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Inverting op amp circuit

  1. Nov 1, 2009 #1
    please look at this circuit

    The transfer function for the the circuit above is

    Vout = (-Rf/Rs)*Vs

    if I were to switch the terminal of the op amp so that Vin is inputted into Vp and Vn =0 instead (everything is the same except that Op amp terminal Vp is now Vn and Vn and is now Vp), how does it change the behavior of the circuit? The transfer function is still the same but I doubt the behavior of the circuit is the same. Or is it?
  2. jcsd
  3. Nov 1, 2009 #2


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    No, it isn't the same and that formula doesn't apply any more.

    It is just the formula for an inverting amplifier and assumes the DC conditions and other connections are correct for an inverting amplifier.

    If you reverse the inputs to the chip or reverse the power connections, the transfer function does not apply.
  4. Nov 2, 2009 #3
    What are the DC assumptions? I still don't understand why if you switch the op amp terminals then the transfer function becomes invalid.

    if Vout = (-Rf/Rs)*Vs is not the right transfer function, then what is the new transfer function? Will the circuit work at all?
  5. Nov 2, 2009 #4


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    You wouldn't expect the transfer function to apply if the opamp had no power would you?
    What if you reversed the power supply?

    The transfer function only applies if you have the circuit functioning correctly as an inverting amplifier.

    If you did reverse the input connections, the transfer function would change. It looks like a Schmitt trigger circuit that way, as you would get positive feedback.


    You might like to read about Schmitt Trigger circuits here:
    Last edited: Nov 2, 2009
  6. Nov 2, 2009 #5

    Say somebody give you the circuit above and you have no idea what it does. How do you know if you can analyze it by writing the transfer function using the assumption vp=vn and In=Ip=0 (op amp terminal current)?
  7. Nov 3, 2009 #6


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    The transfer function for this circuit would be discontinuous and also depend on the output voltage.

    You can analyse it, though. You assign actual values to the resistors and then assume the output is at the negative supply voltage. Suppose you make R2 = 90 K and R1 = 10 K and the supply is +/- 9 volts.

    Then you start increasing the input voltage from the negative supply voltage towards the positive supply voltage.

    The two resistors form a voltage divider between the input and the output.

    At some point, (when the input is greater than +1 volt) the +input will go positive and the output will swing positive. This will swing the +input even more positive.

    If you now reduce the input and keep doing it until the + input goes slightly negative, (when the input is less than -1 volt) the output will go negative again.

    Try it and you will understand this circuit a bit better.
  8. Nov 3, 2009 #7


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    Unfortunately if they have no idea what it does beforehand one doesn't know for certain that they can. But, as it is the simplest case to analyze (and it is often right), you can just assume that you can. If you arrive at a contradiction (say vo depends on more than just vi, or you get two equations which can't both be right, for instance vo=2vi and vo=3vi) then you know your initial assumption was wrong and you can start over knowing that vp!=vn but In=Ip=0 (for the ideal opamp).

    In the schmitt trigger example vp!=vn for all vi.
  9. Nov 4, 2009 #8
    I have a college text book on op amp and it seems like circuits with Vo that feedback to Vn can be analyzed using vp=vn, ip=in=0. Is my observation correct?
  10. Nov 4, 2009 #9


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    Did you understand how the Schmitt trigger worked?
  11. Nov 4, 2009 #10
    Here in thumbnail is a balanced amplifier. R2=R3 and R1=R4. You can use either pos or neg input, and ground the other. This circuit minimizes input bias current offsets.

    Show that if this amplifier is used as a differential amplifier, the output common-mode signal is zero; i.e., Vout = G(Vpos - Vneg)

    Bob S

    Attached Files:

  12. Nov 4, 2009 #11


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    The transfer function for the the circuit above is

    Vout = (-Rf/Rs)*Vs

    Try this:
    Rf = 100 K
    Vs = 2 volts
    Rs = 1 K

    So, Vout = ?

    This is a problem with blindly following transfer functions.
  13. Nov 9, 2009 #12
    So what are you trying to say?
  14. Nov 9, 2009 #13
    Yes I know how it functions but I haven't try to understand why it behaves the way it does.
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