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Finding the transfer function for a difference amplifier

  1. Apr 5, 2017 #1
    1. The problem statement, all variables and given/known data
    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)


    2. Relevant 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)



    3. 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!
     

    Attached Files:

  2. jcsd
  3. Apr 5, 2017 #2
    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...
     
  4. Apr 5, 2017 #3

    gneill

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    Staff: Mentor

    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?
     
  5. Apr 5, 2017 #4
    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?
     
  6. Apr 5, 2017 #5

    gneill

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    Staff: Mentor

    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.
     
  7. Apr 5, 2017 #6
    Yeah, the goal is just to find out which components determine the poles and zero's
     
  8. Apr 5, 2017 #7

    gneill

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    Staff: Mentor

    Okay. So "short out" Vref for the AC model and analyze the circuit.
     
  9. Apr 5, 2017 #8

    Will do, makes this MUCH easier. thanks!
     
  10. Apr 13, 2017 #9

    rude man

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    Homework Helper
    Gold Member

    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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