Is There A Lighting Beginner Guide?

By BeerGeekGamer, in X-Wing Painting and Modification

Hello,

Pretty much what the title says, is there any kind of beginner guide for adding lighting to your models. I have a couple of models that I want to add lighting, but I don't really know anything when it comes to electronics. I have seen people use LEDs and EL Wire/Tape, but have no idea what the power requirements are, and how many LEDs or EL Wire you can run off of one power source and etc.

Chemist says: Ask an electrician.

That aside, it's not overly complicated if you compare it to quantum mechanics. First each color, size, and shape of LED can have a different operational voltage range (for most LED manufacturers the ranges do not differ by size or shape, only color but I have run into a few oddballs). For instance the rectangular diode 5mm blue LEDs I used in my Gozanti Cruiser (post #4 of my repaint & lighting link in my signature) have an operational voltage range of 2.8-3.0V. Now, in order to build a properly powered circuit, you first need to figure out if you're going to set your circuit up in parallel or in series. Parallel circuits have all the negative and positive ends of the diode connected to their like-charged counterparts. You can power a lot more LEDs off of a lot less voltage this way as opposed to in series. LEDs in series are set up so the battery connects to the anode (negative bit of metal) of your LED and then you wire the cathode (positive bit of metal) to the anode of the next LED and so on until your final LED's terminal cathode connects to the battery again. This wastes a lot more voltage but also makes math super easy. Three Blue LEDs, as used in the Gozanti, set up in series will take the sum of their operational voltage range to power (so 2.8-3.0V times 3 = 8.4-9V). In this manner no resistor is needed to "cut back" the voltage across each diode so that you don't blow out the LED. Thus my Gozanti runs on one 9V battery connected, in series, to three blue LEDs.

If you Google "circuit calculator" and you don't do anything fancy (like use multiple colors of vastly different operational voltage ranges) you should be able to figure out what you need based on the diagram and math it does for you.

Good luck!

[Did this help at all?]

So I was thinking of making my Decimator look like this, expect for the blue for the two ports (http://starwars.wikia.com/wiki/VT-49_Decimator?file=SWC_VT-49.png). I was also going to add red for the engines, so a total of 9 red LED's, that I was thinking of getting from here (https://www.adafruit.com/products/1755). Going off of an online calculator the wiring diagram would be the following, down below, with a 3V source. What do you guys think?

Solution 0: 1 x 9 array uses 9 LEDs exactly
+----|>|---/\/\/----+ R = 56 ohms
+----|>|---/\/\/----+ R = 56 ohms
+----|>|---/\/\/----+ R = 56 ohms
+----|>|---/\/\/----+ R = 56 ohms
+----|>|---/\/\/----+ R = 56 ohms
+----|>|---/\/\/----+ R = 56 ohms
+----|>|---/\/\/----+ R = 56 ohms
+----|>|---/\/\/----+ R = 56 ohms
+----|>|---/\/\/----+ R = 56 ohms

The wizard says: In solution 0:

  • each 56 ohm resistor dissipates 22.4 mW
  • the wizard says the color code for 56 is green blue black
  • the wizard thinks 1/4W resistors are fine for your application
  • together, all resistors dissipate 201.6 mW
  • together, the diodes dissipate 360 mW
  • total power dissipated by the array is 561.6 mW
  • the array draws current of 180 mA from the source.

Several tips.

1) Since you want to have 9 LEDs of the same variety, you can just wire them all in parallel, and have a single resistor upstream.

Battery + -> Resistor -> LED1 --> Battery Negative

\->LED2 -/

\->LED3 -/

In which case, your current draw would be 5 mA x9 = 45mA. And purusing the tech specs, I don't see what Voltage they're supposed to be, but red is typically 2.2V in most LEDs, so if you're powering from a 3V source, you'll want a .8V drop. V = IR. .8V = .045A * R. R = 17.78 Ohms. Always round up to the next size, which is 18 Ohms, Brown Gray Black.

2) Though in all honesty, I wouldn't approach it that way. If you're looking for lots of small lights, I would recommend getting yourself some fiber optic cabling, and using that to carry the light from a single 3 or 5mm LED to all of the different locations. It should be cheaper overall, you'll have an easier time fitting it all inside, and it'll draw less current (so your battery run time is longer).

Several tips.

1) Since you want to have 9 LEDs of the same variety, you can just wire them all in parallel, and have a single resistor upstream.

Battery + -> Resistor -> LED1 --> Battery Negative

\->LED2 -/

\->LED3 -/

In which case, your current draw would be 5 mA x9 = 45mA. And purusing the tech specs, I don't see what Voltage they're supposed to be, but red is typically 2.2V in most LEDs, so if you're powering from a 3V source, you'll want a .8V drop. V = IR. .8V = .045A * R. R = 17.78 Ohms. Always round up to the next size, which is 18 Ohms, Brown Gray Black.

2) Though in all honesty, I wouldn't approach it that way. If you're looking for lots of small lights, I would recommend getting yourself some fiber optic cabling, and using that to carry the light from a single 3 or 5mm LED to all of the different locations. It should be cheaper overall, you'll have an easier time fitting it all inside, and it'll draw less current (so your battery run time is longer).

Fiber Optics huh, do you have any resources for that? I know nothing about Fiber Optics, but I like what your suggesting.