Finding the transfer function for a difference amplifier

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Discussion Overview

The discussion revolves around finding the transfer function of a difference amplifier circuit, with participants exploring the theoretical and practical aspects of deriving this function. The context includes considerations of linearity, AC versus DC analysis, and the impact of circuit components on performance.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about deriving the transfer function in the form of Vo/Vin, suggesting that the circuit's behavior may not yield a neat expression due to potential non-linearity.
  • Another participant proposes that the transfer function could be represented as -(Z2/Z1) with an offset, questioning the validity of this approach.
  • A participant raises concerns about the circuit's linearity, suggesting that an open circuit condition at the input could lead to integration effects and saturation issues, complicating the transfer function analysis.
  • One participant references a document that discusses a type II compensator, questioning whether the same neglect of Vref in the transfer function could apply to their circuit.
  • Another participant clarifies that neglecting Vref is appropriate for AC analysis, as it effectively grounds the op-amp inputs for AC considerations.
  • Participants agree on the goal of identifying which components influence the poles and zeros of the transfer function.
  • A later reply notes that the circuit is likely to saturate due to finite offset voltages or mismatches, suggesting limitations in its practical utility.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of deriving a straightforward transfer function, with some suggesting that non-linear behavior complicates this task. There is no consensus on the best approach to analyze the circuit, particularly regarding the treatment of Vref.

Contextual Notes

Participants highlight limitations related to the assumptions of linearity, the impact of DC offsets, and the conditions under which the circuit operates, which may affect the transfer function's applicability.

Dextrine
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Homework Statement


I need to find the transfer function of the attached circuit (this isn't a homework question so I don't know if there's even a solution, but last time I posted this type of question on the EE forum, it was redirected here)

Homework Equations



I know it needs to be eventually of the form Vo/Vin and I know how to solve for Vo but I can't seem to get it as Vo/Vin. If it isn't possible, how would one describe this circuit's transfer function?

Z1=R1
Z2=R2+1/(s*C)[/B]

The Attempt at a Solution



Vo=(Z2/Z1)(5V-Vin)+5V

Vin is the measured signal from the output of a buck converter and 5V is the reference signal, if that helps at all. The closest I've somewhat gotten is
(Vo-5)/(5-Vin)=Z2/Z1

Thanks in advance for all the help![/B]
 

Attachments

  • OPAMPTRANSF.JPG
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The only other thing I can think of would be just letting the transfer function be -(Z2/Z1) with an offset afterwards, but this seems weird...
 
I suspect that the circuit is not going to behave linearly so that a nice neat single expression transfer function for the behavior is not going to be possible. Consider...

Suppose for a moment that the input is open (Vin disconnected: an open circuit) when the power is first applied. What will happen to Vo?

The series capacitor is going to integrate any current through R1 (the top R1; You have two R1's in your circuit diagram) that results from the difference between Vin and 5 V. A continuous input Vin that is not exactly 5 V is going to lead to saturation issues... What happens when Vo hits a power rail?
 
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Hello qneill, thanks for responding to my question. I'm not too sure I follow your suggestions, however, I came upon this http://www.ti.com/lit/an/slva662/slva662.pdf, which has a type II compensator which looks similar to the circuit I have. They kind of just neglect Vref in their transfer function, would I be able to do the same for mine?
 
It looks like they are only looking at the AC transfer function, which is why Vref is ignored (A DC source is just a short circuit to AC). Vref effectively places the two op-amp inputs at ground potential as far as AC is concerned. I suppose you could do the same if your goal is a similar analysis.
 
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Yeah, the goal is just to find out which components determine the poles and zero's
 
Dextrine said:
Yeah, the goal is just to find out which components determine the poles and zero's
Okay. So "short out" Vref for the AC model and analyze the circuit.
 
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gneill said:
Okay. So "short out" Vref for the AC model and analyze the circuit.
Will do, makes this MUCH easier. thanks!
 
The op amp is most likely operated from a single supply, necessitating the 5V input bias. The input is considered "zero" at +5VDC.

This circuit saturates for any finite offset voltage and/or mismatch between dc input voltages (dc input ≠ +5V exactly) so is useless.
 

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