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[teslaalset]

I completely missed this huge post  . It sure has a shock value. Who would have thought that a simple RC circuit could provide this apparent OU.

If your scope allowed to do some more fancy things you could cut on tweaking the parameters a lot. For instance by multiplying the current and voltage channel, taking the mean of that, and dividing that result by the mean of the current squared resistance. This gives you and instant cop value on screen. Since you're dealing with such low capacitance you could make your own simple variable capacitor and adjust it while running or adjust resistance while running. This will probably give some specific RC formula for each frequency that will give maximum cop.

If you don't mind could you please share the above data.

@ Omnibus, Broli,
I followed this thread partly. A few thoughts that I have with this RC example:
- What exactly is the output terminal in this circuit?
- It might be good to do a simulation by performing an 'old school' calculation and compare them with the real measurement results, so we could have a discussion on why there are differences between them.
I could do the calculation, but need to know the output terminal in this circuit case.

[broli]

@ Omnibus, Broli,
I followed this thread partly. A few thoughts that I have with this RC example:
- What exactly is the output terminal in this circuit?
- It might be good to do a simulation by performing an 'old school' calculation and compare them with the real measurement results, so we could have a discussion on why there are differences between them.
I could do the calculation, but need to know the output terminal definition in this case.

Output is measured as current flowing through resistor squared times resistance, Pout=I*I*R. Then the mean of that function can be taken to get the average value.

Input is more straightforward by taking Pin= U*I, and again taking the mean to get an average value. Omnibus shows the work as joules in his graphs. The slope of these graphs is the power. If the slope of the input energy is taking it should give a negative power value. Meaning the source is giving out no energy, it's getting energy from outside.

As for some conceptual analogy. I'm no EE but closing the loop with semiconductors might be tough. My mind is simplistic. What I imagine is a simple strong LC tank tuned at 700kHz with very low resistance. Then the RC OU part is put in parallel with this tank. But the energy gain from the RC circuit has to be bigger than the LC tank losses, mainly due to resistive damping. This should be able to be calculated. At best the LC tank should have an extremely low resistive part. If this can be achieved the tank either remains stable due to some self control or it will runaway increasing voltage as time grows.

[teslaalset]

Output is measured as current flowing through resistor squared times resistance, Pout=I*I*R. Then the mean of that function can be taken to get the average value.

Input is more straightforward by taking Pin= U*I, and again taking the mean to get an average value. Omnibus shows the work as joules in his graphs. The slope of these graphs is the power. If the slope of the input energy is taking it should give a negative power value. Meaning the source is giving out no energy, it's getting energy from outside.

If I understand correctly, below diagram is valid. The resistor acts as an output load.
In that case, voltage across the resistor is missing to make a valid measurement.

[Omnibus]

This is interesting. How did you arrive at that particular value of C ?

I measured it with a RadioShack multimeter which allows for capacitance measurements. The capacitance on the label says 100pF.

[broli]

If I understand correctly, below diagram is valid. The resistor acts as a output load.
In that case, voltage across the resistor is missing to make a valid measurement.

You forgot to include the 3000$ current hall probe :p.

Edit: However since inductance is no longer part of the resistor, it could be a good exercise to do the same calculations using voltage across the resistor and compare for any 1:1 correlation between the hall current probe.