Quantum Energy Generator (QEG) Open Sourced (by HopeGirl)

[Farmhand]

All this reactive power stuff is becoming a real interesting subject. if we look back to the post linked below we see the top left
shot is the fluro traces, and the bottom right shot is the tank traces feeding it.
http://www.overunity.com/14443/quantum-energy-generator-qeg-open-sourced-by-hopegirl/msg414843/#msg414843

Volt x amps in the tank is 84.8 x .922 = 78.18 VA and power factor is (192 cosine) = -0.978,
so 78.18 VA x 0.978 PF = 76.46 VAR.
So 78.14 VA - 76.14 VAR = 1.72 Watts missing from the tank.
With 3.8 Watts to run the AC generator
and a load power of 0.5 W
while the fluro is struck, lit up and conducting out of phase current as well as receiving a spike of positive power and some.

The 3.8 Watts input minus the 1.72 Watts missing from the tank = 2.08 Watts generating losses running the tank at 78.18 VA.
The 0.5 Watts output divided by the 1.72 Watts missing from the tank = 0.29 or 30% efficiency.
But the light is lighting like it's powered by almost 10 Watts = doesn't add up. 

Very interesting.

My setup includes the cost of generating the HF AC power from stored potential "energy". We could think of the
output tank as the grid wall outlet.  And the primary circuit as the HF AC generator, with the DC as the fuel.

A complete system to analyze.

The QEG mob and the other reactive power OU claimants plug in to the grid, which is equivalent to the output of my tank.
They don't take into account the cost of generating the AC in the first place from the stored energy. But they get billed for
the costs in their bill. That's cheating. That cost should be included in the efficiency of running the device.

I can simply use a battery for the 12 volt supply. And charged by a solar system as well. Our grid power is free as well
since our solar array feeds my power supply. Suck it up claimants and include the real costs of running the devices.
Calculate it out or generate your own AC power from stored energy.
..

EDIT: I need to clear up some things on how I confused some people unintentionally, 

1) In the video I show the drawing but I indicate I'm going to scope the tank current when I actually scoped the current out of the
tank to the load. The other two resistors on the board to the left of the right variable cap are for scoping the tank currents.
I moved the reference grounds so as to scope the load to a different spot, but "electrically" the same reference point.

2)On the drawing I use color coding with yellow for the current probe placement, when on the scope the current trace is actually
the blue trace, and on the drawing I use red to show where the voltage is scoped and the voltage trace on the scope is actually
the yellow trace.  Power is the purple trace "V2", the "shown value is divided by 0.1" to get VA.

Video clip of the test referred to. The shots are from a previous test but the values are pretty much the same.
https://www.youtube.com/watch?v=vFOHk_0IDZg

I hope I got all that right.
.

[isim]

@Farmland « Reply #2356 on: Today at 06:49:27 PM »
I need a little reflexion about your circuit and the fluorescent tube. May be I can find a spice model for the tube and simulate your circuit. You will never have a sinusoidal current with a load like this, because a fluorescent tube is a highly not linear device. you need high voltage to initiates the fluorescent tube then the voltage goes down to something like 30V.
- So I don't understand why you have a sinusoidal voltage on the output 2. This is my main problem!

"
Basically the tube itself is the capacitor in this case and discharges it's charge at the beginning of the second half of the cycle.
That upsets the power calculations. The first half power trace is almost symmetrical and the second half is almost all positive.
"
- I don't think so, when the power is "active" the power trace is mainly disymetric. So in the first half trace have only "reactive power and only the second half is "active"

"
And ... If I had an AC primary driver circuit then that spike would not appear I don't think, but the lamp would still light up brightly.
Power would then be symmetrical. Wow again. I need to test that.
"
- If the  power trace is symmetrical with GND, the power is "reactive", no real power.
For an example, look at the curves:

Voltage: yellow
Current: blue
Power: purple
1) phase shift = 0° degre
2) phase shift = 45° degre
3) phase shift = 90° degre


Excuse me but I will study your last post tomorow, its already to late for me!
@+

[Farmhand]

Yeah ok, fair enough, but how does the fluro give off so much light with 0.5 Watts power consumption ? Even if it was consuming
the entire 3.8 watts I think it was still too bright for that input power from the supply. The V x A in the tank is either almost all
active or almost all reactive. But there is real power there in the "Watts range", I think, gone from the tank.

The tank does remain having sine waves but the current trace gets jagged when the fluorescent lights "in any way", so the light
output has a reflection to the tank current. It doesn't distort the tank current or voltage trace because the oscillating power
is so much more in magnitude than the load power I think, If I load the tank with a large capacitor or an inductive load then
the trace can be distorted if the inductive load is not tuned to the tank. 
.
Resonant rise.
With 12.3 volts input I have 85 volts across the load.
Primary turns = 8
Secondary turns = 40
Transformation ratio = 1:5

EDIT: Coupling is not 1  , but I do not know what is, the distance between primary ends and secondary bottoms is
about 12 to 15 mm. End to end coupling distance 12 to 15 mm. I estimate 0.2 to 0.3 "k" to each side, maybe a bit more.


12.3 volts x 5 = 61.5
85v minus 61.5v = 23.4 volts resonant rise still on the secondary tank and load.

The resonant rise is what "strikes" the fluro and the declining Q diminishes V to an equilibrium with the load and input.
The different factors determine if there is resonant rise left applied to the tank and the load or not.
.
The primary reflecting capacitor gets charged to over 40 volts each cycle and that calculates to 7.4 joules per second is put
on the primary "reflecting" capacitor that's across the switch somehow. Exactly double the input. Should be the other way round.
..
It's bizarro world today, everything is back to front and upside down.
.

P.S. Isim if there is more any information/data at all that I can provide to help with a simulation just say and I'll get it as quick as
possible, due to our time differences I need to make an effort to be "in time" for you.

Circuit values measured -
The drawing shows the secondary coils have 112 uH but they turned out to be 119 uH each measured. 1 mm Magnet wire. 40 tuns.
The frequency is now around 420 Khz.
The measured capacitance for resonance with the fluro load in place is 957 pF. Parallel plate air variable capacitor 1350 pF max.
 
Values by estimation and logic. 
Calculator says 119 uH should need 1207 pF for resonance at 420 kHz.
so inter-winding capacitance must be between 200 to 300 pF I estimate. 
The primary inductance now that I've spaced the primary turns does not read on my meter.  2 x 1 mm wire in parallel 8 turns.
The primary switch has a 20 nF capacitor across it and it seems fairly close to resonance so primary should be
in the 5 to 7 uH range.
Fluorescent Lamp is a - F10T8CW also has 40H written on it. Info sheet attached.

..
Coils are all clock wise wound and I'm using the secondary coil at the mosfet drain side of the primary coil.
.
This attached saved  and zipped web page of a solid state Tesla coil calculator models my transformers input power and
output voltage from one side with the fluro load fairly well.

Solid State Tesla coil calculator
http://www.extremeelectronics.co.uk/calcs/index.php?page=sstc_calc.php
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[Farmhand]

These any good for CSR's ? The shunt is a 75 mV one. The other ones come from KillaWatt meters.
..

[MarkE]

The only limitation is the inductance.  The 75mV shunt has lots of inductance and shouldn't be used above mains frequency.  The smaller shunts fro the Kill-a-Watt will be better but depending on their values will probably distort in the high audio range.  The Ohmite WNE's are only about $1.00 each from Digikey and the 0.5 Ohm and 1Ohm will both easily be accurate at 1MHz.