Selfrunning cold electricity circuit from Dr.Stiffler

[DrZLowe7]

Ok, thank you,  then that's encouraging.    Assumptions are nothing bad as long as they're applicable and reasonable.


The break even point for OU in this case would be:

Vdiode =  0.936 watts / (8.6 mA x 52)  =  2.09

So you're assumption is well above this breakeven voltage drop / diode.      I would say you have something significant.

But to be sure, I would measure the DC voltage at the 10 uF capacitor.   If there is quite an AC ripple on it, I would filter the heck out of it with ferrites and include another capacitor if I have to. 

In my previous experimentation, I noticed typical LEDs give out quite a bit of light starting at 1.9 Volts and only get brighter from there.  I'm not insinuating your assumptions are not correct, don't missunderstand, I'm just trying to be carefull when it comes to excess energy measurements.   I made claims like this before and I later retracted.  Hopefully you won't have too.  I hope not.

EM

His LED's are connected in series. If I have 200 volts and place say 20 LED's in series and for math purposes let say each LED has a voltage drop of 2 volts. The voltage drop only changes the 200 volts to now 160 volts for doing equations. So we now have 160 volts with .0086 current going through the all the LED's remember series here. Again Wattage is the total voltage 160 volts X .0086 total current . Again look close at his pictures the resistor is checking total current through all the LED's as the LED's and resistor are in series. If the LED's were parallel as you drew in your versions of the schematic then you would multiply current X 52.

[EMdevices]

*removed* EM

[fritz]

...
Measuring with 100Ohm series resistor would give a nice voltage
reading for our load current. The additional 860mV wont harm the
result because of the current source role of the circuit at this position.
Adding beads at the 10u cap and additional 470n ceramics would give
me the feeling of nice DC loop here.
Calculating the output power as voltage on 10u Cap (- voltage reading on
series shunt) * load current would give me a comfortable feeling.
Measuring the exact resistance of the load shunt with an LCR meter is
a 3 second effort. Same as measuring the load/voltage from the supply
with some trustworthy instrument.

So what is the exact efficency ration of the circuit ??

[DrStiffler]

DRZ,  are you on another wavelegth?

I can see what he is doing clearly and I understand.   I suggest you study the drawings a bit and make yourself a circuit after them.

You might be confused about series and parallel.

He has the LED's in SERIES.   So if each has an assumed 3 Volts drop across them how much of a voltage drop will 52  SERIES connected diodes have?     

Let's do the math    3 + 3 + 3 + ....+ 3 +3 .... + 3   =  52 x 3 =  156 Volts    Got it?

Now the current running through all the diodes is  8.6 mA,   Got it?

So   Power out is    156 Volts x  8.6 mA = 1.34 watts    Got it?

This will be the last time I will walk you by the hand.   And this goes for anybody else that's building up courage and joining in the perceived bashing of Dr Stiffler.   It's one thing to ask intelligent questions and another to be plain silly.

EM

EXCELLENT!

We can carry on a dialog......

Teach me how to ignore, 'well you know' and things may move faster, Yes?

[DrStiffler]

...
Measuring with 100Ohm series resistor would give a nice voltage
reading for our load current. The additional 860mV wont harm the
result because of the current source role of the circuit at this position.
Adding beads at the 10u cap and additional 470n ceramics would give
me the feeling of nice DC loop here.
Calculating the output power as voltage on 10u Cap (- voltage reading on
series shunt) * load current would give me a comfortable feeling.
Measuring the exact resistance of the load shunt with an LCR meter is
a 3 second effort. Same as measuring the load/voltage from the supply
with some trustworthy instrument.

So what is the exact efficency ration of the circuit ??

So Fritz, why don't you take three minutes and get back to us? I have my path charted and it does not include personal requests. I think I must have a great amount of faith in myself to go through this, so help out here and get us a few measurements if your not comfortable?