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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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