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schematic

simulate this circuit – Schematic created using CircuitLab

I am trying to reduce the T feedback network(R2 R3 R4) into a Thevenin equivalent circuit looking into the network from the v- (inverting) terminal. The Thevenin equivalent circuit I get is this:

schematic

simulate this circuit It is easy to see that Voc is just a voltage divider. To calculate the equivalent resistance a test source would be inserted then Isc (short circuit current) would be calculated and then the equivalent resistance is given by $$R_{th} = \frac{V_{oc}}{I_{sc}}$$ This bring me to my question: Is it possible to short the dependent source since $$ A(V_{+}-V_{-})$$ will equal some finite value that is independent of the thevenin equivalent; thus the VCVS is actually independent of this equivalent circuit and can be shorted like an independent source?

The book I'm reading gives the equivalent resistance as $$R_{th} = R_2+R4 || R3$$ Which would suggest that the dependent source has been shorted. To my knowledge we have to use a test source to determine equivalent parameters when there are dependent sources. Is this just a coincidence or can someone explain why we are able to short this VCVS.

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If you think about it that VCVS is not dependent. It depends on some parameters that are not part of the circuit you are analyzing, so you can turn it off (better than saying "short it" in my opinion.

You can also see this from another point of view: if you include the \$V^+V^-\$ part of the circuit and connect a generator to these terminals, then you should turn it off, then \$A(V^+-V^-)=0\$. Or you can just say that if you only have dependent sources in a circuit, then the "all nodes are at 0V" solution works, and some backend math tells us that in a nice network the solution is uniqe if it exists, so also if you don't include the generator on the op amp input terminals you can say well the output must be zero.

Understood this the solution is trivial and it's precisley what your book reports.

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  • \$\begingroup\$ It seems to me that the most logical way to think about it is to consider the dependent source independent in the circuit under analysis for Thevenin's eq. This allows us to use it like we would an independent source, i.e without a test input source necessary since only independent sources are in the circuit at this point. Thus we can use it as a voltage divider for Voc and short it for Rth. Also I am not sure about $$ A(V^{+} - V^{-})$$ being equal to zero. I don't think the amplifier would amplify at that point. I thin the most we can say about ideal op amps is that their gain approaches \$\endgroup\$ – Bwomchickabowow Jan 10 '15 at 23:24
  • \$\begingroup\$ Inifity in the limit and the differential input approaches zero in the limit. Given that, we can say the the out put of the op-amp will be bounded and finite. \$\endgroup\$ – Bwomchickabowow Jan 10 '15 at 23:28
  • \$\begingroup\$ Just stick to the reasoning that fits you better, but keep in mind that \$A(V^+-V^-)\$ can actually be zero, and that's because \$V^+-V^-\$ can be zero with no limits involved, and the limit of \$A\cdot 0\$ with \$A\to\infty\$ is zero. \$\endgroup\$ – Vladimir Cravero Jan 11 '15 at 9:22
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You can consider only a constant voltage source as an "AC short connection" since, when you vary the current through it, its voltage does not change. This means that the source differential resistance is zero (Ri = dV/dI = 0); so it can be thought as of a piece of wire.

Here the voltage source is varying, dynamic, dependent... and you cannot short it. If the voltage divider R3-R4 is low resistive enough, the equivalent resistance of the T-network is Re = R2*(1 + R3/R4), i.e. it behaves as the resistor R2 in the conventional inverting amplifier having the same resistance. See more about the idea behind this clever circuit trick in a ResearchGate question.

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