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

[partzman]

Thanks for the reply partzman

The below pic is the scope and what comes up when I press acquire. No "Menu"

Luc

Gotoluc,

I assume that your model was the same as the 3000 series but I was wrong. I found an online manual and it appears the best resolution would be in the "Average" acquisition mode unless you have high speed captures then you'd use "Peak Detect" but I'm sure you already know all this!

With the cursor button you can select "Time Cursors" to do any measurements that model will allow but they do state that the automatic measurements are more accurate than the cursor measurements and I'm sure you're aware of this as well.

I did not see any math capabilities in the specs but you do have mean, cycle rms, plus other measurements which should allow you to do manual energy and/or power calculations on your waveforms.

partzman

[MileHigh]

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)

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.

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.  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.

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

[verpies]

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.

Below are the two scopeshots superimposed using this online image editor.

[MileHigh]

Below are the two scopeshot superimposed using this online image editor.

Thanks Santa!

[itsu]

I tried the "C2 discharger" using a reed relais, but the instance i remove R2 the contacts fuse together, even when the contacts have a series choke which has a diode/resistor in parallel across it.
The fusing is not permanent, but it makes it impossible to use it.

So i must follow one of verpies his "C2 discharger" designs.

The hall sensor to drive the reed relais did work ok, so i can use that to tune/trigger that setup from verpies.

Itsu