STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[Omnibus]

@lumen,

To reinforce your concern about capacitive coupling, please take another look at the graph I already published here: http://www.overunity.com/index.php?topic=8411.msg242813#msg242813 . As you can see, up to 47kHz the voltage is lagging behind the current which isn't a behavior of inductive load and speaks of a capacitive such. Due to what? It's true the graph shows data for a transformer with a load in the secondary coil. And, then again, at 47kHz the phase shift reverses itself and becomes "proper" (if we consider only active and inductive components of the impedance, as should be the case), decreasing with frequency and tending to disappear around 10MHz.

[lumen]

I would like to understand this argument better. First, as you can see all the voltages, both on the side of the input and on the side of the output coil, are measured against the same ground, as is the correct way of measuring them.

Further, capacitive coupling between the windings of the transformer seems far fetched to me. There may be such a thing but I've never heard of it. A reference or link would help.

Let's, however, suppose that there is such coupling. Understand first that I' calculating both on the input and the output side. The capacitive coupling would affect the phase of the I with respect to V and at some frequencies it might be very favorable. Then, if capacitive coupling brings about the effect let it be more. That not only wouldn't hurt but will favor the results and will explain additionally where the effect might be coming from.

Remember also that at the request of @gyulasun I placed capacitance in parallel with the input and I saw no effect due to this addition.

Now that I'm thinking about this, it is possibly more complicated than I was thinking.

Let me repeat the test first to see if I have it correct.

You put a wave into a transformer of known amplitude or V. Then on the ground side you place resistors of known values so you can monitor the current going to ground. (overly simplified but this is the concept)

This would then let you calculate all the energy going into the device, because you know what is going to ground?
It could be radiating energy or reflecting the wave back into the source like an out of tune radio antenna(standing wave).
This would also reduce the current shown at the output while unknowingly increasing the actual input load.

It seems there may be other problems also. The best test is to some how know the actual input energy put into the device, then compare this to what is going to ground on the output end.

You may be onto something here, but I still think there is a chance something is getting over looked.

[lumen]

Ok, How about this.

Feed the input with a resistor of the same value as on the ground side. Then monitor the input after this resistor in case there is any increased load on the input it will result in lower v on the input side also.

This would correct for the unknown "I" on the input side.

[Omnibus]

Unfortunately, I'm away from my experiment right now. Will be back in Massachusetts sometime Saturday or Sunday and then will a new setup with a new transformer of a different type to check all this.

Now, regarding the ground. The ground is common for both the input and the output. Therefore, on each side it closes the loop on the respective side. So we should just refer to the input circuit as opposed to the output circuit and should abandon the "ground side" concept. It's confusing because you have ground side on both sides of the transformer. Unless you have something else in mind which I can't get.

Once we agree that there are two circuits -- input (of the primary coil) and putput (of the secondary coil) and that these are closed circuits then we should agree that, provided the voltage is measured correctly, the culprit for any overestimation or overestimation of the input and/or output energy would only be the current. Now, the problems with overestimating current due to inductance have already been discussed and it was agreed that measuring it across a metal-oxide resistor (by dividing its resistance into the measured voltage drop) is the most precise way to do it -- inductance of the metal-oxide resistor is negligible even at the studied frequencies. So far so good. As long as current is measured correctly we don't care about radiating energy or reflecting it back to the source because it will be accounted for in the integration. Proper current and voltage measurement is all it takes. If it reduces the current observed at the output that would be detrimental to the effect so, once we observe effect, we shouldn't worry about that loss -- the actual OU will be even more.

Again, it seems that if current and voltage are measured correctly (voltage is, the problem is current; not any more, it seems) then the actual input energy put into the device is precisely known and it can be compared with the output energy. The problems with inductance were discussed. I don't see what the role you mentioned of the eventual capacitance might be other than enhancing the real OU effect.


P.S. Does anyone know of a place to order 1Ohm (or 10Ohm) precision resistor of, say, 0.1% tolerance and possibly 10W max power?

[Omnibus]

Ok, How about this.

Feed the input with a resistor of the same value as on the ground side. Then monitor the input after this resistor in case there is any increased load on the input it will result in lower v on the input side also.

This would correct for the unknown "I" on the input side.

Mind you this -- the resistors on both sides used to derive the two currents are both about 10Ohm, however, the resistance of the primary coil is about 63Ohms while that of the output coil is about 0.6Ohms.