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If you have a resonant circuit, oscillating at its self-resonant frequency, how would you extract the excess power without draining it fully? A high Q resonant transformer impedance matched to the circuit perhaps? Is there another way? I actually need DC output so perhaps rectifying before extraction is a good strategy?

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  • \$\begingroup\$ Try things out using a simulator. It's ideal for theoretical experimentation on this type of idea. Been there etc.. \$\endgroup\$
    – Andy aka
    Nov 30 '21 at 12:03
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    \$\begingroup\$ how would you extract the excess power without draining it fully? By stopping with power extraction before it is fully drained and/or adding more power into the resonator. \$\endgroup\$ Nov 30 '21 at 13:03
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    \$\begingroup\$ Why do I get the feeling the power you WANT is greater than you can skim? The inductor has resistance so it will cause loss and the source has to put power in to deal with that loss. \$\endgroup\$ Nov 30 '21 at 13:36
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    \$\begingroup\$ @StainlessSteelRat It must be that smell, that smell of "free energy" ;-) \$\endgroup\$ Nov 30 '21 at 13:46
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    \$\begingroup\$ This question is an oxymoron. If you're draining power from it, it's not a high Q resonant circuit any more. \$\endgroup\$ Nov 30 '21 at 14:33
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One definition of Q is energy_stored/energy_lost_per_cycle.

Increase the losses, to your pickoff, and you necessarily decrease the Q.

If you have a parallel LC at some point in your oscillator, then a series diode to an output is a good way of getting a DC voltage out, while only loading the circuit at the conduction instants, and leaving it unloaded throughout the rest of the cycle.

For instance

schematic

simulate this circuit – Schematic created using CircuitLab

Note that I've included the sustaining circuit, as resonant circuits don't just resonate by themselves, especially if you skim off the 'excess power'. I've also included its power supply, just in case anybody is so entranced by the idea of resonance they think they can get free energy out of it.

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  • \$\begingroup\$ Thank you Neil, very helpful! \$\endgroup\$ Nov 30 '21 at 22:05
  • \$\begingroup\$ Do be aware that diodes have nonzero capacitance. \$\endgroup\$
    – TLW
    Dec 1 '21 at 4:01
  • \$\begingroup\$ @TLW Do feel free to add your own answer, pointing out how the diode's reverse biassed capacitance changes the resonant frequency of the circuit slightly, and why you should care about this effect, compared to the much larger effect that D1 clamping the positive swing to a diode drop above the peak output DC voltage has. \$\endgroup\$
    – Neil_UK
    Dec 1 '21 at 9:04
  • \$\begingroup\$ It's not the frequency change that I was attempting to draw attention to. It's that you're stating "leaving it unloaded throughout the rest of the cycle". \$\endgroup\$
    – TLW
    Dec 4 '21 at 2:22
  • \$\begingroup\$ @TLW Because the diode residual capacitance is not 'loading' it at all, is not abstracting energy, certainly not compared to the massive loading that the diode conducting into the output capacitor effects. In a hand-wavy circuit like this, I think I've been sufficiently detailed. That's the beauty of a forum like this. One person like me comes in with a simple answer. If you think I've been too sparse with detail, you can post your own. \$\endgroup\$
    – Neil_UK
    Dec 4 '21 at 5:40

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