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

[gotoluc]

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]

@Tinman,

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.

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.

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.

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.

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.

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.

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.

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]

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]

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]

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])