Op-amps can be used as integrators, where the output of the op-amp is the integral of the input voltage, the values of the resistor and the capacitor. The example below shows the op-amp integrator consists of R and C connected in an inverting configuration. The Vout to Vin transfer function is also described. The inverting integrator provides an output voltage that is proportional to the time-integral of the input.


By placing a capacitor in the feedback path and a resistor at the inverting input, we obtain the inverting integrator.
If the initial voltage on C (at t=0) is denoted Vinitial , then the circuit provides an output voltage that is proportional to the time-integral of the input. RC is the integrator time constant. Where Vin and Vout are functions of time, Vinitial is the output voltage of the integrator at time t = 0

Transfer Function

\begin{align} V_{out} = \int_{0}^{t} -\frac{V_{in}}{RC} dt + V_{initial} \end{align}

Frequency Domain
In the frequency domain, the operation of the op-amp integrator can be described alternatively. The bode plot for the integrator magnitude response can be obtained from |Vout/Vin | = 1/(ωRC). It is a straight line of slope -20dB/decade. This line intercepts the 0 dB line at the frequency that makes | Vout/Vin | = 1, which is the integrator frequency ω = 1/(RC) and is simply the inverse of the integrator time constant.


Design an inverting integrator with an input resistance of 10 kΩ and an integration time constant of 1 ms.
What is the value of capacitor? What is the gain magnitude and phase angle of this circuit at 10 Hz and at 100 Hz? What is the frequency at which the gain magnitude is unity? In order to limit the DC signal gain of the integrator circuit to 100, a resistor RF is added in parallel with the capacitor. Find the required resistor value.

Given the transfer function of the integrator: Vo /Vi =-1/(jωRC), ω=2 π f, and the magnitude |Vout/Vin |=1/(ωRC) and phase φ=90°

Integration time constant RC = 10-3s, R = 10kΩ
C = 1m /10 k = 0.1 μf
At f = 10 Hz, |Vout/Vin| = 1/(2 π RC) ≈ 16, φ = +90°
At f = 100 Hz, |Vout/Vin | = 1/(2 π fRC) ≈ 1.6, φ = +90°

Let |Vout/Vin| = 1/(2 π f R C) = 1, we get f ≈ 160 Hz

For DC signal, |Vout/Vin| = R1/R =100, R = 10 kΩ, R1 = 1 MΩ

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