I'm not looking for an answer to a problem here; I'm just giving a particular instance of network theory involving opamps to illustrate an issue that is seldom dealt with. It is of interest to think about.

An opamp circuit appeared on another forum:

If this circuit is solved using the usual "Golden Rules", namely that the opamp open loop gain is infinite, v+ = v- (a consequence of infinite opamp voltage gain) and the inputs draw no current, the voltage gain Vout/Vin is determined to be 5/3 and the output voltage is .12 volts.

If the + and - inputs are reversed and the solution is carried out again using the "Golden Rule" method (without explicitly using the opamp open loop gain, A or Av), the same Vout/Vin is obtained.

In fact, if a solution is obtained with A explicitly involved, such as would be necessary for a solution with a finite A, an expression for Vout/Vin will be obtained. This expression, involving A, can be evaluated in the limit as A approaches ∞, and the same result will be obtained that the earlier "Golden Rule" method gave.

One can obtain an expression for Vout/Vin involving A for the two orientations of the + and - inputs, and the two expressions are different for finite A, but have the same limit as A approaches ∞.

Presumably, one of the orientations is not stable; there is more positive feedback than negative feedback. But, there is no indication of which is which indicated in the solution by the "Golden Rule" method; you get the same solution for both orientations of + and - inputs.

Now, in this rather simple circuit we can see by inspection which is the stable configuration, but what of a more complicated circuit, with several opamps and nested feedback loops? How do we determine stability?

 

The issue you describe applies to any opamp (or comparator) circuit. Applying the assumption that the inputs are at the same potential means that your analysis is indifferent as to which signal is tied to which input because +0=-0.

To determine the stability (at least in the sense that you are talking about), you can do the following (I'm typing this off the cuff, so hopefully I keep things straight in my head): Recognize that the stability goal means that if the output is perturbed in one direction that the circuit responds by telling the opamp to respond in the other direction. So what you want is the gain from the output of the amplifier to the differential inputs. In the case of your circuit, you want

\(
\frac{V_B-V_C}{V_{out}}
\)

Using the same approach as superposition, remove the opamp and zero out all of the independent sources and apply a test voltage between the output of the opamp and ground. Calculate the feedback gain above. If it is positive, then the circuit isn't stable. If it is negative, then it is (again, limited to the notion of "stable" being just whether the + and - inputs of the amp are connected the right way).

In your case, I get a gain of -0.1236, so it is connected the right way.

To see if my musings are valid, replace R5 with a 29kΩ resistor (or anything larger than 19.33kΩ) and perform the analysis using the open loop gain and limiting process that you used previously and see if the opamp doesn't turn out to be the wrong way in that case.

 

I get a similar result for the "gain" if I leave the left end of R1 open, but if the left end of R1 is grounded, my result is that there is no positive value of R5 for which the "gain" changes sign.

My point in this thread so far is that for a simple single opamp circuit, such a calculation is fairly easy, but how can we obtain a similar result for more complicated circuits?

For example, active filters are often built with two opamp GICs. There are 24 ways to arrange the + and - inputs in a GIC:

Which of them are DC stable, and how can we determine which are? This is a question which is rarely asked when students are given these kinds of opamp circuits to analyze.

And, that's not even considering what happens to stability if reactive components are present, or whether a given topology is DC stable with the top node short circuited to ground, or left open.