Hi Singer...

I think you are "off the mark" as you say.  I was okay watching your video until the current measurement part.  I couldn't exactly see what you were measuring, but if you put 1/4A through a neon bulb, it will explode!  Typical neon testers like that draw micro-amps or, at the most, a few milliamps.  You need to test the current and voltage at the same time.  If you are shorting the current meter across the neon tester, then you will find almost zero volts and the neon will not light. 

I think you have a serious measurement error.  Sorry to dampen your obvious excitement!

Linda

Hi Singer,
Linda is right,
you have to measure Volts and Amps at a real ohmical load.

Just try to see, how much voltage and amps you get on a 10 and/or 100 Ohm load
resistor.
(Better use a 10 Watts type, so it will not heat up too fast...)

Regards, Stefan.

Hi Singer,
buy a 1 Ohm
10 Ohm,
100Ohm and
1000 Ohm
all 10Watt type resistors
and use this instead of the neon
bulb as the load.

Then connect each resistor once as the load.
Then measure the voltage across the different resistors
and post each DC voltage at each single resistor.

( Of course put only one resistor of this list to the output and then measure each one by one)

Tested current with 10 ohm 10 watt resistor and got .13 amps with slightly lower voltage.
(Batteries are close to dead- need to buy more)
I'll pick up the 100 and 1000 ohm resistors and batteries tomorrow and post results.

mramos-
the meter was shown up to .29 amps. Is that not close to .33? Mockery is not helpful except to reveal character. And as far as what "was not seen" was obviously me taking the leads off the neon bulb and putting them on the volt meter.

Hi again, Singer,

One thing you said struck me and that was that you were measuring the current "open circuit".  You know, that when you set your meter to any of its current scales and plug into the appropriate jacks, the meter acts as a near short circuit internally.  DMMs are all actually voltmeters.  To get a current reading, they just throw a fat low value resistor across the jacks inside and then measure the voltage drop across it. 

Per Ohms law, if the internal shunt is 0.1 ohm and the current readout says 10A, then there will be a voltage of 1V between the meter jacks.  Just remember that in Voltmeter mode, your meter will have a very high resistance (ideally infinite) and in Ammeter mode it will look like a very low resistance (dead short ideal).  So when you do your resistor tests, measure voltage across (parallel) the resistor and measure current by inserting the meter in series with the test load resistor, not across it.

Probably you already knew all that, but I just wanted to clarify in case you weren't sure.  When you put a current tester directly across the output of your circuit it's a virtual short circuit because of the internal shunt.  I can't count how many times I've blown up the fuse in my meter by forgetting to rearrange the test leads into the right jacks and sticking the meter (still set for a current test) across some beefy power source. 

Anyway...happy testing and don't be discouraged if overunity operation eludes you for a while.  It's managed to hide from the mainstream of practical commercial technology for a very long time!

Linda

No worries mramos, thanks for writing back.

Anybody know the best way to send current back to run the timer once the unit is started? it needs to replace 18V from two 9V batteries.

 

Again, having a little trouble understanding terminology, but what I did was connect the 10 ohm resistor to the positive out on the unit and then connected the positive terminal of the meter to the other side. the negative out of the unit I ran to the negative terminal of the meter. Is this series? I understand series/parallel when it comes to batteries, but with caps and resistors I get confused.

But after reading hartiberlin's post I think I know how to proceed now and will re-test tonight, and post with pix.

A good way to home in on this magic number to extract maximum power is to notice the following:

As you increase the ohmic value of the load resistor, you will most certainly notice an increase in the output voltage.  It will be zero if you use zero ohms (dead short) and will be maximum if you use an open load (no load...open circuit). 

For most (linear) circuits, and I believe yours will probably fall into this category, the ideal load for maximum power transfer will be very near to whatever load causes the voltage to fall to half its open circuit voltage.  Try to find that value by experimenting.  You will then be able to tell if you have hit the maximum power point by raising and lowering the value slightly from there and observing a power falloff in both directions.

Linda