> It appears that the voltage for one of the boards is 2.7 volts. You would
> need 5 of them hooked in series which would give you a little over 13
> volts. With capacitors the voltage goes down as they discharge. With
> that much capacitance they will not go down very fast.
Unfortunately, connecting them in series cuts the capacitance way down. 5
times 2.7F in series is only about 0.5F.
You also need some way to ensure that the voltage divides evenly, which is
usually done with resistors. Which would mean the caps will discharge
fairly quickly.
IIRC (and I probably don't), drawing 1 amp from 1 farad means the voltage
will drop 1 volt per second (1 coulomb of charge).
Definition of CCA is number of amps delivered for 30 seconds with more than
7.2 volts. So if we start with 12.6 volts, and draw 1 amp, a 1F capacitor
should hit 7.2 volts in about 5.4 seconds. (Ignoring the effects of ESR
etc). So it would take over 5F to equal even 1 CCA.
Randall
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