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Overunity Machines Forum



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

Started by gotoluc, November 10, 2015, 07:11:57 PM

Previous topic - Next topic

0 Members and 16 Guests are viewing this topic.

itsu

Quote from: verpies on December 27, 2015, 05:50:26 PM
Almost.
Notice that C2 is not discharging all the way down to 0V.

To discharge C2 all the way down to 0V:
- start discharging C2 sooner
or
- discharge C2 longer. (if I knew Q3's gate waveform then I could decide whether increasing L2's inductance is necessary for this).
or
- discharge C2 faster by decreasing R2 (Warning: don't decrease R2 so much that Q3's maximum pulsed drain current is exceeded)
or
...all of the above or combination thereof

Also, put some small but good bypass caps to ground on the source of Q3 and on cathode of D2, because there is an unwanted voltage spike on the purple trace. While you are at it, decrease the inductance of the entire drain circuit of Q3 including C2 (short leads, small loop area, etc...)

Below is a revised schematic with the correct probe placement and countermeasures against the voltage spike form L2.
If C3 and C4 do not help, then adding R4 will slow down the closing of Q3 and its drain current spike that gets converted to a voltage spike by stray inductances in the drain circuit.  For R4 try anything between 0Ω to 1K.

What are you using for U2 ?  Do you use its inverting input or non-inverting?


Thanks,

i will add the additional components and take the extra scope shots.
U2 is presently an inverting ucc37321, as a non-inverting ucc37322 did not produce the required "discharge C2" signal.

I used a hall sensor feedback system for this signal at first, but the rotor did not want to run properly then allthough i could
tune the sensor anywhere inbetween the L1 pulse train.
I now use the same 555 output for both the Q1 and U2 (inverting)/Q3 inputs.

A 10 Ohm ¼W resistor for R2 lasted only a few minutes before it opened up (invisible damage), so i use a 10W resistor there now,
not sure it was coincidence.

Itsu

gotoluc

Quote from: sm0ky2 on June 17, 1970, 07:20:01 PM
my experience with this matter has led me to believe the "most efficient" motor that mankind has so far created
is  brushless A/C motors, just an inductor, with two large copper links,
and an aluminum rotor with arc-shaped grooves in it.

What you're describing sounds like a shaded pole motor, is this correct?

Quote from: sm0ky2 on June 17, 1970, 07:20:01 PM
When these are operated at a frequency that is compatible (multiple, divisor, or octave of) with the self resonant frequency of the inductive core.
They run at 95+ % efficiency.

Can you provide a demonstration or the details on the 95%+ efficiency test

Thanks

Luc

verpies

Quote from: itsu on December 28, 2015, 04:49:36 AM
U2 is presently an inverting ucc37321,
Note, that the maximum supply voltage of this driver chip is only 16V.
That might seem like enough when you are supplying it from 12V but remember that D2 and ZD12 allow the L2 voltage to swing up 18V higher (depending what ZD2 you have installed) and this gives the total of 30V to ground, that can appear on the output of this driver (minus the voltage drop of R3 and D3) so you are pushing the output transistors of this chip and don't be surprised they break.  ...but if this driver is inexpensive, I would not go to the trouble of adding Q4 since it apparently works well alone.

Quote from: itsu on December 28, 2015, 04:49:36 AM
I now use the same 555 output for both the Q1 and U2 (inverting)/Q3 inputs.
Pretty clever but this way you charge and discharge C2 at the same time.  Q3 should be open and C2 should be discharged, before Q1 opens.
Also, this way you do not have control how long Q3 remains closed. The inductance of L2 determines that for you.

Quote from: itsu on December 28, 2015, 04:49:36 AM
A 10 Ohm ¼W resistor for R2 lasted only a few minutes before it opened up (invisible damage), so i use a 10W resistor there now,
not sure it was coincidence.
I don't think it was coincidence either.
When I was doing the power calculations for putting a small light bulb in place of R2, I calculated approximately 2W.

verpies

Quote from: sm0ky2 on December 28, 2015, 12:22:21 AM
the "most efficient" motor that mankind has so far created is the one we use in nearly every electric device that has a motor in it.... like a stand-alone fan, or a vaporizer, im sure you've seen these things, they are brushless A/C motors, just an inductor, with two large copper links, and an aluminum rotor with arc-shaped grooves in it.
A 3-phase brushless motor with a permanent magnet rotor (PMSM) is more efficient than a shaded pole motor (a type of AC induction motor) because there are no resistive eddy current losses in the PM rotor and in the shaded pole shunt winding.
The resistive conduction of electric current in a copper or aluminum winding/shunt, cannot compete with the atomic bound magnetizing current in a permanent magnet which is lossless and perpetual.

itsu

Quote from: verpies on December 28, 2015, 09:20:17 AM
Note, that the maximum supply voltage of this driver chip is only 16V.
That might seem like enough when you are supplying it from 12V but remember that D2 and ZD12 allow the L2 voltage to swing up 18V higher (depending what ZD2 you have installed) and this gives the total of 30V to ground, that can appear on the output of this driver (minus the voltage drop of R3 and D3) so you are pushing the output transistors of this chip and don't be surprised they break.  ...but if this driver is inexpensive, I would not go to the trouble of adding Q4 since it apparently works well alone.
Pretty clever but this way you do not have control how long Q3 remains closed. The inductance of L2 determines that for you.
Even worse, this way you charge and discharge C2 at the same time.  Q3 should be open and C2 should be discharged, before Q1 opens.
I don't think it was coincidence either.
When I was doing the power calculations for putting a small light bulb in place of R2, I calculated approximately 2W.

OK,  C3 and c4 in place, but it did not make much difference, the ugly spike is still there, going for R4.

I have a 15V 1A zener for ZD2.

L2 has a core which i can move in or out, its now max in (max inductance), which sets the C2 voltage max towards zero.

EDIT, i have removed the screenshots as the setup was wrong so they seem useless.

Screenshot 1 has added point F (compared to B) in the green trace
screenshot 2 is as 1 but zoomed in
screenshot 3 has added point E (compared to B) in the green trace
screenshot 4 is as 3 but zoomed in

Itsu