--- title: "Op-amp properties and summing amplifier" subject: "Basic Electronics Engineering" chapter: 4 key: "semester_2/basic_electronics_engineering:opamp-props-summer-v1-2v2-3v3" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/opamp-props-summer-v1-2v2-3v3 site: BCT Engineering --- # Op-amp properties and summing amplifier Basic Electronics Engineering · Chapter 4: The Operational Amplifier and Oscillator - Asked in: 2071 Bhadra, 2+3 marks - Video: [Summing Amplifiers - Op Amp Circuits](https://www.youtube.com/watch?v=p4FQyOwjhLA) — The Organic Chemistry Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/opamp-props-summer-v1-2v2-3v3 - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-4 ## Question State any four properties of an ideal op-amp. Design a summing amplifier using Op-Amp to get the output voltage V0 = -V1 + 2V2 + 3V3. ## Short answer Ideal op-amp: infinite gain, infinite input resistance, zero output resistance, infinite bandwidth. Two-stage design: first inverting summer for V2, V3 with Rf1=30kΩ, R2=15kΩ, R3=10kΩ gives Vo1 = -(2V2+3V3). Second inverting summer with inputs V1 and Vo1, equal resistors 10kΩ gives Vo = -(V1+Vo1) = -V1+2V2+3V3. ## Answer #### Four properties of an ideal op-amp 1. Infinite open-loop voltage gain. 2. Infinite input resistance (no input current). 3. Zero output resistance. 4. Infinite bandwidth, infinite CMRR and zero offset voltage. #### Design for V0 = −V1 + 2V2 + 3V3 A single inverting summer gives all inputs the same (negative) sign. Here V1 has a negative sign while V2 and V3 have positive signs, so two stages are used. **Stage 1: inverting summer for V2 and V3** V01 = −(Rf1/R2 V2 + Rf1/R3 V3) = −(2V2 + 3V3) Choose Rf1 = 30 kΩ: R2 = 30k/2 = 15 kΩ, R3 = 30k/3 = 10 kΩ. ```text V2 o--[ 15k ]--+ | V3 o--[ 10k ]--+-----[ 30k ]----+ | | | |\ | +-----|- \ | | >------+----o V01 +-----|+ / | |/ GND ``` **Stage 2: inverting summer for V1 and V01** V0 = −(Rf2/R4 V1 + Rf2/R5 V01), with both gains equal to 1: Rf2 = R4 = R5 = 10 kΩ. V0 = −(V1 + V01) = −V1 + (2V2 + 3V3) = −V1 + 2V2 + 3V3 ✓ ```text V01 o--[ 10k ]--+ | V1 o--[ 10k ]--+-----[ 10k ]----+ | | | |\ | +-----|- \ | | >------+----o V0 +-----|+ / | |/ GND ``` **Answer:** Stage 1: Rf1 = 30 kΩ, R2 = 15 kΩ, R3 = 10 kΩ. Stage 2: R4 = R5 = Rf2 = 10 kΩ. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which set correctly lists properties of an ideal op-amp? A. Zero open-loop gain, infinite input resistance, infinite output resistance B. Infinite open-loop gain, zero input resistance, zero output resistance C. Finite gain, zero input resistance, infinite output resistance D. Infinite open-loop gain, infinite input resistance, zero output resistance Answer: D. An ideal op-amp has infinite gain, infinite input resistance (no input current), zero output resistance, infinite bandwidth and CMRR, and zero offset. (Key point: Ideal op-amp properties) 2. Why does V0 = -V1 + 2V2 + 3V3 need two stages? A. The gain of 3 cannot be obtained in one stage B. A single inverting summer gives the same negative sign for all inputs C. An op-amp can take only two inputs D. The resistors would be too large Answer: B. One inverting summer makes every term negative, so a second inverting stage flips the sign of the V2, V3 part. (Key point: Why two stages) 3. With Rf1 = 30 kΩ, what are R2 and R3 in stage 1 so that V01 = -(2V2 + 3V3)? A. R2 = 60 kΩ, R3 = 90 kΩ B. R2 = 10 kΩ, R3 = 15 kΩ C. R2 = 30 kΩ, R3 = 30 kΩ D. R2 = 15 kΩ, R3 = 10 kΩ Answer: D. Gain for each input is Rf/R, so R2 = 30k/2 = 15 kΩ and R3 = 30k/3 = 10 kΩ. (Key point: Stage 1 resistors) 4. What does stage 2 (R4 = R5 = Rf2 = 10 kΩ) do? A. Amplifies V1 by 3 B. Adds V1 and V01 with unity gains and inverts: V0 = -(V1 + V01) C. Subtracts V1 from V01 without inversion D. Acts as a buffer for V01 only Answer: B. Equal resistors give unity gain; V0 = -V1 - (-(2V2+3V3)) = -V1 + 2V2 + 3V3. 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