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

[verpies]

if we charge coil 1, cut the power, dump to cap, how many volts do we have

V=i_L*(L/C)^½, where i_L is the current flowing through the coil when the switch (S1) opens.

L is the inductance of the coil and C is the capacitance of the recovery capacitor and V is the final voltage across that capacitor (providing you do not have any resistor connected across it).
In real world you have to subtract the diode's voltage drop from V, too (usually 0.6V)

...and the time (t) that this transfer of energy from the coil (L) to the capacitor (C) takes, is:
t=π(LC)^½

[gotoluc]

Here is my new test device which uses the coils flyback to do work.

Link to video demo: https://www.youtube.com/watch?v=K8kwdrHCyig

Same observation, it's more efficient with the coil and magnet rotor.

Luc

[verpies]

Here is my new test device which uses the coils flyback to do work.
Link to video demo: https://www.youtube.com/watch?v=K8kwdrHCyig

The video is marked as "Private" and I cannot watch it.
Which circuit and probe positions were used to make these scopeshots?  The current flowing through a coil cannot suddenly jump up like on that annotated scopeshot below !

Also, is Ch2 inverted?

[tinman]

The video is marked as "Private" and I cannot watch it.
Which circuit and probe positions were used to make these scopeshots?  The current flowing through a coil cannot suddenly jump up like on that annotated scopeshot below !

Also, is Ch2 inverted?

Yes-odd how the current shot up as soon as the voltage inverted,which would seem like the switch off time.

Verpies.
If we wound an inductor with say 400 turn's,but at the 200 turn mark we place a 1 ohm CVR in series with the next 200 turns. So we have a series as such-->200 turns of wire on the core,then bring the wire outside the former,through a 1 ohm CVR,then bring the wire back into the former,and continue to wind on our last 200 turns.  Will this CVR now still measure the current flowing through the coil,or will it be seeing the inverted voltage across the winding's when the transistor opens ?.
Im thinking that there should be no voltage reading at the middle point of the windings,but from that mid point to either end of the winding's,we should see a voltage that is 1/2 that of the supplied voltage?.

Brad

[verpies]

If we wound an inductor with say 400 turn's,but at the 200 turn mark we place a 1 ohm CVR in series with the next 200 turns. So we have a series as such-->200 turns of wire on the core,then bring the wire outside the former,through a 1 ohm CVR,then bring the wire back into the former,and continue to wind on our last 200 turns.  Will this CVR now still measure the current flowing through the coil,or will it be seeing the inverted voltage across the winding's when the transistor opens ?.

It will read the current flowing through both coils if you drive it end-to-end.

The midpoint current might only show a difference when the coil's length is comparable to 1/2 of its driving wavelength and standing waves form -  read this.

At frequencies appearing in motors - this is a non-issue as they can be treated as lumped-element systems, in which the current is distributed uniformly along the coil.