I wish to backlight my TI-83+SE.
What all do I need?
It will switch on\off with a tiny switch in the casing that I have.

There are several possible ways to do this, most of them quite invasive to your calculator, all of them risking its health. It's good that you have a TI-83+SE, though; that TI-83 series is much easier to backlight than the TI-84+ series. Are you interested in an LED backlight or an EL backlight? The former is much better understood and easier to find parts for, while the latter is more effective but more complex.

I read somewhere on how to do this, but I forget where. I'll look it up later.

The way you do it is as follows: First, take apart your case. Get some plexiglass, and a LED. Put the plexiglass behind your screen and the LED below it. The LED's light will diffuse through the plexiglass.

Anyway, don't do this based on what I just told you, wait until I remember the site.

That's more or less correct; it sounds like you read the RichFiles page. That's the backlighting method used in the Ultimate Calculators 1 and 2, to varying success. The full steps are much more finicky though, so if you decide to go with the LED method, I'll detail them.

Thanks Kerm, that's the site I was thinking of. Here's a link to the page.

I have been to RichFiles, but it wasn't specific enough. I think what I need would be a piece of plexiglass, a resistor, a switch, an unknown number of LEDs, and more skill than I have . I'm afraid I'll need VERY specific.
Especially considering that tutorial didn't even tell how to remove the screen!

So, what is an EL backlight?

EL stands for "electroluminescent", and is the sort of backlight that is often used in items like digital watches (it produces a distinctive blue/green glow). They're not as easy as LEDs (requiring a high voltage power supply) though do produce a much nicer-looking result. Michael Vincent has some information about adding an EL backlight to a TI-86 on his website.

@Ben: Thanks!

The EL looks very nice, But FRIGHTFULLY expensive?

Edit by Tanner: Fixed the double post
Edit by me: What is that?

One of the main problems that I have seen with EL mods is that the pins from the panel may stick out the side of the calculator. Caleb, I have very specific instructions for taking the LCD apart, but the main problem is that you have to separate the LCD from its ballast, a think metal plate on the back of the LCD. Another option is wrapping the edge of the LCD in EL wire, which I have a long PDF about how to do from a community member.

Kerm, belive me. I have SEPARATED the ballast from the calculator.
But as I said in another thread, in the process I think I SEPERATED my calc from reality. However, I still plan on backlighting it - then try fixing it later.

You've probably separated the LCD module from the mainboard. Are you sure you separated the ballast and silvery reflective layer from the back of the LCD glass?

No, I was too scared to remove the reflective layer. The ballast is off - that was easy with GooGone.

Ah yes, I think you might have even mention that before. The reflective layer is pretty easy; it's basically just silvered paper and more adhesive.

Well, sure. I can probably do that. Anyway, I must leave very shortly, but first I would like to know: How many LEDs would that be? What color and type would they be? What number and type would the resistor be?

What's the "ballast"?

As bright as possible while being as low-current and low-heat as possible. 25mcd indicator LEDs wouldn't work; at least 1000+ mcd bright LEDs would be suitable. 330ohm is usually a safe value for a wide variety of LEDs.

Is that for the resistor?

Yes. LEDs have a fixed voltage drop and will have a maximum rated current (the higher the current, the brighter the LED - until it pops). If you had an LED with a 2V drop and wished to put 10mA through it in a circuit with a 5V power supply your resistor would need to have a 3V voltage drop across it (voltage drops add together in a series circuit, so 2V for the LED and 3V for the resistor comes to 5V to match your power supply) and would need to have 10mA flowing through it (current is the same through all components in a series circuit). Ohm's law tells us V=IR (where V=voltage, I=current, R=resistance) which can be rearranged to R=V/I. For our resistor V=3V and I=10mA which gives us R=300Ω. Resistors are commonly available in certain particular values and 330Ω is the next common value up from 300Ω (a higher resistance will result in less current flow - we're erring on the side of caution).

Caleb, indeed, as Benryves said, I was referring to the resistor for the LED. LEDs are classified by color (often in wavelength), intensity (mcd, or milli-candlepower, where 1000mcd = 1 candlepower), and forward current at nominal voltage.