--- title: "Single op-amp IA with mismatched resistances" subject: "Advanced Electronics" chapter: 4 key: "semester_4/advanced_electronics:single-opamp-ia-mismatched" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/single-opamp-ia-mismatched site: BCT Engineering --- # Single op-amp IA with mismatched resistances Advanced Electronics · Chapter 4: Instrumentation and Isolation Amplifiers - Asked in: 2072 Kartik, 1+1+5 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-mismatched - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-4 ## Question What is an instrumentation amplifier? What are its characteristics? Derive the necessary relation to find the input impedance and output voltage in case of single op-amp instrumentation amplifier with mismatched resistance. ## Short answer An instrumentation amplifier is a high-input-impedance, high-CMRR differential amplifier with gain set by one resistor. For a single op-amp IA with mismatched resistors, output is Vo = (1+R2/R1)·[R4/(R3+R4)]·V2 − (R2/R1)·V1; input impedances are low and unequal: Zin(+) = R3+R4, Zin(−) = R1. ## Answer An instrumentation amplifier is a precision differential amplifier with high input impedance, high CMRR and an accurately set gain. It is used to amplify small differential signals from transducers in the presence of large common-mode noise. #### Characteristics - High and balanced input impedance - High CMRR (more than 100 dB) - Accurate gain set by one resistor, typically 1 to 1000 - Low offset, drift and noise; low output impedance #### Single op-amp IA with mismatched resistors ```text R1 R2 V1 ----/\/\/----+----/\/\/----+ | | (-)\ | >----------+--> Vo (+)/ R3 | V2 ----/\/\/----+ | R4 | GND ``` **Output voltage (superposition):** V1 alone (V2 = 0) is an inverting amplifier: −(R2/R1)V1. V2 alone is a divider followed by a non-inverting amplifier: V+ = V2·R4/(R3 + R4), so the output is (1 + R2/R1)·R4/(R3 + R4)·V2. ```text Vo = (1 + R2/R1)·R4/(R3 + R4)·V2 − (R2/R1)·V1 ``` **Differential and common-mode gains.** Put V1 = Vc − Vd/2 and V2 = Vc + Vd/2, where Vd = V2 − V1 and Vc = (V1 + V2)/2: ```text Vo = Ad·Vd + Acm·Vc Ad = (R1R4 + R2R3 + 2R2R4) / [2R1(R3 + R4)] Acm = (R1R4 − R2R3) / [R1(R3 + R4)] CMRR = |Ad/Acm| ``` With matching R4/R3 = R2/R1: Acm = 0, Ad = R2/R1, and Vo = (R2/R1)(V2 − V1). Any mismatch makes Acm ≠ 0 and limits the CMRR. **Input impedance:** - At the V2 input: the current flows through R3 and R4 to ground, so Rin2 = R3 + R4. - At the V1 input: the inverting terminal sits at V+ (virtual short), so I1 = (V1 − V+)/R1. With V2 = 0, Rin1 = R1. - Differential input resistance (matched, R3 = R1): Rid = 2R1. - Common-mode input resistance (matched, both inputs tied): Ricm = (R1 + R2)/2. The input impedance is only a few kΩ and the two inputs are unequal. This is the main weakness of the one op-amp IA, and the reason for the three op-amp IA with buffered inputs. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With V2 = 0, how does the single op-amp IA treat V1? A. As a non-inverting amplifier B. As an inverting amplifier giving -(R2/R1)V1 C. As a buffer D. It is ignored Answer: B. V1 passes through R1 into the inverting input. (Key point: Superposition for V1) 2. What is the common-mode gain? A. (R1R4 - R2R3)/[R1(R3 + R4)] B. (R1R3 - R2R4)/R1 C. R2/R1 D. Zero always Answer: A. Setting V1 = V2 = Vcm gives this expression. (Key point: Common-mode gain) 3. When is Acm zero, so Vo = (R2/R1)(V2 - V1)? A. R4/R3 = R2/R1 B. R1 = R2 C. R3 = 0 D. R4 = R1 Answer: A. R1R4 = R2R3 means matched ratios. (Key point: Matching condition) 4. What is the main weakness of the single op-amp IA? A. Too high gain B. Low, unequal input impedances (R1 and R3 + R4) C. Needs three supplies D. No inverting input Answer: B. Input impedance is only a few kOhm and the two inputs differ, motivating the 3 op-amp IA. (Key point: Input impedance weakness) ## Related in Advanced Electronics - [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 - [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 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 - [Input impedance of inverting op-amp](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ainverting-input-impedance) — Advanced Electronics, chapter 2 - [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 - [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 other subjects - [CE amplifier input, output resistance and gain (RE1, RE2)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-partial-bypass-numeric) — Electronics Devices and Circuit, chapter 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 - [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 - [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×