🚀 go-pugleaf

The relations needed here are fairly simple.  Capacitance may be defined
by the equation

             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%hpfcla@hplabs.hp.com               |  sentient life-form on this planet.