Hey all,
I've got a strange situation. I'm working on an ERTL Star Destroyer and have 9 LEDs in it. I originally planned on running them off of 3 C batteries stored inside the Destroyer. I made my circuit, tested everything and wired everything up. The only problem is that the 3 C Batteries won't fit inside the ship! At least not easily and safely. What I am wondering is if I can change the 3 C cells to one 9 volt and just add a resistor to the cathode to cut the 9 volts down to 4.5 (which is what all my LEDs are wired to accept). Or should I just change to 3 AA batteries and avoid any possible complications but have a shorter battery life? Please forgive me if this is a "slap you around for being so stupid" question.
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Bobby
I'm assuming youre displaying it in a way that you can't just run a wire down the stand and have the Batteries in the stand? That would seem like the best situation to me.... easier to change, and more stable weight wise....
That said, I cant see it being an issue to drop the level with an extra resistor...
Or you could get a wall-wart and drop that, then power it that way... no changing batts then Thats what I did on my last one, with the help of Zog-the-lighting-master (thanx Zog )
Using a resistor to cut down a voltage is not a good idea - the resistor will take current from the battery and waste it as heat... which means a shorter battery life.
You should first calculate how much current you'll be drawing, then match it with as many smaller batteries that'll provide an equivalent amount of current (without an extra resistor) for a reasonable amount of time. Remember, batteries are rated by voltage and by how much current they can deliver per hour. A 1.5v battery at 100mAh is not the same as a 1.5v battery at 250mAh.
...or just use a wall wart and don't worry about batteries.
Bobby, you can do that with one caveat: if one of your leds fails, the other ones will be getting more current than they should and thus will be brighter/fail sooner. And when another fails, the remaining will get even more current, thus failing even sooner, etc. Also, the 9-volt won't improve things much life-wise over 3 AA batteries, because the extra amp-hours of the extra cells will be used up as heat in the additional resistor.
I'd say use the 3 AA batteries; your circuit doesn't need to be modified and they should last a decent amount of time. If you're really concerned about battery life, you could use two 3-AA battery packs wired in parallel (ideally with a diode on each pack so they aren't directly connected together at both ends).
I've decided to go with 3 AA batteries. I got a 2AA holder and a 1AA holder. Do I solder both positive leads together and both negative leads together? That's what seems to make sense to me, but then my mind plays tricks on me when I think of how you connect batteries end to end.
http://www.starwarsmodels.com <-- I'm always looking for content contributors. Want to lend a hand? PM or E-Mail me!
bobbyfett wrote:I've decided to go with 3 AA batteries. I got a 2AA holder and a 1AA holder. Do I solder both positive leads together and both negative leads together? That's what seems to make sense to me, but then my mind plays tricks on me when I think of how you connect batteries end to end.
So the positive lead from one of the 2AA holder batteries becomes the 4.5V out. The negative lead from that battery goes to the positive lead of the battery next to it. The negative lead from this battery goes to the + of the single battery. The - from this battery becomes the ground for the 4.5 V.
With 3 AA batteries, the LEDs will go dim much quicker than with the 3 C batteries.
At least not easily and safely. What I am wondering is if I can change the 3 C cells to one 9 volt and just add a resistor to the cathode to cut the 9 volts down to 4.5 (which is what all my LEDs are wired to accept). Or should I just change to 3 AA batteries and avoid any possible complications but have a shorter battery life? Please forgive me if this is a "slap you around for being so stupid" question. 
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