Advanced and Delayed magnetic field's.

[picowatt]

Hi Gyula

I would be more than happy to do so,but i dont have an L meter-must get one one day.
The two core's are just being held together with the rubber band-- not glue'd together.

Tinman,

Consider making a scope shot with your external trigger coil positioned right next to the leading edge of your pole pieces so that the trigger pulse is generated from the same magnet passing thru the pole pieces and just prior in time to the pri/sec outputs.

Keep the trigger coil and pri/sec polarities all the same for convenience.  You can temporarily use a channel to view the trigger to ensure this.  Trigger on rising edge.

Thanks...

PW

[tinman]

Tinman here is what is going on:  The windings have a very low coupling factor.  So what the secondary sees is effectively a 10mA p-p current source applied across a roughly 70mH inductor that is in parallel with the potentiometer and 100 Ohm resistor.  Driving at 102Hz:  When the pot is turned down to 2 Ohms the p-p amplitude is 40mV as on your scope, and when the pot is cranked to 1K so that you have 90 Ohms, the p-p amplitude is 800mV as seen on your scope and the phase is advanced ~58 degrees relative to the 2 Ohm case as seen on your scope.

What makes the source look like a current source is a low coupling factor.  (I used a coupling factor of 0.01 in my simulations. ) We know the coupling factor is low from other parts of your demonstrations:  Changing the load on the "secondary" has very little effect on the primary and doesn't affect the current reading on your DMM at all.

So, I think this is a case of mystery solved.

Problem not solved Mark. I think you may be asumeing that more power is being disipated through the secondary coil and resistor when the resistance is turned down to 2 ohm's--> but this is not the case. Im sure you can get an estimation of power disipation from the P/P values over the 90 and 2 ohm resistive load's. The secondary is disipating more power when the resistive load is 90 ohm's,and this is when it is around 58* advanced to that of the primary.

Now,all fancy word's aside,the fact is that a changing magnetic field must be present through the inductor before a current is produced. So how is it that this changing magnetic field appears in the secondary before it dose in the primary?-we are starting from the zero volt line here.

I will also ask you if you know what position the PM's on the rotor will be at relative to the core when the voltage from the inductors is at it's peak?.

[tinman]

Tinman,

Consider making a scope shot with your external trigger coil positioned right next to the leading edge of your pole pieces so that the trigger pulse is generated from the same magnet passing thru the pole pieces and just prior in time to the pri/sec outputs.

Keep the trigger coil and pri/sec polarities all the same for convenience.  You can temporarily use a channel to view the trigger to ensure this.  Trigger on rising edge.

Thanks...

PW

Will do PW.

@TK
I will show a drag test on the rotor in my next video.I am also going to wind some small coils on the top horizontal parts of the two cores-one closest to the rotor will be use as the trigger source,but we can also look at the magnetic field that is traveling across the top of the two coils as well.

[MarkE]

Problem not solved Mark. I think you may be asumeing that more power is being disipated through the secondary coil and resistor when the resistance is turned down to 2 ohm's--> but this is not the case. Im sure you can get an estimation of power disipation from the P/P values over the 90 and 2 ohm resistive load's. The secondary is disipating more power when the resistive load is 90 ohm's,and this is when it is around 58* advanced to that of the primary.

Now,all fancy word's aside,the fact is that a changing magnetic field must be present through the inductor before a current is produced. So how is it that this changing magnetic field appears in the secondary before it dose in the primary?-we are starting from the zero volt line here.

I will also ask you if you know what position the PM's on the rotor will be at relative to the core when the voltage from the inductors is at it's peak?.

As described I have reproduced the relative phase shift and amplitudes in the secondary for the 2 Ohm and 90 Ohm load conditions you demonstrated.  We can divert into the power if you like, but:  It is a new issue and as long as the waveforms in simulation and measurement match for the same component values, then the power matches as well.  I am not sure why you are surprised that the power increases with higher resistance loads.  All that means is that the series impedance is much higher than any of the loads you have tested with so far.  You are on the low load side of the maximum power point.  This is consistent with your other observations that all point to weak coupling.

This is a recurrent intuition problem with AC systems.  It is the rate at which current changes in the primary that induces voltage across the secondary.  As you lighten up the load, it appears more resistive and therefore more in phase with the induced EMF.  The induced EMF is following the derivative of the primary current.  The derivative leads the current.  If the transformer were tightly coupled, the secondary phase lead would shrink and the amplitude would come up. 

The position of the PMs at the secondary peak changes with the load.  Relative to the primary where the relationship is more or less fixed, it will be where the rate of change of flux is highest, so that should be close to when the ferrite gap is half way between two adjacent magnets.

[tinman]

Here is the drag test TK requested,and also the difference between the darg of steel laminated cores and ferrite cores.

https://www.youtube.com/watch?v=uAXtB_7RkEg