Confirming the Delayed Lenz Effect

[DreamThinkBuild]

Hi Luc,

I see your new post, good job. Thanks for taking the time to share your results.

[CRANKYpants]

Hi Thane,

here are a few more Shots to better see the shift. I tuned to Less current and expended Volts Division to Max.

First shot is Normal Probe setup on Primary input and Second Shot I Only moved the Voltage probe across the 10 Ohm bulb.

OKAY IF YOU MOVED YOUR PROBE TO THE 10 OHM LOAD THEN YOU ARE MEASURING THE CURRENT THROUGH THAT LOAD WHICH IS NOT CORRECT.

Anyways, I think that is what is important!... don't you think?
Luc

EVERYTHING IS IMPORTANT BUT LET'S JUST TRY TO FOCUS ON JUST PHASE SHIFT FOR NOW.
PLEASE DO THIS:

1) PUT YOUR CURRENT PROBE ACROSS THE SHUNT RESISTOR.
2) PUT YOUR VOLTAGE PROBE ACROSS THE PRIMARY.

3) DO WHATEVER YOU NEED TO DO (WITH CAPS AND FREQUENCY) TO GET A 90 DEGREE PHASE SHIFT ON LOAD.

4) SHOW NO LOAD...
5) THEN SHOW ON-LOAD (WITH 90 DEGREE PHASE SHIFT) IE ZERO POWER FACTOR! 

6) IF YOU CAN DO #5 WITH REAL POWER THROUGH YOUR LOAD THEN ALL THE POWER IN THE PRIMARY IS REACTIVE IE NO REAL POWER = 0.0 WATTS.

7) EFFICIENCY WOULD EQUAL INFINITY IF YOUR PRIMARY HAD ZERO DC RESISTANCE BUT BECAUSE IT DOES DOES NOT THE HEAT DISSIPATION IS:

I^2 x Rdc

NOW IF SOMEONE (ELSE) IS ABLE TO SHOW A NEGATIVE POWER FACTOR I.E. POWER BEING SENT BACK TO THE GRID THEN WE HAVE A WHOLE NEW BALL GAME HERE! 

ATTACHED IS THE TEST DATA FROM DR. FUSINA OF DEFENSE RESEARCH AND DEVELOPMENT CANADA WHO CAME TO OTTAWA UNIVERSITY TO TEST THE BiTT.

HE FREAKED OUT WHEN HE SAW 0.0 POWER FACTOR SO I TOLD HIM TO APPLY A FUDGE FACTOR - WHICH HE DID.

WE HAVE SINCE FOUND OUT THAT THE PF CAN BE NEGATIVE AND I AM HOPING YOU GUYS CAN ALSO VERIFY THAT?

THIS MIGHT HELP ME GET THAT "LAZY" ENGINEER AT PHILLIPS TO MOVE HIS ASS?

CHEERS
T

[woopy]

Hi Luc and Thane

thank's for advices

I have redo the video to better see the phase shift.

Hope this is what Thane suggested. But my sig gen is very basic and i cannot tune progressively, so it  is very difficult to to get very fine results. So i tested 2 frequency to get with 150 Hz the max phase shift (something less than 180 degree ) and to get 90 degree with 190 Hz  (the cap value is always 12 micro F).

hope this helps

http://www.youtube.com/watch?v=Fa7IsQe0jfc

Laurent

[gotoluc]

OKAY IF YOU MOVED YOUR PROBE TO THE 10 OHM LOAD THEN YOU ARE MEASURING THE CURRENT THROUGH THAT LOAD WHICH IS NOT CORRECT.

You are not understanding what I'm doing. Like I said above the First shot is done like you say:
1) PUT YOUR CURRENT PROBE ACROSS THE SHUNT RESISTOR.
2) PUT YOUR VOLTAGE PROBE ACROSS THE PRIMARY.

3) DO WHATEVER YOU NEED TO DO (WITH CAPS AND FREQUENCY) TO GET A 90 DEGREE PHASE SHIFT ON LOAD.

As I said above, I tuned to minimum Current draw on the Primary and not to 90 Degrees Phase Shift. I can easily tune to 90 Degrees Phase Shift (ON PRIMARY) but the Current draw will go up. So we have to wonder why is this happening in a MOT?... maybe the Primary does not need to be at 90 Degrees Phase Shift for the Secondary LOAD to not effect it if we are working with a Possible LENZ DELAY.

This is what made me think to probe the Secondary with the Voltage Probe but leaving the current Probe on the Primary and use the Primary's Current Phase Reference and compare it's Phase Reference with the Secondary Phase.

As we can see from the Second Scope Shot they are Exactly 90 Degrees out Phase with each other. So my thought is, who knows what is going on in a MOT under these conditions. if the Secondary is 90 Degrees out of Phase when the Current at the Primary is at MINIMUM then maybe this is where this Puppy is Happy.

Just my way of thinking. I know it is not conventional EE but I don't care as long as the load on the Secondary is not reflected on the Primary I'm Happy

Luc

PS I'll be out for a while

[gyulasun]

Just a more practical question that bugs me regarding this MOT experiment:
Why is a single LED surviving in this setup up?
I would expect the voltage at the secondary to be a few KVolt if I look at the difference in inductance between primary and secondary impedance.
Probably due to the high resistance of the secondary winding?

Yes I think also the DC resistance  (86 Ohm or so in Luc's case) serves also as current limiting 'bias' resistor for the LED. 
We can estimate how big the unloaded secondary voltage could be in its unloaded case.  Consider when normal input is designed 120V AC rms primary input and about 2000V rms secondary output, the ratio is 16.6  so for a 10V rms input the open voltage should be around 166V.
Then there is a voltage divider created whenever you connect a load to the secondary: the upper member of the divider is the coil DC resistance, 86 Ohm and the lower member of the divider is the load itself, when this is a LED or anti-parallel LEDs their dynamic AC resistance can be under 10 Ohm as you pointed out, so the significant part of the 166V is dissipated in the coil resistance and the rest can go only to the load.

This latter is an answer also to Luc why he did not see any light from the 12V car bulb: simply the divided voltage is not enough to light it. When I suggested to Luc to use low value resistors as loads I forgot to consider the rather high secondary coil resistance in this case. 
Perhaps a LED lamp of a few Watts designed for normal 120V AC mains could be used? But the output power from the signal generator is limited to half or 1W or so? (its output impedance surely is 50 Ohm).

Gyula