Impossible Op-Amp Circuit....Maybe

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The discussion centers on the feasibility of a particular op-amp circuit, with participants questioning its design and functionality. Key issues identified include the absence of DC feedback, which is crucial for the circuit's operation, and the presence of floating nodes that prevent simulation in pSpice. Participants suggest that adding resistors to ground could help resolve simulation issues, while others argue that the circuit's design inherently leads to saturation and instability. The conversation also touches on the distinction between ideal and real op-amps, emphasizing that real components would behave differently due to non-ideal characteristics. Ultimately, the consensus is that the circuit may not work as intended, particularly with real op-amps, and further analysis is needed to confirm its theoretical viability.
  • #31
jim hardy said:
What do you guys think ?

Well, for REAL opamps I don`t think that a transfer function can be given because there will be no operational point within the linear region of the opamps.
Therefore, the node "NODE" will never be at zero volts (this would be valid for working negative DC feedback for both opamps only).
Instead, I think that both opamps will go into saturation.
 
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  • #32
LvW said:
Well, for REAL opamps I don`t think that a transfer function can be given because there will be no operational point within the linear region of the opamps.
Therefore, the node "NODE" will never be at zero volts

Whenever there's capacitors that can hold charge one must specify among his initial conditions their state of charge. Remember "Constant of Integration" from first semester calculus ?

So, start the thought experiment with initial condition of zero volts everywhere including isolated NODE , and ideal opamps.
Any change in voltage at Ralph's +input will propagate through Ralph, through C2 to Node
which will cause Louie to immediately reach around and undo that change .
So Ralph is held at zero output by Louie, the feedback element. Gain of circuit is zero.

It's a little easier to see if you remove R2, all he does is halve Vt.

Ralph is a unity gain voltage follower , transfer function of 1
Louie is feedback with gain and transfer function of ∞
closed loop transfer function G/(1+GH) is 1/(1+1X∞) = 0

I'd call the circuit a "Frustrated Voltage Follower ".
@Bullington ---- See what Professor thinks ?

but i agree it wouldn't work with real parts.

old jim
 
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