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as some of you may or may not know a few of us have been trying to make the boot garnishes of our r33's light up. see here.

anyway i tried to wire it up in this way

post-30915-1179933652_thumb.jpg

the leds are rated at 2.1v each and have a current draw of 50mA (see here). dunno if tey are the same as mine cos i swear the dude sed 30,000mcd not 20k. but this is the best values i can find.

now here's my problem. i hooked it up to the car the other day and only two and a half letters llit up. took it apart and only four of my leds are working. any dead ones that i replace are immediatly blown. i am getting headaches trying to figure this out. is it my resistor? am i drawing too much current? am i using homosexual components?

if there is no solution can someone confrim that the car runs at 13.4v and 10A? maybe i can figure this out another way

Help!!

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Where are you getting power from? Most things will be 12 volts (lighting etc)

So you have two negative, and one positive going in.. with a resistor on the positive side?

You have 8 LED's at the top, and 8 at the bottom? Is that right? If so, that's a total voltage drop of 8x2.1... 16.8v... just for one side! Way too much for 12v. (you don't want to draw more than 10v at a time)

If you can rig it so that you have four negative points (4 LED's to a circuit = 8.4v), and run one resistor per circuit, you should be fine.

Those four circuits can run in parallel to each other, so there is only ever 8.4 volts being drawn...

Edited by RANDY

thanks for your reply dude. doesnt matter if i connect all the neg points into one wire does it? as long as theres 4 leds per parralell circuit? because what i've done is to connect the two neg lines and run a single positive/neg 2core wire out. so the circuit is actually like this

post-30915-1179959000_thumb.jpg

i thought this technically means im running all 16 in parralell. but it makes sens what you said cos only four leds are working.

Still looks like you're splitting the power AFTER the resistor.

oh oh... i'm still a bit wonky from a nightshift, but you have 8 LEDs in parallel. So you are drawing 2.1v eight times. The resistor is probably calculated for the total voltage, not 2.1

If you run the neg of one LED to the pos of another LED, that would be in series, and you could calculate it that way, but still 16v or so is too much. You need to run two sets of parallels.

I'll reply again when I wake up.

on advice from elsewhere aswell, im doing like you said. i now have 4 lines of power, each with a 120ohm resistor running in parralell. each of these power lines has four leds running off it in parralel.

btw the leds aren't 20k, they are 30k and draw 75mA at 2.4V. anyway it seems to have worked this way from a quick test. now to see if it'll last.

thatnks for your help dude

No worries!

umm... hmmm... You want the 4 LED's to be in a series chain though... so that the voltages add together. If they run in parallel they are still each getting the same power from the one resistor, and each LED is only wanting 2.1v. i.e: if your resistor is dropping 1.6 volts, then each LED is still seeing 12-1.6 = 8.4v EACH.

So... it should go:

Negative power to each series of LED-

Set #1: resistor - neg LED1 pos - neg LED2 pos - neg LED3 pos - neg LED4 pos - positive power.

Set #2: resistor - neg LED1 pos - neg LED2 pos - neg LED3 pos - neg LED4 pos - positive power.

Set #3: resistor - neg LED1 pos - neg LED2 pos - neg LED3 pos - neg LED4 pos - positive power.

Set #4: resistor - neg LED1 pos - neg LED2 pos - neg LED3 pos - neg LED4 pos - positive power.

Positive power to each series of LED+

Each series will draw 2.1x4 = 8.4v.. so you can work out the resistor needed for each series to drop the additional 1.6v, and that resistor will probably be small and not heat up too much. 1/4 watt is probably fine for that voltage. do 1/2watt to be safe?

Well, if your final product looks anything like the above, it should work well and not heat up the resistors too much, and should last forever! =-D

Edited by RANDY

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