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The Q factor of an inductor is given by:

Q=(2*pifL)/R

Is this only the frequency dependent Rf? Or does it also include the series DC resistance?

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  • \$\begingroup\$ The skin effect and proximity effect are embedded into the measurement of Rf that I conducted with a series resonance circuit. So I don't really understand your comment \$\endgroup\$ – Weaverworm Jan 26 '16 at 13:06
  • \$\begingroup\$ It includes that what is in the formula. The ratio of the inductors (imaginary part of the) impedance Z(s) over the real part of Z(s). Usually the real part of Z(s) is equal to the DC resistance but it does not have to be. \$\endgroup\$ – Bimpelrekkie Jan 26 '16 at 14:38
  • \$\begingroup\$ It includes all real resistance, DCR plus AC resistances that are lossy but may be frequency dependent. So Q isn't necessarily a constant. \$\endgroup\$ – John D Jan 26 '16 at 15:33
  • \$\begingroup\$ @John D, Well technically yes, the Q is constant. I mentioned the frequency dependent resistance that is for sure included (I know) so with that given fact the quality factor is constant at a given frequency. Anyhow thanks for the answer. \$\endgroup\$ – Weaverworm Jan 27 '16 at 15:21
  • \$\begingroup\$ My point was that Q isn't constant with frequency, as hysteresis, eddy current, skin effect and proximity losses vary with frequency. Near DC the Q is different from what it would be at 100 MHz. \$\endgroup\$ – John D Jan 27 '16 at 16:33
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In your formula, \$2{\pi}fL\$ repesents the "ideal" purely inductive resistance of the inductor at freuency \$f\$. Thus, by dividing this by the \$R_f\$ measured resistance at frequency \$f\$, you are taking all parasitic resistances into effect (skin effect, DCR, core hysteresis, etc.)

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  • \$\begingroup\$ So you're saying the quality factor is only frequency dependent. As I think about it. You might be right. It is a small signal representation of the inductor. When applying high loads the core losses will join and the quality factor kind of loses its importance.. \$\endgroup\$ – Weaverworm Jan 27 '16 at 15:23
  • \$\begingroup\$ Not quite; the reactance (I called it "purely inductive resistance above) depends only on frequency and inductance. However, in your formula: Q=2*pi_f_L/R_f_ the R_f_ (Resistance at f_requency) is the total resistance/loss measured in an active circuit. By dividing the reactance (_ideal inductor) by the actual, measured resistance, you can account for the overall ratio of energy stored to energy lost in your inductor under the test conditions (frequency & signal level). If you test with your actual usage signal, your calculated Q should account for "real life" losses \$\endgroup\$ – Robherc KV5ROB Jan 27 '16 at 18:41
  • \$\begingroup\$ But if the Rf is low at a certain frequency it means that the Q-factor is high. This does not automatically say that the inductor has a lower dissipation than an inductor with a lower Q-factor. Right? I have an inductor with a Q-factor of 30 and a planar inductor with 55. (not great, I know). The 30 dissipates about 400mW and the 55 dissipates 1200mW. Just because of higher DC and core losses. \$\endgroup\$ – Weaverworm Jan 28 '16 at 9:21

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