Integrator OpAmp Question

Hey All

I have a Question About integrator op amp

First when the Vin signal is high Vout ramp negative ( in the opposite direction)
And vice versa when Vin is low

My Question is the capacitor at the output node charge from where when the signal is high it discharge not charge , i am no explanation for that

Who the capacitor in the output node charge and discharge ???

Thanks

the opamp input has very high impedance and the -verting terminal is at ground (virtual ground) level. so whatever the current coming from vout will go into Vin to charge or discharge so both will be loaded according to the RC condition and input singnal condition.

Thanks venkadesh

what i understand from your replay is that at input node of op amp their is a very high impedance , therefore no current pass in the opamp ( neglected ) , then the capacitor charge from output , my question the feedback from vout to vin how it's reversed to charge the capacitor and what is the effect of the impedance at output node ??

First when the Vin signal is high Vout ramp negative ( in the opposite direction)
And vice versa when Vin is low

Click to expand...

I assume you know that opamp is a differential amplifier have a very high diff gain.

Now consider input is 0 V, not inverting is already in 0 and the output also in 0 V.

Now you are applying some constant voltage in input. so the inverting will be more than non inverting so the opamp will decide to reach the -ve saturation. but when the voltage starts decreasing that will cause current through the feedback path,(cause virtual ground in -) and that will reduce the slope so that it will maintain the integration.

Actually speaking the slope is not limitted by the output impedance but the feed back current going through the input resistance.

I am not completely understand your question, so if you have any doubt please do ask.

eng_ahmed_osama, have you ever heard about the "Miller effect" (which gave this integrator its name)?

The following explanation is based on real opamp properties (large but finite open-loop gain and very small but finite voltage at the inverting input):

As a result, you can treat the whole circuit - referred to the inverting input, which is the only common node for R and C - as a simple RC-combination.
However, because of the Miller effect the capacitor C acts as a capacitor that is multiplied by the open-loop gain of the opamp.
Thus you have an RC lowpass with a very, very low cutoff frequency which can be used as an integrator.
Because of the very low voltage at the inverting opamp input we use instead the opamp output voltage (which is much larger and inverted).

LvW said:

....................
However, because of the Miller effect the capacitor C acts as a capacitor that is multiplied by the open-loop gain of the opamp.
Thus you have an RC lowpass with a very, very low cutoff frequency which can be used as an integrator.
..........................

Click to expand...

The integrator has no cutoff frequency as such and is not the same as an
RC low-pass filter. The integrator goes from DC to an arbitrary high frequency with a constant roll-off slope of 6db/octave.

Thanks For All Of you

So simply i want to summarize your word Venkadesh

The inverting terminal is the main key for the integrator circuit when the inverting terminal ( negative terminal ) goes higher than non inverting ( positive terminal ) Vout goes low

I was just confused about cap role-play

Venkadesh_M said:

Now you are applying some constant voltage in input. so the inverting will be more than non inverting so the opamp will decide to reach the -ve saturation. but when the voltage starts decreasing that will cause current through the feedback path,(cause virtual ground in -) and that will reduce the slope so that it will maintain the integration.

Click to expand...

In your words you say when voltage decrease the Vout will decrease as to maintain virtual ground to the node

So their another reason for Vout t decrease not only when Vin at inverting terminal increase higher than non inverting terminal ??

Thanks

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LvW said:

have you ever heard about the "Miller effect" (which gave this integrator its name)?

Thus you have an RC lowpass with a very, very low cutoff frequency which can be used as an integrator.
Because of the very low voltage at the inverting opamp input we use instead the opamp output voltage (which is much larger and inverted).

Click to expand...

i heard about miller effect c will be multiplied by (1+A)

my Question is related to hoe capacitor charge and discharge

For the point of low pass filter i cannot get the point

Thanks

In your words you say when voltage decrease the Vout will decrease as to maintain virtual ground to the node

Click to expand...

when not voltage decreases, when a constant (+)voltage applied on inverting terminal the Vout will decrease in constant rate until reaching the -ve vsat.

So their another reason for Vout t decrease not only when Vin at inverting terminal increase higher than non inverting terminal ??

Click to expand...

there is no other reason.

If you want to know about capacitors function, i can explain you.

when +ve voltage is applied on input, as you know the opamp will try to drive the output to negative sat, but this causes change in voltage in capacitor, the current equation for cap is

i proportional to dV/dt (change in voltage)

so it will cause a current through both cap and input resistor (same current). because of voltage in input resistor is constant the current also constant so dv/dt also constant and it will remain in same slope.

For beginners understanding a integrator is really difficult (without knowledge of opamp amplifiers).

crutschow said:

The integrator has no cutoff frequency as such and is not the same as an
RC low-pass filter. The integrator goes from DC to an arbitrary high frequency with a constant roll-off slope of 6db/octave.

Click to expand...

Yes - I totally agree with you. However, your comment is true for a pure mathematical definition/description of an ideal integrator only.
As you certainly know, in reality, it is not possible to realize such an ideal analog integrator. As a consequence, each real electronic circuit that is used as a n integrator is, in fact, a lowpass with a small but finite cut-off frequency. An ideal Miller intergator would require an opamp with infinite gain.

LvW said:

Yes - I totally agree with you. However, your comment is true for a pure mathematical definition/description of an ideal integrator only.
As you certainly know, in reality, it is not possible to realize such an ideal analog integrator. As a consequence, each real electronic circuit that is used as a n integrator is, in fact, a lowpass with a small but finite cut-off frequency. An ideal Miller intergator would require an opamp with infinite gain.

Click to expand...

It's true that ideal circuit response requires ideal components but a typical op amp has such a high open-loop gain that it acts very close to that of an ideal integrator at low frequencies. And it's roll-off is a constant 6 dB/octave from near DC to the high frequency limit of the op amp, unlike the flat gain of an RC low-pass filter below it's corner frequency. So you may call an integrator a low-pass filter but it's response is unlike any standard RC or LC low-pass filter.

crutschow said:

It's true that ideal circuit response requires ideal components but a typical op amp has such a high open-loop gain that it acts very close to that of an ideal integrator at low frequencies.

Click to expand...

Very close... yes - that´s exactly what I had in mind writing in post#5 that we can treat the circuit as an "RC lowpass with a very, very low cutoff frequency"
So - we are on the same line again.
On the other hand - I think, it is a nice and interesting view to explain the principle of this integrator using the Miller theorem.

LvW said:

Very close... yes - that´s exactly what I had in mind writing in post#5 that we can treat the circuit as an "RC lowpass with a very, very low cutoff frequency"
So - we are on the same line again.
On the other hand - I think, it is a nice and interesting view to explain the principle of this integrator using the Miller theorem.

Click to expand...

RC low pass filter is not a ideal integrator.

what will be the ouput of low pass filter if you give a square wave input? it should be exactly a ramp for an integrator.

Venkadesh_M said:

RC low pass filter is not a ideal integrator.
what will be the ouput of low pass filter if you give a square wave input? it should be exactly a ramp for an integrator.

Click to expand...

Yes - no doubt about this. I am not sure about the background (meaning) of your contribution.
Just for clarification:
1.) It is not possible to realize an ideal integrator (with infinite DC gain).
2.) Hence, a real integrator always has a finite DC gain. Thus, it is always a first order lowpass which can be used as an integrator for frequencies far beyond the cutoff (3dB) frequency of the lowpass.
This is because an ideal integrator needs a phase shift of 90 deg between input and output. Using real circuit components this can be achieved only for frequencies much larger than the 3dB frequency.

LvW said:

Yes - no doubt about this. I am not sure about the background (meaning) of your contribution.
Just for clarification:
1.) It is not possible to realize an ideal integrator (with infinite DC gain).
2.) Hence, a real integrator always has a finite DC gain. Thus, it is always a first order lowpass which can be used as an integrator for frequencies far beyond the cutoff (3dB) frequency of the lowpass.
This is because an ideal integrator needs a phase shift of 90 deg between input and output. Using real circuit components this can be achieved only for frequencies much larger than the 3dB frequency.

Click to expand...

:wink: It took me atleast half an hour to understand what you guys said. How do you think this can be understood by a OP asking from the basics.

Yeah its really interesting but not easy to understand for everyone.

Hi,

what will be the ouput of low pass filter if you give a square wave input? it should be exactly a ramp for an integrator.

Click to expand...

This is only true if the square wave frequecy is much higher than cutoff frequency of lowpass filter.


If the frequency is much lower than cutoff frequency then it shows a square wave with - i call it - "distorted" rise and fall lines.

And for sure everything inbetween.

Klaus

eng_ahmed_osama said:

i heard about miller effect c will be multiplied by (1+A)
my Question is related to hoe capacitor charge and discharge
Thanks

Click to expand...

Ahmed - based on the Miller effekt, it is rather easy to explain the capacitor charging.
As I have mentioned earlier, you can treat the whole circuit (post#1) as a simple classical passive RC combination - however with a very large capacitor value (enlarged by a factor (1+A) ).
Thus, you have a very large time constant T.
As a consequence, the step response of this lowpass (measured at the common node at the inverting opamp terminal) is, of course, an exponential function with a very, very large time constant.

Now comes the main point: The opamp is in its linear region for very small voltages at the inv. terminal only (µV range because of the very large open-loop gain A).
Hence, only the very first part of this exponential function is used - and available at the opamp output as a much larger signal (multiplied by A).
And this "very first part" can be approximated with very, very good accuracy by a linear rising voltage because of the very large time constant.
It is a well known fact that the quasi-linear part of the charging function increases with the time constant (an infinite time constant would theoreticcaly cause an ideal linear ramp function).

Hi,

how i calculate the capacitance of an integrator:

1) if i have square wave input, the integrator generates an triangle output:
* all the current through Rin is the same as through C: Iin = Uin / R (@ high impedance OPAMP input)
* C = I * t / U in units: F = A * s / V = As/V

* example: Input square wave 1kHz +/-1V, R = 10k; C = 1uF
--> Square wave 1kHz = 1ms total = 0.5V positive 1V + 0.5ms negative 1V
Iin = 1V/10k = 100uA = Ic
t = 0.5s

C = I * t / U --> U = I * t / C --> = 100uA * 0.5ms / 1uF = 0.05V = 50mV
Result: with the positive input the output ravles down 50mV, with the negative input it travels back 50mV upwards. => 50mV pp triangle.
With an integrator the start voltage is not defined. One has to stabilize it by circuit.

2) with sine input the output is sine also.
here i calculate with the impedances.
Uout = Uin * Zc / Rin

example:
1V sine with 1kHz input gives: (again with 10k and 1uF)
Rin = 10k, Zc = 1/ (2 x Pi x f x C) = 1/ (2 * 3.14 * 1kHz * 1uF) = 159 Ohms.

Uout = 1V * 159 Ohms / 10kOhms = 15.9mV
(here it is unimportant if you calculate with RMS, Vpp, Vp, as long as you use on both side the same definition)
1V RMS in ==> 159mV RMS out
1Vpp in ==> 159mVpp out

Also here: Output DC stabilisation circuit needed.


Klaus

Hi LVW

an infinite time constant would theoreticcaly cause an ideal linear ramp function

Click to expand...

Actually i understood whole concept. but i am writing something came to my mind yesterday.

An infinite time constant means infinite C for finite R. then the C would never charge to any voltage whatever the current applied. so whatever the gain we are going to get noting.

Infinite time constant will keep output of integrator at 0V regardless of input voltage. Mathematically ideal integrator will have time constant RC=1.

Venkadesh_M said:

Hi LVW
Actually i understood whole concept. but i am writing something came to my mind yesterday.
An infinite time constant means infinite C for finite R. then the C would never charge to any voltage whatever the current applied. so whatever the gain we are going to get noting.

Click to expand...

Yes - it is a mathematical limiting case. A rising time constant will result in a decreasing slope of the first part of the loading function - and the time, of course, will also increase until the "quasi-final" state is achieved.
And - mathematically spoken - for infinite T the slope is zero. You are right - the last sentence in my former post is somewhat misleading. Thank you.

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Borber said:

Infinite time constant will keep output of integrator at 0V regardless of input voltage. Mathematically ideal integrator will have time constant RC=1.

Click to expand...

Pleasae, can you elaborate and justify the last statement? I can imagine integrators, for example, with T=0.1s or T=100s.