Selfrunning cold electricity circuit from Dr.Stiffler

[DrStiffler]

Very good Stiffler,   not bad.   So you're claiming OU after all.

Now,  how did you measure the Output Power.

1)   How did you get the 3 Volt drop,   from the Spec sheet?    Not a good idea,  it varies with the current.

2)   How did you measure the current, [edit]   Oh I see in the picture, never mind.  I have to analyze it a bit...

You don't have to answer, you don't owe us anything   LOL   

EM

Your a very funny person...........

[DrStiffler]

Prove to you? Who are you? What would YOU ever accept as proof?

No let me show yo a couple of pictures and some very simple math, then we all can listen to what you say is not proof yet. Like did not do this or that, or need to do something else, or faked something, or where are the scope shots. REALLY!!!!

Here is a typical running circuit containing 52 white LEDS.

Now lets see;
52 LED x 0.0086 Series current x 3 volt drop = 1.342 watts

Now difficult, input is 12 volts at 0.078ma = 936mW

MUST BE A REAL MEASUREMENT ERROR HERE!!!

What? Watts = Volts X current . Not Volts X voltage drop. You say you are producing around 200 volts with a 3 volt voltage drop you now have 197 volts . 197 volts X .0086 = 1.6942 watts. However because we are using AC to operate the LED's they are on 50% 0f the time off 50% of the time. So 1.6942 / 2 =.8471 watts a loss of 89mw not over unity

Maybe your using AC, I'm not.............

[DrZLowe7]

Prove to you? Who are you? What would YOU ever accept as proof?

No let me show yo a couple of pictures and some very simple math, then we all can listen to what you say is not proof yet. Like did not do this or that, or need to do something else, or faked something, or where are the scope shots. REALLY!!!!

Here is a typical running circuit containing 52 white LEDS.

Now lets see;
52 LED x 0.0086 Series current x 3 volt drop = 1.342 watts

Now difficult, input is 12 volts at 0.078ma = 936mW

MUST BE A REAL MEASUREMENT ERROR HERE!!!

What Wattage = Volts X Current. Not Voltage drop X current. You say you are getting around 200 volts so 200 volts - voltage drop = 197 volts. 197 X .0086 = 1.6942 watts. Now because you are using AC to drive the LED's they are on 50% of time and off 50% of the time. So divide your results buy 2 = 847mw You now have a loss 89mw. Not over but under unity.

[terry1094]

From a poster on Vortex who is in contact with Dr. Stiffler:

<><><><><><><>

Not the MIB this time ;-)

Apparently he is away with his family for the Thanksgiving Holiday.

As a precaution against a DoS web attack on his server, which has
happened before, he has temporarily taken down the web page and videos
until next week when he returns.

Have a trustworthy turkey-day, all:

http://www.technologyreview.com/Biotech/19741/?nlid=680

<end quote>

Mystery solved.  For our friends in Europe and elsewhere, try some turkey tomorrow!

Terry

[EMdevices]

*removed* EM