--- title: "3 op-amp IA output and common-mode rejection" subject: "Advanced Electronics" chapter: 4 key: "semester_4/advanced_electronics:three-opamp-ia-cm-rejection" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/three-opamp-ia-cm-rejection site: BCT Engineering --- # 3 op-amp IA output and common-mode rejection Advanced Electronics · Chapter 4: Instrumentation and Isolation Amplifiers - Asked in: 2068 Chaitra, 5+2 marks - Video: [Instrumentation Amplifier with 3 OpAmps, How does it work? What are the Gain and Transfer function?](https://www.youtube.com/watch?v=Yq767et8BbY) — Engineering Prof. - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/three-opamp-ia-cm-rejection - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-4 ## Question Find the expression for output voltage of three op-amp instrumentation amplifier. Also, explain how it rejects common mode signal. ## Short answer Output voltage: Vo = (R3/R2)*(1+2R1/RG)*(V2-V1). Common-mode rejection: With V1=V2=Vcm, voltage across RG is zero so no current in R1, thus Vo1=Vo2=Vcm; the first stage passes common-mode with gain 1 while differential gain is boosted, and the difference amplifier cancels the common-mode, yielding zero output. ## Answer The three op-amp instrumentation amplifier has two buffer op-amps (A1, A2) forming an input gain stage, followed by a difference amplifier (A3). #### Circuit ```text V1 --->(+)\ A1>---+--Vo1--[R2]--+--[R3]--+ +->(-)/ | | | | R1 (-)\ | +---------+ A3>-----+--> Vo RG (+)/ +---------+ | +->(-)\ R1 Vo2--[R2]+--[R3]--GND | A2>---+--Vo2 V2 --->(+)/ ``` #### Expression for output voltage **Input stage.** By the virtual short, the two ends of RG are at V1 and V2. The current through RG is: ```text I = (V1 − V2)/RG ``` No current enters the op-amp inputs, so the same I flows through both R1 resistors: ```text Vo1 = V1 + I·R1 = V1 + (R1/RG)(V1 − V2) Vo2 = V2 − I·R1 = V2 − (R1/RG)(V1 − V2) Vo2 − Vo1 = (V2 − V1)(1 + 2R1/RG) ``` **Difference stage (A3).** By superposition: ```text V+ = Vo2 · R3/(R2 + R3) Vo = (1 + R3/R2)·V+ − (R3/R2)·Vo1 = (R3/R2)(Vo2 − Vo1) ``` **Result:** ```text Vo = (R3/R2)(1 + 2R1/RG)(V2 − V1) ``` #### How it rejects common-mode signal Let V1 = V2 = Vcm (pure common-mode input): 1. The voltage across RG is zero, so I = 0 and no current flows in R1. 2. So Vo1 = Vcm and Vo2 = Vcm. The first stage passes the common-mode signal with **gain 1 only**, while the differential signal gets a gain of (1 + 2R1/RG). 3. A3 with matched resistors subtracts its two inputs: Vo = (R3/R2)(Vcm − Vcm) = 0. Rejection happens in two steps. The first stage raises the differential signal relative to the common-mode one by (1 + 2R1/RG), which multiplies the CMRR by this factor. The difference amplifier then cancels what remains. Any residual mismatch in the R2/R3 network is trimmed with a potentiometer in one R3 to maximise CMRR. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For V1 = V2 = Vcm, what is the current in RG? A. Vcm/RG B. Zero C. 2 Vcm/RG D. Infinite Answer: B. There is no voltage across RG. (Key point: Common-mode current) 2. What is the first stage's gain for common-mode signals? A. 1 + 2R1/RG B. Zero C. 1 D. R3/R2 Answer: C. Vo1 = Vo2 = Vcm, so gain is 1. (Key point: First stage CM gain) 3. How does A3 finish the rejection? A. It subtracts Vo1 from Vo2 so equal common-mode values cancel B. It amplifies Vcm C. It adds the two outputs D. It grounds both Answer: A. Vo = (R3/R2)(Vcm - Vcm) = 0 with matched resistors. (Key point: Role of A3) 4. By what factor does the first stage multiply CMRR? A. R3/R2 B. 1 + 2R1/RG C. 2R1 D. 1/RG Answer: B. Differential gain rises by this factor while common-mode gain stays 1. 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