Question:

In the circuit diagram shown below, all OPAMPs are ideal with infinite gain and bandwidth. \(\frac{V_{OUT}}{V_{IN}}\) for this circuit is \(\underline{\hspace{2cm}}\). 

Show Hint

When analyzing OPAMP circuits, break the circuit into stages and calculate the gain of each stage before multiplying them together for the total gain.
Updated On: Dec 24, 2025
  • 5.00
  • 5.33
  • 4.80
  • 6.00
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

To solve for the gain of the circuit (\(\frac{V_{OUT}}{V_{IN}}\)), we need to break down the circuit step by step. This circuit contains multiple operational amplifiers (OPAMPs) with resistive networks that affect the gain. Here are the main steps involved in analyzing the circuit:

Step 1: Analyze the first OPAMP configuration.
The first OPAMP is a non-inverting amplifier. The gain of a non-inverting amplifier is given by: \[ A_1 = 1 + \frac{R_1}{R_2} \] Here, \(R_1 = 4R\) and \(R_2 = R\), so the gain is: \[ A_1 = 1 + \frac{4R}{R} = 5 \]

Step 2: Analyze the second OPAMP configuration.
The second OPAMP is also a non-inverting amplifier, and similarly, the gain is calculated as: \[ A_2 = 1 + \frac{R_3}{R_4} \] Here, \(R_3 = R\) and \(R_4 = 2R\), so the gain is: \[ A_2 = 1 + \frac{R}{2R} = 1.5 \]

Step 3: Analyze the third OPAMP configuration.
The third OPAMP is a difference amplifier, and the gain is given by: \[ A_3 = \frac{R_5}{R_6} \] Here, \(R_5 = R\) and \(R_6 = 2R\), so the gain is: \[ A_3 = \frac{R}{2R} = 0.5 \]

Step 4: Combine the gains.
The total gain of the circuit is the product of the individual gains from all three stages: \[ A_{total} = A_1 \times A_2 \times A_3 = 5 \times 1.5 \times 0.5 = 4.80 \] Thus, the gain of the circuit \(\frac{V_{OUT}}{V_{IN}}\) is 4.80, which corresponds to option (C).

Was this answer helpful?
0
0

Questions Asked in GATE BM exam

View More Questions