I wanted to try a test simulation on LTSpice to understand few things. I have a RLC parallel circuit and I wanted to see how the impedance varies according to each one of the components (very easy but I needed to run the simulation for visualization purposes).


To my surprise, when I plot I(V1) and I(L1)+I(R1)+I(C1) the peak of the currents at the resonance frequency are not the same, and I cannot seem to figure out why. Any ideas?


Thanks in advance :)

  • 1
    \$\begingroup\$ Try making the frequency step resolution smaller to see if it resolves out. It would also help if the result showed the y axis values. \$\endgroup\$ – Andy aka Jan 15 at 10:02
  • \$\begingroup\$ That's not what I get: i.stack.imgur.com/alybD.png. At any rate, there's no need to use those values: from nanoHz until TeraHz is ridiculous. Even the 10g resistor is a lot. You could try changing the colours, the default blue on black is very much obscured by the much brighter green (RClick on the trace's label, in case you don't know). Also, if you press A the small circles from the labels will be gone (if you don't need them). \$\endgroup\$ – a concerned citizen Jan 15 at 16:52
  • \$\begingroup\$ @a concerned citizen, hmm they are equal in your case!? \$\endgroup\$ – Wallflower Jan 18 at 8:13
  • \$\begingroup\$ @Wallflower Use @<TAB> to cycle between the names (there should be no spaces in them). Did you change the default values of some settings in the control panel? It looks like the connection to the left of C1 has a node, which means the wire doesn't stop at C1's pin and goes further than that -- does that influence somehow? Try making a new schematic and add the elements carefully, one by one, then run the simulation. \$\endgroup\$ – a concerned citizen Jan 18 at 12:19

If you use I(V1), you get the magnitude of the current.

So if you plot I(L1)+I(R1)+I(C1) you plot the sum the magnitudes.

Using the magnitude ignores the phase differences between the currents, the imaginary part of the current is removed and converted into the magnitude value.

Only if you would do a "proper" sum of I(L1)+I(R1)+I(C1) that takes into account the imaginary part of the currents and plot the magnitude of that sum, would the result be the same as I(V1).

When you plot I(V1) the summation is done properly (taking the imaginary part, phase etc) into account. So then at the resonance frequency, the inductor and capacitor's impedances cancel each other out and "from the outside" (looking at the impedance of the RLC tank) you only see the impedance of the resistor.

Tip: you're now actually plotting the admittance 1/Z. If you want the impedance plot, print 1/I(V1) or do as I usually do: replace the AC voltage source V1 with an AC current source with AC = 1. Then the voltage at "input" is the same as the actual impedance as V = I * Z = 1 * Z = Z (I = 1 because of the current source with AC = 1).

  • \$\begingroup\$ I don't think this is entirely true. At resonance, the inductor current magnitude and the capacitor current magnitude are not very small - they are both quite significant. The fact that they have entirely opposite phase angles is what makes the net current into the parallel circuit appear to be very small. I'm not saying you are definitely wrong but that your explanation doesn't ring true. \$\endgroup\$ – Andy aka Jan 15 at 11:52
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    \$\begingroup\$ @Andyaka Perhaps I wasn't clear enough and you misunderstood: I never claimed that the inductor and capacitor currents are small. My claim was that at resonance the impedance of the tank (L || C || R) becomes real and equal to R. Indeed the currents through L and C are not zero but exactly each other's opposite so as seen from "outside the tank" they become zero and cancel out. \$\endgroup\$ – Bimpelrekkie Jan 15 at 13:05
  • \$\begingroup\$ I suppose what I'm saying is that if you look at the graph in the OP's question, at 5 GHz the green current is (say) X but, at 500 MHz, you'd expect the sum of the magnitudes to be about 5X but they are only about 1.2X. Now the problem here is that the OP hasn't said what the vertical scale is so it's a bit hand wavey. If the vertical scale is in dB then I think you are right but is the vertical scale in dB? \$\endgroup\$ – Andy aka Jan 15 at 13:13
  • \$\begingroup\$ @Andyaka I simply assumed that the green plot is wrong (I mean, it doesn't show what is really going on regarding the circuit's impedance) so I ignored it and explained why it isn't telling the full story. The blue curve (hard to see) does though. \$\endgroup\$ – Bimpelrekkie Jan 15 at 13:19
  • \$\begingroup\$ Hello guys, I have updated the picture. Thanks for your remarks! \$\endgroup\$ – Wallflower Jan 18 at 8:04

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