I want to add 2 red LEDs to a mask for a costume, but have almost no electrical skills beyond normal household maintenance. Advice from anyone who does know about electricital circuits would be appreciated.
What size battery should I use? Are there different voltages of LEDs available (from Radio Shack or similar store, not specialist dealers) that would affect my setup? Should the LEDs be hooked up in parallel or series? Do I need any sort of resistor in the circuit? Is there any sort of pre-made setup for this sort of thing available?
Well, I’m no expert, but I have experimented a bit with homemade bicycle lights using LEDs. There are tons of variables involved. Your best bet might be taking a prefab cheap bicycle taillight or a similar LED setup – perhaps a flashing tie or some such – and soldering willy-nilly lengths of wire (which will add resistance) to the leads of the LEDs. You’ll have the PCB, the power supply, and all that prefab for you, and it may be that adding a small length of wire won’t cause any problems. Good soldering skills would be a must, I’d think. I’ll be interested to know what the real gurus have to say about this.
However, it’s pretty simple. Red LEDs normally have a forward voltage drop of about 2 volts–check the back of the package for the specific figure, but in general figuring on 2 V is safe. Typically, the maximum forward current is around 20 mA, so use ~10 mA for safety. To calculate the resistor value you need to use for each LED, simply subtract the the forward voltage drop from the supply voltage and divide by the current (10 mA or .010 A). If you are using a 9V battery, for example, you’d take (9 V - 2 V) / .010 which gives you a value of 700 Ohms. The closest standard value would be 680 Ohms. To check if using this value will keep the current below the 20 mA max, just reverse your math: 7 V / 680 Ohms = .0109 A or 10.9 mA. You’re good to go.
Oh, and to answer the rest of your question, the LED’s should be wired in parallel with a resistor in series with each. And if you want to simplify things somewhat, Radio Shack also does sell panel-mound 12 V red LEDs which already have the correct current-limiting resistor incorporated. You need only supply the 12 VDC. These are a bit larger than discreet LEDs, though, so if space is a consideration, the method I’ve outlined will guide you.
Most LEDs will have a forward voltage drop of between 1.5V to 2V when biased at a moderate 10mA. Using 2 in series will drop 3V to 4V, so you’ll need a battery with a higher voltage than that. A PP3 (little 9V battery) will fit into a Halloween costume nicely and contain enough juice to power the LEDs for an evening. Don’t plug the battery in backwards, or the LEDs will pop.
Use a resistor in series to set the current, e.g.: If the 2 LEDs drop 3.5V in total, and you want to get 10mA from a 9V battery, then the voltage dropped across the resistor will be 9V-3.5V=5.5V. As R=V/I, in this case R=5.5/10mA = 550 ohms. A bit of an awkward value to obtain, so use a 470 ohm (gives 11.7mA) or 620 ohms (gives 8.9mA). Precision doesn’t matter too much in this application.
So:
Choose your LEDs, and find out their forward voltage.
Choose your battery voltage.
Choose a current to bias the LED at.
You can calculate the resistor value from everything else you’ve chosen.
Alternatively, there are LEDs that are designed to run on 12V, or a wider range of voltage sources. They contain a little circuit that limits the current efficiently, and so are more expensive. Car accessory stores are a good source of these, but don’t pay too much.
As mentioned by Fridgemagnet, putting the LEDs in series is more efficient than a parallel arrangement. One idea (also mentioned by Fridgemagnet) is to use a 9 V battery and put two LEDs and a resistor in series. It should be kept in mind, however, that while a series arrangement is efficient, there might be a difference between the brightness of the bulbs, as no two LEDs are perfectly matched.
Another idea is to use two AA or AAA batteries in series, and two LEDs in series. The nice thing about this arrangement is that you don’t have to use a resistor, which will make it more efficient. The drawback is that there will only be about 1.5 V across each LED, which means they may not be as bright as you desire.
If you want to use a 9 V battery, can I make a suggestion? Instead of using two LEDs and a resistor in series, why not replace the resistor with three LEDs (i.e. use 5 LEDs and a 9 V battery in series)? In other words, instead of wasting heat in a resistor, why not make some more light?
A note to the other EEs: Yea yea, I know that using a series resistor improves the current regulation in the circuit. I also know that not using a series resistor (e.g. using 5 LEDs with a 9 V battery, or 2 LEDs with two AAs) means that the LEDs will be (more-or-less) voltage controlled, and not current controlled, and that this is usually not a desirable thing. But you’re missing the point: 1) This is not a commercial device, where reliability is critically important. 2) I am assuming that, for this costume, efficiency is more important than reliability. 3) Using 5 LEDs with a 9 V battery, or 2 LEDs with two AAs, will probably work. LEDs are pretty well matched nowadays, especially if they come from the same lot.
Because it would make the current wildly uncontrollable. Seriously, you just wouldn’t do it that way, even for a one-off.
I guess the OP spec is 2 LEDs because they’re going to be spooky glowy eyes. jnglemassiv’s link is a good 'un, and contains everything one needs to know for this application.
True, a series resistor will waste power, but the only efficient alternative is some sort of switching current source, like the little circuit that drives bicycle lamp LEDs. Doing the maths, a 10mA drain on a (cheap) 9V battery with a 150mAh spec will light the LEDs for 15 hours before fading. Plenty long enough for a party.