Impossible Op-Amp Circuit....Maybe

[LvW]

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.

 

[jim hardy]

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