Quote:
|
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.