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What do I need to know? Such as, how would these be best wired up? And what resistance would I need?

It's for a computer case modification I'm planning on doing in the near future, Will be powered from a single 12v source (Computer Powersupply).

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3 Answers 3

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To avoid having to burn too much power in a load resistor, you should put some of the LEDs in series. For example:

LED arrangement

Assume each LED drops 3.5V. Then, the resistors will have (12V - (3.5V * 3)) = 1.5V across them. If you want the current through the LEDs to be 10mA, then the resistor value should be (1.5V / 10mA) = 150 Ohms.

Each resistor in this case would be dissipating (i^2 * R) = (0.01 mA) * (150 Ohms) = 0.015 W, so any size resistor (1/8 W, 1/4 W, etc) would have no problem handling the power.

Note that this diagram has 21 LEDs, while you asked for 20. To keep things simple and symmetric, it would be easiest to just install a 21st LED in the circuit and hide it, if you really only want 20 showing.

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I agree with Jim to place LEDs in series, but not with his strings of three. Jim calculates with the average LED voltage, and that's a Bad Idea™. You should always go for worst case. His 3.5 V drops 1.5 V across the resistors, but when that LED's voltage is 3.8 V that will only be 0.6 V, and that's too little: some variation in LED voltage will give a large variation in brightness. In this case will a LED voltage variation between 3.2 V and 3.8 V give a 4:1 current variation, and that will give a clearly visible difference in brightness.

Now you might say that you bought the LEDs from the same batch, so wide variation is not to be expected. Probably. But how sure are you of the 12 V? A voltage regulator (if the 12 V is regulated in the first place!) may have 5 % tolerance, so worst case that may be 11.4 V, and if you have the bad luck to have 3.8 V LEDs they won't do anything: 3 \$\times\$ 3.8 V = 11.4 V.

Always leave enough headroom. If we only place 2 LEDs in series we'll need a (12 V - 2 \$\times\$ 3.5 V)/10 mA = 500 Ω resistor at the average LED voltage. At 3.8 V LED voltage that 500 Ω resistor will give us 8.8 mA, that's a 12 % deviation, where the three LEDs saw a 60 % deviation. The luminosity ratio between the 3.2 V LED and the 3.8 V LED is then 1.27:1 instead of 4:1.

The drawback is that you'll need 10 strings instead of the 7 you have now, so your total power will be 40 % higher. But you won't need that 21st LED.

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You can run them in parallel, so that vd~3.5V. If you know the typical current consumption, say 10mA ea-- then you can just mult 10mA*20=200mA to get the source current requirement. So you can run a series resistor of (Vdd-Vd)/Id=(12-3.5)/200e-3~42.5 ohms 1/4 watt to 1/2 W should be ok. You can substitute a trim pot or potentiometer of 100 ohms to play with intensity.

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    \$\begingroup\$ A 42.5 ohm resistor with (12-3.5) = 8.5 volts across it would be dissipating 1.7 watts. A 1/4 or 1/2 W resistor will catch fire! \$\endgroup\$
    – Jim Paris
    Commented Aug 28, 2012 at 3:39
  • \$\begingroup\$ How's this? dl.dropbox.com/u/34977901/Computer%20Mods/PCB%20Designs/… \$\endgroup\$
    – Jamie R
    Commented Aug 28, 2012 at 5:23
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    \$\begingroup\$ You'd want one resistor in series with each parallel LED, or small difference in LED parameters can cause fairly large changes in brightness. \$\endgroup\$ Commented Aug 28, 2012 at 13:20
  • \$\begingroup\$ -1 for the reasons given by Scott and Jim. \$\endgroup\$ Commented Aug 28, 2012 at 16:04
  • \$\begingroup\$ Kind of harsh to add all the down votes for a small error given all of the work shared. Incidentally a 2 W resistor will work just fine at the cost of power. \$\endgroup\$
    – pat
    Commented Aug 28, 2012 at 19:04

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