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Author Topic: Sharing ideas on how to make a more efficent motor using Flyback (MODERATED)  (Read 210831 times)

Offline woopy

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Hi Synchro

Very nice explication
Have you some links on the subject for my info

Thank's

Laurent

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Offline gotoluc

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I accept that challenge.
But i can tell you now with some certainty that a low impedance secondary coil will perform better due to lower resistive losses ;)

Glad you accept!

And I would not argue with you that low impedance coils are more efficient since they have next to no resistance but the tradeoff is they have next to no inductance.
I'm suggesting we use the ideal qualities of each (tuned) at the appropriate time to achieve a longer flux holding time to switch a PM flux gate with less input power then just a low or mid impedance coil.

So no need to build a motor, just a flux gate will do. As we know a flux gate will make a motor work and what we need is one that can work with less power input so we get the magnets to do most of the work.

This is what I'm ideally proposing and thought I would make it clear before you start your build.

Thanks for your participation and looking forward to your results

Luc

Offline gyulasun

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....
My beginner's knowledge indicates that, during the pulse, the main (low voltage ) motor coil, get a certain amount of electric energy (voltage and current) from the power station. This energy is dissipated for a small part in  heat due to the impedance of the coil and for the main part in building a magnetic field . During the building of the magnetic field its magnetic strength increases with the increasing current in the inductor and propels the rotor magnet which get kinetic energy.
So it seems to me, that at the end of the pulse, all the electrical input energy should have been transformed in some heat and the kinetic energy of the rotor, yes or not? And  the magnetic field seems to stay there at its max strength until the end of the pulse. yes or not?
Than it seems commonly admitted that at the end of the pulse, the magnetic field brutally collapses and creates the flybackspike.
But if the input energy is totally dissipated at the end of the pulse, how does the collapsing magnetic field create this powerfull flybackspike ? What is the process ?
....

Hi Laurent,

During the time a magnetic field is building up in your coil (input current is ON) and starts to propel your rotor, the field is not fully used up at all in the process (some small part of the field strays away, this depends also on the mechanical construction), just like the field of a permanent magnet is not used up in a normal magnetic interaction.
Of course, normally the action-reaction force rules between the rotor and the magnetic field created from the input energy: if you try to brake the movement of the rotor, input energy consumption increases during the ON time of the input current (Lenz law). 

It is known that an inductance tries to resist to any current change. (A capacitor tries to resist to any voltage change across it). 

In this link I think the "why" i.e. why a coil resists to any current change) is correctly explained:

http://www.allaboutcircuits.com/textbook/direct-current/chpt-15/magnetic-fields-and-inductance/

Quote from the text:

"Because inductors store the kinetic energy of moving electrons in the form of a magnetic field, they behave quite differently than resistors (which simply dissipate energy in the form of heat) in a circuit. Energy storage in an inductor is a function of the amount of current through it. An inductor’s ability to store energy as a function of current results in a tendency to try to maintain current at a constant level. In other words, inductors tend to resist changes in current. When current through an inductor is increased or decreased, the inductor “resists” the change by producing a voltage between its leads in opposing polarity to the change."

Perhaps these can explain better what may happen and helps you.

Gyula

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Offline MileHigh

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With and without the capacitor. It's much the same as to why a vehicle going a set speed can be stopped quicker if you dont lock up the brakes and skid the wheels. Voltage leads current in an inductor,and if the applied voltage pulse is to quick and short,then there will be very little current that follows. However,if that spike is sent to a capacitor where the current leads the voltage,then you can be assured that the inductor will receive all of the !now stored! energy from that inductive kickback that was stored within the cap.

Yes, but you still have the mystery of where the pulse of energy from the drive coil goes and why at least when Luc holds the metal slab in his hands the magnetic attraction is imperceptible.  It doesn't mean it's not there, but at least to his hands it's imperceptible.  When you do a preliminary analysis in your head you would have an expectation that he might feel something.

Quote
Voltage leads current in an inductor,and if the applied voltage pulse is to quick and short,then there will be very little current that follows.

Let's try to be a bit more precise with your explanation.  You have to remember, it's not a voltage pulse, it's a current pulse.

For starters, what happens when coil A discharges into coil B, where the initial conditions are there is some current is flowing in coil A, let's say it's one amp,  and zero current flowing in coil B.  Let's say both coils are one Henry of inductance.  That analogous to what the situation is like in Luc's setup when there is no capacitor.  Anybody want to try to answer that?

Offline MileHigh

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My beginner's knowledge indicates that, during the pulse, the main (low voltage ) motor coil, get a certain amount of electric energy (voltage and current) from the power station. This energy is dissipated for a small part in  heat due to the impedance of the coil and for the main part in building a magnetic field . During the building of the magnetic field its magnetic strength increases with the increasing current in the inductor and propels the rotor magnet which get kinetic energy.
So it seems to me, that at the end of the pulse, all the electrical input energy should have been transformed in some heat and the kinetic energy of the rotor, yes or not? And  the magnetic field seems to stay there at its max strength until the end of the pulse. yes or not?
Than it seems commonly admitted that at the end of the pulse, the magnetic field brutally collapses and creates the flybackspike.
But if the input energy is totally dissipated at the end of the pulse, how does the collapsing magnetic field create this powerfull flybackspike ? What is the process ?

Laurent,

Your analysis is basically correct and I will "fill in some of the blanks" for you.   For starters, this statement, "This energy is dissipated for a small part in  heat due to the impedance of the coil and for the main part in building a magnetic field." should read as, "This energy is dissipated for a small part in heat due to the resistance of the coil and for the main part in building a magnetic field."

The impedance of the coil is associated with its inductance and stores energy, it does not dissipate energy and turn it into heat like a resistance.  So "impedance" and "resistance," although somewhat related because they are both associated with the blocking of the flow of current, in fact they are separate and distinct terms that should not be interchanged with each other.

Quote
So it seems to me, that at the end of the pulse, all the electrical input energy should have been transformed in some heat and the kinetic energy of the rotor, yes or not? And  the magnetic field seems to stay there at its max strength until the end of the pulse. yes or not?

You are essentially correct.  The electrical input energy goes to three places:  1) heat energy, 2) kinetic energy of the rotor, and 3) the BEMF pulse energy.

The magnetic field is not at it's max strength at the end of the pulse.  If the drive coil gives a big kinetic energy push to the rotor, then at the end of the constant-voltage energizing pulse from the reed switch there will be less current flowing through the coil.   In other words, without the rotor in place, then there will be more current flowing through the drive coil at the end of the constant-voltage energizing pulse from the reed switch.

You actually see this effect in your own clip, and I made reference to it in the form of a question but nobody responded.  Please review your clip and see if you can find any evidence for this effect.

Quote
But if the input energy is totally dissipated at the end of the pulse, how does the collapsing magnetic field create this powerfull flybackspike ? What is the process ?

As indicated above, the input energy is not totally dissipated at the end of the constant-voltage energizing pulse from the reed switch.  Some of the input energy remains at the end of the pulse on the rotor and it becomes the flyback pulse.

It goes full circle to what I said before:  How much energy is in the main drive pulse on the rotor from the drive coil?  How much energy is in the flyback pulse?  How do I measure and compare the two?

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Offline synchro1

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Laurent,

Your analysis is basically correct and I will "fill in some of the blanks" for you.   For starters, this statement, "This energy is dissipated for a small part in  heat due to the impedance of the coil and for the main part in building a magnetic field." should read as, "This energy is dissipated for a small part in heat due to the resistance of the coil and for the main part in building a magnetic field."

The impedance of the coil is associated with its inductance and stores energy, it does not dissipate energy and turn it into heat like a resistance.  So "impedance" and "resistance," although somewhat related because they are both associated with the blocking of the flow of current, in fact they are separate and distinct terms that should not be interchanged with each other.

You are essentially correct.  The electrical input energy goes to three places:  1) heat energy, 2) kinetic energy of the rotor, and 3) the BEMF pulse energy.

The magnetic field is not at it's max strength at the end of the pulse.  If the drive coil gives a big kinetic energy push to the rotor, then at the end of the constant-voltage energizing pulse from the reed switch there will be less current flowing through the coil.   In other words, without the rotor in place, then there will be more current flowing through the drive coil at the end of the constant-voltage energizing pulse from the reed switch.

You actually see this effect in your own clip, and I made reference to it in the form of a question but nobody responded.  Please review your clip and see if you can find any evidence for this effect.

As indicated above, the input energy is not totally dissipated at the end of the constant-voltage energizing pulse from the reed switch.  Some of the input energy remains at the end of the pulse on the rotor and it becomes the flyback pulse.

It goes full circle to what I said before:  How much energy is in the main drive pulse on the rotor from the drive coil?  How much energy is in the flyback pulse?  How do I measure and compare the two?

@Milehigh,

This is a quote from you from above:

"Some of the input energy remains at the end of the pulse on the rotor and it becomes the flyback pulse".

This is nonsense! Here's what happens:

The coil is energized by the electrical current. A magnetic field appears in the coil. The current is stopped and the magnetic field collapses. There is no left over current in the coil at this point!

The magnetic field collapse generates a new hi-voltage current. As the field collapses the field is moving in the opposite direction from the expanding field and generates a new current in the opposite direction as it passes inward across the windings. This is the flyback pulse! The faster the current is cut off to the coil the higher the flyback voltage. The magnet rotor can't have any effect on the power coil when the Reed switch is open.

Additionally, the magnetic field collapse generates a longitudinal scaler wave that has an "Impulse Magnetizing" effect. This power is infinite!



Offline minoly

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Glad you accept!

And I would not argue with you that low impedance coils are more efficient since they have next to no resistance but the tradeoff is they have next to no inductance.
I'm suggesting we use the ideal qualities of each (tuned) at the appropriate time to achieve a longer flux holding time to switch a PM flux gate with less input power then just a low or mid impedance coil.

So no need to build a motor, just a flux gate will do. As we know a flux gate will make a motor work and what we need is one that can work with less power input so we get the magnets to do most of the work.

This is what I'm ideally proposing and thought I would make it clear before you start your build.

Thanks for your participation and looking forward to your results

Luc


So what you are proposing is different than what Laurent is doing then? I'm new to this Fluxgate making a motor work can anyone explain this or point me in the right direction?




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Offline tinman

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@Milehigh,

This is a quote from you from above:

"Some of the input energy remains at the end of the pulse on the rotor and it becomes the flyback pulse".

This is nonsense! Here's what happens:

The coil is energized by the electrical current. A magnetic field appears in the coil. The current is stopped and the magnetic field collapses. There's no left over current in the coil at this point!

 This is the flyback pulse! The faster the current is cut off to the coil the higher the flyback voltage. The magnet rotor can't have any effect on the power coil when the Reed switch is open.

Additionally, the magnetic field collapse generates a longitudinal scaler wave that has an "Impulse Magnetizing" effect. This power is infinite!

Quote
The magnetic field collapse generates a new hi-voltage current. As the field collapses the field is moving in the opposite direction from the expanding field and generates a new current in the opposite directions as it passes inward across the windings.

No Synchro,the current dose not reverse direction-->only the voltage across the coil/inductor inverts. You apply a current in one direction to build the magnetic field around the inductor= current to form magnetic field. Then when the supply current is cut off,the magnetic field collapses,and creates a current flow=magnetic field changing in time to create current flow through the inductor. As the magnetic field around that inductor has not inverted (swapped polarities),then the current flow remains in the same direction.


Brad.

Offline tinman

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So what you are proposing is different than what Laurent is doing then? I'm new to this Fluxgate making a motor work can anyone explain this or point me in the right direction?

As the thread is entitled making a more efficient motor using the flyback,then that is what i will do.
The difference between the fluxgate and motor is that you have magnetic fields changing with time with regards to the inductor-this comes from the permanent magnets on the rotor.

I will simply use a generating coil with a variable resistive load to calculate motor torque and RPM. This way we can see how the motor performs under different loads.

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Offline MoRo

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@Milehigh,

This is a quote from you from above:

 "Some of the input energy remains at the end of the pulse on the rotor and it becomes the flyback pulse".

This is nonsense! Here's what happens:

The coil is energized by the electrical current. A magnetic field appears in the coil. The current is stopped and the magnetic field collapses. There's no left over current in the coil at this point! The magnetic filed collapse generates new hi-voltage current. As the field collapses the field is moving in the opposite direction from the expanding field and generates a new current in the opposite directions as it passes inward across the windings. The faster the current is cut off to the coil the higher the fly back voltage.

Absolutely correct.

Additionally, a soft iron core will amplify any field polarity produced by the windings. This is because the 'magnetic domains' align themselves to the external field and add to the sum total of the field.

Voltage is equivalent to pressure, it forces current through the wire. So the higher the voltage, the longer the wire can be through which one can 'move' current.
However, current is NOT movement of anything trough the conductor (wire) end-to-end. It should be thought of as an out-of-phase alignment of conductor atoms by way of force (voltage pressure). The direction of the phase changing pressure relative to the conductor will determine the polarity of the resultant magnetic field around the conductor.

During the time that the phase is changing, the magnetic field will appear to be 'moving'. During the collapse, the out-of-phase atomic condition is springing back into place. So, the field is moving in the opposite direction when collapsing, however the field itself will still be in the same polarity (between 0 and 9 as apposed to -9 and 0).

The amount of out-of-phase atomic alignment is determined by the voltage and length of wire giving rise to sum total resistance to the out-of-phase condition. Once the available voltage causes a maximized out-of-phase condition, the magnetic field no longer appears to be moving. The voltage must be maintained to maintain a certain out-of-phase degree. 

We are looking to get high voltage (pressure) to generate an out-of-phase atomic condition throughout a lengthy piece of wire with many turns rapped around the magnetic domains of a soft iron core, to create a strong usable magnetic field.

I say: increase the back-spike voltage (by way of a step-up transformer) so it works across a longer conductor (multiple electromagnets) (stators).  ::)

MagnaMoRo
 

Offline tinman

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Absolutely correct.

Additionally, a soft iron core will amplify any field polarity produced by the windings. This is because the 'magnetic domains' align themselves to the external field and add to the sum total of the field.

Voltage is equivalent to pressure, it forces current through the wire. So the higher the voltage, the longer the wire can be through which one can 'move' current.
However, current is NOT movement of anything trough the conductor (wire) end-to-end. It should be thought of as an out-of-phase alignment of conductor atoms by way of force (voltage pressure). The direction of the phase changing pressure relative to the conductor will determine the polarity of the resultant magnetic field around the conductor.

During the time that the phase is changing, the magnetic field will appear to be 'moving'. During the collapse, the out-of-phase atomic condition is springing back into place. So, the field is moving in the opposite direction when collapsing, however the field itself will still be in the same polarity (between 0 and 9 as apposed to -9 and 0).

The amount of out-of-phase atomic alignment is determined by the voltage and length of wire giving rise to sum total resistance to the out-of-phase condition. Once the available voltage causes a maximized out-of-phase condition, the magnetic field no longer appears to be moving. The voltage must be maintained to maintain a certain out-of-phase degree. 

We are looking to get high voltage (pressure) to generate an out-of-phase atomic condition throughout a lengthy piece of wire with many turns rapped around the magnetic domains of a soft iron core, to create a strong usable magnetic field.

I say: increase the back-spike voltage (by way of a step-up transformer) so it works across a longer conductor (multiple electromagnets) (stators).  ::)

MagnaMoRo

Quote
Absolutely correct.

Not correct at all-->the current dose not reverse direction when the field collapses around the inductor.

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Offline synchro1

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@Tinman,

Please watch the first portion of this video. Watch where the arrows change direction:

https://www.youtube.com/watch?v=hNHXtIWaaig

Offline tinman

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@Tinman,

Please watch the first portion of this video. Watch where the arrows change direction:

https://www.youtube.com/watch?v=hNHXtIWaaig

You have shown a tank circuit where the current is being rocked back and forth between an inductor and a capacitor--different situation.
When a current is sent to an inductor,the inductor builds a magnetic field around it. When that current supply to the inductor is cut off,the field around that inductor collapses. The voltage across that inductor will invert,but the current flow through that inductor will continue to flow in the same direction. Think about what happens in a water pipe that causes water hammer when the tap that supplies that water flow is abruptly switched off. The pressure (voltage) in that pipe after the tap invert's,but the water(current) wants to keep flowing in the same direction through that pipe.

If we are to help Luc in making a more efficient motor using the flyback,then it is important that this is understood.
Voltage invert's across the inductor,but current keeps flowing in the same direction.

Online Magluvin

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The reason a coil inductance takes time for the field to build and current to eventually fully flow is the fact that the initial expanding field 'generates a reverse current of the input. Due to resistance and losses, the input overcomes the reverse emf. So as Tinman said, when the field collapses, opposite of expanding,  it generates a current in the same direction as the input before it is disconnected.  For me, thats where the term BEMF is not correct or mistaken, and should be called forward emf. As long as the reference is the direction of the input.

Mags

Offline MoRo

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Not correct at all-->the current dose not reverse direction when the field collapses around the inductor.

True. -->> "the current dose NOT reverse direction"
Especially in light of what I have explained 'current' to be: -->> The out-of-phase degree.
However, during field collapse (which is actually: -->> spring-back to normal phase alignment) the detected voltage in the conductor end-to-end will be opposite polarity until the normal phase angle is reached.

MagnaMoRo

 

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