Virtual short circuit and inverting integrator

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

               +------||------+
               |      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.