---
title: "Virtual short circuit and inverting integrator"
subject: "Basic Electronics Engineering"
chapter: 4
key: "semester_2/basic_electronics_engineering:virtual-short-integrator"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/virtual-short-integrator
site: BCT Engineering
---

# Virtual short circuit and inverting integrator

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

- Asked in: 2075 Baisakh, 2+2+2 marks
- Video: [Virtual Short and Virtual Ground in Operational Amplifier | Analog Electronics | Engineering Funda](https://www.youtube.com/watch?v=b7F0nmfveSc) — Engineering Funda
- 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-short-integrator
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-4

## Question

What do you mean by virtual short circuit in OP amp? Draw the circuit diagram of the inverting integrator and show that the output is proportional to the time-integral of the input.

## Short answer

Virtual short: with negative feedback and very high open-loop gain, V+ ≈ V- because Vo = A(V+ - V-) stays finite only if V+ - V- ≈ 0, so inputs are nearly same potential but no current flows. Inverting integrator: input through R, feedback C, virtual ground yields Vin/R = -C dVo/dt, integrating gives Vo = -(1/RC)∫ Vin dt, proportional to time-integral with minus sign.

## Answer

#### Virtual short circuit
In an op-amp with negative feedback, the open-loop gain is extremely large (ideally infinite), so the output Vo = A (V+ − V−) stays finite only when the difference V+ − V− = Vo/A ≈ 0. The two input terminals are therefore at practically **the same potential**, as if they were short-circuited, but no current flows between them (the input resistance is infinite). This is the **virtual short**. If the non-inverting terminal is grounded, the inverting terminal is at virtual ground (0 V).

#### Inverting integrator
```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. What does virtual short circuit mean?
   A. The two inputs are physically connected by a wire
   B. The output is shorted to the inverting input
   C. The two op-amp inputs are at practically the same potential, but no current flows between them
   D. The inputs carry equal large currents
   Answer: C. With negative feedback the difference V+ - V- = Vo/A is about 0, yet input resistance is infinite so no current flows. (Key point: Virtual short)

2. Why is V+ - V- almost zero with negative feedback?
   A. Vo = A(V+ - V-) is finite while A is very large
   B. Because input resistance is zero
   C. Because the output resistance is infinite
   D. Because the capacitor blocks dc
   Answer: A. A finite output divided by a huge gain gives a negligible input difference. (Key point: Why difference is zero)

3. 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)

4. 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)

## Related in Basic Electronics Engineering

- [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
- [Ideal op-amp properties; non-inverting gain](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-noninverting-gain) — Basic Electronics Engineering, chapter 4, asked 3×
- [Ideal op-amp properties; inverting gain](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-inverting-gain) — Basic Electronics Engineering, chapter 4, asked 3×
- [Voltage gain of non-inverting op-amp](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-noninverting-gain) — Basic Electronics Engineering, chapter 4, asked 2×
- [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×
- [Op-amp properties; square wave generator](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-five-props-square-wave) — Basic Electronics Engineering, chapter 4

## Related in other subjects

- [Input impedance of inverting op-amp](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ainverting-input-impedance) — Advanced Electronics, chapter 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 as inverter, integrator and others](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_i%3Aopamp-basic-circuits) — Instrumentation I, chapter 4
- [Output voltage of two op-amp IA](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Atwo-opamp-ia-output) — Advanced Electronics, chapter 4