Real OU-Effect to Share with everyone!!!

[nievesoliveras]

@duff

It seems that I made a mistake with the polarity. I thought that the side of the battery with the 0 was the negative pole.

But if it is the positive it should be tested like this.

This was what I meant with the other two diodes on the wrong polarity.
One diode will charge the cap the other will charge the battery when the cap is discharded. Or that is the hope.

Jesus

[xenomorphlabs]

@Jesus:
What is the purpose of the two diodes actually?
You have mentioned Rectification? But aren`t we having DC in the circuit anyway (at least on cap discharge)?
I don´t see the advantage of using the diodes in contrast to leaving them out, as they open the line in both directions all the time (in contrast to e.g. a TRIAC which is timed), so they can´t control the current direction while cap charge or battery charge.
It would be nice if you could elaborate on their function to help understand the circuit how
you intend it to work, i am sure you have a good reason to use 2 anti-parallel diodes.
Thanks

[nievesoliveras]

@Jesus:
What is the purpose of the two diodes actually?
You have mentioned Rectification? But aren`t we having DC in the circuit anyway (at least on cap discharge)?
I don´t see the advantage of using the diodes in contrast to leaving them out, as they open the line in both directions all the time (in contrast to e.g. a TRIAC which is timed), so they can´t control the current direction while cap charge or battery charge.
It would be nice if you could elaborate on their function to help understand the circuit how
you intend it to work, i am sure you have a good reason to use 2 anti-parallel diodes.
Thanks

I apologize if the capacitor is not charged with ac voltage. I thought that the coil when pulsed was filling the capacitor with ac voltage. So I proposed the diodes in order to rectify the voltage so when the capacitor discharges on the timed rotor the voltage will charge the battery through the second diode.
The circuit was better when I made it on the negative side. There was one only diode to charge the cap and one only diode to recharge the battery. The direction of the diodes were thinking that they were workong on the positive side.

Again I apologize!!!

Jesus

[poynt99]

The question:

Why does it appear that using SW1 AND SW2 together causes the capacitor end-voltage to be almost twice as high when using only SW2 alone and with the diode?

A most probable answer:

RESISTANCE.

Easy to overlook, and not so obvious is the fact that two alligator leads in parallel will have roughly half the resistance of one alone. Where is the electrical path that puts SW1 in parallel with SW2 you ask? Through the capacitor of course!

SW2 connects the coil across the battery directly, and SW1 does so through the capacitor. At the instant a capacitor experiences a change of potential across its terminals it acts as a short, minus a finite internal resistance. This effectively places two connections to complete the circuit for a short instant of time, doubling the initial current. In fact to get optimum results, SW1 should be contacted a fraction of a second before SW2 connects. This most likely explains why the results have not been consistent.

To prove this theory and to consistently obtain better results, use two alligator leads clipped together in parallel to form SW2 and use the diode as per my previous post with the diagram here:
http://www.overunity.com/index.php?topic=6734.msg163425#msg163425


.99

[duff]

The question:

Why does it appear that using SW1 AND SW2 together causes the capacitor end-voltage to be almost twice as high when using only SW2 alone and with the diode?

A most probable answer:

RESISTANCE.

Easy to overlook, and not so obvious is the fact that two alligator leads in parallel will have roughly half the resistance of one alone. Where is the electrical path that puts SW1 in parallel with SW2 you ask? Through the capacitor of course!

SW2 connects the coil across the battery directly, and SW1 does so through the capacitor. At the instant a capacitor experiences a change of potential across its terminals it acts as a short, minus a finite internal resistance. This effectively places two connections to complete the circuit for a short instant of time, doubling the initial current. In fact to get optimum results, SW1 should be contacted a fraction of a second before SW2 connects. This most likely explains why the results have not been consistent.

To prove this theory and to consistently obtain better results, use two alligator leads clipped together in parallel to form SW2 and use the diode as per my previous post with the diagram here:
http://www.overunity.com/index.php?topic=6734.msg163425#msg163425


.99

I performed the test with two leads as SW2. This time I did a total of 60 strikes.

The first 30 I was striking the alligator clips against the copper plate. The average voltage was 105V with a max of 191V.

The next 30 I clipped the alligators to a #14 wire that had one end curled. I struck 1/2 the time with the blunt end and the other 1/2 with the curled end. The average voltage was 150.9 with a max of 319V (one time).

Even though I got one 319V the average is still lower than NRG average which was 204V.

It seems to me the way the strike occurs has significant influence. The #14 wire with the curled end worked best, having more surface area and possibly skidding as the strike was made.

===

I have the commutators functional now but the results are not as good as I had hoped.
I polished the commutator and brushes (AWG 9 with end curled & filed flat on contact area & suspended on independent supports)  with 1600 grit emery cloth. Still there is much noise and I do NOT see the capacitor charging to 300+ volts. I've also noted that I did not make my discharge bar wide enough to fully discharge the cap.


-Duff