# 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.

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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.

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 AAC Fanatic!

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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0
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 AAC Fanatic!

Oct 9, 2007
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Have a look at post #9 in this thread: