Selfrunning cold electricity circuit from Dr.Stiffler

[fritz]

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!!!

Well,

I would expect the tolerance of a multimeter operating in mV range
to be +-5 digits(6% here). Additional you have the contact voltage with the tips, etc...(2%)
Additional 5% resistor tolerance. ->7,53mA ->9,76mA.
52 Leds with Voltage 2.8 - 3.1 give a range from 145,6 to 161.2.
Worst case so far (and my estimations are "nice") 1573.3mW -  1096,4mW

If I estimate the tolerance of your power supply with 10% for the voltage
and 20% for the current reading we got: 1184,0 mW - 716.04 mW

This gives a possible efficiency range from 0.93 to 2.2.
Slightly uncertain result.

BTW:

Do you think you can measure RMS with a resistor, 2 beads and a cap ?
Better using the beads and the caps right at the huge 10u cap.
Makes more sense there - and looks more (dont know how to say that).

[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

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

You do not call RF AC. The RF is rectified buy diodes pulsating DC still on 50% off 50%. If you put the capacitor back in to produce DC then why jump on others who wanted to do this saying it will not give accurate results. My last post got posted twice my computer gave an error message so I re wrote it and sent it again. Sorry

[DrStiffler]

Here we go again, for what now the 10th year and pages and pages on this thread.

NO I AM NOT CLAIMING OU!!!
IT DOES NOT EXIST!!!!

You can not create or destroy energy. YES! I DO BELIEVE THIS.

But you can tap other forms that are not currently understood or utilized.

SO WHAT IS SO HARD HERE?

So if you can not climb out of your boxes and for a split second be open minded and look beyond you bank account (in case your getting paid to bash me) or beyond the trail of follows you might need to prove self worth, then I will not change you, AND YOU WILL NOT CHANGE ME.

Its a take it or leave it, bash me and I am now going to come right back. Guess when Hartmann closes this thing from the worthless banter, them the detractors will be smug and happy. But I will not back down, so those that are working towards something better might get a bit of help.

I'll be back.

[DrZLowe7]

Well, it looks good,  I like what I'm seeing,  my only doubt is the 3V assumption (or is it)

I can see the 10 uF cap, so it shouldn't be hard to measure the DC voltage on it and multiply by the current.


EM

Never mind the 3 volt voltage drop. 3 X .0086 still does not equal over 1 watt.

[Hoppy]

Dr. Stiffler wrote: -

"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!!!"


I'm a bit confused here. The meter is showing 8.6V not 8.6mA.

Hoppy