I'm somehwhat embarassed to say that I've forgotten too much of my college
electronics courses to have a good "feel" for how much juce you can store
in one of these capacitors and how long you can discharge it at a fixed rate
before it's all gone.
I would appreciate it if someone out there could help me get a feel for this,
possiblly by equating one of these supercaps to simple battery terms. Ie:
how many miliamp-hours of capacity one of these supercaps is equivalent to.
My application is a farely simple one...I want to run a blinker-LED off of
a supercap. I was wondering if a 1 farad supercap would have enough juce to
keep a blinker-LED (I'm talking about the kind with the blinker built-in, no
external circuitry) going for 8 hours or so.
If you've got the answer...please write! Thanks!
Greg Corson
19141 Summers Drive
South Bend, IN 46637
(219) 277-5306
{pur-ee,rutgers,uunet}!iuvax!ndmath!milo
Q = CV or C = Q/V
where Q is the charge in coulombs, V is the potential across the capacitor
plates in volts, and C is the capacitance in farads. Remembering that
current is defined in terms of the amount of charge "flowing" past the
point in question over a given period of time: one ampere is one coulomb
per second.
Now, to get a feel for what kind of "battery" a 1F, 5V cap is, let's make
the following assumptions: first, the "LED blinker" requires at least 4.5
volts, and consumes a constant 25 mA regardless of the voltage (a
simplification, but probably not too bad for the case of 5V going to 4.5V).
We'll also assume an ideal capacitor. Now, the question is how long a 1F
capacitor, initially charged to 5V, can supply 25 mA before dropping below
4.5V. We know the capacitance and the change in voltage, so we can calculate
the total change in stored charge as:
1F * 0.5V = 0.5C
One-half coulomb requires 20 seconds to be moved from here to there at
25 mA (0.025A x 20 sec. = 0.5C); therefore, the 1F cap keeps this circuit
going for 20 seconds before the voltage drops below the limit of 4.5V.
You can try to run these same calculations for various voltages, currents,
and capacitances, but the basic answer is that even these "big" capacitors
aren't suitable as battery replacements in relatively high-current
applications. They're much more attractive as supplies for low-current
circuits, such as CMOS RAMs, where the drain can be measured in microamps.
(The 25 mA load used above may be high for this circuit, but you get the idea.)
Hope this helps.
Bob Myers KC0EW HP Graphics Tech. Div.| Opinions expressed here are not
Ft. Collins, Colorado | those of my employer or any other
myers%hpf...@hplabs.hp.com | sentient life-form on this planet.
You can't get enough power out of a supercap to light an LED. As has been
noted before in an earlier thread, the internal resistance is too high. CMOS
RAMs and clocks need only microamps to keep them going. That's why these
things have found a market.
--
David Chapman
{known world}!decwrl!vlsisj!fndry!davidc
vlsisj!fndry!dav...@decwrl.dec.com
In the Linear Databook Vol. 3 from National Semiconductor is a circuit using
a 3909 LED flasher that draws 0.32 mA and continues to run down to 1.15 volts.
You'd probably get a lot more life out of your 1F "battery" using this circuit.
--
Bill McFadden Tektronix, Inc. P.O. Box 500 MS 58-639 Beaverton, OR 97077
UUCP: bi...@videovax.Tek.com, {hplabs,uw-beaver,decvax}!tektronix!videovax!bill
GTE: (503) 627-6920 "The biggest difference between developing a missle
component and a toy is the 'cost constraint.'" -- John Anderson, Engineer, TI