--- title: "Design three op-amp IA with gain 50" subject: "Advanced Electronics" chapter: 4 key: "semester_4/advanced_electronics:design-3opamp-ia-gain-50" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/design-3opamp-ia-gain-50 site: BCT Engineering --- # Design three op-amp IA with gain 50 Advanced Electronics · Chapter 4: Instrumentation and Isolation Amplifiers - Asked in: 2074 Asoj, 2+5 marks - Video: [Instrumentation amplifier using three Op-Amps | IA using three Op-Amps](https://www.youtube.com/watch?v=7nOtMRemsDU) — Electronics for You - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/design-3opamp-ia-gain-50 - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-4 ## Question What are the features of Instrumentation Amplifier? Hence design a three op amp Instrumentation Amplifier having gain 50. ## Short answer For gain 50, with R₂=R₃=10k, R_G=1k, R₁=(50−1)/2×1k=24.5k (matched). Alternatively, split gain: first stage gain 10 (R_G=2k, R₁=9k) and diff stage gain 5 (R₃/R₂=5). Output expression as above. ## Answer An instrumentation amplifier (IA) is a precision difference amplifier with buffered inputs. It has very high input impedance, high CMRR and a gain set by one resistor. It is used to amplify small transducer signals in noisy, high common-mode environments. #### Features of an instrumentation amplifier - Very high input impedance on both inputs - High CMRR (100 dB or more), which rejects common-mode noise - Gain set by a single external resistor RG, accurate and stable - Low offset voltage and drift, low noise - Low output impedance #### Three op-amp IA and gain ```text V1 --->(+)\ A1>---+--Vo1--[R2]--+--[R3]--+ +->(-)/ | | | | R1 (-)\ | +---------+ A3>-----+--> Vo RG (+)/ +---------+ | +->(-)\ R1 Vo2--[R2]+--[R3]--GND | A2>---+--Vo2 V2 --->(+)/ ``` The current through RG is (V1 − V2)/RG. The same current flows through both R1 resistors, so: ```text Vo2 − Vo1 = (V2 − V1)(1 + 2R1/RG) Vo = (R3/R2)(Vo2 − Vo1) = (R3/R2)(1 + 2R1/RG)(V2 − V1) ``` #### Design for gain A = 50 **Option 1, all gain in the first stage:** 1. R2 = R3 = 10 kΩ, so the difference stage gain = 1. 2. 1 + 2R1/RG = 50, so 2R1/RG = 49. 3. Choose RG = 1 kΩ: ```text R1 = 49 × 1 kΩ / 2 = 24.5 kΩ ``` (Or fix R1 = 10 kΩ, then RG = 2R1/(A − 1) = 20k/49 = 408 Ω; use a 390 Ω resistor plus a 50 Ω trimmer.) **Option 2, split the gain (10 × 5):** - First stage: 1 + 2R1/RG = 10, with RG = 2 kΩ and R1 = 9 kΩ - Difference stage: R3/R2 = 5, with R2 = 10 kΩ and R3 = 50 kΩ ```text A = 5 × (1 + 2 × 9k/2k) = 5 × 10 = 50 ``` **Check (option 1):** A = 1 × (1 + 2 × 24.5k/1k) = 1 + 49 = 50. **Design values (option 1):** R1 = 24.5 kΩ (×2, matched), RG = 1 kΩ, R2 = R3 = 10 kΩ (×4, 0.1 % matched), op-amps such as OP07 or LM324. Gain can be changed by adjusting RG only, without disturbing CMRR. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With R3 = R2, which equation gives the three op-amp IA gain? A. A = 1 + 2R1/RG B. A = 1 + R1/RG C. A = 2R1/RG D. A = R1/2RG Answer: A. Unity-gain difference stage leaves A = 1 + 2R1/RG. (Key point: Gain relation) 2. For gain 50 with RG = 1 kOhm and R2 = R3, what is R1? A. 25 kOhm B. 49 kOhm C. 24.5 kOhm D. 50 kOhm Answer: C. 2R1/RG = 49, so R1 = 49 x 1k/2 = 24.5 kOhm. (Key point: Design for 50) 3. In the split option (10 x 5), what sets the factor of 5? A. RG = 5 kOhm B. R3/R2 = 5 in the difference stage C. R1/RG = 5 D. The op-amp type Answer: B. R3/R2 = 50k/10k = 5 multiplies the first stage gain of 10. (Key point: Split gain option) 4. Why are the four difference-stage resistors 0.1 % matched? A. To increase gain B. Mismatch directly reduces CMRR C. To reduce power D. To raise input impedance Answer: B. R3/R2 must be equal on both sides or common-mode signals are not cancelled. (Key point: Resistor matching) ## Close variants set in other exams - [Design three op-amp IA with gain 20](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Adesign-3opamp-ia-gain-20) — Advanced Electronics, chapter 4, asked 3× ## Related in Advanced Electronics - [IA features; 3 op-amp IA output voltage](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aia-features-3opamp-output) — Advanced Electronics, chapter 4 - [3 op-amp IA advantages; electromagnetic isolation](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Athree-opamp-ia-electromagnetic-isolation) — Advanced Electronics, chapter 4 - [Output voltage of two op-amp IA](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Atwo-opamp-ia-output) — Advanced Electronics, chapter 4 - [3 op-amp IA output and common-mode rejection](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Athree-opamp-ia-cm-rejection) — Advanced Electronics, chapter 4 - [Single op-amp IA with mismatched resistances](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Asingle-opamp-ia-mismatched) — Advanced Electronics, chapter 4 ## Related in other subjects - [Ideal op-amp properties; non-inverting gain](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-noninverting-gain) — Basic Electronics Engineering, chapter 4, asked 3× - [Ideal op-amp properties; inverting gain](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-inverting-gain) — Basic Electronics Engineering, chapter 4, asked 3× - [Voltage gain of non-inverting op-amp](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-noninverting-gain) — Basic Electronics Engineering, chapter 4, asked 2× - [Op-amp properties and summing amplifier](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-summer-v1-2v2-3v3) — Basic Electronics Engineering, chapter 4