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gregl
05-15-2010, 09:41 PM
OK, folks, this is really on topic because Iím using my lathe to make the thing that needs this. :) But when it comes to electronics, Iím totally helpless. Cams, gears, levers; I can see those. Electrons are invisible and just not part of my brainís capabilities.

So hereís the situation: I need an indicator light made from a blue LED and run off batteries. I need the light to last for about 40 hours on a battery or set of batteries. Because of the size of the project, battery size is limited to two AA or one 9V, (or smaller-- button cells would work if I can get the hours out of them). The blue LEDs seem to require about 3.5 to 4 volts.

Question 1: Intensity of the light is not a big issue; this is for dusk and night use so the light does not need to be bright enough to be seen in bright daylight. Can I just run a blue LED off two AA alkaline batteries without a resistor? (I know Iíd need one with a 9v. battery.)

Question 2: When I look up AA and 9v. batteries on Wikipedia, I get 3000 mAh for a single AA alkaline, yet only 565 for a 9v. Is this right?

Question 3: If the 3000 mAh for an alkaline AA is correct, am I right in thinking that two of them gives me 6000 mAh? And if the LED draws 20 mAh, Iíd have 300 hours worth of light?

Question 4: And if those numbers are wrong, how many hours would a 20 mAh LED run on two AA batteries (first choice) vs. a 9V battery (second choice)?

Bonus question: I have a little keychain LED light powered by four LR 41 batteries that could be used for the project. At 20 mAh for each battery, does this give me 80 mAh and therefore four hours of light?

THANKS for the help, guys!

J Tiers
05-15-2010, 10:05 PM
Ampere hours are unchanged if batteries are put in SERIES.

Ampere hours ADD UP if batteries are in PARALLEL

"Series" means from common you go say to the "-" of battery #1, connect the "+" of battery 1 to the "-" of battery 2, and the "+" of battery 2 to the load. For you that "load" is the resistor and LED. Series also gives a total voltage equal to 2x one battery for two in series.

"parallel" means you connect the "-" of both batteries together, and the "+" of both together. Normally I don't do that with 'dry cells", but it is common with lead-acid rechargeables. parallel gives the same voltage as one battery, it is just like having one "cell" of a larger size

The 9V battery has a bunch of small thin 1.5V cells in series, and each one has some wasted space around it for insulation, etc. Volume of active material is lower, and I am not surprised by it having less capacity. That's a lot less, are both the same type (carbon-zinc or alkaline)?

Now, the time of run is affected by voltage...... So if the AH is figured with 1.5V start, and a 1.0V final voltage, a series connection starts at 3V, and ends up at 2V. If that is ending at a voltage that won't run your device, you do NOT get the total AH capacity.

AND ampere hours is affected by the drain..... it is specified at a particular current. if you draw LESS you get more AH. if you draw MORE, you will not get to the AH that is quoted. That is due to resistance-based losses in the battery.

For your bonus question...... in the simplest case, where the LED works equally at any voltage (it won't) ...... 0.02AH per cell and 4 cells in series gives 0.02 AH for the series combination. But you get more total power........

Evan
05-15-2010, 10:09 PM
Must it be blue? With red you can get by with just one battery.

I just tested a common ultrabright 5mm blue led. With a 1000 ohm resistor it draws 2 ma on 3 volts and is plenty bright enough to be seen under very bright indoor lighting. At 2.8 volts it draws 1 ma and is still very visible. At 2.65 volts it is on the edge of extinction but is still visible drawing about 300 microamps.

If you use two fresh AA cells with a 1K resistor you can expect at least a couple of thousand hours operation.

gregl
05-15-2010, 10:31 PM
Thanks, J, for the explanation re. the 9v. vs. the AA. Makes sense now. To my brain, the 9v. looks larger and has more volts, so my logic says it should have more electrons inside! After all, a quart of beer has more alcohol than a pint!

And thanks, also, Evan. Yes, the light must be blue. Your answer is just what I need to know. Two AA with a 1k resistor at 1000 hours is quite good enough. What wattage of resistor?

And for the bonus question: I left the keychain light on for an hour and a half just now and at the end of that time the light was almost invisible. However, the batteries, albeit new, are the cheapest ones available and may not be up to snuff, but the failure is enough to show that this won't work for me.

Thanks again.

Evan
05-15-2010, 10:35 PM
Any resistor will do. At 3 volts and 2 milliamps it is using only 6 milliwatts.

gregl
05-15-2010, 10:41 PM
Any resistor will do. At 3 volts and 2 milliamps it is using only 6 milliwatts.


Thanks, again, Evan! Off to Radio Shack!

dp
05-15-2010, 10:41 PM
You can extend the period if the lamp is allowed to blink. Blinking circuitry is pretty cheap and does not appreciably up the current draw. Or you can just buy a blinking LED: http://www.lc-led.com/View/itemNumber/142

darryl
05-15-2010, 10:52 PM
The freshest alkaline battery has shown an open circuit voltage of 1.59 volts for me at least. If two of these are put in series you have 3.18 volts, which I don't think is capable of putting 20 ma through a blue led. You could almost leave out the resistor, with the effect that the led would be quite bright to begin with, but would continue to deliver more light for some time as the battery voltage dropped with use. Using a series resistor, the initial brightness would be less, but then so would the brightness after time passes. At some time as the batteries deplete, you could bypass the resistor and restore some of that 'lost' brightness.

Maybe not an issue in this case, but probably worth mentioning. As Evan suggested, the voltage drop in the batteries is going to result in the led going out at some point. This will be before the full charge has been used. If you powered the led from a 9 v alkaline, you could adjust the current draw with a simple circuit, and the led would remain at the same brightness pretty much for the whole charge available from the battery. At 2 ma current drain, you'd probably get 100 hrs of led life, during which time it wouldn't drop in brightness at all. I'm using a figure of 200 ma/hr for the 9v battery for that calculation, but I don't know if that's still ballpark or if 9v batteries have more or less capacity these days.

The circuit is dead simple- one three lead TO-92 regulator ic, and one resistor, any wattage. 1/4 watt will do fine because it's big enough you can still see it- :)

Evan
05-16-2010, 12:35 AM
Leaving out the resistor will kill the batteries quickly. It will also probably kill the LED since alkalines have a low internal resistance. I'm pretty sure it will exceed the rated current on a new set of AAs. Note the conditions of use. An ultrabright blue LED at 20 ma will be eye blinding bright at night.

There isn't any reason to make it more complicated and a 9 volt battey is a very expensive option per ma hours.