Selfrunning cold electricity circuit from Dr.Stiffler

[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!!!

[EMdevices]

*removed* EM

[plengo]

@Dr. Stiffler

[EMdevices]

*removed* EM

[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? 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