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Op-amp properties and integrator

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.

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.

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

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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