It seems to me that the circuit is only symmetric in a limited sense, namely that if Z5 is removed, the impedance at node 5 is infinite with no load at node 1, and the impedance at node 1 is infinite with no load at node 5, and if the opamp gains are infinite. Otherwise, it would only be symmetric if Z2 = Z3 and Z1 = Z4, which is not the case.

Hello again,

Yes i was assuming you were using a voltage source as drive for the circuit. With a voltage source in the input the impedances appear to be zero.

With nothing connected to the input, then the impedances i get are:
node 5: Z5 (or simply R5)
node 3: -R3*R5/R4
node 3 to node 5: zero

Note yes, one of them is negative.

 

In post#9 the input impedances at the nodes 3 and 5 are wanted.
My answer - as far as node 5 is concerned - was given in my post #14.

Now - here is my answer regarding node 3:
When we remember how Antonious GIC was developed from the basic GIC circuit (see my post #7), it is best to go back to the series connection of the two NIC blocks (jpg file GIC-origin my post #7). Here, both inverting inputs still share a common node which is identical to the node 3 under discussion.
Without any new calculation, again it is possible to give immediately the input impedance:
* It is known that the most right classic NIC block (elements 3,4 and 5) has the negative input impedance Z3,1=-Z3*Z5/Z4.
* Realizing that the grounded element Z5 appears in the numerator, we can say that the input impedance Z3,2 for the most left NIC block (elements 1,2 and "infinite" - no grounded part at the "+"node) will be infinite.
*Hence Zin3=Z3,1||Z3,2 with Z3,2=infinite and Zin3=Z3,1=- Z3*Z5/Z4.

 

Hello again,

Yes i was assuming you were using a voltage source as drive for the circuit. With a voltage source in the input the impedances appear to be zero.

With nothing connected to the input, then the impedances i get are:
node 5: Z5 (or simply R5)
node 3: -R3*R5/R4
node 3 to node 5: zero

Note yes, one of them is negative.

Why would there be any voltage sources other than the one used to determine the impedance (resistance in this case) at a node--the source in the "ohmmeter" so to speak?

OK everybody. Next determine the resistances seen at the five nodes with the opamps remaining ideal
except the internal gain assumes 3 different values, namely 1 , 3/sqrt(5), and 2.

Give the results as 5 element lists; e.g., the result for the question posed in post #9 would be: {15/8 , 0 , -15/4 , 0 , 5}

 

In post#9 the input impedances at the nodes 3 and 5 are wanted.
My answer - as far as node 5 is concerned - was given in my post #14.

Now - here is my answer regarding node 3:
When we remember how Antonious GIC was developed from the basic GIC circuit (see my post #7), it is best to go back to the series connection of the two NIC blocks (jpg file GIC-origin my post #7). Here, both inverting inputs still share a common node which is identical to the node 3 under discussion.
Without any new calculation, again it is possible to give immediately the input impedance:
* It is known that the most right classic NIC block (elements 3,4 and 5) has the negative input impedance Z3,1=-Z3*Z5/Z4.
* Realizing that the grounded element Z5 appears in the numerator, we can say that the input impedance Z3,2 for the most left NIC block (elements 1,2 and "infinite" - no grounded part at the "+"node) will be infinite.
*Hence Zin3=Z3,1||Z3,2 with Z3,2=infinite and Zin3=Z3,1=- Z3*Z5/Z4.

Hi,

And did you calculate the impedance across the two named nodes yet?

 

Why would there be any voltage sources other than the one used to determine the impedance (resistance in this case) at a node--the source in the "ohmmeter" so to speak?

Not sure why i did that to start with, but it did turn out a little interesting just the same in that apparently when there is a voltage source at the input all the nodes show zero impedance to ground. You could try to duplicate that if you like.

OK everybody. Next determine the resistances seen at the five nodes with the opamps remaining ideal
except the internal gain assumes 3 different values, namely 1 , 3/sqrt(5), and 2.

Give the results as 5 element lists; e.g., the result for the question posed in post #9 would be: {15/8 , 0 , -15/4 , 0 , 5}

Ok i'll give it a shot a bit later today i hope.
Did you yourself calculate the other impedances too or no? And if so, did you get similar results?

Here's what i get for the node 5:
Z=((Ap*R3+Ap*R2+Am^2*R2)*R4*R5)/(Ap*R3*R5+Ap*R2*R5+Ap*R3*R4+Ap*R2*R4+Am^2*R2*R4)

where Ap=A+1, and Am=A-1, and A is the internal gain.

So with A=1 for example i get:
Z=R4*R5/(R4+R5)

 

Not sure why i did that to start with, but it did turn out a little interesting just the same in that apparently when there is a voltage source at the input all the nodes show zero impedance to ground. You could try to duplicate that if you like.



Ok i'll give it a shot a bit later today i hope.
Did you yourself calculate the other impedances too or no? And if so, did you get similar results?

The values you and LvW got are correct. Don't you find it interesting that nodes 3 and 5 show finite resistance to ground, yet the resistance between the two is zero?

 

The values you and LvW got are correct. Don't you find it interesting that nodes 3 and 5 show finite resistance to ground, yet the resistance between the two is zero?

Hi,

Yes, in fact, i know there were several times when i had to use feedback to INCREASE the input impedance of an amplifier type circuit, but i cant think of a single time when i had to use feedback to DECREASE any impedance anywhere (assuming an op amp output is low enough of course or else we use a driver, but that's different). In this circuit, we see the circuit react to an excitation in such a way that it keeps the two differential voltage equal across those two nodes, and so we dont get a differential voltage change with any kind of current injection.
That's quite interesting to me and i was thinking of maybe trying to create a new circuit with just one op amp that reduces some impedance to zero like that, maybe the input impedance. That would create a sort of 'ideal' current shunt maybe. Something to think about anyway.
Normally we try to increase the input impedance but that would be for a voltage input not current.
Maybe another challenge
[LATER]
Turns out it is much easier than constructing a temperature compensated shunt circuit. I'll save this for later.

 

Hi,

Yes, in fact, i know there were several times when i had to use feedback to INCREASE the input impedance of an amplifier type circuit, but i cant think of a single time when i had to use feedback to DECREASE any impedance anywhere (assuming an op amp output is low enough of course or else we use a driver, but that's different).

General feedback considerations:
* If the feedback signal is a current (inverting opamp amplifier, BJT stage with RB between C and B) the input impedance is always DECREASED.
* If the feedback signal is driven by (sampled) an ouput voltage (rather than current), the output impedance is always DECREASED.
(in all opposite cases the impedances are INCREASED).

And - yes, I agree that the impedance between nodes 3 and 5 is zero (for ideal opamps) because the voltage difference between both nodes disappears.