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

[verpies]

I setup the circuit as in the below diagram, only i still have all 8 magnets facing north to the coil.
I use a 555 timer to sync the rotor with, presently running at 50Hz @ 31% duty cycle (12V input).

Very clean waveforms - you should collaborate with Gotoluc so he can obtain the same.

I have extended the light blue current ramp-up on your scopeshot without rotor and came to the conclusion that your  ON-pulse lasts for approximately 1.6Tau and that means you have burned 1.5x more energy in the resistance than the energy you had built up in the coil as the magnetic field.
If you decrease the ON time to 0.5Tau than this ratio will fall to 0.37x.  Of course narrower pulse needs to be compensated with higher supply voltage.

The voltage across C2 (yellow Ch1) varies between 4V and 15V so the energy gain of this capacitor is ½C*( (15V)² - (4V)² ). This is how many Joules are recovered per 1 pulse.

At "switch-OFF" (dark blue vertical line) approximately 1.1A flows through the coil, so ½L*(1.1A)² Joules of energy is stored in the coil at that time.

I encourage you to plug the inductance of your coil and the capacitance of your recovery capacitor into these formulas to see how many Joules of energy each one of them represents. (I would do this but I am not sure what L & C are).
Finally, divide these two energies to see what your recovery efficiency is.

If you experiment further you will find out that this efficiency increases if C2 is completely discharged before the "flyback" pulse and no resistor is connected across C2 to burn the recovered energy.

BTW:
The red curve is this curve and the black vertical line is at the break-even-point in time (1.15Tau)

[verpies]

Okay verpies, you should be happy with the below scope shots as C2 is drained to 0v before the next on time.
C2 is now a 60uf motor run cap and has a 47 Ohm 10w resistor across it.

Oh, yes I am very happy.
The current down ramp down has a nice partial sinusoidal shape (voltage waveform too) which is indicative of an efficient resonant power transfer from L1 to C2.

C2 gets charged from 0V to almost 90V peak, which represents ½*60uf*(90V)² = 243mJ of energy recovered from 1 "flyback" pulse.   Energy burned up by R2 is not accounted for.  Actually, you can delete R2 if measuring 1 pulse from the drained C2. 
R2 is not needed to measure the recovered energy (only  the voltage across C2 is).

At switch-off time there is close to 17A flowing through the coil so that represents ½*L*(17A)² Joules of energy.

I am not sure what the inductance of your coil (L) is now but if you plug it into the formula then you will obtain how much energy is stored in the coil at switch-off.  Dividing this energy, by the energy recovered by C2, will give you the efficiency of your recovery process alone.  An interesting number, albeit not the entire story...

P.S.
Your current ramp-up is flattening out similarly to Itsu's (it is not a nice triangle) but the reason for this is different.
I have analyzed Itsu's scopeshot without the rotor but your scopeshot is with the rotor.  Consequently Itsu's current ramp-up is flattening out because he is getting too close to the V/R limit, but your pulse is flattening out because the current induced by the moving magnet in the rotor is getting superimposed on the LR current ramp.  Sorry for comparing apples with oranges, but you gave me only a scopeshot with the rotor's influence.

[itsu]

Fantastic work like usual Itsu.  If I can suggest something it would be to not bother reversing some of the magnets in your rotor.  They are glued in place and there is nothing to be gained and nothing to be learned by doing all of the work to reverse them.

For you scope traces, this time after taking a one-minute look, I cannot see any appreciable differences between the two waveforms.  However, your multimeter is showing slightly higher voltage across the capacitor which is telling us that slightly more current is being drawn by the drive coil when the rotor is stopped.  This is to be expected because there is no influence from any counter-EMF in the drive coil from the moving magnets in the spinning rotor.

With respect to attempting to measure the added power that has to be pumped into the drive coil to make the rotor spin, that could be done.  However, that would require some thought and careful preparation and developing the right measurement regime to detect it.  Right now what is happening is that when the rotor is in place and spinning, the reduced current draw due to the counter-EMF induced into the drive coil is overshadowing the extra power that is added to make the rotor spin.  In other words, the preliminary analysis is that more current draw is reduced due to the counter-EMF than the extra current draw that is required to make the rotor spin.  There are two opposite effects that are happening at the same time with respect to the current draw and it is not necessarily that easy to separate them from each other.

Thanks MileHigh,

i will try to do some power calculations on the voltage and current using the math function, but i do see some strange variations i cannot yet explain,
like the first screenshot (run) blue trace figure says Pk-Pk = 116mV while the second screenshot (stopped) blue trace figure says 113mV, but the second
screenshot blue trace display clearly shows a higher (off the screen) peak.
It probably are some spikes that cause that while running, but i would like to see them.

Itsu

[itsu]

Very clean waveforms - you should collaborate with Gotoluc so he can obtain the same.

I have extended the light blue current ramp-up on your scopeshot without rotor and came to the conclusion that your  ON-pulse lasts for approximately 1.6Tau and that means you have burned 1.5x more energy in the resistance than the energy you had built up in the coil as the magnetic field.
If you decrease the ON time to 0.5Tau than this ratio will fall to 0.37x.  Of course narrower pulse needs to be compensated with higher supply voltage.

The voltage across C2 (yellow Ch1) varies between 4V and 15V so the energy gain of this capacitor is ½C*( (15V)² - (4V)² ). This is how many Joules are recovered per 1 pulse.

At "switch-OFF" (dark blue vertical line) approximately 1.1A flows through the coil, so ½L*(1.1A)² Joules of energy is stored in the coil at that time.

I encourage you to plug the inductance of your coil and the capacitance of your recovery capacitor into these formulas to see how many Joules of energy each one of them represents. (I would do this but I am not sure what L & C are).
Finally, divide these two energies to see what your recovery efficiency is.

If you experiment further you will find out that this efficiency increases if C2 is completely discharged before the "flyback" pulse and no resistor is connected across C2 to burn the recovered energy.

BTW:
The red curve is this curve and the black vertical line is at the break-even-point in time (1.15Tau)

verpies,

great analisys as always, many thanks.


I will shoot a video to show what i have running now, but the setups between GotoLuc and me are way different at the moment.

I will keep the 555 timer on 12V and "up" the voltage on the coil to 24V, hopefully i can narrow down the pulse to 0.5Tau.

C2 is 100uF, so ½C*( (15V)2 - (4V)2 ) =  ½ 100 * 209 = 10450 (uJ?) which is way to low i think (see also http://www.calctool.org/CALC/eng/electronics/capacitor_energy)

L = 38mH, so ½L*(1.1A)² = 19 * 1,21 = 22.99 (mJ?)

Itsu

[woopy]

Hi Luc and all

Yes i have made a small rotor with 2 neomag oriented sideways.

After 2 days of playing with , i have stumbled upon some interesting things.

https://youtu.be/N7S_cbAYp6c

It seems that the timing is very important as well as the distance between the rotor and the core of the coil.

At some "sweet" point the current trace goes smoothly to zero at the end of the voltage pulse and so it seems that the flyback spike disappears completely.

I don't know if it helps, but this is very inteersting

Laurent