I tried to measure the resonance frequency of some multi layer air coils. Typical parameters are 0.6 mm copper wire, 3 to 8 layers, and diameter and height are approx 50 mm. I solder a 1 Ohm resistor to the coil and connect it the shortest way possible to a frequency generator. The output current saturates at low frequencies and the adjusted voltage drops, but I normalize accordingly. When measured with an RLC-meter the coil has about \$L = 5\;\rm{mH}\$ and \$R = 5\;\Omega\$. Measuring \$C\$ does not make sense as the effect is much smaller than \$L\$. So I check the resonance by measuring the voltage over the \$R_0 = 1\;\Omega\$ resistor and the total circuit with a scope. I was expecting a behavior that you get from a simple equivalent circuit, i.e. \$L\$ in series with \$R\$ and \$C\$ parallel to both.


simulate this circuit – Schematic created using CircuitLab

So here is what I get enter image description here

The small dashes are the data. The dotted lines are fits where I fix \$L\$ and \$R\$ and \$R_0\$ as measured from the LCR meter and allow only \$C\$ to vary. This does obviously not fit the data. The dashed line additionally has \$R\$ free to fit, but this doesn't work either, especially not for the 8 layer rectangular coil. The continuous line fits very well but it assumes the resistor \$R\$ to behave as $$ R = R_\rm{const} + \omega \; r_\omega $$ i.e. a resistance changing linear with frequency. The parameters should be pretty far away for significant changes due to skin effect. Losses due to eddy currents are unlikely as well (I checked with and without a big aluminum sheet near by: no difference).

So what am I actually measuring?

Edit Is it radiating? Long wave radio would be up to 300 kHz.

  • \$\begingroup\$ @sstobbe it is the voltage over the 1 Ohm resistor, so at resonance the LRC impedance goes to "infinity" and the drop over R is "zero". (Its normalized by the total voltage) \$\endgroup\$ Feb 20, 2019 at 14:32
  • \$\begingroup\$ Maybe it's the signal generator? Not all like capacitive or inductive loading. \$\endgroup\$ Feb 20, 2019 at 14:35
  • \$\begingroup\$ @aconcernedcitizen I thought about that too, but I am measuring the output voltage over the entire circuit as well. That looks OK on the scope. Moreover, I normalize the voltage by the total output of the signal generator. I still might miss something, but at the moment I don't see how that would work. I can imagine a phase shift or a change in total amplitude but that would be kicked out via the two measurements. \$\endgroup\$ Feb 20, 2019 at 14:40
  • \$\begingroup\$ I see, you are trying to plot tank admittance, I would have preferred Impedance. A frequency dependent R is common, lookup tangent loss/dissipation factor equivalent exists for inductors. \$\endgroup\$
    – sstobbe
    Feb 20, 2019 at 14:44
  • \$\begingroup\$ @sstobbe I found a few equivalent circuits before, showing variable resistance. The parasitic resistance is usually present if it is a coil with ferrite core. e.g. this. Which is copied from here where the dependeny is by square root and not linear. Moreover, it is difficult to imagine a change from 5 Ohm to virtual 150 Ohm by this, especially as it is an air coil. As mentioned in the OP, I tried to provoke eddy losses with a thick aluminum sheet, but no difference has been measured. \$\endgroup\$ Feb 20, 2019 at 15:04

2 Answers 2


I made a little FEM simualtion of an air coil with paramters as you describe (5 cm height and width, 0.6 mm wire diameter copper, L = 5 mH, Rdc = 5 ohm, N = 450 in 6 layers). After 10 kHz the resistance increases significantly due to skin- and proximity-effect, see figure. I guess proximity effect is a main contributor.

enter image description here

  • \$\begingroup\$ Thanks for the effort. While I thought skin effect and proximity could be excluded, I came back to it after checking old papers from Medhurst and this homepage about Tesla coils. Interestingly this should go with sqrt( f ) not f, but fitting my data again with sqrt( f ) shows that it is not much worse. (linear in f fits slightly better, though, especially for small f) \$\endgroup\$ Feb 21, 2019 at 14:14
  • \$\begingroup\$ So at the moment I try to get the paper: Fraga, E., C. Prados, and D.-X. Chen, “Practical Model and Calculation of AC Resistance of Long Solenoids”, IEEE Transactions on Magnetics, Vol. 34, No. 1, January, 1998, pp. 205–212. Maybe it explains even the quantitative values. \$\endgroup\$ Feb 21, 2019 at 14:16
  • \$\begingroup\$ What did you use for simulation? \$\endgroup\$ Feb 21, 2019 at 14:17
  • \$\begingroup\$ I used FEMM, a 2D field simulator, which is free. I made an axis-symmetric model, so this is usually very accurate. \$\endgroup\$
    – UweD
    Feb 21, 2019 at 14:18
  • \$\begingroup\$ femm.info \$\endgroup\$
    – UweD
    Feb 21, 2019 at 14:19

AC resistance of a multilayer coil comprises of skin-effect and proximity-effect resistances $$R_{\rm{ac}}=R_{\rm{skin}}+R_{\rm{proximity}}$$ Depending on the winding structure dominance of the two can vary significantly at higher frequencies. Particularly, if you have very close turns, the proximity effect resistance can be very high at higher frequencies. See the following figure (Source). The x-axis is proportional to the square root of frequency, but the slope is increasing with the number of layers. enter image description here

If you want to reduce the proximity-effect resistance, you may wind the coil with some gap between the layers, it will greatly reduce proximity effect losses. Besides, the accuracy of your measurement technique is highly dependent on many factors like quality of the waveform (because of the presence of harmonics can affect your voltage measurement), the frequency tolerance of \$R_0\$ and even interwinding capacitance is a variable at high frequencies.

  • \$\begingroup\$ Thanks for this additional information and explanation. I found a similar answer, probably the source, in Fraga_IEEETransMag34(1998)205. I don't mind that large R_ac, but I'd like to damp the self resonance, so I need to take R_ac into account, which required to understand what I actually have in the first place. Mission accomplished, I guess. \$\endgroup\$ Mar 1, 2019 at 6:17

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