--- title: "Virtual ground and summer design" subject: "Basic Electronics Engineering" chapter: 4 key: "semester_2/basic_electronics_engineering:virtual-ground-summer" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/virtual-ground-summer site: BCT Engineering --- # Virtual ground and summer design Basic Electronics Engineering · Chapter 4: The Operational Amplifier and Oscillator - Asked in: 2073 Magh, 1+4 marks; 2070 Magh, 2+4 marks (asked 2 times) - Video: [Operational Amplifier: Inverting Op Amp and The Concept of Virtual Ground in Op Amp](https://www.youtube.com/watch?v=AuZ00cQ0UrE) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/virtual-ground-summer - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-4 ## Question Explain the concept of virtual ground. Design a summer circuit using op-amp to get the output voltage as: V0 = -(V1 + 10V2 + 25V3). ## Short answer Virtual ground: in an inverting op-amp with negative feedback and non-inverting input grounded, V- ≈ 0 V due to high open-loop gain, acting as ground without physical connection. Design: inverting summer with Rf=100kΩ, R1=100kΩ, R2=10kΩ, R3=4kΩ yields Vo=-(V1+10V2+25V3). ## Answer #### Virtual ground In an inverting op-amp circuit with negative feedback, the non-inverting input is grounded. Because the open-loop gain is very large (ideally infinite), the differential input voltage is nearly zero, so V− ≈ V+ = 0. The inverting input is therefore at ground potential **without being physically connected to ground**. This is called a **virtual ground**. Since the input resistance of the op-amp is infinite, no current flows into that terminal, so all the input current flows through the feedback resistor. This is why the inputs of a summing amplifier do not interact. #### Design of V0 = −(V1 + 10V2 + 25V3) The inverting summing amplifier gives: V0 = −(Rf/R1 V1 + Rf/R2 V2 + Rf/R3 V3) ```text V1 o--[ R1 ]--+ | V2 o--[ R2 ]--+-----[ Rf ]-----+ | | V3 o--[ R3 ]--+ |\ | +-----|- \ | | >------+----o V0 +-----|+ / | |/ GND ``` Comparing the coefficients: Rf/R1 = 1, Rf/R2 = 10, Rf/R3 = 25. Choose Rf = 100 kΩ: - R1 = Rf/1 = 100 kΩ - R2 = Rf/10 = 10 kΩ - R3 = Rf/25 = 4 kΩ Check: V0 = −(100/100 V1 + 100/10 V2 + 100/4 V3) = −(V1 + 10V2 + 25V3). **Answer: Rf = 100 kΩ, R1 = 100 kΩ, R2 = 10 kΩ, R3 = 4 kΩ.** A resistor Rcomp = R1 ∥ R2 ∥ R3 ∥ Rf (about 2.7 kΩ) placed at the non-inverting input can reduce the offset error. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is a virtual ground? A. The inverting input at ground potential without being physically grounded B. A physical ground connection C. The output at 0 V D. A supply rail Answer: A. Because of the huge gain and negative feedback, V- ≈ V+ = 0 without a direct connection. (Key point: Virtual ground) 2. Why do the inputs of a summing amplifier not interact? A. The summing node is at virtual ground, so each input current depends only on its own resistor B. Each input has its own op-amp C. The inputs are in series D. Rf is zero Answer: A. With the node held at 0 V, each current is Vk/Rk and all currents add through Rf. (Key point: Why inputs don't interact) 3. What is the output of the inverting summer? A. V0 = -(Rf/R1 V1 + Rf/R2 V2 + Rf/R3 V3) B. V0 = +(R1/Rf V1 + R2/Rf V2 + R3/Rf V3) C. V0 = -(V1 + V2 + V3) always D. V0 = Rf (V1 + V2 + V3) Answer: A. Each input is weighted by Rf/Rk and the sum is inverted. (Key point: Summer equation) 4. With Rf = 100 kΩ, which resistors give V0 = -(V1 + 10V2 + 25V3)? A. R1 = 100 kΩ, R2 = 10 kΩ, R3 = 4 kΩ B. R1 = 100 kΩ, R2 = 1 MΩ, R3 = 2.5 MΩ C. R1 = 1 kΩ, R2 = 10 kΩ, R3 = 25 kΩ D. R1 = 10 kΩ, R2 = 100 kΩ, R3 = 400 kΩ Answer: A. Rk = Rf/gain: 100k/1, 100k/10 = 10k, 100k/25 = 4k. 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