Rotating Magnetic Field's and Inductors.

[tinman]

Brad:

Let's assume that the passing rotor magnets induce counter-EMF in the coil and that it exists as some kind of waveform.  You can open-circuit your coil and spin the rotor and see what it looks like.  When I suggest a counter-EMF "battery" in the coil it's just to average out the counter-EMF waveform and turn it into an average DC voltage, that's all.  It's not even a battery that gets charged, it's just an imaginary devices that outputs a DC voltage to emulate the average counter-EMF waveform from the coil.  That will make the average current draw of the coil decrease and chances are you will still have the same amount of power in the back spike.

The next step is to say, "Hey, who needs a DC counter-EMF voltage source inside the coil, I will simply reduce the supply voltage for the coil and I will get the same result."

That's the point, that all of this investigation into the "beneficial effects of a spinning rotor in front of a coil" can be approximately replicated by having no rotor and simply reducing the drive voltage to the coil.  If you can reduce the drive voltage and have no rotor and get the same back spike power as the case where you have the regular drive voltage and a rotor then you can say mission accomplished.

This is just reasonable speculation:  lower drive voltage with no rotor is equivalent to regular drive voltage with a rotor.

Unfortunately, your signals are buried in the "noise" of all of the power lost to heat and it's not necessarily easy to confirm this.  Or perhaps a quick test will confirm this, I don't know.

MileHigh

MH
Reducing the input voltage will reduce the input current. This in turn will reduce the strength of the magnetic field,and that will reduce the P/out when that field collapses,as the field now has less stored energy. Also,to say that the heat is being converted to electrical energy is also not valid. If we reduce the input current,then yes,the heat will be less. But at the same time,we can then raise the input current back to it's original amount,and thus the heat dissipated by the coil will be higher,as the output current would have also gone up,and so the total current flowing through the coil will now be higher-even though the input current is still at it's starting point. The alternating magnetic fields would also introduce further eddy currents in the core,and this would also result in higher core temperatures.

Every example you have given in decreasing the P/in,results in a drop of the P/out.
So i ask you to provide a way where as the P/in is reduced,while the P/out remains the same--without the use of an external alternating magnetic field.
In fact,i invite any of the EE guys to tell me how this can be done-without the use of the external alternating magnetic fields. This dose not include a total rebuild of the coil/inductor-->this is an apples for apples test-->same coil,same frequency,same duty cycle--or a duty cycle of your choice.


Brad.

[NoBull]

For example, at 0.25 Tau the ratio is 6:1 for energy stored vs. energy dissipated when energizing the coil which translates into 85.7% efficiency in energizing the coil.

How did you get from 6:1 ratio to 85.7% efficiency?

Also, what happens to efficiency when I allow the current in the coil to stabilize?  I think they call it the V/R limit.
Finally, how can I see on the scope that I am close to this limit?

[MileHigh]

MH
Reducing the input voltage will reduce the input current. This in turn will reduce the strength of the magnetic field,and that will reduce the P/out when that field collapses,as the field now has less stored energy. Also,to say that the heat is being converted to electrical energy is also not valid. If we reduce the input current,then yes,the heat will be less. But at the same time,we can then raise the input current back to it's original amount,and thus the heat dissipated by the coil will be higher,as the output current would have also gone up,and so the total current flowing through the coil will now be higher-even though the input current is still at it's starting point. The alternating magnetic fields would also introduce further eddy currents in the core,and this would also result in higher core temperatures.

Every example you have given in decreasing the P/in,results in a drop of the P/out.
So i ask you to provide a way where as the P/in is reduced,while the P/out remains the same--without the use of an external alternating magnetic field.
In fact,i invite any of the EE guys to tell me how this can be done-without the use of the external alternating magnetic fields. This dose not include a total rebuild of the coil/inductor-->this is an apples for apples test-->same coil,same frequency,same duty cycle--or a duty cycle of your choice.

Reducing the input voltage will indeed reduce the input current and it will reduce the final current.  That could be critical.  However, nothing is stopping you from measuring the final current with the spinning rotor when the coil discharges, and then you will know if you have "headroom" to operate at a lower voltage and still get the same sized back spike without the rotor in place.  Assume that without the rotor you are free to set any pulse width you want.  And as I mentioned in my previous posting, increasing the drive voltage without the rotor in place will get you to the "desired back spike current level" that much faster with a shorter ON pulse, and that sounds mighty tempting to me.

So i ask you to provide a way where as the P/in is reduced,while the P/out remains the same

Unless I am really screwing up, I think a higher drive voltage and a shorter ON time would do the trick.

If anything, I hope that this discussion is putting some ideas in your head!   For example, one of my favourites is to replace the charging battery with a big fat capacitor and a variable bleeder resistor.  If the voltage across the cap is close to DC you will get a deadly accurate RMS voltage reading with one of your multimeters.  You just have to make a spot-check on the value of the variable bleeder resistor and read off the voltage and get your power out - with added benefit of being able to dial up any "charging battery voltage" you want.  You could prove for yourself that the efficiency will go up as the capacitor voltage goes up.  It sounds like fun to me!

MileHigh

[MileHigh]

How did you get from 6:1 ratio to 85.7% efficiency?

6:1 means six parts magnetic energy to one part resistive dissipation.  So that's 6/7 total which is 85.7%.

[tinman]

MH

Here is your CEMF imaginary battery test i did--just for hoots

https://www.youtube.com/watch?v=J20qEB9Bc0U