Not practical at all
Message #6 Posted by Garth Wilson on 28 Aug 2005, 8:19 p.m.,
in response to message #5 by don wallace
To get as much use between charges as you'd get with one set of alkaline N cells, you'd need somewhere around 1kF-- yes, a thousand farads. That't assuming you used the energy before the capacitor's own leakage discharged it. The supercaps do hold a charge for a long time, but I don't know about years like alkaline batteries manage.
Farads=Amps x seconds / voltage_change
The alkaline N cell gives about a 500mA-hour (not amp-seconds). That's about 1800 amp-seconds, to get from 1.55V down to about 1.2V where the HP-41 low-battery indicator comes on. That's per battery, so multiply by four, assuming the capacitor can handle 6V so you don't have to put them in series. So in the equation above, the "voltage_change" term becomes 4(1.55-1.2), or about 4*.35, or 1.4. The result is almost 1,300 farads-- not microfarads, or even millifarads, but farads. Hardly practical.
To make it worse, suppose the capacitor can't handle the whole six volts. If it can handle three, you'll need two in series, each one of double the capacitance. If you need four in series, each one needs four times the capacitance, or about 5kF. If you say you're only going to charge the set up to 5V instead of 6V, now the "voltage_change" term above becomes smaller, and the required capacitance value further increases-- dramatically!
Edited: 28 Aug 2005, 8:26 p.m.
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