Thane Heins Perepiteia.

[CRANKYpants]

FYI

Wow,
well done Thane!

Does this work just with 60 Hz sinewave AC or are you using some kind of pulsed input ?

YES 60 HZ MAINS POWER.

How do you measure exactly the input power ?

I AM USING THE POWER METER SHOWN IN THE VIDEO AND PHOTOS.
THEN DOUBLE CHECK THE POWER FACTOR ON THE OSCILLOSCOPE.

What kind of meter do you use and do you use it in a range, where not much big error
could occur for the 1 Watt range ?

THE CURRENT AND VOLTAGE READINGS ARE GOOD - THE METER HAS TROUBLE WITH THE Pf THOUGH - SO THAT IS WHY I DOUBLE CHECK WITH THE SCOPE.

P.S: Just saw, that you had a cos phi power factor issue in the other thread...
So is the measurement still valid ? Please clarify
Many thanks.
Regards, Stefan.

NO THE MEASUREMENTS ARE NOT VALID.
WHEN I PLACED THIS TRANSFORMER ON THE SCOPE THE Pf IS ACTUALLY 0.7 OR ABOUT 45 DEGREES. AFTER REWINDING THE PRIMARY - I NOW HAVE HAD IT DOWN TO 0.26.

I AM REWINDING YET AGAIN THIS WEEKEND. 

IDEALLY ONCE WORKING PROPERLY THE PRIMARY CURRENT AND POWER FACTOR WILL NOT INCREASE WHEN PLACED UNDER LOAD BECAUSE THE SECONDARY BACK EMF INDUCED FLUX WILL FOLLOW THE LOWER RELUCTANCE FLUX PATH INTO EACH SECONDARY AND AVOID THE PRIMARY FLUX PATH ALL TOGETHER.

ALSO THE PRIMARY SHOULD BE OPERATED AS CLOSE TO SATURATION AS POSSIBLE.

THIS SCENARIO WILL DO 2 THINGS:
IT WILL SELF REGULATE THE OUTPUT POWER AND THE DIVERTED FLUX WHICH WOULD OTHERWISE CAUSE THE PRIMARY IMPEDANCE TO DROP - CAUSING IT TO DRAW ADDITIONAL CURRENT AND POWER FROM THE SUPPLY - WILL NOT BE REQUIRED.

STILL TRYING TO GET THE WIRE GAUGE TURNS RATIO MIX RIGHT.
.

PLEASE DON'T THINK I AM RUDE BUT I ONLY HAVE TIME TO DO ONE THREAD AT A TIME SO I WILL BE FOCUSING AND POSTING MY RESULTS ON THE OTHER THREAD.


THANKS
Thane

[waterhouse24]

DEAR LEE,

I HOPE THE NEW PHOTOS POSTED BY JUSTME ANSWER YOUR QUESTIONS?
THE AIR GAP IS ABOUT 1/4 INCH.
THERE IS NO AIR GAP BETWEEN THE TOROIDS.
EVERYTHING ELSE YOU GOT RIGHT.

THANKS
Thane

Those pictures speak more than a 1000 words, thanks Thane!

Lee

[CRANKYpants]

FYI Part 2

Do you want to explain this, please?
thanks

THE POWER FACTOR OF AN IDEAL INDUCTOR IS 0 OR VOLTAGE AND CURRENT BEING 90 DEGREES OUT OF PHASE.

IN A TRANSFORMER WITH A PURELY RESISTIVE LOAD THAT HAS A POWER FACTOR OF 1 - THE LOAD POWER FACTOR WILL BE TRANSFERED BACK TO THE PRIMARY AND THE PRIMARY WILL ALSO HAVE A POWER FACTOR OF 1.

THIS MEANS THE POWER DISIPATED BY THE PRIMARY UNDER LOAD IS P = VI or V^2/R..

IF THE POWER FACTOR IS NOT 1 THEN THE POWER DISIPATED BY THE PRIMARY IS
P = VI cos(ANGLE BETWEEN VOLTAGE AND CURRENT) OR
P = VI cos(theta).

SO IN OUR DEMO VIDEO OUR METER SHOWS A POWER FACTOR OF 0.11 AND THE OSCILLOSCPE SHOWS A POWER FACTOR OF 0.7 OR 45 DERGEES OUT OF PHASE.

IN THIS CASE THE METER IS WRONG - BUT ALL IS NOT LOST BECAUSE THE POWER FACTOR SHOULD BE 1 BECAUSE THE LOAD IS PURELY RESISTIVE. THIS IS THE GOOD NEWS AND THE BAD NEWS IS THAT IT IS NOT OVER UNITY - YET.

HOWEVER BY CHANGING THE WIRE GAUGE AND # OF TURNS ETC WE NOW HAVE THE POWER FACTOR DOWN TO 0.26 WITH A PURELY RESISTIVE LOAD - MORE GOOD NEWS.

UNFORTUNATELY OUR PRIMARY IS STILL NOT COMPLETELY ISOLATED - WHICH WE ARE WORKING ON CORRECTING.

CHEERS
Thanelite

[i_ron]

FYI



PLEASE DON'T THINK I AM RUDE BUT I ONLY HAVE TIME TO DO ONE THREAD AT A TIME SO I WILL BE FOCUSING AND POSTING MY RESULTS ON THE OTHER THREAD.


THANKS
Thane

Could someone tell me the addy of the other thread please?

Ron

[gyulasun]

Could someone tell me the addy of the other thread please?

Ron

Hi Ron,

Here you are: http://www.overunity.com/index.php/topic,5159.0/topicseen.html

Gyula