Muller Dynamo

[chrisC]

@MK1

I'm confused. Is it not.........

clockwise going up  =  counter clockwise going down  (same thing)
clockwise going down = counter clockwise going up  (same thing)

wattsup

PS: There are only two ways to wind a coil but there are four ways to describe the wind.
             

Wattsup:

You are confused because you couldn't trace the the fine details of the coil wires in the video.

There are TWO coils in each set. Now how many coil winding variations can you get with 2? It's really simple.  It's 1 plus 1 plus 1 plus 1. Don't need to think so hard.

cheers
chrisC

[yfree]

@EMdevices
@hartiberlin
@JouleSeeker
@wattsup
@All

I posted a reply #3303 yesterday, but it was ignored. Perhaps, because I worded it awkwardly. Here it is again, worded as it should be.
In the first video by Romero (not self-running), there are two digital multimeters connected to the input. One measures the voltage on the battery and the other measures the total current going into the device. When the device reaches a steady state, without load, the input current is 0.94 A and the battery voltage is measured  at 12.59 V. When the load is connected, the current consumption decreases to 0.92 A and the battery voltage drops to12.28 V. The voltage drop under load is 0.3V and the current change is 0.02A. On this basis we can calculate the internal resistance of the battery,
Ri = dV/dI = (0.3V)/(-0.02A) = -15 Ohms.
This means that the battery has negative internal resistance: the more current we draw from it, the higher the voltage on the battery is. For instance, an increase of current draw by 1 A would increase the battery voltage by 15 V. Yes, we are looking for this, but is this really possible with this normally looking battery?

We can also calculate the equivalent resistance of the driver circuits; they draw 0.94A at 12.6V.
R = V/I = 12.5V/0.94A = 13.4 Ohms.
Having this, we can calculate the expected current decrease when the battery voltage drops by 0.3V.
dI = dV/R = 0.3V/(13.4 Ohm) = 0.022A.
This is what we see in the video, within the accuracy of the multimeter. No miracles here. Current consumption decreased because the voltage on the battery dropped.
If, however, we assume that this is a normal battery and has  0.15 Ohms of  internal resistance (a reasonable number for a battery of this size), then we can calculate the value of current change that would cause the observed 0.3V drop under load:
dI = dV/Ri = 0.3V/0.15 Ohm = 2A.
The question now is: where is this unaccounted load on the battery coming from?

Is there a logical explanation to this battery behaviour in this circuit?  Thank you for your contributions.

Best regards to All,

yfree

[Hoppy]

We can also calculate the equivalent resistance of the driver circuits; they draw 0.94A at 12.6V.
R = V/I = 12.5V/0.94A = 13.4 Ohms.
Having this, we can calculate the expected current decrease when the battery voltage drops by 0.3V.
dI = dV/R = 0.3V/(13.4 Ohm) = 0.022A.
Is there a logical explanation to this battery behaviour in this circuit other than the battery with negative resistance?  Thank you for your contributions.

Best regards to All,

yfree

Yes there is. If a battery is partially sulfated, then when a load is applied the terminal voltage can rise for a considerable time. A heavy loading can accentuate this effect.

Hoppy

[nul-points]

nul-points

As i said i really don't know how to use a scope but i took another few shots.

Today i bouth the new coil that are the same as romero's. i'm going to wind some to see how they behave

no problems, marius - they're looking good

i'm assuming all these traces have the same scale vertically?  please let me know if that's not correct


your disconnected single coil is giving approx 67% pk-pk o/p with 10 ohm load compared to o/c - so it's internal impedance could be around 5 ohms - probably not too different from the motor drive load (you'll find that out from your drive i/p tests)

your single-coil connected to FWBR trace is looking pretty much like Romero's single coil into FWBR trace, so that's encouraging!

from your third trace, you can see that under load the combined o/p has quite a different shape to your previous off-load combined o/p trace

IF this trace is at the same scale as the others then you are getting some gain from the combination process - 4 peaks are at 8 grads, whereas your single coil peak BEFORE the FWBR is only 1.5 grads!

also, you can see that the sequential 'increasing' of peaks is not as regular when the FWBR o/ps are on load - maybe some core-rotor gaps need to be adjusted? (just a thought)

i'd say your basic generator o/p is looking pretty healthy - from a regular point of view


do you have a large-value o/p capacitor yet?  you could try adding the capacitor + the 10 ohm load across your FWBR o/p & see what  steady 'DC' voltage you get - this will give you some idea of your system efficiency at the moment, when you compare it with the DC i/p to your drive

please let us know if the scales on all these traces are the same (ie. did you have to adjust the vertical scales between any of the most recent traces?)

how are your existing coils different from the new Romero type ones which you're about to wind?

thanks for sharing!
np


http://docsfreelunch.blogspot.com

[altair]

I will make up another rotor tonight, I have another circle. I will use the magnetite and epoxy mix to fill in holes. I'll post results.
Peace
rawbush

Hi rawbush,
could you tell me please where I could get magnetite powder ?  I have searched the web without results.
Thanks

Altair