--- title: "CM rejection condition; isolation amp applications" subject: "Advanced Electronics" chapter: 4 key: "semester_4/advanced_electronics:single-opamp-ia-cm-rejection" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/single-opamp-ia-cm-rejection site: BCT Engineering --- # CM rejection condition; isolation amp applications Advanced Electronics · Chapter 4: Instrumentation and Isolation Amplifiers - Asked in: 2071 Shrawan, 5+2 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/single-opamp-ia-cm-rejection - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-4 ## Question Determine the condition required to reject common mode signal in a single op-amp instrumentation amplifier with mismatched resistor. List out the applications of isolation amplifiers. ## Short answer Common-mode rejection requires matched resistor ratios: R4/R3 = R2/R1 (typically R3=R1, R4=R2), yielding Acm=0 and Vo = (R2/R1)(V2−V1). Isolation amplifier applications include medical instrumentation, high-voltage sensing, breaking ground loops, industrial process control, and protecting ADCs. ## Answer #### Condition to reject common-mode signal (single op-amp IA) ```text R1 R2 V1 ----/\/\/----+----/\/\/----+ | | (-)\ | >----------+--> Vo (+)/ R3 | V2 ----/\/\/----+---[R4]---GND ``` **Step 1, output by superposition:** ```text V+ = V2 · R4/(R3 + R4) Vo = (1 + R2/R1)·V+ − (R2/R1)·V1 = [R4(R1 + R2) / (R1(R3 + R4))]·V2 − (R2/R1)·V1 ``` **Step 2, apply a pure common-mode input**, V1 = V2 = Vcm: ```text Vo = Vcm [ R4(R1 + R2)/(R1(R3 + R4)) − R2/R1 ] = Vcm (R1R4 + R2R4 − R2R3 − R2R4) / (R1(R3 + R4)) = Vcm (R1R4 − R2R3) / (R1(R3 + R4)) ``` So the common-mode gain is: ```text Acm = (R1R4 − R2R3) / [R1(R3 + R4)] ``` **Step 3, condition for rejection.** Acm = 0 requires R1R4 = R2R3, that is: ```text R4/R3 = R2/R1 (resistor ratios matched) ``` Usually R3 = R1 and R4 = R2. Then: ```text Vo = (R2/R1)(V2 − V1), Ad = R2/R1, CMRR → ∞ ``` With mismatched resistors, Acm ≠ 0 and CMRR = Ad/Acm is finite. Even a 1 % mismatch limits CMRR to about 40 dB. For this reason the resistors are trimmed (for example R4 made adjustable) or laser-trimmed in IC form. #### Applications of isolation amplifiers - **Medical electronics** (ECG, EEG, patient monitoring): protect the patient from leakage and fault currents - **High-voltage sensing** in power electronics, motor drives and power systems - **Breaking ground loops** between remote sensors and the data acquisition system - **Industrial process control:** thermocouples and 4–20 mA loops at high common-mode voltage - Protecting low-voltage control or computer circuits from high-voltage transients. They are also used in nuclear and chemical plant instruments. ## 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 single op-amp IA output? A. Vcm (R1R4 - R2R3)/(R1(R3 + R4)) B. Vcm R2/R1 C. 2 Vcm D. Vcm (R1 + R2) Answer: A. Substituting into the superposition output yields this. (Key point: Common-mode gain) 2. What condition gives zero common-mode gain? A. R1R4 = R2R3, i.e. R4/R3 = R2/R1 B. R1 = R4 C. R2 = R3 D. R1 + R2 = R3 + R4 Answer: A. Numerator of Acm must vanish. (Key point: Condition) 3. What does a 1 % resistor mismatch do? A. Nothing B. Limits CMRR to about 40 dB C. Increases CMRR D. Makes gain zero Answer: B. Acm becomes non-zero, so CMRR = Ad/Acm is finite. (Key point: Effect of mismatch) 4. Which is an application of isolation amplifiers? A. Breaking ground loops and patient protection in ECG B. Audio tone control C. Digital counters D. Clock generation Answer: A. They separate grounds and protect patients and circuits. (Key point: Isolation application) ## Related in Advanced Electronics - [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 - [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 - [Isolation amplifier applications; optically coupled circuit](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aisolation-applications-optical) — 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 - [Practical IA characteristics; optical isolation](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aia-practical-characteristics-optical) — Advanced Electronics, chapter 4 ## Related in other subjects - [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 - [Summing amplifier using op-amp](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Asumming-amplifier-op-amp) — Fundamental of Electrical and Electronics Engineering, chapter 6 - [Cut-off and single-mode range, 4×2 cm](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Awaveguide-4x2cm-cutoff-single-mode-range) — Electromagnetism, chapter 6 - [Single-mode range, air-filled 2×1 cm](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Awaveguide-2x1cm-single-mode-range) — Electromagnetism, chapter 6