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Author Topic: Sharing ideas on how to make a more efficent motor using Flyback (MODERATED)  (Read 347747 times)

gotoluc

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Hi everyone,

I have not been sharing for a while but continued my research in the background and some out of public forum.
I've decided to open all my unlisted video demos to public so anyone can see my research if it interests you.

Here is a link to my videos if you would like to watch what was not public: https://www.youtube.com/user/gotoluc/videos
I have title most of them with "Testing ideas of more efficient Coil Flyback use (test 1 to 9)
There is also a few others and you can tell as they would have a low view count.

Here is the latest video demo where I share what I find most interesting and suggest how to better use Coil Inductive kickback (flyback)

https://www.youtube.com/watch?v=XfLcBD3Fy7M&feature=youtu.be


If you wish to post in this topic please keep it on topic and constructive as I reserve the right to edit or delete any post that are not so.

Luc

Jimboot

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That's awesome Luc thanks. Really interesting. I've only watched the last one that was published so far but I'll be checking out the others soon. It really got me thinking about the secondary in the mot with a cap. I'll be messing with it tomorrow.

wopwops

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I was particularly impressed with the magnetic action you demonstrated from the back EMF. Could you use that principle to pull or push the rotor "for free"? You had that transformer on your bench, and I was thinking: Move that over near the rotor and put it to work...

gotoluc

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I was particularly impressed with the magnetic action you demonstrated from the back EMF. Could you use that principle to pull or push the rotor "for free"?
You had that transformer on your bench, and I was thinking: Move that over near the rotor and put it to work...

Yes, you've got it!... you can re-create a very powerful magnetic field using off pulse (flyback) of a motor coil, and  for free... if you use the correct geometry and timing. And why not put it back to further assist the motor.  In a universal motor when the brushes move from one commutator segment to the other they've designed it to short the flyback so not to arc the commutator as motor designers have been trained that flyback spikes are useless and must be neutralized.

Making modifications to the existing design to allow this useful flyback to come out and maybe redirecting back in the stator coils may assist the motor... no?

There are other things that can be done. Take all these trick and you may end up with something we are looking for.

Keep thinking

Luc
 

wopwops

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I've been following this area of inquiry since the mid 1990s and this last group of videos from you is easily the most impressive thing I've seen so far. The instant I saw the physical result you were getting !!!outside of the duty cycle!!! I thought: Woh. I think Luc's got it.

The strength of that magnetic field seems so strong vs. what was going in. And for free...

It's probably a good sign that hardly anyone is paying attention to this thread. :)

SoManyWires

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very well done there!
seeing the block of metal want to lift that weight shows there is use to this.

ya now to find ways to recycle that into motion maybe a kind of hub motor or some storage.

you are our new leader.

gotoluc

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The strength of that magnetic field seems so strong vs. what was going in. And for free...

It's probably a good sign that hardly anyone is paying attention to this thread. :)

If you think that was good wait till you see the next video 8)

Yes, I agree, it's probably a good thing I'm now mostly ignored. Make things much easier.

Luc

mihai.isteniuc

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Hi gotoluc, hi to all.
 As sometimes happen in life, I was studding for the last couple of weeks the same area you are exploring now.


My testing rig was somehow different but still the same. I was focusing to see how much energy can be recovered as BEMF from an inductor pulsed with DC. The testing circuit was the one from the attachment noted 1. The trace on scope was the one noted 2 from same attachment. I was testing different kinds of inductors (air core, iron core, different number of turns, also different voltages, different pulse width and the list can continue). All test results where noted down into an excel data sheet. Important notice: all tests where done in mW ranges.


By far the best results where obtain from an air core inductor (not saying that this is the way ... air core). Conditions for testing that particular coil where the following: 12V supply voltage, pulse width 7-8% at around 1.5 kHz. With this setup I was able to recover an astonishing 90% of the energy invested in the coil, from back EMF. Better yet, with or without the recovering circuit, the current draw from the source was the same.


 For the rest of the tests (different coils, different working conditions) a media of 50% recovery can be taken into account.


 My understanding of this: you put into an inductor some energy. As a result the inductor exhibit some magnetic effect and also BEMF. Or with another words I put in 1 (one) unit of energy, get some magnetic effect and recover 0.9 units of energy from the same inductor as BEMF.


 Let’s go with the idea one step further. Let’s consider two identical circuits as the one noted 1 in the attachment. For the 1st circuit we need to make it run, 1 unit of energy. One can recovers from that input 50%, or 0.5 units of energy. 2nd circuit is the same. Invest 1 recover 0.5. Now let’s combine the 2 of them using the following logic: the recovered energy from 1st one is half of the required energy for the 2nd one to run. The other half comes from the power supply. The second recovery circuit became the half of the energy requirements for the 1st circuit to run and the other half comes from the second power supply. To better understand what I’m trying to say, see the section 3 from my attachment.


 What’s the gain? One can get twice the magnetic effect for the same input: 1 (one) unit of energy. Things get better and better if the recovery circuit can recover more. This was just an example, to better explain the way I see it. I hope I’m not wrong.


 Some other approach to confirm (at least for me) that BEMF is capable of doing some useful work was the following: take one usual relay that has normal closed contacts. Connect the normal closed contacts in series with the coil of the relay. Connect this setup to a power supply (see 4 from attachement). What u will get is some sort of buzzer that is ringing with some frequency. Nothing new here. Now connect a diode to kill BEMF. The ringing frequency is lower now. This is BEMF expanding the duration of the magnetic field inside the coil. So YES, BEMF can do useful work for us.


 I’ll end my post with a question. Please see this if u have the time: https://www.youtube.com/watch?v=LAtPHANEfQo It is a great animation that explain how a DC motor works, different types of DC motors and so on. At around 3:40 it is explained that BEMF is appearing on a DC motor also. But the explanation (and I saw it so many times over the years) that BEMF is a good thing in a DC motor cause otherwise the motor will overheat and burn is still shocking me every time I hear it. My question: Is this true? What if BEMF doesn’t exist? Or because it is there (and we cannot remove it), then better yet, use it to do some useful work?


 The way I see it is like this: u have a 6 volts lamp and a 12 volts battery. How can u make the lamp work without burning it?
Solution 1: use a series resistor connected with the lamp and the 12 volts battery;
Solution 2: use another 6 volts battery in series with the lamp and the 12 volts battery but connected with the opposite polarity (this is the case of BEMF don’t you think?)
Solution 3: use a 12 volts lamp;
Solution 4: use a 6 volts battery;


 Obviously one of the solution 3 or 4 are the logic (and most efficient) one to be used. Hell now, said the DC motor constructors. Solution 2 is the right way. Huh … really?


mihai

gotoluc

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Hi everyone,

here is a new test video with an accurate way of testing the magnetic flux amplification effect.

Link to video: https://www.youtube.com/watch?v=7dmKEOWOhQA

Luc

SkyWatcher123

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Hi gotoluc, thank you for sharing.
For an easy to build useful test device, the old joule thief with high voltage secondary.
Place bifilar oscillator on one side of toroid, then place high voltage secondary on other side to receive the flyback pulses.
With either a low or high voltage pickup wind over that, depending on desired final output volts needed.
I realize that's not an ideal setup, since it's all on the same core, though it's a start.
It might give us a more efficient jenna led light circuit.
peace love light

woopy

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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

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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=XL/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.   8)

Greetings,
Gyula

MileHigh

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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

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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

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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.