--- title: "Find coefficients of op-amp summing circuit" subject: "Advanced Electronics" chapter: 1 key: "semester_4/advanced_electronics:summing-amplifier-coefficients" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/summing-amplifier-coefficients site: BCT Engineering --- # Find coefficients of op-amp summing circuit Advanced Electronics · Chapter 1: Operational Amplifier Circuits - Asked in: 2070 Asar, 3+2+2 marks - Video: [Operational Amplifier Example implementing a Linear Voltage Combiner](https://www.youtube.com/watch?v=wVEnmBLTA20) — Engineering Prof. - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/summing-amplifier-coefficients - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-1 ## Question For the circuit shown in figure below, the output voltage can be expressed as V0 = a1v1 + a2v2 + a3v3. Now: (i) Find the value of a1, a2 and a3. Also find the value of v0 if (ii) R4 is short circuited and (iii) R4 is removed. [Figure: one op-amp. V1 through R1 and V3 through R3 both go to the inverting input; feedback resistor R4 from the output V0 to the inverting input. V2 through R2 goes to the non-inverting input, which is also connected to ground through R5. No resistor values are given.] ## Short answer a1 = –R4/R1, a3 = –R4/R3, a2 = (1 + R4/(R1||R3))·R5/(R2+R5). If R4 is shorted, Vo = V2·R5/(R2+R5). If R4 is removed, there is no feedback and Vo saturates to ±Vsat depending on input difference. ## Answer Circuit: V1 (through R1) and V3 (through R3) to the inverting input, R4 from output to inverting input, V2 through R2 to the non-inverting input with R5 to ground. Ideal op-amp assumed. ```text R4 +-----/\/\/-----+ V1 --R1--+ | V3 --R3--+---(-) | >-----+---- V0 V2 --R2--+---(+) | R5 | GND ``` #### (i) Coefficients a1, a2, a3 Non-inverting input voltage (voltage divider, no input current): ```text V+ = V2·R5/(R2 + R5) ``` KCL at the inverting input, with V− = V+ = Vp: ```text (V1 - Vp)/R1 + (V3 - Vp)/R3 + (V0 - Vp)/R4 = 0 V0 = -(R4/R1)V1 - (R4/R3)V3 + Vp(1 + R4/R1 + R4/R3) ``` Substituting Vp: ```text a1 = -R4/R1 a3 = -R4/R3 a2 = (1 + R4/R1 + R4/R3) · R5/(R2 + R5) = [1 + R4/(R1 ‖ R3)] · R5/(R2 + R5) ``` So V0 = −(R4/R1)V1 + [1 + R4/(R1‖R3)]·[R5/(R2 + R5)]·V2 − (R4/R3)V3. #### (ii) R4 short-circuited (R4 = 0) Then a1 = a3 = 0 and a2 = R5/(R2 + R5). The output is tied directly to the inverting input and the circuit is a voltage follower of V+: ```text V0 = V2·R5/(R2 + R5) ``` V1 and V3 have no effect (the op-amp output supplies their currents). #### (iii) R4 removed (R4 = ∞) There is no feedback, so the op-amp works open-loop as a **comparator** with gain A (very large): ```text V- = (V1·R3 + V3·R1)/(R1 + R3) V+ = V2·R5/(R2 + R5) V0 = +Vsat if V+ > V- V0 = -Vsat if V+ < V- ``` The output saturates near the supply rails (about ±13 V for ±15 V supply); the linear relation V0 = a1v1 + a2v2 + a3v3 no longer holds. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are a1 and a3? A. a1 = 1 + R4/R1, a3 = 1 + R4/R3 B. a1 = −R4/R1, a3 = −R4/R3 C. a1 = R4/R1, a3 = R4/R3 D. a1 = −R1/R4, a3 = −R3/R4 Answer: B. V1 and V3 use inverting inputs, so each is multiplied by −Rf/R with Rf = R4. (Key point: a1 and a3) 2. What is a2? A. R4/R5 B. R5/(R2 + R5) C. −R4/R2 D. [1 + R4/(R1‖R3)]·R5/(R2 + R5) Answer: D. V+ = V2 R5/(R2 + R5) is amplified by the non-inverting gain 1 + R4/R1 + R4/R3. (Key point: a2) 3. With R4 shorted, what is V0? A. −V1 − V3 B. Saturated C. V2 R5/(R2 + R5) D. 0 Answer: C. Output is tied to the inverting input, so it is a follower of V+. (Key point: R4 shorted) 4. With R4 removed, how does the op-amp behave? A. Integrator B. Linear summer C. Comparator saturating near ±Vsat D. Voltage follower Answer: C. No feedback gives open-loop operation, so output goes to ±Vsat. 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