Sharing ideas on how to make a more efficent motor using Flyback (MODERATED)

[tinman]

Okay so I see the crisis has been resolved.

Old data:

A) 17.745mJ  Energy expended by the power supply during one ON-pulse, to bring coil's current from 0 to ~600mA,
B)   6.804mJ  Calculated energy stored in the 37.8mH coil at the end of the ON-pulse,
C) 49.005mJ  Recovered energy in C2 after one ON-pulse.

New data:

E₁ = 5.382W * 4.32ms = 5.382 * 0.00432 = 0.02325024J = 23.25024mJ
E₂ = ½ * 37.8mH * (824mA)² = ½ * 0.0378 * 0.824² =  ½ * 0.0378 * 0.678976 = 0.0128326464J = 12.8326464mJ
E₃ = ½ * 1μF * (99.6V)² = ½ * 0.000001 * 9920.16 = 0.00496008J = 4.96008mJ

E₃ / E₂ =  4.96008mJ / 12.8326464mJ = 0.38652 = 39%  (stored to recovered energy)
E₃ / E₁ =  4.96008mJ / 23.25024mJ = 0.21333 = 21%      (expended to recovered energy)



Now to get to my main theme, the differences in the current waveform between no rotor and the rotor in place.  I made a quick attempt to see if the "Dia" program would allow me to make black a transparent colour on one of the two scope captures to superimpose one on the other but quickly gave up.  I attached the two close-up graphics again.

Why is the current a lower value when the coil is driving the rotor?  I bet that many people would believe that when the coil is driving the mechanical load of the rotor that the coil should draw more power and therefore draw more current.  So why is it backwards?

The answer lies in the nature of the way the coil works and what it means when it is exporting power to the outside world to make the rotor turn.

When you apply voltage across the coil you are overcoming the electrical inertia of the coil and the current slowly rises.  The voltage is "pulling the current up" in the coil and we know that the applied voltage is a constant DC value.

When the coil is driving the rotor, a portion of that voltage is not "pulling the current up" in the coil.  Instead, some of the voltage is being "consumed" to drive the rotor.  That means there is less voltage available to "pull the current up" in the coil and therefore there is less current flowing through the coil.  The instantaneous "missing voltage" times the instantaneous current is the instantaneous power that is driving the coil.  (I am simplifying and ignoring other losses.)

If you look carefully at the two current traces, and temporarily ignoring the effects of the wire resistance, the slope of the current curve is telling you the approximate amount of voltage "available" to increase the current flow.  The slope of the current curve when the rotor is in place is lower than the slope when there is no rotor.  Therefore less voltage is "available" to increase the current flow in the coil when the rotor is in place.  So, in a manner of speaking, you are indirectly "seeing" how much voltage is being "consumed" to make the rotor spin.
That's why the DSO shows more energy put into the coil during the energizing cycle than the calculated 1/2 L i-squared value at the end of the energizing cycle.  The difference is the energy "exported to the outside world" to make the rotor turn.

MileHigh

So, Brad, I think it's appropriate to rub it in this one time:  You jumped the gun.  I am not trying to be nasty or anything like that.  Let this be a lesson for you.  Your mocking comments about "the books" ring hollow.

Lol-right on que 

Your mocking comments about "the books" ring hollow.

Well if your explanation below is from books MH,then i my self now have to have a laugh.

Voltage that is consumed to make the rotor spin is not available to increase the amount of current in the coil and therefore the slope of the rising current waveform is less compared to when there is no rotor.

MH
We all know that voltage alone is not power,so this part makes no sense at all. Also,as this !! voltage !! that is being consumed by the rotor is after the measuring equipment,then any power the rotor is suppose to be consuming must go through the measuring equipment first,unless you have found a way to some how go straight from the source,bypass the measuring equipment,and jump straight into the rotor. There is also the fact that even in your own words,there is more energy being put back into the system from the rotor than was used to drive it.

If you look at the two scope shot's below,and can understand what it is you are looking at,you can see where some of the energy is coming from to spin the rotor--i will also attach the schematic, with probe locations. Every thing above the red line is indicating a current flow through the charge battery !! and what !! ? When the yellow line is flat during the off time,is showing you the supply battery voltage. So can you understand what these two scope shots are showing you?.

This thing about the rotor stealing battery voltage is nothing more than fairy tails MH. The rotor is generating current flow through the coil before the transistor switches on,and so this electrical inertia has already been taken care of by the rotor,and thus the supply voltage dose not have to work so hard to get this electrical inertia flowing-so to speak.

So MH-you too jumped the gun with your analogy,as it is wrong.


Brad

[picowatt]

If you look at the two scope shot's below,and can understand what it is you are looking at,you can see where some of the energy is coming from to spin the rotor--i will also attach the schematic, with probe locations. Every thing above the red line is indicating a current flow through the charge battery !! and what !! ? When the yellow line is flat during the off time,is showing you the supply battery voltage. So can you understand what these two scope shots are showing you?.

Tinman,

I disagree with your "red line' comment.  It looks to me as though no current can flow into the charge battery until the voltage at the coil/collector junction exceeds the voltage of both batteries in series plus one diode drop (approx. 25V).  This would mean that current only flows into the charge battery at and during the mostly squared off top of the voltage waveform.

PW

[picowatt]

Tinman,

Also, during the off portion of the cycle with the rotor installed, where we see the sinusoidal waveform produced by the passing magnets, no current flow is occurring anywhere (other than leakage) so we are only seeing an unloaded voltage waveform being generated by the rotor magnets during that time.

PW

[picowatt]

It might be interesting if someone could edit Tinman's "'with rotor" scope capture by copying and pasting the sinusoidal waveform seen during the off time onto and aligned with the left side of that sine wave so that we can see if its phase/frequency remains constant during the on time.

In other words, if a copy of the observed sine wave is shifted to the left, does it match up, is it time continuous, with where the preceding sine wave stops at the beginning of the on time?

This may provide evidence as to whether the rotor's speed is constant or not.

Trying to do this by counting scope divisions is a bit difficult at the sweep rate used, but it does look like there may be one minor division or so missing, which would indicate rotor acceleration during the on time.

PW

[MileHigh]

Brad:

I think you had a tough day and are giving me some excess push-back.

MH
We all know that voltage alone is not power,so this part makes no sense at all. Also,as this !! voltage !! that is being consumed by the rotor is after the measuring equipment,then any power the rotor is suppose to be consuming must go through the measuring equipment first,unless you have found a way to some how go straight from the source,bypass the measuring equipment,and jump straight into the rotor. There is also the fact that even in your own words,there is more energy being put back into the system from the rotor than was used to drive it.

If you look at the two scope shot's below,and can understand what it is you are looking at,you can see where some of the energy is coming from to spin the rotor--i will also attach the schematic, with probe locations. Every thing above the red line is indicating a current flow through the charge battery !! and what !! ? (http://overunity.com/Smileys/default/wink.gif) When the yellow line is flat during the off time,is showing you the supply battery voltage. So can you understand what these two scope shots are showing you?.

This thing about the rotor stealing battery voltage is nothing more than fairy tails MH. The rotor is generating current flow through the coil before the transistor switches on,and so this electrical inertia has already been taken care of by the rotor,and thus the supply voltage dose not have to work so hard to get this electrical inertia flowing-so to speak.

So MH-you too jumped the gun with your analogy,as it is wrong.

I think what I wrote is pretty clear and I suspect that you are not getting it.  I am not going to reword it so I suggest that you read it again and think about it.  You are off the scent of the trail but I am sure you can get back on track.  I am quite confident about what I am saying.

There is also the fact that even in your own words,there is more energy being put back into the system from the rotor than was used to drive it.

No, no, no, no, no.  I don't know how you see that in what I write and it looks to me like you are still projecting your wishes onto this situation as if they are real.  I am on the edge of a "going bonkers" breakdown over the rotor.

This all goes back to schematics and timing diagrams.  In the realm of electronics, when you are trying to truly analyze a circuit, the schematics and timing diagrams rule.  Sometimes the timing diagram does not or cannot show you all the information you are looking for.  You still can use your knowledge of the circuit and electronics and your knowledge of what the pulse motor itself is doing in the physical realm to extract "hidden" information in the timing diagram.

MileHigh