Thevenin of an op amp

Discussion in 'General Electronics Chat' started by johnk, Dec 1, 2008.

  1. johnk

    Thread Starter New Member

    Dec 1, 2008
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    I need some serious help regarding thevenizing this circuit. I am struggling to get the answer in the back of the book so I need someone to explain to me what I am doing wrong.
    I am trying to calculate the open circuit voltage and short circuit current which should give me the Thevenin resistance.
    Help would greatly appreciated at this point. It almost as if I am solving for Ro by having circular definitions.

    :mad::mad:
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Are you sure you got the book values for all those resistances correct? Ro at 200k seems a little high for a voltage output opamp. I would expect 200 ohms to be more typical. Is this a current output opamp by any chance?
     
  3. johnk

    Thread Starter New Member

    Dec 1, 2008
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    Yes, I think it is more of an exercise than a real application. I know op-amps are supposed to have low (<100Ω) even in open loop.

    I have been working at this circuit for a few hours and it is driving me crazy. :mad:

    I rewrote my equations and I now calculate 40kΩ which at least is lower than the 200kΩ. I would expect the Thevenin impedance to be lower than the 200kΩ when the gain is greater than 1.

    I know there are several ways to tackle Thevenin/Norton equivalents, but I find the open circuit voltage and short circuit current the easiest method.

    Any ideas?
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Try this method; apply 1 volt to terminal a (with respect to ground) and calculate the current. The output resistance will be the reciprocal of that current.

    It's important to set Vs to zero; in other words, ground the left end of Rg.

    Proceed like this:

    Rg and Rd are in parallel and that combination is in series with Rf, so the current through the Rf path is 1volt/(Rf + ((Rg*Rd)/(Rg+Rd))).

    The voltage at the top of Rd is a simple voltage divider calculation:

    V1 = 1volt*((Rg*Rd)/(Rg+Rd))/((Rg*Rd)/((Rg+Rd)+Rf))

    The voltage at the left end of Ro is -A*1volt*((Rg*Rd)/(Rg+Rd))/((Rg*Rd)/((Rg+Rd)+Rf))

    The current through Ro is (1volt+A*1volt*((Rg*Rd)/(Rg+Rd))/((Rg*Rd)/((Rg+Rd)+Rf)))/Ro. Don't forget that the voltage at the right end of Ro is 1 volt.

    Add the currents in Rf and Ro and take the reciprocal. I get 2032.86786359 ohms.
     
    Last edited: Dec 2, 2008
  5. The Electrician

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    Oct 9, 2007
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    Another way to do this is to write the admittance matrix by inspection:

    Code ( (Unknown Language)):
    1. [ 1/Rg+1/Rd+1/Rf   -1/Rf   ]
    2. [   A/Ro-1/Rf    1/Rf+1/Ro ]
    then invert the matrix. The output resistance will be the (2,2) element of the inverse. I get the same value, 2032.86786359 ohms.
     
  6. johnk

    Thread Starter New Member

    Dec 1, 2008
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    Well that certainly helps. I appreciate the explanation. Going back through my calculations it looks like I would need to use determinants to solve for Rth.

    The test voltage source was a lot easier. Also, I am not familiar with the admittance matrix; is that two port analysis?
     
  7. The Electrician

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    Oct 9, 2007
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    Have a look at post #9 in this thread:
    http://forum.allaboutcircuits.com/showthread.php?t=16454

    When you do a circuit analysis using the nodal method, you end up with as many equations as nodes, as in that post.

    The coefficients of the node voltages, on the left hand side of the equals signs, can be viewed as the elements of an n x n matrix, with the excitation vector on the right hand side. That n x n matrix is the admittance matrix for the circuit, and, with practice, it can usually be written just by inspection.
     
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