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Integrator operation

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zxcv2201

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I wonder how it will work if there is a reference voltage as shown in the following figure.

1674877491580.png
 

Of course, a voltage Vref at the non-inv. opamp input also acts as a signal voltage.
The corresponding gain expression (ideal opamp) is

Acl+=1+(1/jwC)/R=1+1/jwT (with T=RC).

When a signal Vs at "IN" is available the opamp output is (superposition):

Vout=Vref*[1+(1/jwT)] - Vs*(1/jwT)=(Vref-Vs)*(1/jwT) + Vref.

Comment:
This is an idealized expression. The circuit will not work as a "stand-alone" unit due to opamp non-idealities (offset). However, it will work well when part of an overall negative DC feedback loop.
 

It will still work.
If the reset switch is periodically closed, then the circuit acts as a voltage follower when the switch is closed, and the integration output starting voltage will be equal to Vref when the switch opens.
 

It will still work.
If the reset switch is periodically closed, then the circuit acts as a voltage follower when the switch is closed, and the integration output starting voltage will be equal to Vref when the switch opens.
Yes - that is obvious. My comment applies to the circuit without such a switch. I think, there are only a few applications which allow/require such a switch.
 

Yes - that is obvious. My comment applies to the circuit without such a switch. I think, there are only a few applications which allow/require such a switch.
It may not be obvious to the TS.
The posted circuit does have such a switch, and I didn't understand why you ignored that, hence my post.
It's called a boxcar integrator/averager, and has often been used to recover periodic signals from noise.
 

It may not be obvious to the TS.
The posted circuit does have such a switch, and I didn't understand why you ignored that, hence my post.
It's called a boxcar integrator/averager, and has often been used to recover periodic signals from noise.
I am very sorry that I have "ignored" the switch in my comment at the end of my contribution (post#3). You have corrected me - and I have confirmed that you are right (post#5).
So - what is the problem?
Because you are mentioning the TS (your first sentence): Perhaps you have noticed that I did not ignore the question from the TS regarding the reference voltage (my answer in post#3).
 

It may not be obvious to the TS.
The posted circuit does have such a switch, and I didn't understand why you ignored that, hence my post.
It's called a boxcar integrator/averager, and has often been used to recover periodic signals from noise.
I had always heard of this but never investigated, fascinating.



Regards, Dana.
 

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  • zi_whitepaper_principles_of_boxcar_averaging_latest.pdf
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I am very sorry that I have "ignored" the switch in my comment at the end of my contribution (post#3). You have corrected me - and I have confirmed that you are right (post#5).
So - what is the problem?
You confirmed my post by saying it was "obvious", which is dismissive of what I said.
Now you take exception to my reply to that with another dismissive comment.

So the problem is, I don't tolerate condescending/patronizing replies.
 

The posted circuit ......It's called a boxcar integrator/averager, and has often been used to recover periodic signals from noise.
I will now be very careful with the wording of my question:
If the circuit shown is to work as a "boxcar integrator/averager", doesn't we require Vref=0 ?
I think, otherwise the task to recover periodic signals from noise is disturbed, isn't it?
(See the expression for Vout in my post#3).
Regards
LvW
 

If the circuit shown is to work as a "boxcar integrator/averager", doesn't we require Vref=0 ?
I think, otherwise the task to recover periodic signals from noise is disturbed, isn't it?
Yes, that's a good question.
It would still recover the signal but the output would have the Vref offset.
Without knowing the actual use of the circuit we can only speculate.
Perhaps the output receiver circuit needs the integration output to start at other than 0V for some reason, such as to increase the dynamic range if the signal were unipolar.
 

Hi,

An integrator without feedback needs to be "reseted" from time to time to prevent from output saturation.
Maybe this is the function of the switch.

For sure the voltage at +In creates some offset at the output.
But maybe the idea is to compensate for an offset at the signal input.

Example:
Some current transducers have a DC offset of VCC/2. So if one wants to integrate the current signal one needs to put VCC/2 to +In. Especially for 0V/5V supplied circuits this could make sense.

Klaus
 

For sure the voltage at +In creates some offset at the output.
But maybe the idea is to compensate for an offset at the signal input.
I rather think that Vref does not only create a (fixed) offset, but it causes a rising DC value at the output. This is because Vref is integrated according to
Vout=Vref*[1+(1/jwT)] - Vs*(1/jwT)=(Vref-Vs)*(1/jwT) + Vref.
 

Hi,

See my example:
If the output of a current sensor has an unavoidable offset of 2.5V, then VRef also needs to be 2.5V ...especially to avoid a rising (integrating) output.

This way you have a 2.5V offset at the output and the current becomes integrated both directions. Pretty clean solutions for an ADC with 0V..5V input range.
--> Close the switch, do the ADC offset calibration. Open the switch, do the measurement.

I´ve used them before:

Klaus
 

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