The Ossie motor

[mscoffman]

Hi Captain

thanks for answer  have you some shematic as how to transfer the cap energy back to the run battery or how to use and what for a Zener diode value ?

Is it a way to calculate the power going in the big cap along the time . As the time from about 12.5 to 50 volt was  30 minutes is it possible to deduce some power from those datas ?

Or what would be the datas to register to get the power of those Flyback spikes ?

regards

laurent

@laurent

Question A-
On the zener...it is not efficient for discharging the 10Kuf capacitor.
The zener to use, would be the voltage you want to see, lets say
12.5volts. You put a resistor in series with the zener so that at
12.5 volts you get your required .007 amps = 7ma. pass-current.
The resister has to be of the appropriate wattage. The problem is
that same resistor at 50.0 volts will be conducting 28ma so at the high
voltage the power wasted is very high. P=E*I. where E= (max V - base V)
where base V is 12.5 volts. A series pass DC regulator will have this same
wasted power.  What you need is a switching regulator for efficient
energy conversion.

Question B -
I wondered the same thing. What happens if you turn all of
the charge energy of the 50V 10kuf capacitor into 12.5 volts
at .007amps to run the motor:

o) Well 1C=1amp for 1sec. which means C= .007amps * 1800 seconds
or about 12.5Coulombs charge required by motor to run for 1/2 hour.

o) Also equals 1C=1Farad at 1Volt which means the capacitor
is C = .01F * 50Volts or only .5Coulombs charge stored in the
10Kuf capacitor.

So the capacitor is fairly light duty for running the motor for so long.
This is not suprising given the narrow width of the back EMF pulse
in ratio to the motors total pulse cycle time. You'll need to look
at a supercap in this application to have a shot at success.

:S:MarkSCoffman

[woopy]

@Mscoffman

Thanks a lot for the very comprehensive explanation

Hebeh ! I am far from the goal.!

If i have right understood ,  if i need 0.007 amps to run the motor for 30 minutes (1800 sec), the stored energy in the cap should be at minimum 12.6 Coulombs.

And as the actual energy after 30 minutesa is only 0.5 coulombs, i am about 25 time under any attempt for a self runner by this way.

Even if i only consider the pure motor current needed that is 0.0014 amps (without the Hall circuitery) i am always 5 time under the goal.

OU is not to be searched in the Flyback this way. Unless there is a mean to get far more of them  and  or stronger ones. Any idea ?

Can i state that in this case with 0.007 amps motor current by geting back all the fly back spike power i could  get back 4 % energy or improve the motor  efficiency of 4 % ?

And if yes if we could reduce the Hall circuitery to zero (ideal case), In this case with 0.0014 amps the efficiency could be improved of 20 %?

Just for thinking

Regards

Laurent

[Jimboot]

Ok I put the gen motor aside for now. I'm attaching pickup coils. What sort of circuit would be best to capture current through the coils?
mmm just had a thought. What if I used the extra rotor & coils in series with the one below? The mags on the top rotor are opp polarisation to the one below. Could this cause some sort of phase cancellation?

[captainpecan]

After banging my head against the wall a bit more, I stepped away from the project for a bit.  After thinking it all through again, I came up with an interesting variation that could be of use.  I'm thinking of replacing the bridge rectifier with a voltage doubler circuit.  In theory, it eliminates 2 diodes so there will be less voltage drop on the energy captured from the induction.  Also, it would allow the turn of the rotor to generate less voltage than the run battery, and still charge the run battery anyway.

A quick rundown of how it works.  AC current comes from the rotor induction and feeds into the voltage doubler.  Each half wave charges it's own cap through only one diode.  When the reed switch is tripped, both caps are hooked in series to double the voltage and dump into the run battery.  Pretty simple, just wanted to share what I was cookin' up in my head.  Haven't tried it at all, just kicking it around.  But in theory it may be a way to suck a bit more energy from the turn of the rotor without adding more pickup coils yet.

[Jimboot]

After banging my head against the wall a bit more, I stepped away from the project for a bit.  After thinking it all through again, I came up with an interesting variation that could be of use.  I'm thinking of replacing the bridge rectifier with a voltage doubler circuit.  In theory, it eliminates 2 diodes so there will be less voltage drop on the energy captured from the induction.  Also, it would allow the turn of the rotor to generate less voltage than the run battery, and still charge the run battery anyway.

A quick rundown of how it works.  AC current comes from the rotor induction and feeds into the voltage doubler.  Each half wave charges it's own cap through only one diode.  When the reed switch is tripped, both caps are hooked in series to double the voltage and dump into the run battery.  Pretty simple, just wanted to share what I was cookin' up in my head.  Haven't tried it at all, just kicking it around.  But in theory it may be a way to suck a bit more energy from the turn of the rotor without adding more pickup coils yet.

I like it CP, except the bit about tuning 3 reed switches. The simpler te better tho I think.