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Op-amp as inverter, integrator and others

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.

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

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

                 +----[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

 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)

                 +----[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

                 +-----||-----+
                 |      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

                 +-----[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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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