Can anyone give an explanation about general impedance converter? This question might not be specific enough, but I will appreciate if someone could share as much as possible about this circuit.

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    \$\begingroup\$ Can you post a schematic? \$\endgroup\$ – Martin Petrei Sep 24 '14 at 14:34
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    \$\begingroup\$ Re: the down votes, And ... how is this NOT EE? I don't see any other similar questions on this stack, so this is an opportunity to flesh out the knowledge base here. And it's not overly broad as there are real limits to how these can be applied. \$\endgroup\$ – placeholder Sep 24 '14 at 15:06
  • \$\begingroup\$ Try searching for gyrator circuit analysis. Here's a nice link, mysite.du.edu/~etuttle/electron/elect66.htm Try working through it and tell us where you get stuck. \$\endgroup\$ – George Herold Sep 24 '14 at 15:26
  • \$\begingroup\$ oops, that link has a mistake. I think it's the voltages at nodes a,c and e (not d) that are equal (via feedback) \$\endgroup\$ – George Herold Sep 24 '14 at 16:11
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    \$\begingroup\$ So what exactly do you want here? An explanation of what a general impedance converter is, or a proof about the derivation of the impedance formula of the Antoniou circuit as shown in this picture? If it's the former you want you can google it and wikipedia results is on top. If it's the latter i can prove it to you. \$\endgroup\$ – Nikos Sep 2 '15 at 11:01

As George noted in his comments, if you label the nodes 1-5 from left to right, nodes 1,3 & 5 are all the same voltage due to the op-amps forcing enough current until those nodes are equal.

Node 1 would also be known as the input voltage. Notice that the output current is Vin / Z5 then. Which is also equal to the current through Z4.

Note that the current through Z2 and Z3 have to be equal to eachother. From the information so far, you can determine the voltages at nodes 2 and 4 with respect to Vin. Once you know node 2 voltage, you know the input current (Vin-V2)/Z1. The output current is in terms of Vin. If you divide the output current by the input current, you'll know the difference in impedance because the equations will be true over all Vin. That is, Vin/Iin / Vout/Iout = impedance_conversion. Vin=Vout in this case, so Iout/Iin = impedance=conversion.

Working through what I just described you should end up with this:
$$ Iout=Vin/Z5 $$ $$ V4=Vin+Vin/Z5*Z4 $$ $$ V2=Vin + (Vin-V4)/Z3*Z2$$ $$ (Vin-V2)/Z1=Iin$$
$$ \frac{Iout}{Iin} = \frac{\frac{Vin}{Z5}}{\frac{Vin-V2}{Z1}} = \frac{Z1}{Z5}\frac{Vin}{Vin-V2}$$ $$ = \frac{Z1}{Z5}\frac{Vin}{\frac{-Vin+V4}{Z3}Z2} = \frac{Z1Z3}{Z5Z2}\frac{Vin}{-Vin+V4}$$ $$ = \frac{Z1Z3}{Z5Z2} \frac{Vin}{\frac{Vin}{Z5}Z4} = \frac{Z1Z3}{Z2Z4} $$

$$ \frac{Zin}{Zout} = \frac{Iout}{Iin} = \frac{Z1Z3}{Z2Z4} $$

If the impedances are just resistances I can't imagine it being terribly interesting as it's just a resistor made up by 4 resistors.

If the impedances are capacitive and resistive and you place them right, you could potentially get an impedance that looks like an inductor. This has an interesting potential usage. In integrated circuits, capacitors are much prefered to inductors due to their small size. Now you can make a circuit that acts like an inductor without an inductor. It can also be tuned and altered using the resistors and capacitors so that sizing can be optimal for a chosen inductance value. This has applications in integrated filters and probably other integrated circuits.

This should be enough to get you started on this topic. You should probably read up on gyrators if you're interested in this more. Hopefully I didn't make any typos in all of my equations.

Another way of looking at the result is this as Z5 = Zout: $$ Zin = \frac{Z1Z3}{Z2Z4}Z5 $$ So Z5 can be used as just another knob to adjust the impedance you're creating.

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I think, this is not the right place to explain how the GIC works - nevertheless, some general remarks:

  1. The GIC topology as shown was invented by A. Antoniou in 1969 (!). For a specific dimensioning some non-ideal opamp properties cancel each other.

  2. The working principle of the GIC can be derived from a series connection of the two different forms of a negative impedance converter (NIC).

  3. The GIC das been proven as one of the most versatile active circuits and is extensively used in active filter and oscillator circuits. GIC applications include active (lossless) L-simulation as well as FDNR realizations (frequency-dependent negative resistor). For this purpose, the various impedances Z are chosen as R or 1/sC, respectively.

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  • \$\begingroup\$ It's an easy derivation, (assuming ideal opamps) labeling the nodes a,b,c,d,e from left to right. Feedback makes Va=Vc=Ve= Vin. And then solve for the various currents and unknow voltages, starting at Z5 and working to the left. I5=Ve/Z5=Vin/Z5, I4=I5... etc. \$\endgroup\$ – George Herold Sep 24 '14 at 18:31
  • \$\begingroup\$ Something wrong with my GIC contribution? The questioner has asked us to "share as much as possible about this circuit". \$\endgroup\$ – LvW Sep 25 '14 at 7:19
  • \$\begingroup\$ I learned about the Antoniou inductor in school. My mind was blown at that moment when the prof derived that Zin = sCR^2 (assuming all resistors were the same). Amazing stuff. \$\endgroup\$ – efox29 Nov 24 '14 at 9:24
  • \$\begingroup\$ Did your prof tell you also about some other interesting applications? For example: FDNR (frequency-dependent reistor)? Zin=-1/(R*C1*C2*s^2) \$\endgroup\$ – LvW Nov 24 '14 at 10:21

Adding to what Horta wrote...

If you make Z4 a capacitor, then Z4 becomes 1/sC. So the impedance becomes:

Zin = s (Z1 Z3 Z5) / Z2

Which is an inductor of value Z1 Z3 Z5 / Z2

If we replace Z5 with a trim pot, then we have a variable inductor...

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