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When studying a system that has the same behavior of an RLC parallel circuit, the slope of the impedance before the resonance corresponds to the value of the equivalent inductance, the slope after the resonance to the value of the equivalent capacitance and the value of the impedance at the resonance frequency to that of the resistance (phase = 0).

My question is, let's say we study an RLC circuit in real life -instead of modeling the studied system with RLC parallel circuit with constant values-, the parameters R, L and C are frequency dependent (R(f), L(f) and C(f)), then to what value of inductance (what frequency) exactly does the slope before the resonance correspond to (same for the slope after the resonance aka the capacitance)?

Thanks in advance

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  • \$\begingroup\$ You can greatly change the slope below resonance without changing the inductance : the slope is much more strongly dependent on Q than on inductance. So the premise of the question looks incorrect to me : some clarification needed? \$\endgroup\$
    – user_1818839
    Dec 7, 2020 at 12:26
  • \$\begingroup\$ I don't understand why the slope wouldn't change when the value of the inductance changes? \$\endgroup\$
    – Wallflower
    Dec 7, 2020 at 12:33
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    \$\begingroup\$ I didn't say it wouldn't. I said it'll change more when Q changes. For example keep LC constant and vary R to sweep Q between 1 and 100. \$\endgroup\$
    – user_1818839
    Dec 7, 2020 at 12:36

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If a resistance has a dependency on the frequency, it's no longer a resistance, but an impedance. That said, frequency-dependent components implies an alternative, equivalent circuit, and then you may no longer have only one resonance, since now you have additional (equivalent) RLCs. However it may be, the same basic rules apply as to any RLC: positive slope means derivative (parallel inductance, series capacitance), negative means integral (series inductance, parallel capacitance), no slope means resistive.

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    \$\begingroup\$ It's the presence of reactive components, not frequency dependence, that characterise an impedance as opposed to a resistance. Skin effect modifies a wire's resistance as a function of frequency; the value changes but it's still a resistance (real valued impedance) \$\endgroup\$
    – user_1818839
    Dec 7, 2020 at 12:28
  • \$\begingroup\$ @a concerned citizen, I agree with you, especially when the resistance and inductance's frequency dependent behavior is taken into account using an equivalent circuit. What I can't get my head around is when the system is supplied by say for instance a sinusoidal 50Hz voltage. All the parameters are to be computed at 50Hz to get the correct distribution of the voltage across the system. In this case, no frequency dependent behavior is needed. So the impedance of the system would be, if unable to measure it, the equivalent impedance of all components with their values at 50Hz. \$\endgroup\$
    – Wallflower
    Dec 7, 2020 at 12:37
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    \$\begingroup\$ @BrianDrummond Some books might disagree, some might not, and discussions on this theme have been carried by far more knowledgeable people than me. I merely settled to one side. For example, here is one discussion that happened in the LTspice group. I guess by now it's a matter of semantics. \$\endgroup\$
    – a concerned citizen
    Dec 7, 2020 at 12:38
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    \$\begingroup\$ @Wallflower Also, using any other waveform except a sinusiodal one will discard the "single frequency" part, since you are now talking about harmonics. Which means every element will react accordingly to each harmonics. This is no different than performing an AC analysis over a range of frequencies, sampled at each harmonic. \$\endgroup\$
    – a concerned citizen
    Dec 7, 2020 at 12:44
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    \$\begingroup\$ @Wallflower Now I see what you mean. Yes, analyzing at only one frequency partly discards the frequency-dependent part. I say "partly" because you should still account for the "equivalent circuit" that I mentioned: after all, no matter the frequency at which you're measuring, there will be an equivalent circuit given by the fact that all the elements are not fixed RLC. So this eqivalent circuit will be dictated by the frequency dependency of the consituent elements, though at measuring, there's no frequency changing. (I hope my English made it safe) \$\endgroup\$
    – a concerned citizen
    Dec 7, 2020 at 12:51

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