Integrator OpAmp Question

Hi,

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

Click to expand...

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

KlausST said:

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

Click to expand...

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."

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.

Click to expand...

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.

Hi,

yes, OPAMP integrator...

the response of the Integrator will be more alike ideal

Click to expand...

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

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.

Borber said:

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

RC=1 means that integrator has gain=1.

Click to expand...

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.

Hi,

RC=1 means that integrator has gain=1.

Click to expand...

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...

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.

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.

Borber said:

equivalent in frequency domain which is 1/f.

Click to expand...

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

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.