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I've been working on a proof for University and I've been able to understand most of what I was asked, but I'm having trouble understanding why is it that the equivalent inductor in an Antoniou Gyrator has inductance L = (R1*R3*R5*C4)/R2, referring to the image below.

Antoniou Gyrator

I've understood the parts of the proof concerning the differentiator and the operational amplifiers and I have almost closed my reasoning, but I don't precisely understand why that equation works out to be that way. From what I've understood, R5*C4 is the gain of the differentiator, but other than that, I don't know where the equation comes from.

Can you guys help me on this? Any tips? I looked at the other question on here about this type of gyrator, but it didn't answer my specific question. Thank you!

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schematic

simulate this circuit – Schematic created using CircuitLab

The figure shows two NIC circuits (NIC=Negative Impedance Converter).

  • NIC 1 is short circuit stable (a low source resistor at "in1" makes the circuit stable; negative feedback dominates). It is relatively simple to derive the input impedance at "in1" as Zin1=-\$\frac{Ro.R2}{R3}\$.

  • NIC 2 is open-circuit stable (100% negative feedback; a large source impedance at "in2" is required.) The input impedance at "in 2" is Zin2=-\$\frac{R4.R6}{R5}\$.

Now, if you replace the grounded resistor Ro by the input resistance Zin2 the whole circuit remains stable because the negative input impedance Zin2 is "far beyond" infinite. A mathematical proof is possible. That means: The inverting node of NIC1 is now connected with the inverting node of NIC2 (and Ro is removed).

Therefore, the resulting input impedance "in1" will now be:

Zin1=Zin=+\$\frac{R2.R4.R6}{R3.R5}\$.

This is the basic form of a "Generalized Impedance Converter GIC".

This two-opamp block can be used to create an active grounded inductor (replace R3 or R5 by a capacitor). In many cases, such a GIC is also used to realize an FDNR (Frequency-Dependent Negative Resistor) .

A better circuit arrangement (Antoniou`s circuit):

The discussed circuit will work - however, it was shown, that a modification (proposed by Antoniou) has better properties because the opamps non-idealities cancel each other up to a certain degree (higher frequencies possible). The new circuit (Antoniou`s GIC block) can be explained as follows:

For quasi-ideal op-amps both non-inverting input nodes (of the basic GIC) have the same potential because the voltage across the op-amp´s input nodes is assumed to be zero. Hence, the positive input nodes can be exchanged against each other. That`s all. When you redraw the circuit (without any line crossings) you will arrive at the circuit arrangements as shown in the original question. Of course, the expression for the input impedance remains unchanged.

Replacing R3 or R5 with a capacitor (Z:1/sC),thus, allows you to realize a grounded inductor. The value of the inductance can be selected with 4 resistors and/or one capacitor.

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    \$\begingroup\$ @ LvW Quite a fine answer this is. \$\endgroup\$ – analogsystemsrf Jun 26 '18 at 14:58
  • \$\begingroup\$ Thank you so much, I've been trying to understand where it came from for a while. Your explanation was superb, I'm finally able to close the reasoning. Thanks! \$\endgroup\$ – Arrigo Lupori Jun 26 '18 at 16:02
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    \$\begingroup\$ Arrigo, I must admit that some years ago I had the same problems. And it was really a large problem to find the explanation for Antoniou`s modification. There is no textbook which explains how to derive this circuit. Good luck. \$\endgroup\$ – LvW Jun 26 '18 at 21:47

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