Question:

The $\dfrac{V_{OUT}}{V_{IN}}$ of the circuit shown below is _____________

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When you see a final op-amp with $R_3$ into the inverting node and $R_4$ in feedback, it’s an inverting stage. If the preceding block presents $V_{IN}$ across $R_3$, the overall gain is simply $-R_4/R_3$.
Updated On: Aug 28, 2025
  • $-\dfrac{R_4}{R_3}$
  • $\dfrac{R_4}{R_3}$
  • $1+\dfrac{R_4}{R_3}$
  • $1-\dfrac{R_4}{R_3}$
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The Correct Option is A

Solution and Explanation

Step 1: Identify the rightmost stage.
The rightmost op-amp has its non-inverting input at ground, input resistor $R_3$ to the inverting input, and feedback $R_4$ from output to inverting input $\Rightarrow$ inverting amplifier: \[ V_{OUT}=-\frac{R_4}{R_3}\,v_x, \] where $v_x$ is the signal applied through $R_3$. 
Step 2: Evaluate $v_x$ from the left block.
Each of the two left op-amps has its inverting input fed back from its own output (via $R_2$) $\Rightarrow$ both act as voltage followers. Top follower outputs $v_a=V_{IN}$; bottom follower (non-inverting at $0$ V) outputs $v_b=0$. The resistor $R_3$ feeding the right stage is connected between these two follower outputs, so the voltage impressed across the input resistor is \[ v_x=v_a-v_b=V_{IN}-0=V_{IN}. \] Step 3: Overall gain.
Substitute $v_x=V_{IN}$ in Step 1: \[ \frac{V_{OUT}}{V_{IN}}=-\frac{R_4}{R_3}. \] \[ \boxed{-\dfrac{R_4}{R_3}} \]

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