--- title: "Summing amplifier V0 = 3V1 + 2V2 + V3" subject: "Basic Electronics Engineering" chapter: 4 key: "semester_2/basic_electronics_engineering:summing-amplifier-3v1-2v2-v3" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/summing-amplifier-3v1-2v2-v3 site: BCT Engineering --- # Summing amplifier V0 = 3V1 + 2V2 + V3 Basic Electronics Engineering · Chapter 4: The Operational Amplifier and Oscillator - Asked in: 2068 Bhadra, 6 marks - Video: [Design an OP-AMP circuit for given output expression.](https://www.youtube.com/watch?v=EK5G8Gd_vXc) — Engg-Course-Made-Easy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/summing-amplifier-3v1-2v2-v3 - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-4 ## Question Design the summing amplifier using Op-Amp to get the output voltage V0 = 3V1 + 2V2 + V3. ## Short answer Use a two-stage circuit: an inverting summer with Rf=30 kΩ, R1=10 kΩ, R2=15 kΩ, R3=30 kΩ gives V01 = -(3V1+2V2+V3). Follow with a unity-gain inverter (two equal resistors, e.g., 10 kΩ) to obtain V0 = 3V1+2V2+V3. ## Answer A summing amplifier gives an output equal to the weighted sum of its inputs. A single inverting summer gives a negative sum, so the required positive output V0 = 3V1 + 2V2 + V3 is obtained by following an inverting summer with a unity-gain inverter. #### Stage 1: inverting summing amplifier ```text V1 o--[ R1 ]--+ | V2 o--[ R2 ]--+-----[ Rf ]-----+ | | V3 o--[ R3 ]--+ |\ | +-----|- \ | | >------+----o V01 +-----|+ / | |/ GND ``` For an ideal op-amp (virtual ground at the inverting input): V01 = −Rf (V1/R1 + V2/R2 + V3/R3) We need the weights 3, 2 and 1, so Rf/R1 = 3, Rf/R2 = 2 and Rf/R3 = 1. Choose Rf = 30 kΩ (assumed value): - R1 = Rf/3 = 10 kΩ - R2 = Rf/2 = 15 kΩ - R3 = Rf/1 = 30 kΩ Then V01 = −(3V1 + 2V2 + V3). #### Stage 2: unity-gain inverting amplifier ```text +-----[ R ]-----+ | | | |\ | V01 o--[ R ]--+-----|- \ | | >-----+----o V0 +-----|+ / | |/ GND ``` With both resistors equal (R = 10 kΩ), V0 = −(R/R) V01 = −V01. #### Result V0 = −V01 = 3V1 + 2V2 + V3 **Answer: R1 = 10 kΩ, R2 = 15 kΩ, R3 = 30 kΩ and Rf = 30 kΩ in the first stage, followed by an inverter with two equal resistors (10 kΩ each).** The two resistors of the second stage can be any equal pair. A 741 op-amp with a ±12 V supply can be used, as long as the output stays within the supply limits. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the output of the inverting summer (stage 1)? A. V01 = +Rf(V1/R1 + V2/R2 + V3/R3) B. V01 = -(R1V1 + R2V2 + R3V3)/Rf C. V01 = -Rf(V1/R1 + V2/R2 + V3/R3) D. V01 = -(V1 + V2 + V3)Rf/R1 only Answer: C. With virtual ground the input currents add and flow through Rf, giving a negative weighted sum. (Key point: Summer output formula) 2. For weights 3, 2, 1 with Rf = 30 kΩ, what are R1, R2, R3? A. 10 kΩ, 15 kΩ, 30 kΩ B. 90 kΩ, 60 kΩ, 30 kΩ C. 30 kΩ, 15 kΩ, 10 kΩ D. 10 kΩ, 20 kΩ, 30 kΩ Answer: A. Each weight is Rf/R, so R1 = 30k/3, R2 = 30k/2, R3 = 30k/1. (Key point: Choosing resistors) 3. Why is a unity-gain inverter added after the summer? A. To triple the gain B. To remove the virtual ground C. The summer output is negative, -(3V1+2V2+V3), so it must be inverted to get a positive sum D. To increase input resistance Answer: C. An inverting summer gives a negative sum; an inverter with equal resistors restores the positive sign. (Key point: Need for second stage) 4. What condition makes the second stage a unity-gain inverter? A. Both resistors are equal, so V0 = -(R/R)V01 B. Feedback resistor is twice the input resistor C. Non-inverting input is open D. Both resistors are zero Answer: A. Gain is -Rf/Rin, which is -1 when the two resistors are equal. 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