"VARs can be converted to real power but not at a rate of more than the supply can provide it."

Almost right. You can get all the energy in the tank out in microseconds if you want, or sometimes if you don't want. But it will then take time to fill the tank back up, since you have collapsed your resonance.

For example, take the circuit I am working with at the moment. You have seen the tank measurements and the input power measurements. If I cut the coil while the thing is operating, all the stored energy in the tank will be released -- in the mosfets -- in a matter of microseconds. This is a high power level, and I have the blown-apart mosfets to prove it.

Let's amend your statement to say "you cannot get real power out of your VARs _continuously_ at a faster rate than the supply can provide it to keep the tank full."

"VARs can be converted to real power but not at a rate of more than the supply can provide it."

Almost right. You can get all the energy in the tank out in microseconds if you want, or sometimes if you don't want. But it will then take time to fill the tank back up, since you have collapsed your resonance.

not, you are mistaken because i showed you that the resonance may not to destroy, but you continuing to you insist the opposite

Sorry...had the lower opto diode reversed. This one is correct.

Barry

"VARs can be converted to real power but not at a rate of more than the supply can provide it."

Almost right. You can get all the energy in the tank out in microseconds if you want, or sometimes if you don't want. But it will then take time to fill the tank back up, since you have collapsed your resonance.

For example, take the circuit I am working with at the moment. You have seen the tank measurements and the input power measurements. If I cut the coil while the thing is operating, all the stored energy in the tank will be released -- in the mosfets -- in a matter of microseconds. This is a high power level, and I have the blown-apart mosfets to prove it.

Let's amend your statement to say "you cannot get real power out of your VARs _continuously_ at a faster rate than the supply can provide it to keep the tank full."

Quite right Tinsel, I should have said continuously, that's what I meant.

Xname, show us how you can convert the VAR's into Watts without reducing the VAR's.

Both Tinsel and I can easily produce oscillating powers (VAR's) to input power in the order of over 100:1 and we can convert that to useful output
and we as well as many others that tell the truth say that when you convert a VAR to a Watt the VAR no longer exists that was
converted, and so that VAR of reactive power cannot be returned to the supply or continue to oscillate, to keep the same VAR's
the supply must provide more power. Any power converted to Watts is real power and is supplied by the power supply.

Forest, energy brought in from outside if it is, has nothing to do with reactive power and if there is an external input it should be
easy to identify and would be the more important part of any investigation into the device. Finding where the extra energy comes
from and how it enters the system would be important.

When people talk reactive power with a grid supplied system they mean power that is supplied by the grid that goes unused by
certain loads and therefore gets returned to the supply that same power can get power factor corrected and consumed as real
power and it does quite often, when that happens the consumed power is real power because it is consumed and cannot return
to the supply.

The instant power is consumed by a load it is real power and only real power is consumed. Reactive power cannot be consumed
by definition.

..

Xname, show us how you can convert the VAR's into Watts without reducing the VAR's.

of such like you even if show you such a thing, will again are you looking about what to when he messed. May not to please him of such like you, of a curve rocket space and interferes

Ok the confirmation :

"VARs can be converted to real power but not at a rate of more than the supply can provide it."
I feel much safer about some things that had said in the previous post.

" You can get all the energy in the tank out in microseconds if you want, or sometimes if you don't want. But it will then take time to fill the tank back up, since you have collapsed your resonance."

Take time ... How long it takes to fill the tank circuit? Have you sure that tank circuit couldn't fill at same rate or fastest than the supply ?

It takes a time interval that can be calculated and measured. It's not "much" time because even in a large tank like the FTW QEG, the actual energy circulating in the tank isn't all that large. How long does it take to deposit four Joules in a tank when your supply is providing, say, six watts? Two-thirds of a second. And the delays of that order can be seen in tank circuit oscillators when they are started from zero voltage and zero stored energy, in sims and in real devices.
How am I sure that it can't fill at the same rate or faster than the supply? 1: I have measured it and at no time have I ever seen a tank filling faster than it was being supplied. 2: it is logically incoherent. If the tank is filling, where is the filling coming from, if not from the supply???

I see you circuit and i think that understand what you try to show but i think that your circuit does not illustrate the phenomenon in the same way that other people are testing.

Depends on which people, I think. You may note that once a resonant tank is filled, its oscillations are sinusoidal. In fact that is one way you can tell that you are indeed in resonance. So once I have a resonant tank filled and oscillating at resonance, how is what happens after that, in any way dependent on what comes before that? That is, if I am resonating a tank with a sine wave oscillator instead of a pulsed squarewave drive... can the downstream load, feeding off the tank, tell the difference?

https://www.youtube.com/watch?v=DaHXLZ6QJss
You can see in this video, that the output power generated is greater than the input used;
I say that because i use only a pulse in input , and at least in my understanding i think that is a gain . or not ?

Pulsed inputs have their own problems which must be addressed. It is very likely that you are underestimating the power in your input. But I can't tell what is going on in that video, except I am waiting for someone to electrocute themselves. What, you are showing a neon glowing a little bit for a few seconds from the energy stored in a capacitor from the single backspike of a huge inductor as you  make and break the connection to the battery or supercap that is the green cylinder? . OK, fine. I don't see any measurements in there that indicate OU. If you are impressed by lighting a neon from a lower voltage source using inductive collapse spikes, take a look at the JT in the video below. A single depleted AAA battery lights up _four_ NE-2 neons in series. But nobody seems to think That is OU. It's hard for me to find better performances though. Unless... of course.... it's done wirelessly.
http://www.youtube.com/watch?v=zYhISYeWGTo
 

About you say : "But it will then take time to fill the tank back up, since you have collapsed your resonance."
If you can reduce the damping in cycles and makes the inductor feed the capacitor, and vice versa, you can maintaining the same resonant current in the circuit, and converting all the current into useful work.

You can maintain the same resonant current in the circuit, yes. In the video above you see multiple chained examples of just that. Converting all the _input_ current into useful work, _after_ passing through the various tank circuits in the demonstration. What goes in, comes out the other end, minus losses. Where is any magic extra energy supposed to come in?

I think is what happens in the video , because after you remove the input power you see that oscillations maintain the circuit running in resonance.

I doubt it. I think that what you are seeing in your video is the Neon running on straight DC from the charged up capacitor. Strip the neon out of your circuit tester so you can see both electrodes. Are they both glowing equally, or is only one glowing? If you are resonating, they will both be glowing equally. If you are running on DC, only the negative polarity electrode will be glowing.

TinselKoala a like your reviews and videos.
 Iḿ  your youtube subscriber at least 3 years and i like your dedication at this subject,   but i don't like your temper and tone of your reviews.

I have no tolerance for people who insult my education and my work, without having corresponding work of their own to demonstrate their claims. And when people say idiotic things or misrepresent my work, I will let them know about it. I am not in a popularity  contest, I am doing science.

I ask you to be gentle in your reply in case you want to consider comment on what I said.
I'm always open to healthy criticism

I hope you have found me gentle enough. Thank you for being a longtime subscriber and I hope I don't lose you. But please, do the little test I suggested above: Get your hands on a bare NE-2 neon bulb and repeat your MOT test with that, and tell us which electrode glows, or if they both do.

TinselKoala
see now in the patent clearly states: Reactive power input =33,5VA, Active power out =25,3W, power consumption from grid =1,16W, Is not that this proof?