Second Stage Joule Thief Circuits

[gadgetmall]

@groundloop . Here is a replication of the Ou controler . Its just waiting for the parts missing a few diodes and two scr's and a few connections

It will be a Nice little package indeed !!!

[Groundloop]

PaulLowrance,

I agree with you. And that means that the only thing we need to know is
the voltage a capacitor has, to be able to find the energy stored, as long
as we know the capacitor value.

Alex.

[jadaro2600]

I'm not sure how you came up with those numbers.

I think there's been a misrepresentation of the statistics of the 650F capacitor @ 2.7v: at maximum, it can hold 1755 coulombs of charge since 1 farad = 1 coulomb per volt, if you were to short this, you would dissipate 1755 coulombs instantly. As we all know, an amp is a measure of coulombs per second, so, if it were capable of discharging 1755 amps instantly, it would, but not 3500.

As you probably know, a joule is a unit of energy, and coulomb is electric charge. One coulomb equals 1A * 1s. The energy contained in a capacitor is 1/2 * C * V^2.  So 1/2 * 650F * 2.7^2 = 2369 joules.

Paul

Yes you are correct, i've apparently hobbled the math - as we WERE discussing power and energy, however, I did go back and edit my some of my ignorance out and thank you for the comparison / correction.

What I was interested in is the potential amps which could be produced.  Hopefully I've gotten the rest of the math right ( but probably not ).

Gadget, could you resize that image, it's simply massive!

[Groundloop]

@gadgetmall,

It looks nice. The missing parts is being shipped by Santa right now. :-)

I have made a little change to the circuit. I have made room for a larger BY255
instead of the 1N4007 (charge current going back to AA battery) by your request.
So the new pcb (when done) has room for this diode. (There are some BY255 in your parcel.)

Groundloop.

[PaulLowrance]

I'm just waiting for someone to say this is not topic of advanced JT. But gadget's claim is so big, that it encompasses all threads. 

If gadget can charge the 650F cap to like 2.5V, then assuming the manufacturer is correct about it being 650F, then one can *efficiently* discharge the cap and get very close to 1/2 * 650F * 2.5V^2 = 2031 joules. That's money in the bank, no?

Paul