Will an 8-hole AA battery pack work with only six batteries?

[jkiker]

Hi all,

So I think I have settled on using a 9 volt battery pack as well as a 9 volt regulated DC wall wort for the alternative power sources for my lighted Y-wing project. I want to use rechargeable AA batteries rather than the rectangular 9 volt battery, since I believe the AA batteries will give a longer run time. I have also picked 9 instead of 12 volts as I think it will mean less heat from the resistor'd LED's.

So in looking at battery packs, Radio Shack and others have a 4 hole and an 8 hole AA battery holder. At 1/5 volts, I will need six of the AA batteries to yield 9 volts. Can I load the 8 hole battery pack with just six batteries and still have it produce the desired amount of power, or does it need to be full? And have I missed seeing a battery pack for six AA batteries?

Thanks, Jim

[Ziz]

Nope. All battery holders are wired in series - the positive of each battery connects to the negative of the one next to it so that it all adds up by the time you get to the other end. If you break the chain - no power at all.

You need to either find a six spot AA holder or build one out of some combination of 1, 2 and/or 4 spot holders. The only other choice would be to determine where either the first or last two batteries are in the loop and re-wire the holder to bypass them.

[Joseph Osborn]

Here's a 6x AA battery holder

Also note that rechargeable AA batteries usually are rated at something like 1.2 to 1.3 volts, not the regular 1.5 that you get from good alkaline batteries.

[jkiker]

Hi Joseph,

Thanks for the "point" to the battery holder, but especially for the note about rechargeable batteries! I'd never have guessed that. So, what happens if I use 9 volts to figure out the required resistors for my LED's, and then give them 7.2 volts instead of 9? Do they simply get less bright? Would that damage anything, or cause the lights to last a shorter amount of time?

Thanks, Jim

[Ant]

As an aside, if you're using just LEDs why are you going all the way up to 9v only to have to drop it back down to the LED voltage?

4.5v should be fine, maybe 6v tops if you have some exotic white or blue devices with a high forward voltage. If you're looking for longevity, move up from AA to C cells - they will run for months!

[jkiker]

Hi Ant,

The reason I had settled on 9V is because I am a total noob, which is illustrated by the several general and specific questions I have been posting lately. I will be using two bright white LED's for the engines rated at around 3.2V, and two smaller white LED's (less bright, but still likely around 3V since they are white). No timing, no blinking, no sound, no card; just an appropriate resistor for each LED, with both battery and wall wart power running through a three-way switch. So six volts is plenty? I plan on wiring the LED/resistors in parallel to the power lines, and run the power down the model's stand to where all the power stuff will be located.

So to confirm, I could use 6 volts, with the applicable resistors, and still be fine?

TIA, Jim

[naoto]

You need to either find a six spot AA holder or build one out of some combination of 1, 2 and/or 4 spot holders. The only other choice would be to determine where either the first or last two batteries are in the loop and re-wire the holder to bypass them.

A kludgy solution is to create a "cheater" -- basically a cylinder the same size as a battery, that has a wire between the + and - ends -- essentially a short that you put in place of the missing cell.

[Ziz]

Hi Ant,

The reason I had settled on 9V is because I am a total noob, which is illustrated by the several general and specific questions I have been posting lately. I will be using two bright white LED's for the engines rated at around 3.2V, and two smaller white LED's (less bright, but still likely around 3V since they are white). No timing, no blinking, no sound, no card; just an appropriate resistor for each LED, with both battery and wall wart power running through a three-way switch. So six volts is plenty? I plan on wiring the LED/resistors in parallel to the power lines, and run the power down the model's stand to where all the power stuff will be located.

So to confirm, I could use 6 volts, with the applicable resistors, and still be fine?

TIA, Jim

OK, crash course in power flow. Ready?

All power sources and devices that use them have two ratings/power needs - voltage and amperage. In layman's terms, voltage is how much power is going to the device, amperage is how fast it's getting there.

Depending whether you run your circuits in series or parallel changes the power the circuit needs to run. Likewise, connecting batteries in series or parallel changes how much power you're supplying to your circuit.

Series is when you connect positive to negative to positive to negative in one continuous chain. Break the chain anywhere along the length and the whole thing fails. In series, more voltage is needed while amperage required remains consistent.

Parallel is when you connect all the positives together and all the negatives together so that each device effectively has its own connection to the power source. In parallel, voltage remains consistent while more amperage is needed.

Batteries work the same way. Six AA batteries in series adds up to 9v but will only give the average amperage rating of a single AA battery. Six batteries in parallel only gives 1.5v of a single battery but more total amperage.

This is why you have to design your circuits carefully and make sure you have the right power supply - in both voltage and amperage - to handle the needs of the circuits. Too much of one and not enough of the other will either burn out your circuit or fail to get it operating in the first place.

Don't just pick an arbitrary voltage because it's convenient. Do the math and figure out what you truly need.

[jkiker]

Hi Ziz,

Well, this is easier if one already knows what to do first. I am not trying to arbitrarily pick a voltage, but I have been guessing based on what I have read; as I study more, my sense of the right amount of power has dropped. I thought I needed 12 volts at one time because I was adding up the voltage required, which was wrong for parallel wiring.

So let me imagine I just found a 3mm clear white LED for my two cockpit lights. One I found just this afternoon is 1100mcd, 3.5 volts and 20mA, for two bulbs (two seats, instrument panels plus some indicator lights on the consoles). The engine lights I have, two 5mm LED's making 7000mcd, 3.3- 3.6 Volts and 25mA, also for two bulbs. The resistors will be wired in series to each LED. If I connect each light/resistor pair in parallel as I have seen done in several "how to" articles, then theoretically I would only need 3.6- call it 4- volts of power, and I have a total amperage of 90mA, correct?

Therefore, a 6 volt power supply is more than enough. For batteries I could use AA's or C's, in a battery holder for 4 batteries. For the wall wart, a 6 volt regulated DC adapter, rated at .5A would be fine, correct?

For resistors, the engine LED's need 100-120 ohms and I think a 1/4 watt resistor would do for each engine. For the cockpit lights, the resistor needs to be 150 ohms, and again 1/4 watt should do fine.

Have I done the math correctly, and what have I missed?

So, can I de-rate the brightness of the two cockpit lights further (and should I?), given that I plan on using each cockpit LED for some direct light behind the instrument panels, plus using some fiber optics for the console lights, likely a about 12 strands per LED? And to do that, could I use a pair of 27 ohm resistors to knock the voltage down to 4 volts?

I have to say, I've spent some hours searching for the "right" lights, and the white LED's do not seem to come in relatively low brightness unless I want to down into the 20-50mcd range. That seems far too low to me, but then, I'm still on a steep learning curve!

Thanks in advance for all your inputs and for sharing your knowledge!

Thanks and cheers, Jim

[Ziz]

Basically you've got it correct, just don't round off. Work it out with the decimal places because that might be the difference in getting, say, a 270 ohm resistor vs a 330 (just picking random numbers for example sake)

.5A = 500mA so that's more than strong enough so you're good there.

Either way, wire up your circuits before you put them in the model to make sure not only that you've got the connections right but that all your other issues are good - power supply not getting too hot, LEDs are correct brightness, etc. Solve every problem you can think of while you're still outside the model.

[jkiker]

Hi Ziz,

Not to worry, I didn't round off when doing the calculations; I just wrote that because I do not think one would go looking for/find a 3.6 volt power source!

Thanks again, Jim

[en'til Zog]

Hi, again

3 x 1.2 V = 3.6 V, if you use rechargable batteries.

Just say'in is all.

[jkiker]

Hi,

And I'm hearin', too!

Thanks, Jim

[Mr. Engineer]

Just in case you're wondering, naoto's 'cheater' is a dummy battery which you can get in most shops, I think.

But the guys are right. You do not need a 9volt. In layman's terms, a lot of 'extra' unwanted voltages would be dissipated by that current limiting resistor to the LEDs. For me, 4.5v to 6v are the best and this is what I am going to use for my future designs. (Apart from the 1/350 Enterprise modules, that is *ahem*)

[MillenniumFalsehood]

Something I wanted to mention, in case you weren't aware.

MCD is millicandles, and is the measurement of the amount of light a single LED produces. Unless you're looking for specific brightness requirements, like looking for a very bright light for a photon torpedo circuit or a hull spotlight, then you can safely ignore this specification.