A General Impedance Converter (GIC) looks like this:

enter image description here
Source: http://www.ee.nmt.edu/~wedeward/EE212L/SP15/ImpedanceConverters.html


$$Z_{tot} = \frac{Z_1*Z_3*Z_5}{Z_2*Z_4}$$

But I need a circuit that gives me \$ \frac{1}{Z_{tot}}\$

How would I create a circuit that gives me \$ \frac{1}{Z_{tot}}\$ for Zin?

(Or something like \$ Z_{in} = \frac{Z_2*Z_4}{Z_1*Z_3*Z_5}\$)

The problem is if I put an inductor on this circuit, I can't select a series of impedance's Z_1 through Z_5 using only resistors and capacitors to end up with a combined impedance of 1.

  • \$\begingroup\$ Hmm interesting. \$\endgroup\$ – Andy aka Jul 29 at 22:49
  • \$\begingroup\$ So you want a general conductance converter (GCC?) \$\endgroup\$ – jonk Jul 30 at 0:23
  • \$\begingroup\$ @VoltageSpike So you just need to insert a transconductance -- something that presents current at the output given voltage at the input. Like a BJT. ;) \$\endgroup\$ – jonk Jul 30 at 18:45
  • \$\begingroup\$ @LvW I want to be able to reverse the impedance of an inductor, using only resistors and capacitors. So if I had an inductor on Zin, I could match the impedance with an inverse impedance. \$\endgroup\$ – Voltage Spike Jul 31 at 19:21
  • \$\begingroup\$ ....and what is the unit for the "inverse impedance"? Is it capacitor? \$\endgroup\$ – LvW Aug 1 at 8:16

In theory, the inverse of the inverseconverse of the converse is the original impedance which was a filter with a bunch of RC's.

NIC Negative Impedance Converters give the negative of impedance , not the inversion.

enter image description here

The performance as shown would be highly vulnerable with stray coupling. Yet the desired or expected performance was never specified.

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  • \$\begingroup\$ So, if I put an inductor on Vin can I get 1/Z_l from the circuit? \$\endgroup\$ – Voltage Spike Aug 5 at 20:37
  • \$\begingroup\$ You have it backwards.. on the top half you put C on Vin+ then Vin appears as an inductor yet current limited... \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 5 at 20:44
  • \$\begingroup\$ The input resistance at the node Vin is Z=-Z1Z3Z5/Z2Z4. Hence, there is only a sign inversion. Is that the solution the OP was asking for? \$\endgroup\$ – LvW Aug 6 at 7:19
  • \$\begingroup\$ Yes @LvW Thankyou. The conversion is just the opposite polarity not an inversion. my bad choice of words. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 6 at 7:28
  • \$\begingroup\$ I rather think that the OP has made a bad choice of words resp. symbols. He did mention Zin=1/Ztot.....whatever this could mean... \$\endgroup\$ – LvW Aug 6 at 9:02

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