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

[itsu]

Very clean waveforms again and they appear how they are supposed to.  No inversion changes necessary.

The double M waveform in the trivial circuit is caused by the fringing flux of the magnets. The 2 peaks occur when the magnet is half way eclipsing the coil's core and the dip between these peaks occurs at TDC.

The R2 is significantly distorting the voltage measurements across C2 !!! This happens because now you have higher voltage across C2 (~57V) and the peak discharge current increased to 57V/120Ω = 475mA ...and before it was 15V/120Ω=125mA, if the previous R2 was the same.

If you do not want to go to the trouble of making an active C2 discharger, at least increase R2 so much that the voltage across C2 falls down to 0V just before the MOSFET opens at the end of the next energizing pulse....like I have drawn below.

If you do this, then the average DMM voltage reading will increase significantly, but it is the peak voltage calculated by the scope, that will be your true indicator of the recovered energy, according to E=½CV². 
And if you zoom in and give me the rise time of the voltage across C2 I should be able to calculate how much higher the voltage peak would be if R2 was deleted altogether.




P.S.
If you had an active C2 discharger circuit, then the average voltage shown by the DMM would become equal to the peak voltage across C2, while the yellow waveform would become almost rectangular ...and the voltage peak itself would be higher, too.

Ok, great.

I changed the flyback diode to a 0.4V forward voltage one and put a 5K potmeter as R2 and had to put it to the max. (5K), then the voltage across the C2 looks like what you had drawn,  see screenshot 1.
The DDM now shows around 20V.

Screenshot 2 is somewhat zoomed in.

Both situations are with a running rotor.

Itsu

[verpies]

I changed the flyback diode to a 0.4V forward voltage one and put a 5K potmeter as R2 and had to put it to the max. (5K), then the voltage across the C2 looks like what you had drawn,  see screenshot 1.
The DDM now shows around 20V.

If L₁=38mH and peak i=880mA then E₁=½L₁i² = ½*0.038*0.880² = ½*0.038*0.7744 = 14.71mJ
If C₂=10μF and V₁=0V and peak V₂=106V then E₂=½C₂(V₂² - V₁²) =  ½  * 0.000010 * 106² - 0² =  ½  * 0.000010 * 11236 = 56.18mJ

So the recovery efficiency calculates to: E₂/E₁ = 56.18mJ / 14.71mJ = 3.82 or 382%   
...and that is before the C2's voltage sag due to the D1 and R2 wasting energy during the first 114.8μs, is compensated for.

All right, who divided by zero?

[woopy]

Hi Luc

Very impressive work as usual

Fantastic conclusion in your video

So if we use a main coil,  the flyback of which is transfered to an assistant coil, so:

1 when the torque is increased on the rotor shaft, then the current in the main coil will increase
2 as the current is increased, the flyback spike is also increased
3 that  increased flyback spike is recovered by the assistant coil .
4 the assistant coil is therefore more assisting the main coil to support the increased torque .

Seems very good all this stuff

Bravo and thank's for sharing

Laurent

[gotoluc]

Hi Luc

Very impressive work as usual

Fantastic conclusion in your video

So if we use a main coil,  the flyback of which is transfered to an assistant coil, so:

1 when the torque is increased on the rotor shaft, then the current in the main coil will increase
2 as the current is increased, the flyback spike is also increased
3 that  increased flyback spike is recovered by the assistant coil .
4 the assistant coil is therefore more assisting the main coil to support the increased torque .

Seems very good all this stuff

Bravo and thank's for sharing

Laurent

Merci Laurent,

you have the same understanding as I do ... I think feedback is the way to go for what we are looking for.

I did not demonstrate this but in my previous build it was much more apparent then in this one when using the high torque low rpm DC motor to assist instead of the assist coil. Something in the magnitude of around 3 times more torque (pressure on the rotor) to the input power to increase on the drive coil compared to just powering the DC motor directly with DC.

This gave me the idea of, what if we take an off the shelf brushless DC PM motor which are controlled by 3 hall effect sensors which tell the switches (mosfets or transistors) to turn on and off at the appropriate time ... I'm quite sure in these control circuits they do nothing with the flyback right?... so all we do is collect the flyback and with it power a second motor tied to the same shaft and there you go, a motor that once under load the assist motor will become stronger which should help maintain shaft torque.

I think Tesla had a device that would operate on this same principal.

What do you think of this instead of trying to re-invent a motor, we just add a turbo assist to it and we may see it go over 100%

Let me know what you think

Luc

[verpies]

Quoting myself:

If L₁=38mH and peak i=880mA then E₁=½L₁i² = ½*0.038*0.880² = ½*0.038*0.7744 = 14.71mJ

We cannot trust the L₁ in the formula E₁=½L₁i² because L₁ might not be constant during the ON-pulse, when the rotor is influencing it.

However, if we use the fool-proof, albeit difficult method (V_inst *I_inst * t) of calculating the energy expended by the 12V power supply to bring L₁'s current up to 880mA, we get approximately 45mJ (from ½ * 12V * 1A * 7.5ms, when we assume ideal sawtooth current fitted symmetrically into the real curved-down current waveform - see below).
A Math trace of Ch2*Ch3 integrated between cursors set to the width of the ON-pulse, would give us a more precise number.

We can trust that the capacitance of C2 is constant, so we can trust the formula E₂=½C₂V²

Nonetheless the energy recovery efficiency form L1 into C2 is still E₂/E₁ = 56.18mJ / 45mJ = 1.25 or 125%   


So the first task for Itsu is to confirm, that the constant supply voltage applied to L1 through the MOSFET is really 12VDC and that PEAK current flowing through L1 is really 880mA  and that C2=10μF and that the CSR = 0.1Ω and that scope probe attenuation is set correctly (e.g.: x10).
The second one is to do the Math trace Ch2*Ch3 and to integrate it for the duration of the ON-pulse, to confirm the energy delivered to L1 by the power supply.