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

[MileHigh]

Recycling unused energy that would normally be wasted is a good thing.  But how many times do you want to recycle that energy is a valid question.  For example, when you look at Laurent's clips you notice when he dumps the drive coil pulse into the capacitor and then the capacitor discharges through the second drive coil it completely discharges and you are done.  That's arguably an "elegant" solution - do a single recycle, burn off all of the unused recycled energy and you are done.  There is arguably little to gain by adding a second energy recycling stage.  It makes the circuit more complex and at each stage there are losses.  More stages equals more losses.

There is a somewhat ironic situation happening.  This thread is about a "more efficient motor" and only myself and Tinman have mentioned that the high-voltage coil has a relatively high resistance and that means more losses.  I have challenged you guys about measuring the losses in the high-voltage coil and nobody wants to touch it.  Also, nobody really knows what a true "Goldilocks" pulse would be for the high-voltage coil.  Perhaps a longer push-pull pulse that subtends the angle between the limits of the "U" core of the high-voltage coil?  (Relative to the passing rotor magnet.)  What about a shorter stronger pushing pulse that is focused on one end of the "U" core only where the magnetic field is stronger?  It's an unknown.

On the side of the good news, you are free to move the high-voltage coil around and hunt for a sweet spot.  That is your very easy variable timing system.  You don't actually need to scope the timing or make an LED timing gun.  What I can say through is that if I was a builder, I would add this to my bag of tricks:  In the TK MHOP clips he over-drives a "regular" LED and has a circuit to flash it at the start and end of the drive coil ON time.  To make it simpler, I would get a big mean very high-power LED, and use the op-amp to switch on and off a power transistor.   Then I would paint little white dots on top of the rotor magnets, and voila, you have your LED strobe timing gun to see the conduction angle for the drive coil.  You could make a little jig and then use your LED strobe timing gun on any pulse motor that you build.

So here is a variation on what you see in Laurent's clip to consider:  You need to have a high-voltage coil with multiple taps.  Then you can experiment with different capacitors, and different tap points on the high-voltage coil.  Can you shape the pulse to get more rotor bang for your pulse that comes from the drive coil?  Here is the big question:  When I change taps on the high-voltage coil, I lower the resistance of the coil.  Will that improve my overall setup because there is less resistance in the secondary coil?  The presumption is that you will get a shorter length pulse.  But don't forget that you are dealing with less turns, and therefore a weaker generation of a magnetic field for the same amount of current.  It's hard to know, but it could be an interesting investigation.

[MileHigh]

Now I am going to play devil's advocate.  Sometimes in life you arrive at a zero sum game, and that may apply to the multiple-tap high-voltage coil.   In one case you have higher turns, higher resistance, and a higher strength of magnetic field generated.  In the other case you have lower turns, lower resistance, and a lower strength of magnetic field generated.  What happens to the resistive losses in the coil when we compare the two and we want the same strength of magnetic field generated?  Well, we know that low current, high turns, high resistance will be equal to high current, low turns, low resistance for the same strength of magnetic field.  So what if we look at the resistive losses in these two cases?  You are comparing (low current x high resistance) losses to (high current x low resistance) losses for the same strength of magnetic field generation.  That sounds like a zero sum game to me.  You can crunch the numbers on paper without even building anything to check this.

If you assume that it is indeed a zero sum game with respect to the resistive losses, (and we can't be 100% sure of that in the real world actual build) then the key to extracting the maximum rotor bang for your pulse buck may me more directly related to pulse strength, pulse timing and pulse duration.  Let's not even discuss the pulse strength because you have limited control over it.  So, to adjust the timing is trivial, because you have the luxury of physically being able to move the high-voltage coil around to effectively change the pulse timing.  Pulse duration is controllable with the size of the capacitor with the caveat that the larger the capacitor, the longer the pulse duration, and the weaker the pulse strength.

As you can see, it's an interesting study in trade-offs when you play with the various parameters.  This is all presuming that you have a fixed amount of pulse energy to work with.   And I will leave you with another question to ponder:  You are trying to make the most efficient pulse motor by recycling the pulse energy that comes from the drive coil when the reed switch (or transistor/MOSFET) switches off.  But perhaps a big pulse of energy from the drive coil is not as good as optimizing the drive coil pulse first and foremost, and then working with a smaller pulse for the high-voltage coil.  What is the best energy mix between the low-voltage drive coil and the high-voltage secondary coil?  70-30?  50-50?  30-70?  It's hard to know that one but at least being conscious of it is worthwhile.

[Magluvin]

I had looked through my vids to see if i had one, but dont....  I remember trying to get a bemf spike into a higher henry coil and the higher H coil seemed to block most of the spike rather than take advantage of the full potential. Like a subwoofer crossover coil, it blocks out the high frequencies.  So the capacitor across your higher inductance coil probably loads up first then delivers it charge to the parallel coil?

Mags

[MileHigh]

One other comment that just occurred to me.  We notice in Laurent's clips that the capacitor simply discharges through the high-voltage coil and we observe the voltage decreasing in an almost linear fashion.  That means there is a fairly even pulse of current going through the coil because i = C dv/dt.  This is somewhat strange, because the circuit is an LC tank, and normally it's supposed to resonate.  My assumption is that between the EMF induced into the high-voltage coil from the passing rotor magnet, and the relatively high resistance of the coil, you "stumble" upon this favourable capacitor discharge curve for the pulse motor operation.

However, if the secondary coil resistance starts to get lower because you change to a tap on the coil with a lower number of turns and a lower resistance, then you might start to see the LC circuit start to resonate.  It's pretty safe to assume that a resonating LC tank circuit will NOT help in the operation of the pulse motor.

[tinman]

@Tinman,

Pay attention Bub! The current from the primary coil generates a magnetic field that collapses and reverses current polarity and travels to the right at the junction when the Reed switch opens. When the Reed switch is closed the current from the primary coil travels to the left toward the negative ground! The yellow marker's pointing at the general area in the schematic below!

Bub

No, the current dose not revers polarity in the primary inductor-only the voltage dose.
You have confused current flow direction with current flow path.

OK,i will try one more time.
Please see attached pic below.
Diagram at top shows conventional current flow when reed switch is closed.
Diagram at bottom shows conventional current flow when reed switch opens.
As you can see by the red arrows,the current continues to flow in the !same! direction through the primary coil when the reed switch opend. You only have to look at the diode direction in the picture you posted of Woopy's circuit to see that this is true. If the current flow reversed as you say,then the diode would have to be turned around in order for that flyback current to flow into the cap/inductor combo on the secondary.

The cap will charge first with the flyback current,as it has a lower series resistance than that of the high turn coil. The cap will then start to discharge into the secondary coil as it reaches peak volatge. This is the reason for the !almost! linear discharge through the secondary coil,and the absence of resonant oscillations.

Hope that clears things up.