Equivalent Non-Ideal Opamp

Discussion in 'Homework Help' started by EE_Bob, Aug 18, 2009.

  1. EE_Bob

    Thread Starter Member

    Jul 21, 2009
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    This is an opamp circuit that is easily dealt with when assuming ideal opamp constraints. However when you are dealing with a non-ideal opamp the analysis becomes a bit more complex. My problem is that I cannot successfuly convert this circuit into one involving the actual components of the opamp.

    [​IMG]
    Can anyone draw the equivalent circuit with opamp components (input resistance, output resistance, voltage controlled voltage source with gain) to help solve my dilemma.
     
  2. bertus

    Administrator

    Apr 5, 2008
    15,645
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    Hello,

    Just take a look at the attached PDF.
    It handles opamp parameters.

    Greetings,
    Bertus
     
  3. EE_Bob

    Thread Starter Member

    Jul 21, 2009
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    I appreciate your suggestion and I did read through most of the pdf file.

    The opamp in question is a 741 model with gain of 2x10^5, input resistance of 2x10^6 Ohms, and output resistance of 50 Ohms.

    I'm required to find the gain of the opamp and can do so using convential nodal analysis under ideal opamp constraints. It is when having to deal with the nonideal model that requires use of the of the mentioned values, that stops me in my tracks.

    The gain found using the ideal opamp model provides only an approximate value, and while this is acceptable for my level of studies I would like to calculate the actual gain and output current of this circuit.

    I will include a photograph of my redrawn circuit and you will see my area of trouble.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well, you can apply this model
    [​IMG]
    But in yours circuit we have positive feedback.
     
  5. EE_Bob

    Thread Starter Member

    Jul 21, 2009
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    Is this circuit correct?
    [​IMG]
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes, circuit is correct, but I believe that this model is incorrect for positive feedback amplifier.
    And what are the results?
     
  7. EE_Bob

    Thread Starter Member

    Jul 21, 2009
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    The model was incorrect as that circuit came from an older edition of a text and was misprinted.

    It should have a negative feedback loop as you said.

    Im getting a gain of 8.99959118707
    and current Io=449mA

    However, I think this is incorrect. The book specifies that the gain should be 9.00041 and Io should be 657mA.
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well, the both answers are "almost" correct, it depends on a feedback.
    So recalculate the circuit once with negativ and then with positive feedback.
    And I think it that current should be in "uA" not in "mA".
     
  9. EE_Bob

    Thread Starter Member

    Jul 21, 2009
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    End up getting close to 9 for the gain and 649 micro amps for the current.

    The book uses an approximation for their calculation of V1. They round down a number 451 to 450.

    I have to assume its correct with the negative feedback. I've spent too much time on this to go through sign changes for the "positive feedback route"

    You must be getting different values if you say they're both almost correct? If so what are your values and how did you determine them.
    I used nodal analysis and calculated V1 in terms of Vs and Vo in order to find the gain. For the current I used the original circuit and used nodal analysis on node Vo.
     
  10. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    If I use the positive feedback arrangement shown in your first post, and ignore the fact that the circuit would latch up, just going through the math gives a gain of 9.000407384 and a current of 650.0123 μA.

    But, with negative feedback (the two cases can be calculated with the same equations by just choosing A = 2E5 or A = -2E5), I get gain = 8.99959 and Io = 649.9709 μA.
     
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