PhysicsProf Steven E. Jones circuit shows 8x overunity ?

[TinselKoala]

hi Steven

i think you have an incorrect method for calculating Ein

you should calculate the Energy stored in C for each voltage, start & end

THEN subtract to get total

hence:-

40mF
1.385V => 38.37mJ
1.225V => 30.01mJ
               Ein 8.36mJ

Pin = 8.36/0.5 = 16.72mW

another potential issue to note - the cap value can be +/- 10-20%

when doing these calcs, it's wise to measure the cap value

of course, for a 'ball-park' calculation it's not necessary! 

hope this helps
np

PS  i admire your other 'extra-mural' work , investigating & providing low-cost solar cooking solutions for developing countries!


http://docsfreelunch.blogspot.com

Vfinal is 1.255, not 1.225 as you have it. The algebra is correct, both methods give the same answer, but you've got to use the same input numbers !!

(CVinitVinit)/2 - (CVfinalVfinal)/2 = (C/2)(ViVi)-(C/2)(VfVf) = (C/2)(ViVi-VfVf)

[yssuraxu_697]

BTW When using very large capacitors, is capacitive reactance considered?
For example 40000uF has Xc=10e-7ohms at 4Mhz.

[xee2]

delta-E = 1/2 C(Vi**2 - Vf**2) ,  Pin = deltaE / delta-T  , 30 seconds.  C = 40mF.

Start, Vinitial = 1.385V  , Vfinal =  1.255V

So delta-E = 6.8 mJoules.
and Pin = 6.8/o.5min = 13.6 mW,   pls check my math.

It is a good thing you are a physics professor. I would have gotten this wrong on a test. I get:

( 500 ) * ( 40e-6 ) * ( 1.385^2 - 1.255^2 )

=  6.864 milli Joules total energy

therefore over 30 seconds =  6.864/30  =   0.2288 mW

note > watt = Joule/sec

What did I do wrong?

[nul-points]

[...]
therefore over 30 seconds =  6.864/30  =   0.2288 mW

note > watt = Joule/sec

What did I do wrong?

LOL - if you're wrong then at least two of us are!

[...]
a Joule is a Watt-second - i see you've divided Ein by units of minutes;
[...]
Steven, apologies - your Ein calc method is good, but Pin needs units of seconds, not minutes]

Pin = 6.87/30 = 0.23mW
[...]
np
[...]

...fortunately, i suspect we're both correct! 

[Tudi]

call me a simple guy BUT, we are debating over this power usage / gain issue for days and days. If you would have left that LED(better more then 1) on the device running and some of the output to loop back, at least we would have a very vague idea if it runs for a short time or very long time. ( yes, i realize that these high frequency ringer circuits are tricky regarding the light emiting consumers = same visual brightness if close to 30 fps or continues operation )
Better yet, if this circuit is so easy to replicate, just make a new one and put the new one to run on a button cell battery with a big as possible consumer that you estimate it should hold. And put another battery without the circuit with similar load. Yes, these are very barbaric tests, not even close to an 50% precision. But if both seem to die off in close "year" then maybe the gain is very small. What do you have to loose ? 2 button cells and 3 hours ?
What can you gain ? realize that some specific part of this circuit is very important to know to replicate it ( maybe key component)