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Let us consider the circuit above (R3 and R4 should actually be \$1\$ ohm each)

Using time-domain analysis, we reach that \$V_o=-6V_2\$.

Using laplace transforms (Block Diagrams), we would just have a negative transfer function.

What I cannot understand is this:

1) Intuitively, this system must oscillate since we are making a signal greater by a constant factor, returning it back, then multiplying again, so where did I go wrong in calculations?

2) Can we reach that this system oscillates algebraically without any intuition whatsoever?

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    \$\begingroup\$ "R3 and R4 should actually be 1 ohm each", then update your schematic. \$\endgroup\$ – Harry Svensson Apr 3 '18 at 17:54
  • \$\begingroup\$ I don't think it would matter much (if we had an ideal op-amp) to be honest since I'd still have the same \$A\$ and \$\beta\$ which is my point. It's just that I used 1 ohms while simulating \$\endgroup\$ – Hasan Saad Apr 3 '18 at 17:56
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    \$\begingroup\$ When you use the CircuitLab button on the editor toolbar the editable schematic gets saved inline in your post. This avoids the need to take a screengrab (with the grid in it) and is editable so your component errors can be fixed. Double-click a component to edit its properties. "It's just that I used 1 ohms while simulating." If the simulator is in any way realistic it won't work with such a low impedance load. \$\endgroup\$ – Transistor Apr 3 '18 at 18:54
  • \$\begingroup\$ Even if it weren't realistic, I'm concerned with the understanding of the ideal case. Also, thanks for the tip on the editor toolbar \$\endgroup\$ – Hasan Saad Apr 3 '18 at 18:55
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    \$\begingroup\$ By "ideal case" do you mean "infinite voltage range at the output" and "infinite gain?" \$\endgroup\$ – jonk Apr 3 '18 at 20:05
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There are two possible states. In both cases, \$V_2\$ is irrelevant since it is finite.

  1. If \$V_O=+\infty\$ then \$V_2\$, being finite, can be ignored. \$V_+=\frac{+\infty}{2}\$ and \$V_-=\frac{+\infty}{3}\$. Since \$V_+\gt V_-\$, the opamp holds \$V_O=+\infty\$.
  2. If \$V_O=-\infty\$ then \$V_2\$, being finite, can be ignored. \$V_+=\frac{-\infty}{2}\$ and \$V_-=\frac{-\infty}{3}\$. Since \$V_+\lt V_-\$, the opamp holds \$V_O=-\infty\$.

The above follows because:

$$\begin{align*} V_+&=\frac{V_O}{2}+V_2\\\\ V_-&=\frac{V_O}{3} \end{align*}$$

It's a binary state that depends on the initial condition of the output.


If you limit the output to a pair of voltage rails you provide, \$V_\text{CC}\$ and \$V_\text{EE}\$, then the situation is different. You have a range of hysteresis. Starting \$V_2\$ from well below \$\frac{-V_\text{CC}}{6}\$, and raising it upwards, the output will transition from \$V_\text{EE}\$ to \$V_\text{CC}\$ at \$V_2=\frac{-V_\text{EE}}{6}\$. Starting \$V_2\$ from well above \$\frac{-V_\text{EE}}{6}\$, and lowering it downwards, the output will transition from \$V_\text{CC}\$ to \$V_\text{EE}\$ at \$V_2=\frac{-V_\text{CC}}{6}\$.

In the somewhat more realistic case where also the open loop gain is large, but finite, then there will be a very tiny linear transition slope nearby the above-mentioned transition points. (And assuming an even more realistic opamp with BJT inputs, there will be substantial input loading for the input nodes when the voltages at the nodes are more than a few tens of millivolts apart from each other, since the differential amplifier inputs only present a tiny load in the case where the inputs are very close to each other in voltage.)

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  • \$\begingroup\$ How did you actually conclude that \$V_o\$ is infinite in the first place? I mean, we know it oscillates, but it doesn't appear here. The only explanation I could reach is that this circuit is not realistic since we just assumed that there is no delay in the paths and that somehow this system will become stable (so we could use Kirchoff's equations) I suppose with delay, it will appear to oscillate. Otherwise, nothing makes sense. \$\endgroup\$ – Hasan Saad Apr 15 '18 at 8:22
  • \$\begingroup\$ @HasanSaad I started out with ideal amplifiers which means their outputs are unbounded. I thought that approach might help. Obviously, it didn't help you at all. But I also dealt with the case where they have finite limits to their output. \$\endgroup\$ – jonk Apr 15 '18 at 8:37
  • \$\begingroup\$ I understand your point, but due to the application of feedback on the inverting part of the ideal op-amp (with infinite gain \$A\$), the gain would become finite as is obvious by applying the finite model then taking limits. What I mean is, why would there be any oscillation at all? Where did I go wrong in my derivation from Kirchoff's law? \$\endgroup\$ – Hasan Saad Apr 15 '18 at 8:42
  • \$\begingroup\$ @HasanSaad I don't know what you mean by the term, "oscillation." I think there is a language barrier. It doesn't oscillate, so far as I know. So you must be using it in a way I haven't experienced. \$\endgroup\$ – jonk Apr 15 '18 at 18:45
  • \$\begingroup\$ It doesnt? Doesnt this work in the same principle as the Wein Bridge Oscillator except that there isnt any resonant frequency? \$\endgroup\$ – Hasan Saad Apr 15 '18 at 20:40

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