Thane Heins Perepiteia.

[LarryC]

Hi Thane,

I had run a test with a small coil/core against the motor shaft length to test the voltage. The core was 1MM thick by 10MM wide by 40MM long. The coil was wound so that the 40MM sides were the poles. It did produce voltage as expected by your N S flipping statement at bottom/top. Unless you allready know and if you have the time, could you measure both sides at the same time with your oscilloscope to see the phase difference and if N S is occurring at the same time?

Still working on my theory and I can't get my stupid version of this song out my head.

Everybody went Flux fu fighting
Those kicks were fast as lighting
In fact it was a little bit frightning
But they fought with expert timing

Thanks,
Larry

[ramset]

NICE tune LAR    Chet

[OilBarren]

Okay, I'm not going to go wacko yet. Having a hard time making out the decimal place on the ampmeter. If .01 then .035W out and .019W in. If so, how much for the big honking 5000 watt version?
Regards,
Larry

HERE IS SOMETHING TO CONSIDER HOWEVER...

POWER FACTOR IN A TRANSFORMER PRIMARY IS LOAD DEPENDANT, I.E. A PURELY RESISTIVE LOAD - POWER FACTOR OF 1 PRODUCES A PRIMARY POWER FACTOR OF 1 AND IF THE LOAD POWER FACTOR CHANGES THE PRIMARY ALSO CHANGES.

IN OUR TRANSFORMER IT IS NOT - WITH A PURELY RESISTIVE LOAD Pf = 1 OUR POWER FACTOR IS JUST 0.26 AND THERE IS NO SCIENTIFIC REASON  (THAT I KNOW OF) WHY THIS CAN CHANGE JUST BECAUSE THE INPUT POWER DECREASES. LOAD POWER FACTOR IS ALWAYS TRANSFERRED BACK TO THE PRIMARY VIA MUTUAL COUPLING.

WE HAVE ALWAYS ASSUMED THE POWER FACTOR IS 1 (WORST CASE)

LET'S ASSUME FOR ARGUEMENT SAKE (AND I AM BEING FACETIOUS BECAUSE WE LIKE TO ARGUE HERE) THAT IT IS 0.26 IN THE LAST PHOTO WITH LUC'S NEW 175 STRAND LITZ WIRE. IF THIS HOLDS OUT TO BE TRUE THEN:

INPUT POWER   = 1.921 volts x 0.01 amps x 0.26 Pf
                          = 0.005 watts

OUTPUT POWER   = (5.94 volts)^2 / 1000 ohms
                             = 0.035 watts

EFFICIENCY  = 0.035 watts out / 0.005 watts in
                      = 705.7%

IF ALL GOES WELL WE SHOULD KNOW SUNDAY...

Thane

[polarbreeze]

INPUT POWER Ã,  = 1.921 volts x 0.01 amps x 0.26 Pf
Ã,  Ã,  Ã,  Ã,  Ã,  Ã,  Ã,  Ã,  Ã,  Ã,  Ã,  Ã,  Ã,  = 0.005 watts
...
Thane

Could you help me understand why it's necessary to take these measurements at such very low inputs currents? There's still that difficulty of the 0.01 amps not being a reliable number because the clamp meter has a +/- 3 digits resolution (ie a reading of 0.01 could actually mean as high as 0.04A). Couple of ways around that:

- run it at a higher input voltage to bring the input current up to, say, 0.5A (or even higher if possible), which would at least get the accuracy into a range of about +/- 10% or so.

and/or

- use the same power meter (DW-6090) as you're using in the latest series of experiments, which has better resolution than the clamp meter (1 digit rather than 3 digits) - or better yet, use an ammeter with better resolution than that.

PB

[OilBarren]

INPUT POWER   = 1.921 volts x 0.01 amps x 0.26 Pf
                          = 0.005 watts
...
Thane

Could you help me understand why it's necessary to take these measurements at such very low inputs currents? There's still that difficulty of the 0.01 amps not being a reliable number because the clamp meter has a +/- 3 digits resolution (ie a reading of 0.01 could actually mean as high as 0.04A). Couple of ways around that:

- run it at a higher input voltage to bring the input current up to, say, 0.5A - higher if possible, to get the accuracy into a range of about +/- 10% or better.

and/or

- use the same power meter as you're using in the latest series of experiments, which avoids the clamp meter problem (but of course I don't know if its resolution is any better: what does it go down to?)

PB

TOMORROW WE WILL CHECK THE NEW PRIMARY AT HIGHER CURRENTS.
THE PROBLEM IS THE PHYSICAL SIZE OF OUR PRIMARY CORE IS 5MM SO HIGHER CURRENTS (HIGH ENOUGH TO USE THE Pf METER DON'T WORK WELL AND WE HAVE 200 FEET WOUND ON OUR PRIMARY NOW BUT WE SHOULD HAVE ABOUT 1000 TO GET THE IMPEDANCE DOWN TO A POINT WHERE WE CAN INCREASE THE VOLTAGE WITHOUT MELTING OUR PRIMARY WIRE. - WE ARE USING 175 STRAND 46 GUAGE LITZ - BUT WE WILL GIVE IT A GO AND SEE WHAT HAPPENS.

ASSUMING YOU ARE RIGHT ABOUT THE CURRENT BEING 0.04 amps THE EFFICIENCY ONLY DROPS TO 177.3% IF THE POWER FACTOR IS 0.26

SURELY EVEN YOU MUST BE TAKEN ABACK BY A PRIMARY POWER FACTOR OF 0.26 WITH A LOAD POWER FACTOR OF 1? DEFINATELY NOT NORMAL TRANSFORMER BEHAVIOR.

THIS IS CLOSE TO DESIGN SPECS WHERE THE PRIMARY POWER FACTOR WILL BE 0 ON NO LOAD AND REMAIN AT 0 WHEN PLACED ON LOAD AND ALL THE CURRENT IN THE PRIMARY WILL BE REACTIVE.

Thane