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Old 07-02-2007, 01:55 AM   #1
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Default how to add LED's??

ok ive done senior physics and am studying engineering at university but havnt done a whole heep of electronics i know V= I*R and al that basic stuff but im trying to work out how much resistance i need to drop 12v down to 3.5v for an led on a power supply?
do you work of the current ie
3.5v .2amps =17.5 ohms?
or dodo i takes that 17 ohms and divide 12 by 17 to give me .685 amps?
so basicly how do you work out voltage drop through resistance?
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Old 07-02-2007, 02:20 AM   #2
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you should just go to radio shack (if your in the usa) and get a 12v LED

the work and stuff to make the other way work isnt worth the hassle
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Old 07-02-2007, 02:27 AM   #3
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You require a voltage drop of 12-3.5 = 8.5V

The current you want flowing through the LED's will be around 20mA, therefore you want 20mA flowing through the resistor too.

Then just use V = IR:

R = V/I = 8.5/0.02 = 425ohms, then choose the nearest preferred value which is higher than that value, so 430ohms should do it.
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Old 07-02-2007, 03:09 AM   #4
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Josh try this site.
You actually want to limit the current with the resistor as it's the current that will kill the LED. Don't forget most of our RC power supplies are 13.5-13.8V not 12V. (unless you are using a PC power supply).
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Old 07-02-2007, 05:27 AM   #5
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Here is how i did it (the Blue 350z is my show drifter shell)

I posted a full tut on the theory and how to make neons.

U can apply the voltage theory to 12v that i used to step down from 9v.

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Old 07-02-2007, 09:53 AM   #6
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Heh, I find running them in series is easier for me, though I used a 3.2 voltage drop LED and I ran them in series from the receiver to get them to turn on with the ESC. Don't have any pics of the car at night :P, though, I used 10mm LED's :-D... I had to make a test light yesterday, with the 3.2 drop LED's I was able to use a 220ohm resistor, just remember, series wastes the voltage, parallel distributes it evenly :P
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