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Resistor Help


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#1 silvia Ks

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Posted 09 November 2004 - 12:56 AM

im wiring 8 LEDs facing inwards the jewel and a separate 10 leds facing outward. the metku calculator says a 10 ohm resistor and 12 ohm. i go to rat shack and i dont know what wattage a should get? (1/4, 1/2, 1)

also instead of putting a resistor in front of the leds can i just get a bunch of 100 ohm (im using 5v)and wire them to each LED so i dont have to go through the hassle of getting different resistors since im also lighting the sides and controller vents? thnx

#2 silvia Ks

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Posted 09 November 2004 - 03:24 AM

bump

#3 laxdrummer04

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Posted 09 November 2004 - 03:34 AM

well to do wat u want to do, u would have to solder a wire to the 5v power, then solder an end of ALL the resistors to the other end of the resistors... so it wouldnt be pretty

Also, please wait 24 hrs before you bump... someone will reply... dont worry

#4 silvia Ks

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Posted 09 November 2004 - 03:42 AM

alright i decided to ditch the resistor per LED idea. now how do i know what resistor wattage i need? ( like in ben999s led tut, did he use a 1/4 or 1/2 watt resistor

#5 knappster

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Posted 09 November 2004 - 05:14 AM

First of all I'd like to mention that 18 LEDs seems a little excessive. But I don't know what you're shooting for, so maybe it will be ok for you. Instead of listing a a bunch of equations that probably won't be helpful for you I'll try to explain. But if you could list some of the specs of the LEDs and the size of the resistor you're using, it will be easier to determine. Basically, the resistor you need depends on:

a) how many resistors you have in parallel(which is how it appears that you will be connecting them)
cool.gif how much voltage each LED is rated at

Since the LEDs are in parallel, there is an equal voltage drop across each LED(You probably used it to determine the size of the resistor). Let me call this VL for clarity. So the voltage drop across the resistor will be the remaining voltage, Vr = 5v - VL.

Example:
1.5V LED with a 5v source
Vr=5v-1.5V = 3.5V

Once you find that, there is more math to do. The current,IL, through each LED is going to be equal as well, but the current through the resistor, Ir, will be X times as high, where X is the number of resistors in series.

Example:
8 LEDs rated 10 mA
IL = 10 mA
Ir = 8 x 10 mA = 80 mA

Now the power dissipated by the resistor is simply Vr x Ir = Pr. This is measured in Watts and will tel you if you need a 1/4, 1/2, or 1 Watt resistor.

Example
Vr = 3.5V
Ir = 80 mA = .08A
Pr = 3.5V x .08A = .28W, which is slightly too much power for the 1/4W resistor but should be fine for the 1/2W.

In the case that there are no 1/2W resistors of that type, you could always buy 2 resistors with half the resistance and attach them in series to get the same resistance necessary. I hope this helps you, I'll check the board for a little while if you're still unsure.

-Knappster




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