Running a LED on 6v?

[Chriszuma]

I know this isn't exactly related to FIRST, but I don't know of any other place to get the opinion of hundreds of electrical engineers.
I'm going to mod my calculator, and light up the screen with a blue LED and some fiber optic cables. I can get the fibers from a cheapo Ikea lamp of something, but I'm having trouble finding a suitable led; the calc runs off 4 AAAs (6v, in case you're anti-multiplication). I really don't want to pay $3.29 for Radioshaft's 6v-max-rated led, but I can't seem to find any other 6v ones (and i'm beginning to suspect that they are liars). What would you suggest? Possibly only wiring it to 3 of the batteries somehow, or putting a voltage regulator on there? Maybe wiring two leds in parallel (I'm en electronics newb, if you can't tell).

Thanks for your help.
~Chris Hammond

[jgannon]

Originally Posted by Chriszuma

I know this isn't exactly related to FIRST, but I don't know of any other place to get the opinion of hundreds of electrical engineers.
I'm going to mod my calculator, and light up the screen with a blue LED and some fiber optic cables. I can get the fibers from a cheapo Ikea lamp of something, but I'm having trouble finding a suitable led; the calc runs off 4 AAAs (6v, in case you're anti-multiplication). I really don't want to pay $3.29 for Radioshaft's 6v-max-rated led, but I can't seem to find any other 6v ones (and i'm beginning to suspect that they are liars). What would you suggest? Possibly only wiring it to 3 of the batteries somehow, or putting a voltage regulator on there? Maybe wiring two leds in parallel (I'm en electronics newb, if you can't tell).

Thanks for your help.
~Chris Hammond

Use Ohm's law to figure out what size resistor you should put in the circuit. I assume that the LED wants 10mA, and the drop over the LED is somewhere around 2.5V, leaving 3.5V for the resistor. Thus, you need a resistor that is 3.5/.01 = 350 ohms, which you would put in series with the LED. If you have actual ratings for the LED, I would suggest using those, but this should be a good enough approximation. Also, a word of advice from someone fairly involved in the calculator scene: be very careful when you're doing this. The inside of graphing calculators is very delicate, and while you're trying to remove the adhesive foam between the screen and the circuit board, you may very well rip off the ribbon cable that connects the two, and there is no easy way to repair this. Once you get past that stage, removing the reflective backing from the screen is also a pain, and you'll never quite be able to get all of the adhesive off of the screen, so it will always look smeary. I gave up, and ended up buying a backlit TI-86 from someone who knew what he was doing. Nonetheless, good luck with your project.

[Alan Anderson]

LEDs don't exactly run off voltage; they run off current. Being a diode, they have a characteristic voltage drop across them, so you do need to have at least that much voltage supplied. But the important part is a current-limiting resistor.

jgannon's analysis is fine. A blue LED at 2.5 volts and 10 milliamps sounds about right. Red or green LEDs usually have a much lower voltage drop, and often like much less current, though some are perfectly happy at up to 25 milliamps or more. If you can, you should look at the specific ratings for the LED you're using.

[Chriszuma]

Originally Posted by jgannon

Use Ohm's law to figure out what size resistor you should put in the circuit. I assume that the LED wants 10mA, and the drop over the LED is somewhere around 2.5V, leaving 3.5V for the resistor. Thus, you need a resistor that is 3.5/.01 = 350 ohms, which you would put in series with the LED. If you have actual ratings for the LED, I would suggest using those, but this should be a good enough approximation. Also, a word of advice from someone fairly involved in the calculator scene: be very careful when you're doing this. The inside of graphing calculators is very delicate, and while you're trying to remove the adhesive foam between the screen and the circuit board, you may very well rip off the ribbon cable that connects the two, and there is no easy way to repair this. Once you get past that stage, removing the reflective backing from the screen is also a pain, and you'll never quite be able to get all of the adhesive off of the screen, so it will always look smeary. I gave up, and ended up buying a backlit TI-86 from someone who knew what he was doing. Nonetheless, good luck with your project.

Okay, I understood around 80% of that, didn't quite extract what kind of resistor to use (please explain in newb-understandable terms). Also, when you say running it off 3.5v, do you mean altering the voltage with a regulator, or have I completely misconstrued what you were trying to tell me?

Pertaining to the delicacy of internal electronics: I have already cracked it open and examined it. I found a place to mount the led and run the fibers with a minimal of dremeling and wiring. Also, I plan on simply running the fibers up the back of the board and wrapping them around the front of the screen to get a nice front-lighting effect, since backlighting would require a lot more involved and breakage-likely operation.

[Kevin Sevcik]

More information than you ever wanted on LEDS and their usage:
Dan's Data LED Caselight article

[Chriszuma]

Originally Posted by Kevin Sevcik

More information than you ever wanted on LEDS and their usage:
Dan's Data LED Caselight article

thank you, that site was incredibly informative. I worked this out:
i can solve the voltage difference between my calculator batteries and the led rating by putting two in series, but then if i have too much current flowing over them i'll get thermal runaway (because their resistance goes down with temperature increase) and they'll burn up, since i can't fit a resistor in there without lowering the voltage too much.
alternately, i can use 1 led and a resistor (which i calculated to be 96 ohms, or the closest standard size)

The single-led solution seems best, but i don't know if one will be enough to light up the screen very well. Would i have a significant decrease in light if i ran them at lower voltage?

[Kevin Sevcik]

I'd go with the single LED, but I think you should be able to fit a resistor in there for the double LED.....

[Chriszuma]

will it affect the brightness if its run at, say, 2.5 volts instead of 3.6?

[Kevin Sevcik]

it's always running at 3.6V. The voltage drop across an LED is a fairly fixed thing. The resistor affects how much current will be passing through the LED. Basically, if it's a 3.6V LED, that means the resistor will HAVE to have 6V-3.6V of voltage across it. So the resistor has 2.4V across it. Then you just use Ohm's law to set the resistance of the resistor to get the desire current through the circuit.

Now, the current is what actually affects the brightness of the LED. there's probably a graph of the LED's brightness vs. current somewhere if you look hard enough.

[Chriszuma]

i already did the ohm's law thing. But if i were to use two leds in series to double the voltage drop to 6 volts, then i wouldnt need a resistor for voltage, but the current would still be too high.

You know what, i just realized that if i wired them in parallel I would just need a resistor to drop to 3.6v and 50mA. duh. I guess that's what the stress of winning a regional does to you.

[Alan Anderson]

Originally Posted by Chriszuma

...You know what, i just realized that if i wired them in parallel I would just need a resistor to drop to 3.6v and 50mA...

Putting diodes in parallel isn't usually a good idea. If one of them has a slightly lower voltage drop than the other, it will carry the lion's share of the current. If you want the LEDs' brightness to match, give each its own current limiting resistor (or put them in series).

[Chriszuma]

to give each its own resistor, would i split the power supply wire, then put a 100-ohm resistor before each led, or would that not have the desired result?