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LED Tail Light 12v to 9v

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JdotFite

New Member
Hi guys, I wanted to learn more about LEDs so I began making an LED board for a tail light on my scooter. I am using Cree P4 LEDs (2.5 forward voltage, 70mA forward current). I didnt have a lot of room in the casing, so I wired them in rows of the three and used 22 ohm resistors and hoped I could figure out how to get the input voltage down to 9v.

I have everything soldered and it works great when I test everything with a 9v wall wart. How should I handle dropping my bikes 12v input down to 9v?

I bought a bunch of Sharp 9v regulators (PQ09RD21) from digikey, and they get very hot very fast. I found a couple of heat sinks today that I could use to help with the heat, but I wasn't sure if there was a better option. I don't have a lot room to put a bunch of heat sinks. I was hoping I could fit everything into the tail light casing. Is there a better way to handle this?

Thanks for any and all help. I am just getting my feet wet, so please assume I am a noob!

Original Board:
8532-DSC07178-1.jpg


Soldering:
8533-DSC07180-1.jpg


Done:
8534-DSC07193-1.jpg
 
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colin55

Well-Known Member
Firstly put 15mA through each set of 3 LEDs and see how bright they are. Most LEDs give very good output at 15-17mA.
This is your starting point.
 

JdotFite

New Member
Thank you for the response, but would you mind elaborating? I am not sure if I understand what you are asking me to do. I apologize if I sound like an idiot!
 

bountyhunter

Well-Known Member
I have everything soldered and it works great when I test everything with a 9v wall wart. How should I handle dropping my bikes 12v input down to 9v?///Is there a better way to handle this?
Just drop the voltage in the resistor in series with each string.

With 12V source, three LEDs drop about 7.5V, that leaves 4.5V. If you want 20mA per string:

R = V / I = 4.5/.02 = 220 Ohm (1/4 W)


At that current, each string of LEDs will dissipate about 150mW, with 16 strings that makes a total power dissipation on the board of about 1.9W. It looks like that would be OK from the picture.

If you reduce the resistance and crank up the current, the power dissipation goes up. I think the additional brightness above the 20 - 30 mA range is small for the amount of power it takes to get it.
 
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RODALCO

Well-Known Member
15 mA

I second colins comment.

Reduce the LED current to about 15 mA's and your problem is solved.

The 70mA quoted is peak impulse current. I doubt that those LED's stand 70 mA's continuous.
 

mneary

New Member
70 mA is the maximum rating of the LEDs, there's no need to run them at that current if they are bright enough at (for example) 15-20 mA.

Re-calculate your series resistor for lower current, and the regulator shouldn't get hot. The regulator is a nice idea since the LEDs wouldn't fluctuate with RPM etc..

Even a small heat sink makes a huge difference in the regulator temperature. If the tab is fully insulated (data sheet offers no clues) you can bolt it to anything (vital anyways in a vibrating environment).
 

JdotFite

New Member
I second colins comment.

Reduce the LED current to about 15 mA's and your problem is solved.

The 70mA quoted is peak impulse current. I doubt that those LED's stand 70 mA's continuous.


Thank you for pointing out that the 70mA is their max. I didn't even realize that. So you are saying I should reduce the LED current to 15 mAs, and just use the 12v from the battery? So I would need a 330 ohm resistor for each set of three? Thats basically what bountyhunter is suggesting, just running them at 20 mA with 220 Ohm resistors, right?

Thanks again for everyones help!
 
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Diver300

Well-Known Member
Most Helpful Member
Looking at the data sheet for the LEDs, there is quite a bit of variation in forward voltage. That could alter the voltage across the resistor a lot if you have a 9 V regulator.

I suggest that you measure the voltage across the resistor when running, and see how close to 1.5 V it is. You choice of a 9V regulator means that small variations in the LED voltage will give a big current change.

Linear regulators will always get hot. Between the resistors and the regulator the voltage drop is 4.5 V with a 12 V supply, and 6.5 V with a 14 V supply. With 16 strings of LEDs the total current is 1.12 A so the power dissipation is 7.28 W. You didn't say how many PQ05RD21 regulators you uses, but their limit is 1.4 W with no heat sink, so you would need 5 or more without heatsinks, or quite a good heat sink with just one regulator.

If you want to have less heat, you need a switch mode regulator.
 

kpatz

New Member
I'm not sure exactly which LEDs you're using, but assuming it's these:
**broken link removed**

70mA is the maximum continuous forward current, and typical at 2.5V. So these LEDs are designed to run at that current. 15-20 mA is more typical for lower-power LEDs.

What I would have done was run 5, or better yet, 6 LEDs in series instead of 3. Then you have 2.5 x 5 = 12.5 volts, or 2.5 x 6 = 15 volts (consider that the electrical system is running closer to 15 volts when the engine is running). With 6 LEDs in series, you could get by without a resistor or regulator.

Otherwise, you should use a switching regulator, or a microcontroller providing PWM to the LEDs so they aren't running at 100% duty cycle. Another benefit to the microcontroller approach is you could have 2 brightness levels, for taillight and brake light.
 

audioguru

Well-Known Member
Most Helpful Member
I made a night-light with Luxeon Super-Flux LEDs that look like yours and have the same spec's. Their max current is 70mA so I run them at "only" 53mA. There are 36 of them in the case for a compact assette and there are many cooling holes drilled in the case.

It gets much too hot. The plastic case melted and turned foggy. The clear plastic case on the LEDs turned yellow.

The LEDs average 2.5V each and with 36 at 53mA the total heat dissipation is 4.77W plus a little more for the current-limiting resistors. It is too hot for a small case.

Your LEDs are not 2.5V. They are any voltage from 2.0V to 3.0V. Are you going to measure each one so you can connect a low voltage one in series with a high voltage one for matching?
 

Diver300

Well-Known Member
Most Helpful Member
What I would have done was run 5, or better yet, 6 LEDs in series instead of 3. Then you have 2.5 x 5 = 12.5 volts, or 2.5 x 6 = 15 volts (consider that the electrical system is running closer to 15 volts when the engine is running). With 6 LEDs in series, you could get by without a resistor or regulator.
If you run 6 LEDs in series, the current will be uncontrolled. Small variations in the LED or battery voltage will cause huge variations in the current.
6 LEDs would mean your tail light wouldn't work at all if the engine was off. 5 LEDs would be very dim with the engine off.

Otherwise, you should use a switching regulator, or a microcontroller providing PWM to the LEDs so they aren't running at 100% duty cycle. Another benefit to the microcontroller approach is you could have 2 brightness levels, for taillight and brake light.

If you do that, make sure that off period is less than about 1 ms or the flashing is visible.
 

tytower

Banned
How to power up 24 ?
And have them flash of course
 

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JdotFite

New Member
thanks again to everyone who is helping me out.

If I could drop the whole 9v idea, that would be great. If I do decide to run them at 12v, should I use a 12v regulator to make sure the voltage never gets to high? I have had the meter on the battery and it was staying ~12v while I was riding, but I know it does fluctuate a little.

I will be using pulse width modulation, I have a small board made for that.

I stuck with three in a series because of the amount of space I had to solder everything together. Not sure how I could do it differently without making a custom circuit board or using less LEDs. I was trying to stuff as many as possible in there.

There are actually 27 Cree P4 Red Leds in the center and 9 Cree P4 Amber Leds on each side for the signals.

audioguru, thanks for explaining that the forward voltage isn't 2.5, I see what your saying. I had no idea I was supposed to measure each one. It sounds like I will be removing all these LEDs or buying new ones, so I can measure them either way. So you are saying I should match them up so they average out. Beyond matching them up, based on your experience with
your night-light, what do you suggest? BTW, I do have a bunch of Luxeon Super-Flux Leds here too, you are right, they look almost identical and the specs are similar.

Diver300, thanks for explaining the 5 or 6 LEDs in a series thing. I wondered why the calculators online always kept the amount in a series low, like 3 or 4 max.

Alright, based on what everyone is saying....if I ditch the 9v input idea (would like to!) and use the scooters 12v battery as input and kept the LEDs in series of three @ 20mA I would need to use 1/4W 270 ohm resistors according to the calculator I am using online. Seems like the resistors would be dealing with a lot of left over. Is this a bad idea? Would the resistors get too hot?

Thanks again everyone, I really appreciate the help!
 

bountyhunter

Well-Known Member
Thank you for pointing out that the 70mA is their max. I didn't even realize that. So you are saying I should reduce the LED current to 15 mAs, and just use the 12v from the battery? So I would need a 330 ohm resistor for each set of three? Thats basically what bountyhunter is suggesting, just running them at 20 mA with 220 Ohm resistors, right?

Thanks again for everyones help!
I would try a 330 and see if it is bright enough. Remember, tail lights have to be seen in daylight and about 99% of the LED's out there are not bright enough for use as a tail light and would be an illegal modification. I have been through this since I put LED tail and brake lights on my motorcycle. A good reference for this info is:

https://www.superbrightleds.com/

I used the 3W Luxeons for mine, none of the others were bright enough.
 
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crutschow

Well-Known Member
Most Helpful Member
What I would have done was run 5, or better yet, 6 LEDs in series instead of 3. Then you have 2.5 x 5 = 12.5 volts, or 2.5 x 6 = 15 volts (consider that the electrical system is running closer to 15 volts when the engine is running). With 6 LEDs in series, you could get by without a resistor or regulator.
kpatz: If you run the LEDs without series resistors you will likely end up with the LEDs kaput.;)
 

bountyhunter

Well-Known Member
I made a night-light with Luxeon Super-Flux LEDs that look like yours and have the same spec's. Their max current is 70mA so I run them at "only" 53mA. There are 36 of them in the case for a compact assette and there are many cooling holes drilled in the case.

It gets much too hot. The plastic case melted and turned foggy. The clear plastic case on the LEDs turned yellow.
I have a flashlight with similar LEds. In stock form, they ran about 10mA through each of 24 LED's and it was OK. I modified it so I could kick it up to about 30 mA per LED and it is brighter, but not 3X brighter. You get into diminishing returns rapidly above about 10 mA.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
I have a flashlight with similar LEds. In stock form, they ran about 10mA through each of 24 LED's and it was OK. I modified it so I could kick it up to about 30 mA per LED and it is brighter, but not 3X brighter. You get into diminishing returns rapidly above about 10 mA.

Isn't it a log scale like sound?, so you need ten times the current to be twice as bright.
 

audioguru

Well-Known Member
Most Helpful Member
You don't know what "twice the light" or 'half the light" is.
Two LEDs shining on a white ceiling at night light the room dimmly. One LED also lights the room dimmly, maybe a little less bright but the light power is half.

Three LEDs shining on the ceiling light the room dimmly at night and one LED also lights the room dimmly with a little less light.

I think there will be a big difference when 10 LEDs shine on a ceiling at night then only one LED does. Is the brightness of the light doubled with 10 LEDs when compared to one LED?
 

JdotFite

New Member
bountyhunter, you are suggesting using the 12v as input and using 330 ohm resistors in strings of three?

To address your comment about safety, I agree with you, but here is a photo of the original crappy LEDs that were in the tail light (this isn't the stock tail light). These aren't the best comparison shots...

Before:
8541-DSC06742-1.jpg


Here is the Cree P4 Leds (had them running at 70mA when I took this picture, not sure how bright they will be at 10-20mA, but hopefully will still be much brighter)

8542-DSC07189-1.jpg


Left Turn signal:

8543-DSC07199-1.jpg


So does everyone think I will be ok running them in series or three @ 12v and just using 330 resistors? Will the resistors get too hot? I don't want them to melt the plastic casing. They are kinda close to the plastic.

Thanks guys, I have learned a lot from this thread already!

Justin
 
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Hero999

Banned
Isn't it a log scale like sound?,
Probably.

so you need ten times the current to be twice as bright.
I doubt that because that would make my 150W floodlight only twice as bright as the 15V light in my fridge - it just doesn't add up.
 
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