3
\$\begingroup\$

A General Impedance Converter (GIC) looks like this:

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

Where

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

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.

\$\endgroup\$
  • \$\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
0
+50
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.