COP 20.00 (2000%) Times, Reactive Power Energy Source Generator,

[picowatt]

A bit of a correction and more observation:

In the left side capture, there appears to be approx 4 amps peak during the first positive cycle.  If Rload is indeed 10 ohms (as surmised from the right side capture data) and that 10R load is across the transformer output, then Vpeak on the left side graph would have also been similar to the 40volts seen on the right side graph (40V/10R= 4 amps).

However, if indeed Vpeak was 40V in the left side capture, the math trace should only show 160 watts or so, but instead it shows closer to 500 watts during that peak.

Possibly this indicates that the transformer is being used for more than just isolation.  Alternately it may be that the current shown in the left side capture does not truly reflect the 10R load resistor, but instead indicates the current thru the reactance of any caps being charged or discharged (and possibly Rload as well).   

PW

[x_name41]

here s my schematic for BiToroid transformer as a reactive power energy source generator , use simple, a reliable and efficient resonant oscillator with self-excitation based on Mazilli ZVS circuit but here use bipolar transistors, This oscillator can operate both with one or two transistors and will be called the "x_name41_ZVS oscillator"



$ 1 5.0E-9 2.9224283781234943 50 5.0 50
r 464 416 464 464 0 22000.0
w 640 400 608 400 0
c 720 400 720 304 0 4.7000000000000005E-7 20.311132933747196
w 624 432 608 432 0
w 720 304 816 304 2
w 816 304 816 320 0
w 816 384 816 400 0
w 720 400 816 400 1
w 624 432 624 496 0
w 624 496 368 496 0
v 368 400 368 320 0 0 40.0 3.7 0.0 0.0 0.5
w 368 400 368 496 0
r 368 224 368 320 0 0.1
w 368 224 432 224 0
w 432 224 464 224 0
t 560 416 608 416 0 1 -13.994493455539088 0.15599672830778347 1000.0
l 816 224 816 272 0 9.999999999999999E-5 0.036129450108768606
l 784 224 784 272 0 9.999999999999999E-5 -0.036129450058019645
w 784 224 816 224 0
w 784 384 816 384 0
w 560 464 672 464 0
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w 464 416 512 416 0
w 464 224 672 224 0
w 736 224 784 224 0
w 784 272 784 336 0
w 672 224 736 224 0
w 784 336 784 384 0
w 816 272 816 304 0
d 720 400 688 400 1 0.805904783
w 640 400 688 400 0
d 720 304 672 304 1 0.805904783
l 816 320 816 384 0 6.799999999999999E-6 -1.0626308848043475
x 68 649 142 655 0 24 battery
x 241 651 399 657 0 24 inductor 6,8uH
x 429 650 601 656 0 24 capacitor 470nF
x 643 650 784 656 0 24 choke 100uH
x 911 652 963 658 0 24 base
x 486 83 751 89 0 24 x_name41 ZVS oscillator
x 562 385 660 391 0 24 2SC3987
x 652 286 734 292 0 24 1N4007
x 689 436 771 442 0 24 1N4007
o 10 64 0 298 5.0 0.0125 0 -1
o 10 64 0 297 2.5 0.1 0 -1
o 10 64 1 299 0.3125 9.765625E-5 0 -1
o 36 64 0 298 40.0 3.2 1 -1
o 36 64 0 297 20.0 6.4 1 -1
o 36 64 1 291 80.0 9.765625E-5 1 -1
o 2 64 0 298 40.0 3.2 2 -1
o 2 64 0 297 20.0 6.4 2 -1
o 2 64 1 299 80.0 9.765625E-5 2 -1
o 16 64 0 298 20.0 0.1 3 -1
o 16 64 0 297 10.0 0.2 3 -1
o 16 64 1 299 2.5 9.765625E-5 3 -1
o 23 64 0 298 1.25 3.90625E-4 4 -1
o 23 64 0 297 0.625 7.8125E-4 4 -1
o 23 64 1 299 7.62939453125E-5 9.765625E-5 4 -1


and see my successful experiment of a single wire energy transmission  http://realstrannik.ru/forum/80-temy-xname41/117136-re-generatory-kapanadze-obshhaya-tema.html#228052

[mscoffman]

One more note.

As the left side capture is labeled as transformer output, it is possible the variac is driving the primary of the transformer and that the secondary (transformer output) is connected to the circuit.  This arrangement allows for both isolation and input V adjustment.

I must again address my concern, however, that the data collected from the transformer output is not necessarily going to reflect similar data measured directly at the AC line.  It is difficult to believe that anyone claiming to be able to draw real power from the AC line while returning the same or more power back to the AC line, would not include AC line measurement data.  A variac and that transformer must surely represent a fair amount of inductance between the AC line and the points in the circuit used for input power measurements.

I will remain skeptical until AC line input measurements are released that reflect what is happening directly at the AC line, and that also includes any power drawn from the driver circuitry.  The total AC line input power measured directly at the AC line, compared to that simultaneously made across Rload, would paint a more accurate picture.

PW

I happen to think that a high efficency 97% HV Inverter 100-600Vdc input with MPP impedance matching 120-240VAC 50 or 60Hz
output should be part of the instrumentation tool kit of a power supply professional.   This inverter used in conjunction with a bridge
rectifier capacitor filter can operate as a nearly pefect "VARS Stop". This will keep VARS from transistioning in either direction across
a power link. This is because there is no way to represent VARS on a stable DC link and so will power readings can be taken with low cost
instruments there, if the inverter in operation remains stable. If the inverter is destablized by the load,  then a Corcom filter can be added
as a rotating power VARS resiviour on the output side of the inverter. These things will make sure real power can always be measured
on the DC link without the interference in either direction by virtual AC power. This is what is required for OU self looped operation.
The diode bridge will not have a PF power factor of zero but that should not distort virtual power from the utility net because non-zero
power factor is something that the utility network can handle and you will not be billed for.


---


This will be easily checked in this case, I suspect that the lamps plugged into the output of the inverter and the
transformer plugged into the inverter will show nearly the same power values on the DC link with excess virtual power
rotating in conjunction with the inverter if the inverter can remain in stable operation. While the killawatt meter may
show different.




:S:MarkSCoffman

[x_name41]

Look like don't work for me here...
I've made a simulation and don't seem to have excess.
The logic is like this.
0-90° --> charge in parallel.
90-180° --> discharge series.
then same thing but for the inverted polarity.
Putting cap to series/parallel doesn't give excess (this a well know fact).
When you charge cap 1/2 of energy is lost no matter what kind of resistance you use. It can be the parasitic resistance of the wire, capacitor ESR, or a bulb.
This is also true at discharge cycle 1/2 is again lost.
So for one unit of energy, 1/2 get "lost" and at discharge 1/2 of 1/2 get "lost" again.
So it give that 3/4 is lost/burned and only 1/4 return to source... 3/4+1/4=1 so energy is conserved in the whole process, I don't see unfortunately OU in this setup at least.
---------------------------------------------------------------------
Here the code:

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o 12 64 0 291 640.0 6.4 0 -1
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o 26 64 0 35 10.0 9.765625E-5 5 -1

---------------------------------------------------------------------------------

or

$ 1 5.0E-6 19.867427341514983 50 5.0 50
w 304 240 304 160 0
w 304 160 352 160 0
r 352 160 448 160 0 1000.0
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w 560 208 560 160 0
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178 496 208 496 288 0 1 1.0E-9 -1.383E-321 0.05 1000000.0 1.0E-9 1000000.0
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w 560 304 672 304 0
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w 656 400 656 480 0
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w 832 320 832 304 0
w 832 304 672 304 0
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o 10 64 1 291 80.0 9.765625E-5 0 -1
o 2 64 0 291 320.0 1.6 1 -1
o 10 64 1 299 229.11123134779967 9.765625000000002E-255 2 -1
o 2 64 1 299 214.30172143725346 2.4414062500000007E-305 3 -1
o 10 64 0 298 559.9361855444511 0.17498005798264096 4 -1
o 24 64 0 299 10.0 9.765625E-5 5 -1
o 6 64 1 291 320.0 9.765625E-5 6 -1
o 7 64 1 291 320.0 9.765625E-5 7 -1

[wayne49s]

Look like don't work for me here...

---------------------------------------------------------------------------------

I think you should consider the resistor (light bulb) as the output in this device. It is not a loss. It is the device to extract the overunity power. I would think it would be more like a battery is what I would be interested in lookng into.


/Wayne