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1Step 1
Locate an appropriate power supply and check out its spec label. I lucked out; mine has all the info I need:
The five voltages listed are the standard outputs for an ATX supply; in my case, I decided to pull all five out. The most important thing to note here is minimum current rating; switchmode supplies will only put out a stable voltage if you give them a minimum load. In this case, I need to hang a 1-amp load off the +5vdc and +12vdc lines, and a 0.3-amp load off the +3.3vdc line.
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2Step 2
Time to apply Ohm's Law to determine proper resistor values. Pretty simple: V=IR, and since the resistor is connected between ground and each supply rail, V = supply voltage. They give us current, so we can calculate the three resistor values:
3.3V rail: 11 ohms
5V rail: 5 ohms
12V rail: 12 ohms
Simple enough. But remember, resistors work by converting electrical energy into heat; as such, they all have wattage limits. P=IV, so:
3.3V rail: 1W
5V rail: 5W
12V rail: 12W
Too much power through a resistor = overheating = bad.
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3Step 3
Calculations in hand, head to your local surplus store. In the midst of getting lost in a sea of ammo cases, CB radios, unidentifiable motors, and assorted pieces of self-adhesive foam, collect some parts:
You'll note that they didn't have any large wattage resistors, so I picked up a handful (at 10 cents apiece with a baker's dozen discount) of various sizes to build up the values and wattages I need. Also note the awesome organ-style pull switch and the handy red insulating bushings for the binding posts.
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4Step 4
Chop the connectors off of your power supply, leaving enough room to use them in another project ten years from now. Just kidding, you'll store them in a shoebox for ten years but then forget about them when you actually need them and end up buying new ones. Might as well just throw 'em out and use the shoebox for something else.
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5Step 5
Measure the binding post bushing diameter, rip the supply's cover off, and start peppering the front panel with holes, hopefully in a somewhat straight line:
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6Step 6
Measure the LED housing. Grow overly confident in your ability to accurately drill through steel plate without a drill press and decide to skip the pilot hole this time. Also skip clamping a backing board on to the plate. Immediately regret both decisions:
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7Step 7
Get some help from the Internet to determine how much current your wires can handle. That will dictate how many of each color wire you'll need to run to each binding post. I threw in an extra wire or two for good measure; in my case, the 12v and 5v lines each took six 18-gauge wires.
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8Step 8
Crank that soldering station up for the crimp connectors. In my case, I'm using RoHS stuff so I'm up to 850 F. It's a good idea to solder the wires together first, let them cool, trim them back a bit, then shove them into the crimp connector with a bunch of flux before heating the whole thing up and hitting it with some more solder:
Probably not a bad idea to clean up the 'no-clean flux'. I skipped that step so the joint is a bit discolored, but the actual connection will work.
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9Step 9
Heat shrink time. Fortunately, the tubing is big enough to slip over the connector, unlike last time when you forgot to thread it on prior to soldering.
Measure, cut, and repeat 'til you've covered all the voltages and ground.
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10Step 10
Thank goodness for needle nosed pliers! Bend that messed up hole back as best you can. Don't worry, your mistake will be mostly covered by the LED bezel.
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Thanks for the guide!
Regarding Step 12: "I'm running a bunch of 2-watt resistors in parallel; remember, wattages stack in this configuration if the resistance values are the same."
The effective resistance of a bunch of resistors in parallel is
R_effective = 1 / ( 1/R_1 + 1/R_2 + 1/R_3 + ... + 1/R_n ),
or just
R_effective = R_common / number
if they all have the same resistance.
Then, each resistor gets part of the current, with a share equal to 1/R_i. For example, if all resistors have the same voltage, then they split the current evenly (in theory). Or if there's a 1-Ohm and a 2-Ohm, then the 1-Ohm gets twice as much current as the 2-Ohm.
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