Noninverting Amplifier with series RC network in the Feed back Path

Hari,

I concur with Michael.  If the capacitor in the feedback path were shorted out, then you would see the output go to the V+ rail.  As it is, there is no DC path in the feedback.

If you were to actually build it as is, I would expect the amp to be always stuck at V+ anyways.  But, I doubt that's what anyone wants from this.

Regards,

Daniel

Hi Hari,

can you show the complete schematic?

Kai

Hi Kai,

Please find the complete schematic

Regards

Hari

Hi Michael,

I did not put the complete schematic. It was a mistake from my side. Now I have added the complete schematic. This will give you a clear picture.

As Kai asked.

Regards

hari

Hi Daniel,

I put the complete schematic.

Could you please check it.

Regards

Hari

Hi Kai,

Thank you very much.

The input,DVVD_IN is a DC voltage of value 3.5V

I am struggling to understand the circuit in the schematic.

if you don't mind Could you please tell me how it works.

Regards

Hari

Hi Hari,

Invertering input of amp will be equal to non-inverting input of amp due to negative feedback.  Due to R2 and R3, output of IOP2 will be 2x the value of VG1.  Since IOP2 is in a buffer, VF2 will be equal to ~2 x VG1.  The output of IOP1 depends on whether the transistor is on or off.  These are the two scenarios presented by Kai above.  The output of IOP1 will rise to ensure that VF2 = 2 x VG1.

Note this behavior may change by frequency, as Kai has shown above.

Regards,

Daniel

Hi Daniel,

Thank you very much.

Here the input is a constant DC voltage of Value 3.5V.

Then the output will be always at Supply voltage .In DC C1 will be open and opamp output will be at Supply rails.

Please correct me if I am wrong.

Regards

Hari

Hello Hari,

For the first circuit with no DC feedback path, this is true.

However, for the more complete circuit which does have a DC feedback path, the output of the first stage will be 2 x VG1 = 7V.  You can verify this by going to Analysis -> DC Analysis -> Calculate Nodal Voltages and running the DC sim for Kai's TINA file.

Regards,

Daniel

Hi Daniel,

Thank you very much. I got it.

May I know that is the use of that series RC network.

Regards

Hari

Hi Hari,

in what application is this circuit used?

Kai

Hi Kai,

This is used in the Evaluation of an ASIC.DVDD_DUT is the Asic supply voltage.

I think the first OPAMP with series RC is a PID controller.Correct me if I am wrong

Regards

Hari

Hi Hari,

Which series RC?  You mean the one in the first stage?

I'm not sure that it would a PID.  Why are you inclined to think that?  Because of the component arrangement or the transfer function?

Regards,

Daniel

Hi Daniel,

Thank you for your support.

Yes, I mean RC in the first stage.

I would like to know its role.

To get a gain of 2 it is not required.

Would like to know the role of first series RC.

Regards

Hari

Hi Daniel,

Thank you very much.

I read the compensator document which you provided.

While deriving the transfer function of Compensator, may I know why Vref is not considered?

Regards

Hari

Hi Hari,

not every circuit you find in a schematic needs to be a good one...

By the way, it's no fun to analyze a schematic which isn't complete. Your schematic says nothing about the used amplifiers :-(

Kai

Hi Kai,

Thank you very much:)

Regards

hari

You are welcome :-)

Kai

As always, thanks for your help with this one Kai!

Regards,

Daniel

Hi Kai and Daniel,

Thank you for your help.I got a chance to talk with designer. Please see his response below.

"U12A buffers the input signal. The DUT current is measured over R32 and buffered by U13. U12B compensates for the voltage error that R32 introduces and feeds the error signal into the summing point of U12A. 

R33 and C50 adds some AC enhancements to the feedback loop. My best bet is AC enhancement or maybe stability."

I am attaching Schematic for your reference

Regards

Hari

Hello Hari,

I'm not sure what type of component U13 is.  However, if it is an op amp, then I don't see how it is acting as a buffer since there is no feedback loop.  Similarly, the R33 and C50 components would not be in the feedback loop.  I'm guessing this is a different type of component, such as a dedicated buffer.

Also, can you please clarify whether you have a question here?  I am not sure what you need help with.

Regards,

Daniel Miller

Hi Daniel,

I agree with you. We would need to know what chip U13 is :-)

Kai

Dear Kai and Daniel,

U13 is an instrumentation amplifier.Used as unity gain.

Below is the statement the Designer told

"Input voltage (DVVD_IN) is DC but the output current (the Current Drawn by the ASIC from DVVD_DUT) is AC and U12A needs to compensate for that.

The load current varies between 0.5mA and 1mA and you will see that there will be a voltage over R32 that U12A needs to compensate for via the feedback path through U12B.

R34 and C49 is for compensation.

R33 and C50 make a small AC current path from U12A to the DUT" 

My Questions are below

But I don't understand how that compensation is taking via U12A.

For ca current path R32 itself is sufficient correct?

Regards

Hari

Hi Hari,

the phase stability analysis shows that the circuit is not very stable with the LM358B or OPA2197. The phase margin is too low:

hari_rc_feedback_2.TSC

hari_rc_feedback_3.TSC

Kai

Hello Hari,

I don't believe that I understand your questions.  But, I have answered as best I can with my limited understanding of the desired circuit functionality.

U12A has two feedback paths.  The more direct one will block DC voltages and pass AC signals.  Based upon the designer's comments, the other seems to be some sort of feedback to account for current draw at the output.  Whether or not this is accurate, I cannot tell.  I think you will need to have the designer himself explain, because I cannot sort out what the intention of this overall circuit is or what the designer was thinking when adding things in.

If U13 is an instrumentation amplifier, then I am not sure as to why an "AC path" with R33 and C50 is needed.  I would think R32 would be the necessary component here, based upon my limited experience working with instrumentation amplifiers.  Again, we can guess.  But, the designer may be the only one that knows for sure.  Only he will know the intended purpose.

My recommendation would be to build the circuit in a simulator and strip out the portions that you do not think are necessary.  See if the circuit can perform the desired function.  If it cannot, then start adding in the designer's extra portions.  This will help to determine the function of each segment.

Regards,

Daniel

Hello Hari,

I hope you have found this information useful, even though you may have further questions regarding the circuit for your designer.

Please feel free to reach out if you would like further assistance.

Regards,

Daniel