# Impossible Op-Amp Circuit...Maybe

Gold Member
Could anyone verify if this is an impossible circuit, I put it in pSpice and it didn't work, I have no op-amps or voltage generators to build the circuit otherwise I would have built it. What do you all think, is it possible to find the transfer function of the far right op-amp? The schematic is attached

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• Op-amp.jpg
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NascentOxygen
Staff Emeritus
There is an absence of low-frequency (DC) feedback, so I expect the outputs will in short time be pegged at the supply rail.

Jeff Rosenbury and Bullington
meBigGuy
Gold Member
The 5 from the opamp to gnd is useless since OUT is 0 impedance.

There needs to be a resistance fron the center node to ground or spice won't converge (possibly PSPICE puts it there automagically. Try 100MegOhms tp ground.

No idea what "does not work" means. What was it supposed to do?

Bullington
Gold Member
Sorry, pSpice says there are two floating nodes, one at the node where C1 and R2 meet, and another where R3 and C2 meet. This is actually a circuit that was done in class, my professor came up with a solution for the transfer function; he said that the node where C1 meets R2 is not zero but it is an actual voltage.

NascentOxygen
Staff Emeritus
The op-amp inverting inputs cannot be supplied with bias current, they are floating.

Bullington
Did you create the circuit by yourself? I doubt if you have found this circuit in any documentation.
HOWEVER: This circuit is very similar to Antonious very well known GIC circuit.
This GIC block plays a major role in active filter circuits (electronic (active) inductors, FDNR realizations, impedance transformation,...)

Bullington
meBigGuy
Gold Member
They are not opamps. They are ideal circuit elements, so there is no such thing as bias current.

Problem is that pspice will not simulate with floating nodes. For pspice, you need to add 100MegOhm resistors to ground on the floating nodes, other wise it can't find a quiescent solution to begin simulation.

Bullington
NascentOxygen
Staff Emeritus
They are not opamps.
They're not? What's that label beside each triangle in the schematic?

Bullington
Gold Member
I really wanted to get your opinions on this circuit without changing it, although I think if I add resistors it won't make much of a difference, the problem I see is the branch with C1 and R3 on it; this is an ideal op-amp so the voltage across the input terminals of an op-amp must be zero and C1 and R3 lay right in-between those terminals. Did I assume that correctly? Now that I have seen the GIC circuit, I do believe my professor actually meant to draw that one, or at least something similar to it.

meBigGuy
Gold Member
They're not? What's that label beside each triangle in the schematic?

I misspoke, kinda. They are ideal differential voltage sources, not "real" opamps. No output impedence, no bias current, no offset, infinite gain, etc etc.

Actually, using real models would solve his problem since those node would not be floating. Ironically that would change performance by more than the .000005% the resistors would add

Bullington
Now that I have seen the GIC circuit, I do believe my professor actually meant to draw that one, or at least something similar to it.
Yes - thats very probable. There are several alternatives within the GIC block for placing resistors and capacitors. There are two basic versions for active inductors and three versions for realizing an FDNR. More than that, active filter stages can also be realized directly. But in any case, it is absolutely necessary that each opamp has a negative DC feedback (DC path between output and inv. input), otherwise the circuit cannot work.

Bullington
meBigGuy
Gold Member
But in any case, it is absolutely necessary that each opamp has a negative DC feedback (DC path between output and inv. input), otherwise the circuit cannot work.
Can you explain why in the context of a perfect amplifier with infinite input impedance, infinite gain, 0 offset, etc.

Bullington
Can you explain why in the context of a perfect amplifier with infinite input impedance, infinite gain, 0 offset, etc.
What is the question? ("Why in the context of...." ?)

Bullington
meBigGuy
Gold Member
Can you explain why? ( in the context of a perfect amplifier with infinite input impedance, infinite gain, 0 offset, etc.)

Bullington
I am really sorry but I dont know what I should explain?

Bullington
meBigGuy
Gold Member
Where the DC content comes from that requires DC feedback path, for starters?

If there is no DC input, why is DC path needed?

Bullington
OK - when I see an opamp symbol, immediately, I think of a real opamp and if the whole circuit will work as expected.
More than that, in the first post the questioner has mentioned that he couldn`t build the circuit because of some reasons.
Hence, my answer is related to real world conditions only. Was that your question?

Bullington
Bullington, the attached image shows a GIC-based lowpass filter.
As you can see, it looks very similar to your circuit.
By the way - the necessary negative DC feedback for the most right opamp is established via R5, the other opamp, R3.

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Bullington
Gold Member
So while I did try to build it, I still don't see how it is even theoretically possible; how could the branch where C1 and R3 are located do anything? Theoretically the voltage between the positive and negative inputs of an op amp are equal, so the current through C1 and R3 must be zero and if the current there is zero then the only place current can be measured is between the resistors R1 and R2 and the voltage out of the U2 op-amp is zero. Is this wrong?

Bullington, are you still speaking about "your" circuit ? Why do yu want to analyze a circuit that makes no sense?

Gold Member
Well, has anyone actually confirmed it doesn't make sense on both a physical and theoretical scale? I have my reasons for it being a bad circuit but I wanted to see if you all agree.

Bullington, when I am asked to confirm if a certain circuit makes sense or not, at first I need to know what the purpose (the task) of the circuit is.
So - what is it?

Gold Member
is it possible to find the the gain or transfer function?

Ys, here is the transfer function between input and output of the most right opamp.
The FINITE opamp gain values are E1 and E2, respectively. Good luck.

( + E1 C2 R2 R4) s

( + E1 C2 C1 R2 R3 R4) s^2

------------------------------------------------------------------------------

( + E2 C1 R2 R4 + E2 C1 R1 R4 + C1 R2 R4 + C1 R1 R4 + E2 E1 C2 R2 R4 + E2 E1 C2 R1 R4 + E1 C2 R2 R4 + E1 C2 R1 R4 + C2 R2 R4 + C2 R1 R4) s

( + E2 E1 C2 C1 R2 R3 R4 + E2 E1 C2 C1 R1 R3 R4 + E1 C2 C1 R2 R3 R4 + E1 C2 C1 R1 R3 R4 + C2 C1 R2 R3 R4 + C2 C1 R1 R3 R4 + C2 C1 R1 R2 R4) s^2

Gold Member
Wow! How could that be if the voltage across C1 and R3 is zero? And, if the voltage across those components are not zero, then how? For an op amp the voltage across the positive and negative inputs are zero; so if a branch is connecting the two then that branch is zero, and if that branch has zero voltage then the current is zero; is this correct?