ENERGY AMPLIFICATION

[Farmhand]

No offence Br 549 but I think you simply have a measurement error, considering the tiny power level it would make sense that the power would be incredibly difficult to measure properly, you should try it with 10 Watts or more input. At least then if any extra energy was produced it would an obvious result and enough extra power to measure properly and do something with. No one will be convinced by apparent OU at such low power levels.

I have also measured OU from just a regular transformer using the scope and considering the phase angles and power factor and with a pure sine wave at very low power levels although more than you used I think. It is caused by measurement error in the equipment, the nature of the currents and the power level is so low a very small deviation or measurement error would produce a large error in the results.

I urge you to try the experiment with over 10 Watts and I bet you will see the anomalous measurements disappear, and get more proper results.

The tiny power levels are no use to obtain proper measurements. In my opinion. I'm sure many would agree.

Did you correct for power factor ?

Yes Gyula is correct, there is no load, therefore there is no output. No load no output, simple. efficiency is zero. In my opinion.

Cheers

[br549]

gyulasun - The meter measuring the input current is directly in series with the input circuit (between the H-Bridge and the  Transformer primary), nothing else. The output is directly connected to the the inputs of the meter measuring the output current, and that is it. One channel of the scope is measuring the input RMS voltage, and the other channel is measuring the output RMS voltage. The input frequency and the output frequency are the same.
Madddann - I have tried many differnt combinations, and so far this one seems to be the best (for the components that I am currently using). If you are interested, I can repeat and show some of the tests and results that lead to this point. If the flux bypass coils and the way they are connected make all the difference. The strength of the magnets make a difference.
Farmhand - I don't realy know how to make the measurement process more simple, and accurate (with the equipment on-hand) than what I am currently doing (since the input and the output signal are very simular, and I am measuring the the both the same way. Voltage RMS X Current = power. I have tried to eliminate any measurement error. My equipment is not the best, but it is accurate, so I'm certain the readings are correct, but like you say, there can always be unforseen errors not readly seen. I can present (as time permits) some of my earlyer tests with results.

[br549]

gyula - The meter that is directly connected to the output is rated at 3 amps. It is the load. (I did this to try and eliminate induced error due to any thing placed between my transformer output and my measuring device).

[profitis]

@farmhand,a working perpetual motion device doesnt get energy from nowhere,it sucks in ambient heat on one side and spits it out on the other,chek out the theory page.

[gyulasun]

gyula - The meter that is directly connected to the output is rated at 3 amps. It is the load. (I did this to try and eliminate induced error due to any thing placed between my transformer output and my measuring device).

Hi Br549,

Thanks for clarifying the load: it is the inner resistance (impedance) of the HP3478A multimeter when it is in AC amper mode. I looked up the manual on the web for this multimeter but the inner resistance (impedance for AC) was not specified for the two current ranges neither for DC nor for AC,  lower range is 300mA, higher is 3A (fused).  If I accept the 1.51V rms voltage drop across the ammeter input and I accept the current is 50mA rms then the inner resistance (impedance for AC) is R=1.51/0.05=30 Ohm   So this means your load, willy-nilly, was a 30 Ohm resistor, right?  This 30 Ohm sounds a bit high to me for a current meter. Agree with this?  I also assume that such a current meter (HP) does not have an inductive inner impedance for current measurements, so the AC current and voltage more or less are in phase, right? 
Are you not curious if the 30 Ohm inner resistance could be measured from the outside by a normal handheld digital Ohm meter? Well I am...      Just for a simple cross check, probably the DC ammeter input is the same two terminal as the AC ammeter input, so by choosing the AC current mode from the DC current mode on the HP the input reistance will not change I suppose. You could see this resistance on the hand held DMM hooked to the ammeter input.

One more thing: would you mind checking the output voltage across the ammeter with the other HP multimeter, switched to AC voltage mode?  Also a simple cross check.  Another thing: if you have a 100 or 200 or 300 Ohm (non wire wound) resistor (wattage is 0.25 or 0.5W)  and used it as the load across the transformer output and also measure the AC voltage across this resistor, you could see the rms voltage either on the scope or the HP multimeter (no need to check the current in this resistor load case).

Very interesting setup, have you considered to use a tuning capacitor at the output like at the input? This way the output would be more sinusoid too.

rgds, Gyula