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I have done some impedance measurements on an unknown 1 port device. I have measured its impedance at different frequencies (from 10 MHz to 600 MHz) and I have seen it on the normalized (on 50 Ohm) Impedance Smith Chart:

enter image description here

At Low frequencies we are at the short circuit point (left). Then, the device offers an inductive impedance, until, at about f0 = 470 MHz, we arrive at the real axis (right). It seems it arrives at the open point but it is not: if you see the following graph you see that the impedance is, obviously, purely real, but not infinite. It is equal to about 471 Ohm (absolute value).

enter image description here

You may see these results also in the following graphs (the first one contains real and imaginary parts of the impedance, while the second one its absolute value).

enter image description here

enter image description here

So, at f0 the devices resonates and offers a purely real impedance. Now my question is: which is the equivalent RLC circuit of this device according to this analysis?

Precisely:

  • from the Smith Chart I may get the value of L: I should take the Imaginary Part of Zin at low frequencies (in which the device is almost purely inductive) and divide by 2*pi*f.
  • from the Smith Chart I may get the value of R (it is 471 Ohm, as told before)
  • from the Smith Chart I may get the value of C (at high frequencies, where the device is almost purely capacitive)

So, which is the equivalent model? A parallel RLC, a series RLC? Moreover, at f0, does the device offer a parallel resonance, or a series Resonance? I have been told that if I measure R,L,C through the Impedance Smith Chart, they are the values of the RLC series model, but I am not sure about it because I can also do my measurements on the Impedance Smith Chart for a device which is not a RLC series circuit. For instance, something like that:

enter image description here

Moreover, I'd say it cannot be an RLC series circuit, because at low frequency we are at short circuit point, and not open circuit point.

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  • \$\begingroup\$ You should just play around with some calculations or with a simulator, and see what kind of trace your circuit makes on the Smith chart. Then you'll know right away whether it matches your measurement or not. \$\endgroup\$
    – The Photon
    Jan 19, 2020 at 6:02
  • \$\begingroup\$ Good Idea. I have done now a simulation with the impedance of a Series RLC circuit (Zin = R + jomegaL - j(1/omegaC)) but the path of the Smith Chart is different (in fact it starts at the open position and arrives at the short position). But you have made me think of a possible solution. \$\endgroup\$
    – Kinka-Byo
    Jan 19, 2020 at 9:28
  • \$\begingroup\$ Maybe the solution is that with this kind of measurements we model the device with a series RLC circuit, but if the original device is not a series RLC circuit, we will get some frequency terms in R, L, C (I am thinking now at the equivalence formulas with Q factors, which contain omega). So this may explain the fact that the starting point is not the open point as an RLC circuit with R,L,C constant \$\endgroup\$
    – Kinka-Byo
    Jan 19, 2020 at 9:30

1 Answer 1

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That characteristic looks awfully like a parallel RLC circuit. At resonance the LC part becomes infinity and, because it is shunted with R, the effective impedance is R, possibly a 470 ohm resistor. At DC and low frequencies the inductor dominates with its low impedance and, if you went significantly higher than 1 GHz the parallel capacitance would dominate and start to act as a low impedance shunt.

Below is just a convenient graph I found for a parallel RLC circuit: - enter image description here

The difference between series and parallel resonance is shown side-by-side here: -

enter image description here

Picture from here.

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