Muller Dynamo

[mariuscivic]

Hi Mike!

I can't do that experiment becouse i have changed my setup once again. The bolt ads drag to the rotor and the rpm will be much more hyer without the bolt.

[mariuscivic]

Hi mariuscivic,

use you meter on AC to measure the current between the 220vac and you transformer. First measure how much current the transformer uses or wastes (not connected to motor) then start motor to calculate how much it uses and then connect LED and diode to see if current goes down. Please also remove all the other coils so we know they are not being used. We don't need to see if the motor accelerates since we can hear it.

Please also tell us the DC output voltage from your transformer when the motor is operating.

Please do this new test; remove the LED and see if you get even more acceleration using only the Diode. I believe you will as I have also found 2 year ago that if you recirculate the flyback (BEMF) back in the coil there's a better push on the magnet.

Please make a video of the new tests.

Thanks for sharing

Luc

Hi gotoluc!

I made another video . I measured the input curent of the transformer with and without the LED conected to the BEMF.

GREAT SUCCES!!!       http://www.youtube.com/watch?v=DuE3p8gUFMA

[gotoluc]

Hi gotoluc!

I made another video . I measured the input curent of the transformer with and without the LED conected to the BEMF.

GREAT SUCCES!!!       http://www.youtube.com/watch?v=DuE3p8gUFMA

Thank you mariuscivic for making a new video.

Interesting that if you remove the LED but keep only the diode connected you lose the effect. I would of liked to see that as it's hard to understand why it would do that.

Can you measure the DC voltage used to the motor and let us know. Also, maybe you can give some details on your coil. You said it is an air core but how many turns, wire size, length and resistance.

Does your meter have a Freq. setting?... if so you could measure and give us the number and the amount of magnets on your rotor so we could calculate the RPM.

Current (when diode Led added)  of 8mA @ 220v = to about 1.75 watts.  Did you measure how much current is used when the motor is not connected?  since all transformers use current even if nothing is connected to the secondary.

Thanks for making this new video and sharing your findings.

Great job

Luc

[Magluvin]

Hey Marius

Did you say in the vid that when adding the diode alone, without the led, that the speedup didnt happen?  Seems odd.

Maybe the diode is bad. Shorted?  And the led was the only diode in the circuit.  Have you tried with just the led?

Having had my share of pulse motor projects, I believe the current going lower at higher speeds is due to the shorter pulses to the coil. The longer the coil is energized the more current till peak as time goes by. If you could run your setup again, measuring the same current, add another motor coil separately on the other side of the rotor to make the rotor go faster than the original coil could and see if the current goes down even further without the diode.

Seems as though if we add more coils....

Say 1 coil is 5v 100ma  example.
And if we speed up the rotor say 10% and the current reduces to 90ma..

Now 2 coils will be at 5v 90ma    tot 180ma but a lot faster.

4 coils will be at 5v 81ma  tot 360ma   way faster..

And add the diodes  even faster..  And it should be putting out more rotation energy per power in as compared to just one.

Here is an experiment.  Test rpm/ voltage(say 10v) / current with 1 stator, then use 2 coils(stators) and give each half the power(half the voltage 5v) each and input tot will be the same as was used in the first test by putting them in series and align them with the rotor. I bet the rotor goes faster using the same total input, where the 2 coil version uses only half each. It works with speaker systems also.  ;]


Sorta shows how inefficient using 1 stator can be. but 2, 4, 8, 16 32 64?  How much can we save like this and does it approach OU ?

;]

Mags

[Magluvin]

Correction

Say if 1 coil has 10v  coil 10 ohm(pete sake) is 1 amp = 10w

We want 10 watts with 2 coils  so each coil will consume 5watts at the same time.

10/1.414(constant) =7.072v for each coil   

7.072 / 10ohm = .7072A    7.072v x .7072A= 5w each  tot 10w   ;]

sorry bout that.   now me sleeps. 

Mags