---
title: "Op-amp properties and integrator"
subject: "Basic Electronics Engineering"
chapter: 4
key: "semester_2/basic_electronics_engineering:opamp-props-integrator"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/opamp-props-integrator
site: BCT Engineering
---

# Op-amp properties and integrator

Basic Electronics Engineering · Chapter 4: The Operational Amplifier and Oscillator

- Asked in: 2069 Poush, 2+4 marks
- Video: [Op-Amp Integrator (with Derivation and Solved Examples)](https://www.youtube.com/watch?v=OPvs7A554Rw) — 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/opamp-props-integrator
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-4

## Question

State four important properties of ideal operational amplifier. Draw the circuit diagram of an integrator using op-amp and show that output is the integration of input signal.

## Short answer

Ideal op-amp properties: infinite open-loop gain, infinite input resistance, zero output resistance, infinite bandwidth. Integrator: input through R, feedback capacitor C. Virtual ground: Vin/R = -C dVo/dt => dVo/dt = -Vin/(RC). Integrating gives Vo(t) = -(1/RC) ∫ Vin dt + Vo(0). Output is integral.

## Answer

#### Four properties of an ideal op-amp
1. Infinite open-loop voltage gain.
2. Infinite input resistance (zero input current).
3. Zero output resistance.
4. Infinite bandwidth, infinite CMRR and zero offset voltage.

#### Integrator using op-amp
```text
               +------||------+
               |      C       |
               |     |\       |
 Vin o--[ R ]--+-----|- \     |
                     |   >----+----o Vo
               +-----|+ /
               |     |/
              GND
```
The input is applied through R to the inverting input; a capacitor C is connected from the output to the inverting input; the non-inverting input is grounded.

**Proof**
- Virtual ground: V− = 0.
- Input current: i = Vin/R. No current enters the op-amp, so the same current flows through C.
- Capacitor current: i = C d(0 − Vo)/dt = −C dVo/dt.

Equating: Vin/R = −C dVo/dt

dVo/dt = −Vin/(RC)

Integrating both sides:

**Vo(t) = −(1/RC) ∫₀ᵗ Vin dt + Vo(0)**

So the output is proportional to the time integral of the input, with a minus sign. Examples: for a constant (dc) input, the output is a ramp (a linearly changing voltage) with slope −Vin/RC; for a square wave input, the output is a triangular wave; for a sine input Vm sin ωt, the output is (Vm/ωRC) cos ωt, a cosine wave that leads the input by 90°.

**Practical note:** a resistor Rf across C (about 10R) limits the dc gain, preventing output drift and saturation.

## 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. Infinite open-loop gain, infinite input resistance, zero output resistance
   B. Zero open-loop gain, infinite input resistance, infinite output resistance
   C. Infinite open-loop gain, zero input resistance, zero output resistance
   D. Finite gain, zero input resistance, infinite output resistance
   Answer: A. 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. In the inverting integrator, why is the inverting input taken as 0 V?
   A. The capacitor is shorted for dc
   B. The input resistor R is zero
   C. The non-inverting input is grounded and with negative feedback the two inputs are at the same potential
   D. The op-amp output is always 0 V
   Answer: C. Negative feedback with huge gain forces V- = V+ = 0, so the inverting input is a virtual ground. (Key point: Virtual ground)

3. What is the capacitor current in the integrator, with V- = 0?
   A. i = -C dVo/dt
   B. i = C Vo/R
   C. i = +C dVin/dt
   D. i = Vo/(RC)
   Answer: A. Voltage across C is (0 - Vo), so i = C d(0 - Vo)/dt = -C dVo/dt. (Key point: Capacitor current)

4. Which expression is the integrator output?
   A. Vo = -RC integral of Vin dt
   B. Vo = +(1/RC) integral of Vin dt
   C. Vo = -(1/RC) integral of Vin dt
   D. Vo = -RC dVin/dt
   Answer: C. Equating Vin/R = -C dVo/dt and integrating gives Vo = -(1/RC) integral Vin dt, with a minus sign. (Key point: Result of integration)

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## Related in other subjects

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