Partnered Output Coils - Free Energy

[tinman]

Tinman,

You might consider trying to drive your primary directly from your FG using sine waves. Try sweeping the FG output between 30KHz and 60KHz, or there about, while watching the inner secondary with your scope.  Try it with the secondaries unloaded at first.  Remember to reset your FG's offset back to zero before doing so.

That steel putty you used, does it retain magnetism noticeably?  That is, if you place a magnet up against it, does that spot retain magnetism sufficiently to lift small objects.  Do you have a scrap piece of the hardened steel putty to experiment with?  I am just wondering how hard or soft, magnetically, that material is.

As for your earlier post regarding your "1:1" or "100%" comments, I am afraid I am a bit confused as to what you meant, and/or how you determined that.  Perhaps you would be so kind to explain that to me a bit further.
PW     

https://www.youtube.com/watch?v=WBgJfdaGLBY

[tinman]

@Tinman - I would advise you to not release any Figure's on IN/OUT

Not here, the GOON Squad will do nothing but ridicule the absolute most ridiculous things...

"The sun was setting so it must be fake", "Diode was too far from the Fett, it must be fake", "Your Scope was the wrong colour, it must be fake", you get the idea... If I were you, I would share privately between trusted individuals.

Not here. Not with the Trolls.

Of course, the decision is yours, and I know you can make your own decisions...

   Chris Sykes
       hyiq.org

Chris
I know you must have your reasons,but i see most people here are here to help.
I have made no claims of any OU device,so i dont see why posting measurements here would call up your goon squad.

I am posting results piece by piece,and am still undecided as wether to posting my P/in and P/out measurements anywhere at all. I am well aware of what comes after that,and this is the very reason for carrying out all requested test-along with my own.

I could just throw up a video !of say! two identical caps-one the input with X amount of volts in it,and one on the output with 0 volts in it. We switch on the FG,get the fets switching,and then hook up the input cap to the system. We let the system run until all energy stored within the input cap has been consumed,and we then see that we have a higher voltage value on the output cap(which is identical to the input cap in every way)at the end of the run. We then swap the caps around,and run the system again,only to find at the end of that run our input cap now hase 3.2 volt more in it than what we started out with in the first place.

Now see-something like that would get the juices flowing all over the place,but what if there was power coming into the system from some other source?.
This is why i go piece by piece,test by test,and make no such claim to the test above.

[MileHigh]

Brad:

Al explained in the next video. But to try here--> when L1 is powered up for that 5% of the cycle,we get say 1 watt output from the secondary during this 5%. Then L1(the primary) switches off,and we get a further 1 watt from the secondary during this 95% of the cycle.
Now,the average voltage over the whole cycle is 0 volt's,and the average DC current over the whole cycle is also 0 amps. This means that the total power flowing in one direction(during the 5% on time) is exactly the same as the total power flowing in the opposite direction(during the 95% off time) Now people are calling this the inductive kickback(the 95% off time)power,but when have you ever seen an inductive kickback produce 100% of the power of the forward on state power-(our 5% on time)with inductive coupling-transformer effect.

I am not really involved in this discussion, but I will make some comments for you to think about.  The text above is out of kilter because your basic concepts and methods might be at least in the ballpark, but they are not truly in the game.

I'm assuming you are talking about a MOSFET pulsing a coil, and the coil has a fly-back diode or a fly-back diode in series with a load resistor or a lamp bulb.  Even if you are talking about a load on a secondary, what follows applies in a general way.

When the MOSFET is on, this is the power sucking up phase for the coil.  So instantaneous power goes into the coil, and that becomes a certain amount of energy by the end of the ON time for the MOSFET.  There is also instantaneous power being dissipated in the resistance of the coil, and in the D-S resistance of the MOSFET, and that can be measured as total dissipative energy at the end of the ON cycle.

I think your scope is a DSO with math functions, so this could be measured.

Then when the MOSFET switches off, the stored energy in the coil does a "burn" through the coil resistance, the diode, and the load resistor or lamp bulb.  You can then add all those instantaneous power values up (integrate) and at the end of the cycle have energy measurements.

End of Input cycle:  a) Resistive energy burnt in the coil, b) resistive energy burnt in the MOSFET, c) energy stored in the coil.

End of Output cycle:  a) Resistive energy burnt in the coil, b) resistive energy burnt in the diode, c) resistive energy burnt in the load resistor or lamp bulb.

That is the general frame of reference for the sort of stuff you are talking about.  You have to discuss instantaneous power or (integrated) energy and not interchange the two concepts.

The moment the MOSFET switches on, instantaneous power starts to flow into the circuit.  What happens and where it goes is a great challenge to investigate.  But you need to "keep your concepts well grounded."

A final comment would be that there is always the possibility in a pulsing coil setup with a fly-back diode that current never actually stops circulating in the coil.  So that should always be investigated because in most cases presumably this is not a desirable thing and it represents unnecessary wasted power.

MileHigh

[MarkE]

OK
so now i would like to do some power calculations,and would like some input in case i screw this up lol.
First the P/in.
I am going to calculate P/in using instantaneous measurements,and then averaging that out-as my DMM amp meter dosnt like the instant current. So i am using a 10 watt .33 ohm resistor,and scopeing across that resistor to get the instantaneous voltage(below is the scope shot)
So besides the ringing after switch off,i calculate the voltage across that resistor(.33 ohms) to be an average of around 1.7 to 1.8 volts. So our instantaneous current would be 5.15 amps. We average this current out by multiplying it by the duty cycle,which is 5%.
So our average current is 5.15 x 5%=257.5mA. With a large cap between my DMM and the circuit,my DMM reads 251.6mA--so we are close.

Now,the voltage we use to calculate P/in is ?

It is the voltage applied across the branch you are measuring the current through.

The voltage of the battery,or the voltage across the coil(L1) during the on time?-->so as we eliminate power disipation of the rest of the circuit.
If it is the voltage across the coil,do we then use that instantaneous voltage value,RMS value,Mean value ,or average value?.

Use P_INSTANT = V_INSTANT*I_INSTANT  P_AVERAGE = 1/t*integral(P_INSTANT). 
If you have a constant voltage and a variable current:  P_AVERAGE = V_CONSTANT*I_AVERAGE

[MarkE]

Which is, as I have already proven many times, perfectly fine to use, if one eliminates the possible small measurement defects encountered!

your continued argument here is TOTAL RUBBISH MarkE!!!

you're full of crap and as helpful as a bag of bricks!

if only you understood the absolute basics of "Power Generation"

LOL, it was not so long ago that you made the false claim that AC coupling is necessary to accurately evaluate AC signals.  Given that you have recently acknowledged that claim of yours was wrong, now comes the issue of when AC coupling is does not excessively distort a measurement.  I've listed the correct criteria.  All you have to do is copy and paste it.

BTW:  When do you plan to retake all those measurements you erroneously took using AC coupling when it wasn't appropriate?