I would like to create a simple circuit where a couple double aa's charge a capacitor then use the capacitor to briefly power a 3v led. I guess i need to know how much capacitance I need and also a basic idea of how you charge both side of the capacitor with batteries.
2007-06-14 10:11:00 · 4 answers · asked by Zak G 1 in Science & Mathematics ➔ Engineering
So long as the capacitor is rated for more than 3 volts (6.3 WVDC is common), simply connect the battery '+' terminal to the capacitor '+' terminal, and '-' to '-'. Leave it for at least a couple of seconds and the capacitor will be charged (unless it is one of those 'super' capacitors with a value of 1 Farad or more, then you need to leave the battery on for about 1 minute). It won't hurt the battery or the capacitor if you leave them connected indefinitely.
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2007-06-14 10:20:02 · answer #1 · answered by tlbs101 7 · 0⤊ 0⤋
you probably need more than 100 microF, (maybe 1000 microF) to light up a LED for more than a sec.
I'd recommend putting a resistor in series to limit the charging current since the capacitor w/o charge can act as a short circuit.
make sure that you use a capacitor having proper voltage rating
2007-06-14 12:18:00 · answer #2 · answered by jebin 2 · 0⤊ 0⤋
The LED will discharge your capacitor in probably less than a second. It's impractical; just use the batteries.
2007-06-14 10:27:12 · answer #3 · answered by Gene 7 · 0⤊ 0⤋
Connect positive to positive and negitive to nagetive.
C=100 micow F
2007-06-14 10:23:49 · answer #4 · answered by Anonymous · 0⤊ 0⤋