New white LED 3rd brake light MOD
#1
Three Wheelin'
Thread Starter
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Since a few of my bulbs were burned out in the 24 that are in my 3rd brake light, I though I'd replace them with superbright LED's. I just ordered a bunch (60) white LEDS from ebay, as that was WAY cheaper then getting them from an online place like mouser.. I bought 60 for $16!! (- it was close to $150 form mouser.. jeez..) so the plan is to desolder the existing lamps out and solder in the 5mm LED's in place. the holes are 6mm and the factory bulbs are 5.6mm.. I need to place a resistor in series with the anode end.. question: since I have 24 LED's in parallel that need to be knocked down to 20mA, what is the value of the resistor? can I use a 470ohm as that would be for a single LED, would that work for the bunch?? or do I have to multiply it by the consumption of the group? (c'mon, I was a design major, not an EE)
Coleman
Coleman
Last edited by Coleman; 06-01-2004 at 08:30 PM.
#3
Burning Brakes
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Let me know how you get on with this mod
(if you get it working) ![hiha](https://rennlist.com/forums/graemlins/roflmao.gif)
It is one I may be interested in as I have 29 5000mcd super white leds lieing around.
![Smilie](https://rennlist.com/forums/images/smilies/smile.gif)
![hiha](https://rennlist.com/forums/graemlins/roflmao.gif)
It is one I may be interested in as I have 29 5000mcd super white leds lieing around.
Last edited by GabiOsz; 05-21-2004 at 08:41 AM.
#4
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Sorry to dampen your high spirits, but you should have ordered RED leds.
The white led's will look orange and dim through a red lens.
You would be better modifying the PCB so that you can run groups of 4 LEDs in series, then wire the whole lot in parallel, that way you'll only need a small resistor and that will keep the heat down.
Regards
Alan
The white led's will look orange and dim through a red lens.
You would be better modifying the PCB so that you can run groups of 4 LEDs in series, then wire the whole lot in parallel, that way you'll only need a small resistor and that will keep the heat down.
Regards
Alan
#5
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Maybe you know but .......
Make sure you wire the LEDs' anodes (long leg) and cathodes (short leg)the right way round - could make for an interesting light pattern if not!
On the sibject of LEDs, when I got my p-car not all the lights above the two banks of centre console switches were working. I replaced all the bulbs with clear LEDs, and now I have lights that won't blow.
Make sure you wire the LEDs' anodes (long leg) and cathodes (short leg)the right way round - could make for an interesting light pattern if not!
On the sibject of LEDs, when I got my p-car not all the lights above the two banks of centre console switches were working. I replaced all the bulbs with clear LEDs, and now I have lights that won't blow.
#6
Three Wheelin'
Thread Starter
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Alan, have oyu tried this?? I can always get superbright reds.. they're almost free! the original bulbs are white, they look petty good... the white LED's come off orange?? hmm I'll experiment before I solder..
I understand the polarity of the LEDs, thanks.. and nice tip on the grouping on 4 in series.. if I have 6 series groups of 4 and those are in parallel, what would be the final R value??
Coleman.
I understand the polarity of the LEDs, thanks.. and nice tip on the grouping on 4 in series.. if I have 6 series groups of 4 and those are in parallel, what would be the final R value??
Coleman.
#7
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assum R1 is the resistance of 1 Led
for one series group totalR = 6*R1
for the entire setup then the total resistance is
1/Rt = 1/(6R1) + 1/(6R1) + 1/(6R1) + 1/(6R1)
1/Rt = 4/(6R1)
so the total resistance in your led network should be
Rt = 6/4 * R1
If you add a resistor in series to your network then
newRt = 6/4*R1 + yourResistor
I think you want newRt to equal the resistance of your original light.
I'm just a mechanical engineer so I could be wrong about the equations.
But in case you want to check for yourself, do a search on Kirchoff's Laws
and how the laws are used for network reduction.
In general, for 'n' resistors in series the equivalent resistance is
Req = R1+R2+R3+....Rn (sum the resistors)
for 'n' resistors in parallel
1/Req = 1/R1 + 1/R2 + 1/R3+.....1/Rn
hope that helps
for one series group totalR = 6*R1
for the entire setup then the total resistance is
1/Rt = 1/(6R1) + 1/(6R1) + 1/(6R1) + 1/(6R1)
1/Rt = 4/(6R1)
so the total resistance in your led network should be
Rt = 6/4 * R1
If you add a resistor in series to your network then
newRt = 6/4*R1 + yourResistor
I think you want newRt to equal the resistance of your original light.
I'm just a mechanical engineer so I could be wrong about the equations.
But in case you want to check for yourself, do a search on Kirchoff's Laws
and how the laws are used for network reduction.
In general, for 'n' resistors in series the equivalent resistance is
Req = R1+R2+R3+....Rn (sum the resistors)
for 'n' resistors in parallel
1/Req = 1/R1 + 1/R2 + 1/R3+.....1/Rn
hope that helps
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#8
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The reason white bulbs are OK behind a red lens is this:
White light is made up of all the colours of the spectrum, and when a lens is placed in front of it, in this case red, all the colours are blocked except the red light, which is free to pass through.
An LED is different in that the white light is made up of mostly blue light. As the red lens will only allow through red light, the output is considerably reduced (because there is little red light in a white led)
If you use RED leds, the light will pass straight through and work very efficiently.
Once you have your LED's specifications, I will calculate the resistor reqd.
Alternatively, purchase some 12V LEDs, these will be a direct replacement.
Regards
Alan
White light is made up of all the colours of the spectrum, and when a lens is placed in front of it, in this case red, all the colours are blocked except the red light, which is free to pass through.
An LED is different in that the white light is made up of mostly blue light. As the red lens will only allow through red light, the output is considerably reduced (because there is little red light in a white led)
If you use RED leds, the light will pass straight through and work very efficiently.
Once you have your LED's specifications, I will calculate the resistor reqd.
Alternatively, purchase some 12V LEDs, these will be a direct replacement.
Regards
Alan
#9
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Hmm, I need to replace a bulb in my hoop spoiler so while it's off the car, I might as well replace the bulbs with LEDs. Assuming it's essentially the same mechanism inside as the top-level brake light, how many 12v LEDs would I need to buy?
#10
Three Wheelin'
Thread Starter
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I'm doing it on a 95, so it is the same as yours.. (basket handle) you need 24 Ultra-red. 5mm. Online they are a lot cheaper.. I just got 100 Ultra-red's for 15.00 You can take the light/lens off while the basket handle is still on the car... 2 screws under the handle.
Coleman.
Coleman.
#12
Three Wheelin'
Thread Starter
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So.. using all the input that I had received on my previous posts, I completed the MOD! Works like a charm... Okay.. after much thinking about the way I was going to wire it (the resistor setup had me sweating a bit..) I figured out that if you put 6 in series, each consumes about 2 volts, it's about 12v total.. and that is what we have at the source voltage! no resistors-!! this is done by dremeling the traces on the PCB so that the 4 groups run in series.. with these 4 groupings, and like Alan/Acepatrs suggested, just run those in parallel. then some wire across the (+) and (-) of the 4 groups, tying all the (+)'s together and all the (-)'s together, and then close her back up!! It's not dark yet, I'll post a photo of the LED's in action later.... tested them and it works....
Coleman.
Coleman.