Author Topic: Sharing ideas on how to make a more efficent motor using Flyback (MODERATED) (Read 354598 times)

gotoluc · « **Reply #90 on:** November 19, 2015, 11:47:23 PM »

Quote from: minoly on November 19, 2015, 10:09:42 PM

Luc,
I don't understand why you keep deleting my posts. I'm trying to understand more clearly exactly what you are doing. You even go on to answer my question yet you still delete my post???

The only posts I deleted were your first posts relating to J Bedini. Please tell me which post you believe to be deleted and why I would answer your question to a post you believe to be deleted?

BTW, you never replied to my post to you on page 2: http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466026/#msg466026

Now on page 7 and still waiting

Luc

tinman · « **Reply #91 on:** November 20, 2015, 12:15:18 AM »

Quote from: synchro1 on November 19, 2015, 03:51:11 PM

@Tinman,

Quote

There's no path back to the positive electrode for the reversed current with the Reed switch open.

1st,there is no reverse current.
2nd,i have not drawn a current path with the reed switch open back to the source.

Quote

The destination determines the current polarity.

No it dose not. The current flow direction through the primary inductor is determind by the inflow current direction.
Once again,the current flow through the primary inductor will remain the same once the reed switch is opened.

Quote

The negative ground is the destination of the power pulse; Then the reversed current polarity seeks a positive ground!

No.
The current flow will seek the path of lowest resistance once the reed switch is opened. In this case,it is the capacitor,and this forms a current loop-->first through the capacitor and primary inductor,and the secondary loop once the cap is charged is from the cap through the secondary inductor--as i have drawn.

Quote

You caused this same kind of confusion in that Universal motor schematic you passed to Chris Sykes on the bucking coil thread.

Only those that do not understand that the current continues to flow in the same direction through the primary coil once the current supply to that coil is switch off,are the only one's that get confused.

Quote

Charged capacitors resist change in voltage just like inductors. The sitting charge on the capacitor would determine the destination of the flyback, either through the inductor or to the cap.

If you look at Woopy's scope shot's,you can see that the cap is discharged each cycle,and thus the current would flow into the cap first in the next cycle.

Quote

One the other or both, the flyback causes the capacitor to discharge into the Hi-voltage coil to accelerate the magnet rotor with a power pulse.

The flyback current charges the capacitor,and the capacitor then discharges via the secondary inductor. This you can see in the linear discharge scope trace in Woopy's video's.

Quote

You're trying to maintain that a current can exist with a reversed voltage polarity. This is impossible!

This is not impossible at all,and is exactly what happens across the primary inductor when the reed switch opens-->the voltage across that inductor inverts,but the current flow continues in the same direction.

Quote

The current changes direction headed towards an opposite ground albeit by a circuitous path. The path is illusory, the direction of the current is switched from negative to positive inside the primary after the magnetic field collapse.

No again.
The current dose not change direction,but the voltage polarity dose invert across the primary coil.
So when the reed switch is closed,the positive voltage side of the inductor is the same as the positive side of the supply(battery/power supply). When the reed switch opens,the current will continue to flow through that primary coil in the same direction,but the voltage across that coil has now inverted-->so what was the positive voltage side of the primary coil when the reed switch was closed in now the negative(ground) side once the reed switch opens.

minoly · « **Reply #92 on:** November 20, 2015, 12:39:53 AM »

Quote from: gotoluc on November 17, 2015, 06:58:24 PM

Please post the link to your video experiments so we can see your work.

Have I not shared the above to woopy and anyone else who reads this topic?... I don't appreciate your tone or unfounded accusation.
Let's see what you have shared over the years

Maybe you need to do the experiment?

Regards

Luc

this post was gone, I checked from two computers and 2 different browsers, now it's back... http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466152/#msg466152
My mistake.

you answered indirectly when you replied to MoRo's attempt to reply to my question:
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466166/#msg466166
"The next thing I would recommend is a device that needs a long flux holding time. This is why I recommend a PM flux switch.[/size] [/size]I have already designed and built it. I'll call it the GTL (short for "go to luc") Gate. The GTL Gate uses the same design principle of my mostly magnet motor, where both inner and outer fields of the coil is used which gives you double the magnetic field and obviously double the core surface area."[/size]
I still do not understand it though.

You know me from past. I've often commented in your vids. I've always had good things to say about your work. https://www.youtube.com/channel/UCrIHB1lv7XElAD_ztL-1FPw

You might not appreciate my tone, but I did not appreciate the smack when all I did was share my opinion in the first post about JB.

I hope I'm understanding this correctly, it is not discharging into a cap that is in parallel to a coil that will aid in spinning a rotor (to put it simply) that you are interested in; however, your interest is in creating a fluxgate using only high voltage/impedance coils that will hold/pull with more force using the spike? This GTL sounds very interesting. I have lot's of ideas how to use the spike to provide more torque, I cannot think of any that would aid in a mostly magnet motor - very interesting indeed.

Magluvin · « **Reply #93 on:** November 20, 2015, 12:54:31 AM »

Quote from: tinman on November 20, 2015, 12:15:18 AM

1st,there is no reverse current.
2nd,i have not drawn a current path with the reed switch open back to the source.

No it dose not. The current flow direction through the primary inductor is determind by the inflow current direction.
Once again,the current flow through the primary inductor will remain the same once the reed switch is opened.

No.
The current flow will seek the path of lowest resistance once the reed switch is opened. In this case,it is the capacitor,and this forms a current loop-->first through the capacitor and primary inductor,and the secondary loop once the cap is charged is from the cap through the secondary inductor--as i have drawn.

Only those that do not understand that the current continues to flow in the same direction through the primary coil once the current supply to that coil is switch off,are the only one's that get confused.

If you look at Woopy's scope shot's,you can see that the cap is discharged each cycle,and thus the current would flow into the cap first in the next cycle.

The flyback current charges the capacitor,and the capacitor then discharges via the secondary inductor. This you can see in the linear discharge scope trace in Woopy's video's.

This is not impossible at all,and is exactly what happens across the primary inductor when the reed switch opens-->the voltage across that inductor inverts,but the current flow continues in the same direction.

No again.
The current dose not change direction,but the voltage polarity dose invert across the primary coil.
So when the reed switch is closed,the positive voltage side of the inductor is the same as the positive side of the supply(battery/power supply). When the reed switch opens,the current will continue to flow through that primary coil in the same direction,but the voltage across that coil has now inverted-->so what was the positive voltage side of the primary coil when the reed switch was closed in now the negative(ground) side once the reed switch opens.

Thinking about it more...

If I recall correctly, my coils in my vid are 2mh .45ohm. They wont do much with a spike from another coil because it wont let high freq through. So you are most likely correct to try a coil like mine in the circuit with the cap in parallel.

Mags

MileHigh · « **Reply #94 on:** November 20, 2015, 01:06:14 AM »

Gyula:

Excellent post about the power measurements and quite similar to what I was thinking. I don't have the real bench measurement skills that you have but permit me to give an outline of my line of thinking and analysis for this task.

Just for fun, I am going to keep a running tally of all of the measurements and assign each one a capital letter.

For the input power I was thinking of just putting a one-ohm CVR before the reed switch and measuring across that with a quality multimeter to get the average voltage to compute the average input current for the entire circuit [A]. A non-inductive CVR would be preferable but not necessarily critical in this application. This assumes that the voltage output from the power supply is constant and then we can measure the input power for the whole circuit.

Then, if you measured the RMS voltage across the same CVR it would be giving you the RMS current through the drive coil . Therefore you have the power dissipation in the 0.5 ohm resistance of the drive coil [C].

The average input power would be the average input current times the input voltage [D].

Therefore you are already in a position to estimate the power in the drive coil that is exported to the "outside world." The outside world includes two components, 1) the mechanical power that is imparted on the spinning rotor, and 2) the power that goes through the diode and drives the LC circuit. So the "exported power from the drive coil" [E], would be the average input power [D] minus the power dissipation in the resistance of the drive coil [C].

You can make an accurate measurement of the pulse frequency [F], therefore you can measure the input pulse energy based on the average input power [G].

You can also measure the "exported pulse energy from the drive coil" [H].

Here is where we are so far:

A - Average input current for entire circuit
B - RMS current through the drive coil
C - Power dissipation in the resistance of the drive coil
D - Average input power
E - Exported power from the drive coil, ([D] - [C])
F - Pulse frequency
G - Input pulse energy ([D]/[F])
H - Exported pulse energy from the drive coil ([E]/[F])

citfta · « **Reply #95 on:** November 20, 2015, 01:15:49 AM »

For Synchro

I think what is confusing you about the current not reversing after the reed opens is you are not realizing the coil becomes the source of current and not a load.

You can easily prove this to yourself if you have a 2 channel scope. Simply add a low value resistor like 1 ohm or so to the circuit Tinman has provided in series with only the coil. Now watch the scope channels as you turn the coil on and off. You will see the signal across the resistor only goes up as the coil is charged and discharged. But the channel across the coil will show a reversal of polarity. But obviously if the channel across the resistor doesn't change polarity then the current does NOT reverse.

If we use the old school idea of electron flow you can look at it this way. The negative terminal of the battery has a surplus of electrons since electrons are negatively charged. So when the reed switch is turned on electrons move from the negative of the battery into the bottom of the coil. And electrons flow out the top of the coil back to the battery. If we put a meter on the coil we see the top as positive and the bottom as negative because the excess of electrons is at the bottom of the coil.

Now when we turn off the reed switch the collapsing magnetic field wants to keep the electrons flowing in the same direction. Since the coil is now the source and the current is still flowing in the SAME direction as before we now see the top of the coil as being more negative than the bottom. The current has not changed direction but the polarity of the voltage has changed.

Respectfully,
Carroll

MileHigh · « **Reply #96 on:** November 20, 2015, 01:25:54 AM »

Power measurements continued...

For the LC circuit by looking at Laurent's waveforms we can see that essentially all of the drive coil's pulse energy first goes into charging the one-uF capacitor. This is what was expected because the secondary coil will be blocking the initial current flow from the pulse of current from the drive coil.

So, we have initial maximum voltage measurement across the capacitor . From that we can calculate the pulse energy that goes into the capacitor [J]. If Laurent has a capacitance meter it should be used to get a more accurate value of this capacitor.

Now, we know the "exported pulse energy from the drive coil" [H], and we know the maximum energy that is transferred into the one-uF capacitor [J]. Therefore we can make a reasonable inference for how much pulse energy gets imparted onto the rotor from the drive coil [K], which is ([H]-[J]).

Here is where we are:

A - Average input current for entire circuit
B - RMS current through the drive coil
C - Power dissipation in the resistance of the drive coil
D - Average input power
E - Exported power from the drive coil, ([D] - [C])
F - Pulse frequency
G - Input pulse energy ([D]/[F])
H - Exported pulse energy from the drive coil ([E]/[F])
I - Maximum voltage across the 1 uF capacitor
J - Maximum energy in the 1 uF capacitor
K - Main drive coil pulse energy that is imparted onto the rotor ([H]-[J])

Magluvin · « **Reply #97 on:** November 20, 2015, 01:27:03 AM »

Thinking a bit more, and on topic....

I reposted Woopys circuit for reference below.

If you see the red x labeled 4, if you put a diode there, same direction as the one to the left of the x, the cap will discharge all of its charge through the coil only once.

Now for possibly an even better effect. Havnt tried but I think it would be a good thing if it works....

When the cap charges, then discharges through the coil with the added diode, the cap will fully discharge, and then some due to the flywheel effect of the coil, meaning the cap will end up with a charge reverse of what it started with. The possible advantage is, when the next spike comes from the primary drive coil, it will be facing a cap with a charge that is well willing to accept the spike as the polarities will be in series. So the spike enters the cap with help of the charge already in the cap. The possibility is that the cap may charge higher than without that added diode.

Im pretty sure of it from my experience in this stuff.

Mags

Magluvin · « **Reply #98 on:** November 20, 2015, 01:40:31 AM »

Thinking more, again. I need to try what I just wrote in my last post. There might be something really good there. possibly more than I imagined.

If the cap does get an increased voltage level, the next pass may get even bigger and so on. Just went through it in my head. It just might be so.

Mags

gotoluc · « **Reply #99 on:** November 20, 2015, 01:48:47 AM »

Quote from: minoly on November 20, 2015, 12:39:53 AM

this post was gone, I checked from two computers and 2 different browsers, now it's back... http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466152/#msg466152
My mistake.

Yes, your mistake as it's impossible for me to delete a post and then re-post it using your user name.
That's the second time you accuse me of something that is not true. So please chill out and stop jumping to conclusions.

Quote from: minoly on November 20, 2015, 12:39:53 AM

you answered indirectly when you replied to MoRo's attempt to reply to my question:
http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg466166/#msg466166
"The next thing I would recommend is a device that needs a long flux holding time. This is why I recommend a PM flux switch.[/size] [/size]I have already designed and built it. I'll call it the GTL (short for "go to luc") Gate. The GTL Gate uses the same design principle of my mostly magnet motor, where both inner and outer fields of the coil is used which gives you double the magnetic field and obviously double the core surface area."[/size]
I still do not understand it though.

You didn't add the part I wrote saying if you don't understand you will once I demonstrate the GTL Gate.

Quote from: minoly on November 20, 2015, 12:39:53 AM

You know me from past. I've often commented in your vids. I've always had good things to say about your work. https://www.youtube.com/channel/UCrIHB1lv7XElAD_ztL-1FPw

So you're cool Joule

Quote from: minoly on November 20, 2015, 12:39:53 AM

You might not appreciate my tone, but I did not appreciate the smack when all I did was share my opinion in the first post about JB.

My first post has this warning: "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"

So how can this be a shock when you post about J Bedini stuff and then more posts arguing with others that it's the same thing?... I think you're over-dramatic. And right now me taking the time to write this is taking away from experiments and also adding pages with useless posts. Do you not see this?

Quote from: minoly on November 20, 2015, 12:39:53 AM

I hope I'm understanding this correctly, it is not discharging into a cap that is in parallel to a coil that will aid in spinning a rotor (to put it simply) that you are interested in; however, your interest is in creating a fluxgate using only high voltage/impedance coils that will hold/pull with more force using the spike? This GTL sounds very interesting. I have lot's of ideas how to use the spike to provide more torque, I cannot think of any that would aid in a mostly magnet motor - very interesting indeed.

Yes correct, my experiment results lead me to believe a flux gate would be best use of this effect but that doesn't mean there is no other use for it.

The mostly magnet motor was a good learning experiment. It has taught me many things that will be incorporated in the GTL Gate.
I see no reasons you have not started experiments as we I can see you have plenty of stuff to use and make a video of your findings.

Looking forward to seeing ways you can think of using this and your experimenl results

Regards

Luc

gotoluc · « **Reply #100 on:** November 20, 2015, 01:55:38 AM »

Quote from: citfta on November 20, 2015, 01:15:49 AM

For Synchro

I think what is confusing you about the current not reversing after the reed opens is you are not realizing the coil becomes the source of current and not a load.

You can easily prove this to yourself if you have a 2 channel scope. Simply add a low value resistor like 1 ohm or so to the circuit Tinman has provided in series with only the coil. Now watch the scope channels as you turn the coil on and off. You will see the signal across the resistor only goes up as the coil is charged and discharged. But the channel across the coil will show a reversal of polarity. But obviously if the channel across the resistor doesn't change polarity then the current does NOT reverse.

If we use the old school idea of electron flow you can look at it this way. The negative terminal of the battery has a surplus of electrons since electrons are negatively charged. So when the reed switch is turned on electrons move from the negative of the battery into the bottom of the coil. And electrons flow out the top of the coil back to the battery. If we put a meter on the coil we see the top as positive and the bottom as negative because the excess of electrons is at the bottom of the coil.

Now when we turn off the reed switch the collapsing magnetic field wants to keep the electrons flowing in the same direction. Since the coil is now the source and the current is still flowing in the SAME direction as before we now see the top of the coil as being more negative than the bottom. The current has not changed direction but the polarity of the voltage has changed.

Respectfully,
Carroll

Excellent post and test suggestion Carroll

Thanks for taking the time to help

Luc

gotoluc · « **Reply #101 on:** November 20, 2015, 01:58:34 AM »

Quote from: Magluvin on November 20, 2015, 01:27:03 AM

Thinking a bit more, and on topic....

I reposted Woopys circuit for reference below.

If you see the red x labeled 4, if you put a diode there, same direction as the one to the left of the x, the cap will discharge all of its charge through the coil only once.

Now for possibly an even better effect. Havnt tried but I think it would be a good thing if it works....

When the cap charges, then discharges through the coil with the added diode, the cap will fully discharge, and then some due to the flywheel effect of the coil, meaning the cap will end up with a charge reverse of what it started with. The possible advantage is, when the next spike comes from the primary drive coil, it will be facing a cap with a charge that is well willing to accept the spike as the polarities will be in series. So the spike enters the cap with help of the charge already in the cap. The possibility is that the cap may charge higher than without that added diode. Im pretty sure of it from my experience in this stuff.

Mags

Sounds very interesting Mags

Thanks for sharing

Luc

synchro1 · « **Reply #102 on:** November 20, 2015, 02:08:04 AM »

Quote from: citfta on November 20, 2015, 01:15:49 AM

For Synchro

I think what is confusing you about the current not reversing after the reed opens is you are not realizing the coil becomes the source of current and not a load.

You can easily prove this to yourself if you have a 2 channel scope. Simply add a low value resistor like 1 ohm or so to the circuit Tinman has provided in series with only the coil. Now watch the scope channels as you turn the coil on and off. You will see the signal across the resistor only goes up as the coil is charged and discharged. But the channel across the coil will show a reversal of polarity. But obviously if the channel across the resistor doesn't change polarity then the current does NOT reverse.

If we use the old school idea of electron flow you can look at it this way. The negative terminal of the battery has a surplus of electrons since electrons are negatively charged. So when the reed switch is turned on electrons move from the negative of the battery into the bottom of the coil. And electrons flow out the top of the coil back to the battery. If we put a meter on the coil we see the top as positive and the bottom as negative because the excess of electrons is at the bottom of the coil.

Now when we turn off the reed switch the collapsing magnetic field wants to keep the electrons flowing in the same direction. Since the coil is now the source and the current is still flowing in the SAME direction as before we now see the top of the coil as being more negative than the bottom. The current has not changed direction but the polarity of the voltage has changed.

Respectfully,
Carroll

@Citfta,

That sounds pretty scientific. What both you and Tinman are proposing is that it's possible for electrical power to share current and voltage of opposite polarities. This is physically impossible!

There is no "Left Over" old current in the primary coil after the violence of the magnetic field collapse; There is only new current and voltage of opposite polarity. There is no cause and effect between the old current and the new current. The old current is in the past behind the event horizon. The violence of the field collapse has utterly and with absolute finality obliterated any trace of the old current along with any trace of previous electron paths for eternity. The new current can go in either direction depending on the pathway. Given a pathway of less resistance it will travel that way to it's newly biased ground. That includes an opposite direction. There is no force tendency of any kind at work on the new current from the old current. That's just superstition.

Magluvin · « **Reply #103 on:** November 20, 2015, 02:09:52 AM »

just tried on circuit sim. The reverse charge on the cap feeds back to primary. So we end up with a alternating oscillation. Unless the timing of the proceeding spikes are at the point where the cap is in full reverse charge before it is fed back to the pri. Just solid state at the moment to get the hang of it Will play with it more to see what I can do to fix that. Love that sim for times like this. Real time saver on simple stuff.

Mags

MileHigh · « **Reply #104 on:** November 20, 2015, 02:27:44 AM »

I will try to finish off the measurement discussion....

The only thing left to do is look at the capacitor discharging through the secondary coil. I agree with Gyula that changing the resistor at position #4 from one ohm to 10 ohms would be a good idea. When you look at Laurent's existing scope shots at position #4 you see a linearly decreasing ramp of current flow. That looks just like a resistor. That would suggest that the 210-ohm resistance of the coil is dominating over the inductance, but I am not sure how much secondary coil pulse energy will be imparted onto the rotor. The measurements and the number crunching should give us a handle on that.

I am going to assume a quality multimeter or Laurent's scope will give an accurate measurement for the RMS voltage across a 10-ohm CVR at position #4. When in doubt, do the research yourself to check these things otherwise you might be a victim of GIGO.

This time I am not going to give a description because it would just be a repeat of what was stated for other components.

Here are the new measurements:

L - RMS current through 10-ohm resistor at position #4
M - Average power dissipated in the resistance of the secondary coil
N - Pulse energy dissipated in the resistance of the secondary coil ([M]/[F])
O - Secondary coil pulse energy that is imparted onto the rotor ([J]-[N])

Here is the whole measurement shebang with the more interesting stuff highlighted. Some of the measurements are derived from other measurements, and you can easily determine other efficiency metrics, etc.

A - Average input current for entire circuit
B - RMS current through the drive coil
C - Power dissipation in the resistance of the drive coil
D - Average input power
E - Exported power from the drive coil, ([D] - [C])
F - Pulse frequency
G - Input pulse energy ([D]/[F])
H - Exported pulse energy from the drive coil ([E]/[F])
I - Maximum voltage across the 1 uF capacitor
J - Maximum energy in the 1 uF capacitor
K - Main drive coil pulse energy that is imparted onto the rotor ([H]-[J])
L - RMS current throuugh 10-ohm resistor at position #4
M - Average power dissipated in the resistance of the secondary coil
N - Pulse energy dissipated in the resistance of the secondary coil ([M]/[F])
O - Secondary coil pulse energy that is imparted onto the rotor ([J]-[N])
P - Percentage of input pulse energy that becomes resistive losses (([C]/[F] + [M}) /[G]) * 100
Q - Percentage efficiency (total rotor pulse energy/input pulse energy) * 100 == (([K]+[O])/[G]) * 100
R - Ratio of secondary coil pulse energy imparted to rotor to drive coil pulse energy imparted to rotor [O]/[K]

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Author Topic: Sharing ideas on how to make a more efficent motor using Flyback (MODERATED) (Read 354598 times)

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