Integrator OpAmp Question

[KlausST]

Hi,

I can imagine integrators, for example, with T=0.1s or T=100s.

for a low pass filter t = the point where the output voltage reaches 63% of it´s end voltage level (after infinite time)
it is t = R x C

for an integrator seen as a lowpass i find it difficult to recognize a t = 0.1s or T = 100s,
because from the graphical view it can not reach the 63% threshold, because there is no end voltage level.

But...

for regulation loops (PID) one defined the I parameter as a time.
Here the T = 0.1s or T = 100s is defined as:
* Starting an integrator with zero output voltage
* apply voltage x on the input
* wait until the output voltage (here negative) reaches the same voltage as the input voltage.
* this delay is defined as the time constant of the I part of an PID regualting loop.
(this time is independent of the absolute input voltage)

example:
Rin = 10k, C = 100nF
Applying 1 volt at the input causes a current of U/R to charge the capacitor. = 1V / 10k = 100uA
C = I * t / U ==> t = C * U / I = 100nF * 1V / 100uA = 1ms.

And now we have the same time constant as with a low pass filter: t = R x C: 10k Ohms x 100nF = 1ms.


Klaus

 

[LvW]

for an integrator seen as a lowpass i find it difficult to recognize a t = 0.1s or T = 100s,
because from the graphical view it can not reach the 63% threshold, because there is no end voltage level.
Klaus

Yes - of course.
However, in post#19 Borber spoke about a "Mathematically ideal integrator" which - according to his opinion - "will have time constant RC=1."

 

[Venkadesh_M]

Rin = 10k, C = 100nF
Applying 1 volt at the input causes a current of U/R to charge the capacitor. = 1V / 10k = 100uA
C = I * t / U ==> t = C * U / I = 100nF * 1V / 100uA = 1ms.

And now we have the same time constant as with a low pass filter: t = R x C: 10k Ohms x 100nF = 1ms.

The formula C = I * t / U cant be used here, because I is not constant for a RC circuit having a step input.

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Oh its ok you have used it on a opamp integrator, right ?

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Summary :

For an Integrator the time constant should be as much as possible to get a nearly Ideal response, but it should not be a infinite.

When the time constant increases the response of the Integrator will be more alike ideal but there will be a reduction in amplitude.

The selection of Time constant is based on maximum frequency of the input signal and amplitude of the output of the same.

 

[KlausST]

Hi,

yes, OPAMP integrator...

the response of the Integrator will be more alike ideal

I used an integrator to generate a reference triangle out of an square input (0V..1.8V @10kHz with LT1639).
From my measurements the triangle signal had linearity errors of less than 1%.
This mostly depends on OPAMP, and there are better OPAMPs than the used LT1639.

I´d expect no visible distortion in the output of an integrator. (with good selected devices. No "Z" or "Y" ceramic capacitors)

Klaus

 

[Borber]

https://en.wikipedia.org/wiki/Op_amp_integrator

Above words "Practical circuit" there is equation for vo as function of vi. Before integral it has a constant 1/RC. If RC=infinite then vo is zero for any vi. RC=1 means that integrator has gain=1.

 

[LvW]

https://en.wikipedia.org/wiki/Op_amp_integrator

RC=1 means that integrator has gain=1.

At which frequency? An integrator has a gain that drops continuously with 20dB/dec.
It has a unity gain of 0 dB at one single frequency only: wo=1/RC.

 

[KlausST]

Hi,

RC=1 means that integrator has gain=1.

How can this be?
Integrator output is frequency dependent.
Twice the input freuquency gives half the output voltage.

In my understanding gain = 1 means 1V in --> 1 V out.
But for an integrator this is true only for one specific frequency.

Klaus

P.S: too late.. i had a break before sending my post...

 

[Borber]

Integrator expression is simple vo=integral vi dt and it determines equivalent in frequency domain which is 1/f. Practical realisation with OPA has it's limitations such is corner frequency or zero in transfer function like low pass filter.

 

[crutschow]

I'm not sure if it has been stated explicitly in this thread, but here's a simple relationship between the input and output of an integrator. For a DC input of V volts, in one time-constant, the output voltage of an integrator linearly changes by -V volts. From that it's easy to calculate an integrator output change due to any DC input. Thus for an RC op amp integrator value of 1 second and a +/- 1V square-wave input with a 2 second period, the output will be a 1Vpp triangle-wave.

 

[LvW]

equivalent in frequency domain which is 1/f.

Transfer function (ideal) is H(s)=1/sT

 

[Borber]

Look here: https://www.rapidtables.com/math/calculus/laplace_transform.htm what is Laplace transform of integral and if f(t)=1. It is 1/s.