In the previous tutorials we have seen circuits which show how an operational amplifier can be used as part of a positive or negative feedback amplifier or as an adder or subtractor type circuit using just pure resistances in both the input and the feedback loop.

But what if we were to change the purely resistive ( Rƒ ) feedback element of an inverting amplifier to that of a frequency dependant impedance, ( Z ) type complex element, such as a Capacitor, C. What would be the effect on the op-amps output voltage over its frequency range.

By replacing this feedback resistance with a capacitor we now have an RC Network connected across the operational amplifiers feedback path producing another type of operational amplifier circuit commonly called an **Op-amp Integrator** circuit as shown below.

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As its name implies, the **Op-amp Integrator** is an operational amplifier circuit that performs the mathematical operation of **Integration**, that is we can cause the output to respond to changes in the input voltage over time as the op-amp integrator produces an *output voltage which is proportional to the integral of the input voltage*.

In other words the magnitude of the output signal is determined by the length of time a voltage is present at its input as the current through the feedback loop charges or discharges the capacitor as the required negative feedback occurs through the capacitor.

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When a step voltage, Vin is firstly applied to the input of an integrating amplifier, the uncharged capacitor C has very little resistance and acts a bit like a short circuit allowing maximum current to flow via the input resistor, Rin as potential difference exists between the two plates. No current flows into the amplifiers input and point X is a virtual earth resulting in zero output. As the impedance of the capacitor at this point is very low, the gain ratio of Xc/Rin is also very small giving an overall voltage gain of less than one, ( voltage follower circuit ).

As the feedback capacitor, C begins to charge up due to the influence of the input voltage, its impedance Xc slowly increase in proportion to its rate of charge. The capacitor charges up at a rate determined by the RC time constant, ( τ ) of the series RC network. Negative feedback forces the op-amp to produce an output voltage that maintains a virtual earth at the op-amp’s inverting input.

Since the capacitor is connected between the op-amp’s inverting input (which is at earth potential) and the op-amp’s output (which is negative), the potential voltage, Vc developed across the capacitor slowly increases causing the charging current to decrease as the impedance of the capacitor increases. This results in the ratio of Xc/Rin increasing producing a linearly increasing ramp output voltage that continues to increase until the capacitor is fully charged.

At this point the capacitor acts as an open circuit, blocking any more flow of DC current. The ratio of feedback capacitor to input resistor ( Xc/Rin ) is now infinite resulting in infinite gain. The result of this high gain (similar to the op-amps open-loop gain), is that the output of the amplifier goes into saturation as shown below. (Saturation occurs when the output voltage of the amplifier swings heavily to one voltage supply rail or the other with little or no control in between).

The rate at which the output voltage increases (the rate of change) is determined by the value of the resistor and the capacitor, “RC time constant“. By changing this RC time constant value, either by changing the value of the Capacitor, C or the Resistor, R, the time in which it takes the output voltage to reach saturation can also be changed for example.

If we apply a constantly changing input signal such as a square wave to the input of an **Integrator Amplifier** then the capacitor will charge and discharge in response to changes in the input signal. This results in the output signal being that of a sawtooth waveform whose output is affected by the RC time constant of the resistor/capacitor combination because at higher frequencies, the capacitor has less time to fully charge. This type of circuit is also known as a **Ramp Generator** and the transfer function is given below.

We know from first principals that the voltage on the plates of a capacitor is equal to the charge on the capacitor divided by its capacitance giving Q/C. Then the voltage across the capacitor is output Vout therefore: -Vout = Q/C. If the capacitor is charging and discharging, the rate of charge of voltage across the capacitor is given as:

But dQ/dt is electric current and since the node voltage of the integrating op-amp at its inverting input terminal is zero, X = 0, the input current I(in) flowing through the input resistor, Rin is given as:

The current flowing through the feedback capacitor C is given as:

Assuming that the input impedance of the op-amp is infinite (ideal op-amp), no current flows into the op-amp terminal. Therefore, the nodal equation at the inverting input terminal is given as:

From which we derive an ideal voltage output for the **Op-amp Integrator** as:

To simplify the math’s a little, this can also be re-written as:

Where ω = 2πƒ and the output voltage Vout is a constant 1/RC times the integral of the input voltage Vin with respect to time. The minus sign ( – ) indicates a 180^{o} phase shift because the input signal is connected directly to the inverting input terminal of the op-amp.

If we changed the above square wave input signal to that of a sine wave of varying frequency the **Op-amp Integrator ** performs less like an integrator and begins to behave more like an active “Low Pass Filter”, passing low frequency signals while attenuating the high frequencies.

At 0Hz or DC, the capacitor acts like an open circuit blocking any feedback voltage resulting in very little negative feedback from the output back to the input of the amplifier. Then with just the feedback capacitor, C, the amplifier effectively is connected as a normal open-loop amplifier which has very high open-loop gain resulting in the output voltage saturating.

This circuit connects a high value resistance in parallel with a continuously charging and discharging capacitor. The addition of this feedback resistor, R_{2} across the capacitor, C gives the circuit the characteristics of an inverting amplifier with finite closed-loop gain of R_{2}/R_{1}. The result is at very low frequencies the circuit acts as an standard integrator, while at higher frequencies the capacitor shorts out the feedback resistor, R_{2} due to the effects of capacitive reactance reducing the amplifiers gain.

Unlike the DC integrator amplifier above whose output voltage at any instant will be the integral of a waveform so that when the input is a square wave, the output waveform will be triangular. For an AC integrator, a sinusoidal input waveform will produce another sine wave as its output which will be 90^{o} out-of-phase with the input producing a cosine wave.

Further more, when the input is triangular, the output waveform is also sinusoidal. This then forms the basis of a Active Low Pass Filter as seen before in the filters section tutorials with a corner frequency given as.

In the next tutorial about Operational Amplifiers, we will look at another type of operational amplifier circuit which is the opposite or complement of the **Op-amp Integrator** circuit above called the Differentiator Amplifier. As its name implies, the differentiator amplifier produces an output signal which is the mathematical operation of differentiation, that is it produces a voltage output which is proportional to the input voltage’s rate-of-change and the current flowing through the input capacitor.

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super stuff

If four is add to one that answer is five.

Hi Wayne! This site is really great!

I wonder what happens in the circuit in the last picture (“The AC Op-amp Integrator with DC Gain Control”) if you apply only DC to the input and suddenly you change that to 0V. What will the R2*C cause to the output? I am asking this because I also read your article about “Discharging circuits” and MAYBE that comes into play here but am not sure how.

Prove that, Vo = -RfCi(d e in /dt)

The course is interesting and allows to understand the use of an integrator efficiently. But i came to this site to find out if there was a solution for a question in a text book regarding the crossover frequency of an op amp with three corner frequencies.the only data given is a frequency response curve and i am been asked to compute the crossover frequency. without a phase angĺe plot i am lost and do not know how to begin and my sseveral attemps to solve the equation have been in vain.i would much appreciate if Electronics Tutorials could lecture on the subjects with a few examples so that i may try again at the equations.by the way, the book is the work of american author R.A Gayakwad and the question is on page 233 no.6.9.hoping thhat you can help and being much obliged to you.i am yours sincerely

Eric Lamarque

I do not own the book to which you refer so can not make any comment about its contents.

…Oops! I meant inverting amplifier in my previous suggestion.

This is nearly all good, but I am having a little trouble with a small part of your explanation of ac op-amp integrator. Can you please review the section under ‘The AC or Continuous Op-amp Integrator’ ?

‘…This circuit connects a high value resistance in parallel with a continuously charging and discharging capacitor. The addition of this feedback resistor, R2 across the capacitor, C gives the circuit the characteristics of an inverting amplifier with finite closed-loop gain of R2/R1. The result is at very low frequencies the circuit acts as an standard integrator, while at higher frequencies the capacitor shorts out the feedback resistor, R2 due to the effects of capacitive reactance reducing the amplifiers gain.’

In the 3rd sentence, it would be clear to me if it read something like this ‘…The result is that at very low frequencies the circuit acts as a standard non-inverting amplifier with finite close loop gain of R2/R1, while at higher frequencies low capacitive reactance of C shorts out feedback resistor R2 due to…’

Do you agree?

For the Op-amp Integrator Ramp Generator, the input square wave should be between -1v and +1v instead of -1v and 0v or else you would get an output wave that would just reach saturation and remain there.

You are right.

That depends if you are integrating using a single or dual op-amp supply. The point is that for a step input the output magnitude is a constant ramp with a slope in the opposite direction because the voltage output is taken from the capacitor. Then we can use this effect to convert a rectangular or square wave waveform at the input into a triangular waveform at the output with the same frequency as the input.

Hey. I do not understand how the Vout formula was simplified

the charge at virtual earth

Vc =Vi-Vout

but Vi =0

then we get

Vc=-Vout

the charge on capacitor

Q=CfVc

=> Q=-CfVout

At the end of the time t say

0 to t/i dt = =-CfVout

we know that

i=Vi/R1

substitute in above equation we get

0 to t/(Vi/R1) =-CfVout

=>Vout = -1/CfR1(0 to t/Vi dt)

Hello guys, please help me out, I have working on integrator circuits both active and passive for more than one week, but I have not recorded any success, I know the circuit is easy to make, but It has frustrated me. please help me

You really don’t expect anybody to help you by posting this question, right? You should show us your circuit diagram, your physical construction, explain what’s not working and then you can hope that somebody will bother to answer to you. Otherwise, the answer is – take a break, read again about integrators and opamps, double check your circuit, take measurements in various points and find the problem.