Joule Thief

[freepow]

@  Gadgetmail

I thought of something...

1x AA              = 2500 mA for  1 Hours use

1x Cap @ 650F =  approx  3000 Amps.

Does'nt that mean 3000 Amps for 1 sec of use
and 3000/60/60 = .833 Amps for  1 Hour of use

Which means  1x AA  has 2500 mA for 1 Hour
which means  1x cap has  833 mA  for 1 Hour

?

[dasimpson]

it could be m8 but my maths and understanding ant upto working that out i think the true way to know is my putting loads on them running them side by side to compare the battery to the cap to get best indication to how long they will last that way we will start to work out a mah rating for them ethern if it is just here it is ideal information to know

[dasimpson]

well thats my jt now running below 25ma
line filter out of a psu
bc337-25
5k resisters
4 super bright leds in parallel from a one of them sticky back lights that go under the worktops

now the longivity test if im right it should last for about 100 hours give or take

im not expecting it to run for 100 hours as to do that at 25ma draw it would have to run the battery to 0

only problem is there is a whining noise hopefully when the 2n3904 arrive they may be in a diffent hurts range that or i just have good hearing

[TinselKoala]

@  Gadgetmail

I thought of something...

1x AA              = 2500 mA for  1 Hours use

1x Cap @ 650F =  approx  3000 Amps.

Does'nt that mean 3000 Amps for 1 sec of use
and 3000/60/60 = .833 Amps for  1 Hour of use

Which means  1x AA  has 2500 mA for 1 Hour
which means  1x cap has  833 mA  for 1 Hour

?

Why the puzzlement? ENERGY is the important parameter for longevity, not "Amps" or mA-H.

The energy stored in a capacitor in Joules is given by (C * V * V)/2, where C is capacity in Farads and V is the voltage it's charged to. NOT the rated voltage of the cap, but the charge that is ACTUALLY on the capacitor.

The energy in a battery, if the battery's capacity is given in Amp-hours, is
(V * Capacity) = Watt-hours
Watt-hours * 60 * 60 = Watt-seconds or Joules.

So a 1.3 volt rechargeable battery of 2500 mA-H capacity has
1.3 * 2.5 * 60 * 60 = about 12000 Joules, assuming a relatively flat discharge curve.

And a 650 Farad supercapacitor charged to that same 1.3 volts has
650 * 1.3 * 1.3 * 0.5 = about 550 Joules.

The capacitor charged to the same voltage as the battery contains less than 1/20 the energy of the battery.

The capacitor has much lower internal resistance than the battery and so can deliver its ENERGY much faster: that is, in a pulse that has greater amperage. In fact the cap can deliver essentially ALL the stored energy in one very quick pulse. The battery, with an internal resistance of several tenths of an ohm, can't come close to the current delivery capability of a capacitor. But it takes a BIG capacitor to store anything like the total energy in a decent battery at the same voltage.



Note well: you get more "bang for your buck" by increasing the voltage on the capacitor. The energy on the cap goes up as the square of the voltage but only linearly with capacitance. Double the voltage, you get 4 times the energy, at a given capacitance. Double the capacitance and you only double the potential energy storage at a given voltage.

For your amusement:
http://www.youtube.com/watch?v=cj5T0zRALKc

[Pirate88179]

TK:

Thank you very much for that very detailed explanation.  As most on here know, I love the caps but I also somehow knew that an AA had more "power" But, as you have stated so eloquently, it all depends upon what you are after.  I have seen my supercaps charge up "right away" and run for a long time where a battery takes quite a while to do so.

Anyway, thanks again for keeping us based in real numbers and science.

Bill