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# Modding LEDs on a Antec 900 - Page 2

Ok so I figured out what I'm going to do given the values in the table, which was exactly what I needed btw. The calculator gave me one set up but I didn't like how it put a resistor (9 total) in series with each LED when you could just stick a single one in series with the whole array.

I'm going to put all of them in parallel with each other, wired to the switch, then that will be in series with a 8ohm resistor. Should give 3.6V and 20mA to each LED. Now I just need to solder and wire it all up. Hopefully my math is right
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no, it recomends 1/4w resistors
9*1/4 > 2w

10w 10ohm resistor will be close enough

i hate shopping at radioshack, but here ya go
I don't think that I need a 10W resistor.

Using 20mA through each LED gives 180mA through the single resistor. Also each LED will need 3.6V across it based on the table.

So setting up a KVL equation around a loop in the circuit with a 5V source gives 5=(.18)R + 3.6. Solving for R gives 7.77ohm, which is why I said 8.

P=VI and V=IR, giving P=R(I^2).

Going with a 10ohm resistor gives us P=(10)(.18^2) = .324 W. So a 1/2 watt resistor should be enough.
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Quote:
 Originally Posted by Xazen I don't think that I need a 10W resistor. Using 20mA through each LED gives 180mA through the single resistor. Also each LED will need 3.6V across it based on the table. So setting up a KVL equation around a loop in the circuit with a 5V source gives 5=(.18)R + 3.6. Solving for R gives 7.77ohm, which is why I said 8. P=VI and V=IR, giving P=R(I^2). Going with a 10ohm resistor gives us P=(10)(.18^2) = .324 W. So a 1/2 watt resistor should be enough.
yea 1/2 watt resistor is more than enough. A 10w resistor is extreme overkill for what you are doing. I think RadioTrash Usually stocks a pretty good assortment of 1/2 watt resistors and anything from 8-12 ohm should suffice if you cannot find a 10 ohm.
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Quote:
 Originally Posted by Xazen I don't think that I need a 10W resistor. Using 20mA through each LED gives 180mA through the single resistor. Also each LED will need 3.6V across it based on the table. So setting up a KVL equation around a loop in the circuit with a 5V source gives 5=(.18)R + 3.6. Solving for R gives 7.77ohm, which is why I said 8. P=VI and V=IR, giving P=R(I^2). Going with a 10ohm resistor gives us P=(10)(.18^2) = .324 W. So a 1/2 watt resistor should be enough.
After Re - Examining this I decided to simulate your array. Considering the Source voltage is 5v and the draw of Each led is 3.6v The system needs to be set up to have an ~82ohm 1/4w resistor/ Per LED.

If you use a 12v supply you could use 3 arrays of 4 LEDS with a 68ohm 1/4watt resistor.

an 8ohm resistor will work for a short period of time, considering the LED could probably handle 5v for between 2 seconds to 48hours. But to bring down the source voltage to the proper range you will need more resistance and making a series array of leds is out because it would so under-power the leds they would not light up at all.
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That is what I was considering at first, as that is what the calculator gave me. However the same result can be achieved by removing each resistor that is in series with the LEDs, and instead add a single resistor after the 5V source but before all of the LEDs that are in parallel. To lower the current to something that the LEDs can handle over a long period of time all that I would need is a 8ohm resistor in this configuration. I already worked out the math. I will probably use a 10ohm just because.

Anyway its coming along. I already cut the wires to/from each leds, soldered wires to them, and put them all in parallel. Now I just need to fit the switch to the back of the case somewhere and solder the power and wires to the fans to it.

Thanks for the help cook, +rep.
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Quote:
 Originally Posted by Xazen That is what I was considering at first, as that is what the calculator gave me. However the same result can be achieved by removing each resistor that is in series with the LEDs, and instead add a single resistor after the 5V source but before all of the LEDs that are in parallel. To lower the current to something that the LEDs can handle over a long period of time all that I would need is a 8ohm resistor in this configuration. I already worked out the math. I will probably use a 10ohm just because. Anyway its coming along. I already cut the wires to/from each leds, soldered wires to them, and put them all in parallel. Now I just need to fit the switch to the back of the case somewhere and solder the power and wires to the fans to it. Thanks for the help cook, +rep.
Thanks!

no problem

I was having a hard time visualizing what you where saying. After the simulation, Right on. With an 8 ohm resistor you would be running 3.2v which is more than good. Here is what I tested just for giggles and +rep for your math equation goodness.

Oh yea and here is The Circuit Simulator I used in the event you want to test further. Make sure you have Java installed on your system as this is a .jar.
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you could just wire all the leds to one negative and have a switch on that one neagtive
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Quote:
 Originally Posted by cook Thanks! no problem I was having a hard time visualizing what you where saying. After the simulation, Right on. With an 8 ohm resistor you would be running 3.2v which is more than good. Here is what I tested just for giggles and +rep for your math equation goodness. Oh yea and here is The Circuit Simulator I used in the event you want to test further. Make sure you have Java installed on your system as this is a .jar.
Awesome simulation btw. Will defiantly keep that site in my bookmarks.

Quote:
 Originally Posted by RIFOLWANTS2BUY you could just wire all the leds to one negative and have a switch on that one neagtive
Yea that's pretty much what I did. However the resistor was needed in order to keep limit the current to something the LEDs could handle.

Anyway I spent most of the day today wiring everything up. It took a looong time, there were a ton to wires to solder. Not to mention the wires to the anode and cathode of each LED were absolutely tiny and impossible to work with. I'll post a pic or two later.
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Everything is finished. Updated first post.
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