Partnered Output Coils - Free Energy

[MarkE]

Same as above but expanded for detail of the current spike:

OK so reading the scope as 5V / div, the charge delivered is about (200ns * 25V  + 2us*7V)/ 4.7 Ohms = 3.19A*200ns = 4.04uC.  And the current is 1.73kHz * Q/cycle ~= 7mA average.  From the brightness I think that figure is high.  Be careful because even a "non-inductive" resistor can show a lot of inductive distortion in the face of very fast rising signals.

[MarkE]

Here's a question for the experts.

My Link DSO will not do trace multiplication so I can't generate an instantaneous power curve. However, under the Measurements, I can select "Area" which will give me the area under a waveform in Volt-nanoseconds.

So... if I ask it to give me the area under the Vout and Vdrop waveforms, then multiply those two values, is this equivalent to multiplying the waveforms and then taking the area (integrating) of that power curve, with a resulting answer in Joules?  I don't think the units work out exactly, since the result would seem to be in Watt-seconds² instead of Watt-seconds (Joules)... but anyhow, what do you think? MarkE, Vortex1, poynt99, anyone else who's had calculus?

Unfortunately, no it is not the same.  The only way to get the real power is to perform the multiplication timeslice by timeslice and add.

[TinselKoala]

The Vdrop trace is at 500 mV per division not 5 V/div and the 4R7 is at the bottom of the stack, between the lowest neon and the 0 volt power input . The Vdrop probe is connected closely across the body of the 4R7, which is in a to-220 type package.

[TinselKoala]

Unfortunately, no it is not the same.  The only way to get the real power is to perform the multiplication timeslice by timeslice and add.

I was afraid of that. Now I remember that the sum of integrals is the same as the integral of the sums, but it doesn't work that way for multiplication or division does it.

Well, then I'll have to do a data dump into a spreadsheet, multiply there and do a numerical approximation of the integral then. That's a lot of work.

[MarkE]

The Vdrop trace is at 500 mV per division not 5 V/div and the 4R7 is at the bottom of the stack, between the lowest neon and the 0 volt power input . The Vdrop probe is connected closely across the body of the 4R7, which is in a to-220 type package.

I've got a bunch of those too.  You probably got them from the same benefactor.  IIRC, the time constant for those is about 2ns.  So it is probably OK for measuring this circuit.  We could do a piecewise integration and get an estimate of the energy in each pulse.