(R\) is
\[ I=\frac{V_{in}}{R} \]
Because the input pins draw no current, this same current \(I\) must also flow through both feedback resistors \(R_f\), one on the top op-amp and one on the bottom op-amp.

Step 3: Write the two output voltages.
The current \(I\) flowing away from one inverting node through its \(R_f\) raises that op-amp's output above its inverting-input voltage by \(IR_f\), while at the other op-amp the same current flowing the opposite way lowers its output by \(IR_f\) below its inverting-input voltage. Adding the two contributions, the differential output across the two op-amp outputs works out to
\[ V_{out}=I\!\left(R+2R_f\right)=V_{in}\left(1+\frac{2R_f}{R}\right) \]
This is the standard gain formula for this two-op-amp front end.

Step 4: Plug in the numbers.
\[ V_{out}=0.1\left(1+\frac{2\times10}{1}\right)=0.1\times(1+20)=0.1\times21 \]
\[ V_{out}=2.1\ \text{V} \]

Step 5: Check this is inside the supply rails.
The op-amps run off \(\pm15\) V supplies, and \(2.1\) V is well inside that range, so neither amplifier saturates. Options (A) \(-15\) and (B) \(15\) would only be correct if the circuit railed against a supply, which does not happen here at this small input. Option (D) \(1.1\) V comes from using the wrong, single-op-amp gain formula \(1+\dfrac{R_f}{R}=11\) instead of the correct two-op-amp formula \(1+\dfrac{2R_f}{R}=21\), forgetting that both feedback resistors add to the gain.

Final Answer:
\[ \boxed{V_{out}=2.1\ \text{V}} \]