--- title: "IA features; 3 op-amp IA output voltage" subject: "Advanced Electronics" chapter: 4 key: "semester_4/advanced_electronics:ia-features-3opamp-output" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/ia-features-3opamp-output site: BCT Engineering --- # IA features; 3 op-amp IA output voltage Advanced Electronics · Chapter 4: Instrumentation and Isolation Amplifiers - Asked in: 2074 Chaitra, 2+5 marks - Video: [Instrumentation Amplifiers Explained (Amplifiers #7)](https://www.youtube.com/watch?v=VfJ2AHO2aHg) — Aaron Danner - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/ia-features-3opamp-output - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-4 ## Question What are the features of Instrumentation Amplifier? Derive the expression of output voltage of 3 op-amp Instrumentation Amplifier. ## Short answer IA features: very high input impedance, high CMRR (>100 dB), gain set by one resistor R_G. Derivation: current in R_G is (V₁−V₂)/R_G; then V₉₂−V₉₁ = (V₂−V₁)(1+2R₁/R_G). Difference amp output: V₉ = (R₃/R₂)(V₉₂−V₉₁) = (R₃/R₂)(1+2R₁/R_G)(V₂−V₁). ## Answer An instrumentation amplifier (IA) is a closed-loop differential amplifier with buffered inputs. It has very high input impedance, high CMRR and a precisely set gain, and it is used to amplify low-level transducer signals. #### Features of an instrumentation amplifier - Very high input impedance (hundreds of MΩ), so it does not load the transducer or bridge - High CMRR (more than 100 dB), which rejects common-mode noise and hum - Gain set accurately by one resistor RG, typically 1 to 1000 - Low offset voltage, low drift and low noise - Low output impedance and good gain stability with temperature #### Three op-amp instrumentation amplifier ```text V1 --->(+)\ A1>---+--Vo1--[R2]--+--[R3]--+ +->(-)/ | (a) | | | R1 (-)\ | +---------+ A3>-----+--> Vo RG (+)/ +---------+ | +->(-)\ R1 Vo2--[R2]+--[R3]--GND | A2>---+--Vo2 (b) V2 --->(+)/ ``` #### Derivation of output voltage **Step 1, input stage (A1, A2).** By the virtual short, the inverting inputs of A1 and A2 are at V1 and V2. So the voltage across RG is V1 − V2, and the current through it is: ```text I = (V1 − V2) / RG ``` The op-amp inputs draw no current, 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) ``` Subtracting: ```text Vo2 − Vo1 = (V2 − V1) + (2R1/RG)(V2 − V1) = (V2 − V1)(1 + 2R1/RG) ``` **Step 2, difference amplifier (A3).** For a difference amplifier with matched resistors: ```text Vo = (R3/R2)(Vo2 − Vo1) ``` **Result:** ```text Vo = (R3/R2)(1 + 2R1/RG)(V2 − V1) Ad = (R3/R2)(1 + 2R1/RG) ``` **Example:** R1 = 10 kΩ, RG = 1 kΩ, R2 = R3 = 10 kΩ gives Ad = 1 × (1 + 20) = 21. A common-mode input (V1 = V2) gives no current in RG, so Vo1 = V1 and Vo2 = V2 with gain 1, and A3 cancels it. The gain can be changed with RG alone. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. By the virtual short, what is the current through RG in the three op-amp IA? A. (V1 + V2)/RG B. (V1 - V2)/RG C. V1/RG D. Zero always Answer: B. The two ends of RG sit at V1 and V2. (Key point: Current through RG) 2. Why does the same current flow through both R1 resistors? A. The op-amp inputs draw no current B. R1 are in parallel C. A3 supplies it D. Because RG is zero Answer: A. No current enters the inverting inputs, so I passes through RG and both R1. (Key point: Same current in R1) 3. What is the final output of the three op-amp IA? A. Vo = (R3/R2)(1 + 2R1/RG)(V2 - V1) B. Vo = (R2/R3)(1 + R1/RG)(V2 - V1) C. Vo = (1 + R3/R2)(V2 - V1) D. Vo = (R3/R2)(V2 + V1) Answer: A. Vo2 - Vo1 = (V2 - V1)(1 + 2R1/RG) and A3 multiplies by R3/R2. (Key point: Result) 4. What happens to a pure common-mode input (V1 = V2)? A. It is amplified by 1 + 2R1/RG B. No current in RG, so it passes with gain 1 and A3 cancels it C. It saturates A1 D. It doubles at output Answer: B. I = 0, so Vo1 = Vo2 = Vcm and A3 gives zero. 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