---
title: "Op-amp as inverter, integrator and others"
subject: "Instrumentation I"
chapter: 4
key: "semester_4/instrumentation_i:opamp-basic-circuits"
source: https://www.bctengineer.com/q/semester_4/instrumentation_i/opamp-basic-circuits
site: BCT Engineering
---

# Op-amp as inverter, integrator and others

Instrumentation I · Chapter 4: Electrical Signal Processing and transmission

- Asked in: year not given, 8 marks
- Video: [Operational Amplifiers - Inverting & Non Inverting Op-Amps](https://www.youtube.com/watch?v=idJEMYhrIfs) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_i/opamp-basic-circuits
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_i/chapter-4

## Question

Explain with circuit diagram, how does an op-amp act as: a. Inverter b. Non inverter c. Subtractor d. Integrator e. Differentiator.

## Short answer

Inverter: Vo = −(Rf/R1) Vi. Non-inverter: Vo = (1+Rf/R1) Vi. Subtractor: Vo = (R2/R1)(V2−V1). Integrator: Vo = −1/RC ∫ Vi dt. Differentiator: Vo = −RC dVi/dt. All use negative feedback and virtual ground.

## Answer

An ideal op-amp has infinite gain, infinite input impedance and zero output impedance. With negative feedback the two inputs are at the same voltage (virtual short, V+ = V−), and no current enters the input terminals. When the (+) input is grounded, the (−) input is a virtual ground.

#### a. Inverter
```text
                 +----[Rf]----+
                 |            |
 Vi ---[R1]------+---(-)\     |
                         >----+---- Vo
          gnd -------(+)/
```
The current through R1, Vi/R1, also flows through Rf, so **Vo = −(Rf/R1) Vi**. For Rf = R1 the gain is −1 (a pure inverter).

#### b. Non-inverter
```text
 Vi -----------------(+)\
                         >----+---- Vo
             +-------(-)/     |
             |                |
             +------[Rf]------+
             |
            [R1]
             |
            gnd
```
V− = Vi = Vo·R1/(R1 + Rf), so **Vo = (1 + Rf/R1) Vi**. The gain is ≥ 1 and the input impedance is high. With Rf = 0 (and R1 removed) it is a voltage follower (gain 1).

#### c. Subtractor (difference amplifier)
```text
                 +----[R2]----+
                 |            |
 V1 ---[R1]------+---(-)\     |
                         >----+---- Vo
 V2 ---[R1]------+---(+)/
                 |
                [R2]
                 |
                gnd
```
Using superposition: **Vo = (R2/R1)(V2 − V1)**. With all four resistors equal, Vo = V2 − V1.

#### d. Integrator
```text
                 +-----||-----+
                 |      C     |
 Vi ---[R]-------+---(-)\     |
                         >----+---- Vo
          gnd -------(+)/
```
The current Vi/R charges C: Vi/R = −C dVo/dt, so **Vo = −(1/RC) ∫ Vi dt**. A square-wave input gives a triangular output. In practice a large resistor is added across C to limit DC drift.

#### e. Differentiator
```text
                 +-----[R]----+
                 |            |
 Vi ----||-------+---(-)\     |
        C                >----+---- Vo
          gnd -------(+)/
```
The current through C, C dVi/dt, flows through R: **Vo = −RC (dVi/dt)**. A triangular input gives a square-wave output. A small resistor in series with C limits high-frequency noise.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The output of the op-amp inverter is:
   A. Vo = (1 + Rf/R1) Vi
   B. Vo = -(Rf/R1) Vi
   C. Vo = -(R1/Rf) Vi
   D. Vo = Vi
   Answer: B. Using virtual ground, the input current Vi/R1 flows through Rf, giving Vo = -(Rf/R1)Vi. (Key point: Inverter gain)

2. What is the gain of a non-inverting amplifier, and what happens when Rf = 0?
   A. -Rf/R1; becomes an inverter
   B. Rf/R1; gain becomes zero
   C. (1 + Rf/R1); it becomes a voltage follower with gain 1
   D. (1 - Rf/R1); gain is -1
   Answer: C. V- = Vi = Vo R1/(R1+Rf) gives Vo = (1+Rf/R1)Vi; Rf = 0 gives unity gain. (Key point: Non-inverter)

3. An integrator with a square-wave input gives which output, and what is its relation?
   A. Square wave; Vo = -RC dVi/dt
   B. Sine wave; Vo = Vi
   C. Spikes; Vo = -Rf/R1 Vi
   D. Triangular wave; Vo = -(1/RC) integral of Vi dt
   Answer: D. Current Vi/R charges C, so Vo = -(1/RC)∫Vi dt; integrating a square wave gives a triangle. (Key point: Integrator)

4. Which pair is correct?
   A. Differentiator Vo = -RC dVi/dt; subtractor Vo = (R2/R1)(V2 - V1)
   B. Differentiator Vo = -(1/RC)∫Vi dt; subtractor Vo = V1 + V2
   C. Differentiator Vo = RC Vi; subtractor Vo = (R2/R1)(V1 + V2)
   D. Differentiator Vo = Vi/RC; subtractor Vo = 0
   Answer: A. The current C dVi/dt through R gives -RC dVi/dt; superposition gives the difference amplifier result. (Key point: Differentiator and subtractor)

## Related in other subjects

- [Op-amp properties and differentiator](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-differentiator) — Basic Electronics Engineering, chapter 4, asked 2×
- [Transfer function of op-amp integrator-type circuit](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-opamp-r1-r2-c) — Control Systems, chapter 2
- [Op-amp properties and integrator](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-integrator) — Basic Electronics Engineering, chapter 4
- [Transfer function of op-amp phase-shift circuit](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-opamp-allpass-r1-c-r2) — Control Systems, chapter 2, asked 2×