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

[woopy]

Hi Luc

Thank's very much for sharing your experiment, and it seems to work very well.
What puzzle me is the fact that the very sharp fly back spike normally has almost no energy, it has very high voltage peak, but very low current, and if you connect a diode and a capacitor, it takes a lot of pulsse to charge it to obtain some power. But in your design the cap get a real power at each pulse.
So i think that we have to think of what happen in the fly back spike itself.
How does the flybackspike "capture " this energy ?

https://youtu.be/tag5OlvPi54

Thank's

Laurent

[gyulasun]

Hi Laurent,

Very clever experiment, thanks for showing it!
To explain your question, the first thing to consider is that the actual load for the sharp flyback pulse is not only your 1 uF capacitor but the 1.8 Henry coil,  i.e a parallel LC combination. So they have an AC impedance and although they do not seem to be in resonance with the frequency  the rotor RPM establishes, they cannot represent as a low impedance for the spike as a solely (discharged) capacitor normally would.

So the "trick" is to load the flyback spike with a relatively high impedance component or circuit like a multiturn coil or a parallel LC circuit brought near or very near to the spike's frequency,  these insure a high impedance load. An empty capacitor is just the opposite: it is a very low impedance load first (when fully discharged) and then it represents an increasing impedance as it charges up.

Addition: a HV coil has many number of turns so a small current a flyback can insure is still able to perfom a decent magnetic force.  Amper times turns is involved, the relatively low current is backed up by the many fine turns of the HV coil.

To widen the pulse width of the captured energy in the HV coil, try to use a 2.2 uF capacitor instead of the 1 uF and see which direction it changes. Or you already tested this and arrived at the 1 uF value as the best choice?

The 2.2 uF with the 1.8 H coil could resonate pretty close to the rotor speed what I took from the scope shot as cca at 77 Hz.
This would be  4620 RPM, but we have to divide this by 3 (number of magnets) and we get 1540, pretty close to your tacho metered 1551.   (With the 1 uF capacitor the 1.8 H coil resonates at 118.6 Hz.)  Notice that I did not consider that the input coil "sees" the HV coil during the ON time of the flyback diode, this may mean a different value capacitor, so it needs to check some other values besides the 2.2 uF. Nevertheless, the tuning of the LC circuit need not be very sharp in this given case because the 210 Ohm DC resistance of the HV coil is a loss and reduces the loaded Q (figure of merit, Q=X_L/R) of the LC circuit.  So the voltage amplitude is limited by this transformed loss across the LC circuit, and the lower the DC resistance of the HV coil is, the higher the resonant impedance is hence the higher the voltage amplitude could be across it.

Perhaps the reed switch could be placed a little bit further away from the rotor magnets to shorten the ON time. This would probably involve less RPM but this may be compensated with placing the LV coil a bit closer to the rotor magnets, and this may be true for the HV coil too.  I suggest these,  if you feel like doing such refinements.   

Greetings,
Gyula

[MileHigh]

I only looked at Laurent's clip, and it's pretty clear that the strategy proposed by Luc is working.   I am not going to disagree with Gyula, but instead give a different take on describing what is happening.

For starters, when the capacitor discharges into the high-voltage coil, you notice that there is no LC tank resonance taking place due to the 210 ohm resistance of the coil, creating an "over damped" condition.  The nice linear decreasing ramp voltage is a bit of a surprise, but that is a surprise in favour of the experiment.  It's telling you that a nice even pulse of current is flowing through the coil with a more or less flat top.  So that means that the high-voltage coil is giving you a nice ON-OFF pulse of current and an ON-OFF pulse of attraction-repulsion to drive the rotor.  This could easily be verified by putting a one-ohm current sensing resistor in series with the drive coil.

Here are the circuit dynamics:  The reed switch switches ON and current ramps up in the low-voltage drive coil.  When the reed switch switches OFF, the drive coil has to discharge its stored energy.  You can never forget that this energy is discharged in the form of a pulse of current at a to-be-determined voltage, and NOT as a pulse of voltage.

The pulse of current first flows through the diode.  We know that when you try to pulse the current through the high-voltage coil, it acts like an open circuit and refuses to let current flow though it.   We know that when you try to pulse the current into the capacitor, it acts like a short circuit and readily accepts the current flow.  Therefore, when the reed switch switches OFF, essentially all of the current pulse flows into the capacitor, and none of the current pulse flows into the low-voltage coil.

When the capacitor is at its peak voltage, that means that the low-voltage drive coil has completely discharged its current pulse energy.   Then the capacitor starts to discharge through the high-voltage coil in a nice linear fashion as described above.

How much energy is in the pulse from the low-voltage coil?  That's easy, you can calculate it based on the peak voltage of the capacitor and the value of the capacitor.

Where does that pulse energy go?  It goes to two places in and around the high-voltage coil:  1)  some of the energy pushes on the rotor, and 2) some of the energy is lost in the resistance of the high-voltage coil.

So, that leads up to the question, how much of the pulse energy pushes on the rotor and how much of the pulse energy is lost due to the resistive losses in the high-voltage coil?

That's an interesting question because it tells you how efficient your setup is in recovering the recycled pulse energy from the low-voltage drive coil.  I will leave that as an exercise for the enthusiasts to work out.

Finally, Laurent did not try changing the angle of the high-voltage coil to look for a sweet spot.  It would have been a worthwhile exercise to try that.   In fact, I would have moved the reed switch angle first to find the sweet spot for the reed switch only (no high-voltage coil) and then I would have added the high-voltage coil and then looked for the sweet spot angle for that.

[minoly]

Hi Luc

Thank's very much for sharing your experiment, and it seems to work very well.
What puzzle me is the fact that the very sharp fly back spike normally has almost no energy, it has very high voltage peak, but very low current, and if you connect a diode and a capacitor, it takes a lot of pulsse to charge it to obtain some power. But in your design the cap get a real power at each pulse.
So i think that we have to think of what happen in the fly back spike itself.
How does the flybackspike "capture " this energy ?

https://youtu.be/tag5OlvPi54

Thank's

Laurent

I'm glad people are starting to look at this using the many different methods to harvest and use the spike. This is what John Bedini and others have been talking about for years. Try it on the window motor with the full or even half bipolar Bedini/Cole switch or even the zero force... - you will all be amazed!
Cheers - Patrick

[tinman]

Hi Luc

Thank's very much for sharing your experiment, and it seems to work very well.
What puzzle me is the fact that the very sharp fly back spike normally has almost no energy, it has very high voltage peak, but very low current, and if you connect a diode and a capacitor, it takes a lot of pulsse to charge it to obtain some power. But in your design the cap get a real power at each pulse.
So i think that we have to think of what happen in the fly back spike itself.
How does the flybackspike "capture " this energy ?

https://youtu.be/tag5OlvPi54

Thank's

Laurent

Very nice experiment Laurent.