I am starting my first LED project--I am wiring eight white LEDs for an ambient footwell lighting project in my 4Runner. Two per side under the front dash panel and two per side under the front seats pointed towards the rear footwells. I plan to use the standard 3.6V 20 mA white LEDs.

This would be one way to do it:


However, battery voltage (measured at the dome light, which will trigger the footwell lighting) can vary from 11.9 volts to 14.8 volts. Picking my resistor based on an average of 13 volts means I am pushing only 16 mA with the engine off and overdriving the LEDs at 26 mA with the vehicle running. If I recall correctly, LEDs are pretty dim at 16 mA.

Has anyone ever used a current-limiting regulator like the LM317? Supposedly all I need to balance the circuit correctly is R = 1.25 / I (I being the current in amps). So R = 1.25 / 0.020 amps (two 20 mA LEDs in series per array), which means a 62 ohm resistor and I'm set? This seems too easy. Anyone familiar with this stuff?

I played a little with similar similar devices, but that was a few years ago for an automotive electronics class (A6). I haven't done anything with them since, but if I recall correctly, they really were that simple.

I say buy some red LEDs with similar current requirements and give it a shot (red LEDs are cheap). Use an adjustable power supply to give it up to 17v and down to 9v. Those should be just outside the extremes of expected operation for your rig. If it works for those, it should work great with with your white LEDs. <img src="/forums/images/graemlins/kewl.gif" alt="" />

Ken, you should have taken a look at my rocklights at mammoth. It's almost exactly what you're doing, except I used luxeon 1W LEDs.

http://yotatech.com/showthread.php?t=89847





I think you found the same webpages I found when I did the research. Have you figured out where that 1.25 value comes from?

1.25 is the minimum output of the LM317 regulator and it is capable of regulating power up to 37 volts. So basically, the 1.25 is the "overhead" required to run the regulator.

Did you use a regulator, or just current-limiting resistors? If the latter, how much difference in brightness have you noticed with the engine off versus on?

Oh ok, that 1.25 had been bothering me forever.

I used the LM317. The brightness changed dramatically with only resisters and considering the expense of the luxeons I didn't want to burn them out.

But with normal LEDs I'd recommend just resisters if you can get away with it...it's just a simpler circuit. Run some tests and see how bright they are at lower voltage.

Oh good, I was hoping to find someone that had experience using the LM317.

My last LED project was illumination of a custom boost gauge in the Supra. I was using 3 white LEDs in series with a resistor, much like the diagram in my first post. Problem was, the guage was barely illuminated with the engine off and the brightness would get noticeably brighter with an increase in engine RPMs. This is what I was hoping to avoid by using the LM317. If you take the voltage, subtract the voltage drop from the LEDs, then divide by the resistor rating in ohms, you can calculate how many milliamps you are running the LEDs. At 11.9 volts (engine off), you get 16.2 mA, well below the recommended 20 mA for this LED. At 14.6 volts (engine on), you're at 26 mA and overdriving the LEDs by almost 28%. This is why their brightness varied so much.

Did you simply use the R=1.25/I formula and throw whatever resistor this equation calculates for each array? (I can't see the LM317 in your pics.)

When you say the brightness varied dramatically, do you mean the LEDs were brighter when you started using the LM317?

In essence, that's what I did, but I ended up changing it a little for the luxeons. If they were normal LEDs, yeah that would have worked fine. But with luxeons they like 350mA so there's a lot more power going through them. I wired up 2 lux's in series to use up some of the power, but there was still a lot going through the LM317 so both the LEDs and the LM317 heated up pretty good. Near 100C. I ended up throwing another 10 watt resister in series to dissipate the power to more managable levels. That's what the big yellow thing is in the pics. The LM317 is the black thing on the right. Did you look at the link to YT in my first post? There's more pics there.

When I said the brightness changed dramatically, I meant the circuit that used only resisters and no LM317. With the LM317 the LEDs are steady as a rock. There's no noticable change in brightness even when starting the engine.

Just gonna throw this in as a suggestion... it's a good idea to put a heat sink on those voltage regulator chips because they can get quite hot otherwise. Those look to me like a TO-220 package, and those are so common that Rat Shack would have a heat sink for them. Otherwise, check Mouser or Digikey.

EDIT: Wooooo, 1000 posts! <img src="/forums/images/graemlins/smile.gif" alt="" /> <img src="/forums/images/graemlins/cheers.gif" alt="" />

Yeah, I should probably be using a heat sink but so far so good. With the extra resister in there the temp of the LM is down to 60ish deg C.

On a side note, anybody know how I can wire up 3 watt luxeons? I want to make some LED turn signals that only need 1 LED for each side and I think the 3 watt ones will do.

You should have used something like this for your 1-watt luxeons:
R = 1.25/0.35 (350 mA) = 3.6 ohm resistor. Should be located where it says "3.9 ohms" on the diagram below.



I haven't seen the specs for the 3-watt LEDs, but hooking up will be the same as with the 1-watt LEDs. Assuming the forward voltage is 7 volts at 700 mA:
R = 1.25/0.70 = 1.8 ohm resistor is needed.

The part that gets me is forward voltage does not come into play in these calculations. So you ask "how many LEDs can I run in this series?" Two LEDs in series will pull the same current (700 mA) as one LED provided you double the supply voltage. And there's the key, you need to have an adequate supply voltage. With a forward voltage of 7 volts, plus the 2-3 volts of "overhead" required by the LM317 regulator, you can only run one of these 3-watt Luxeons per circuit. You would need about 17 volts to run two 700 mA LEDs. (17 volts - (7 volts X 2 LEDs) - 3 volt LM317 drop).

Given this, looks like you could barely run 3 of the 1-watt Luxeons as they generally have a 3.6 volt forward voltage.

Yeah, that's the diagram I started out with, but modified it as I did testing. With only 1 lux everything got really hot really quick. So instead of one luxeon I used two and found that everything still heated up but it was better. So then I threw in a big resister in with the 2 LEDs to dissipate more power and that brought down the temps to a resonable value.

Forward voltage for 1w lux is 3.4V and the 3w is 3.6V. I've done the calcs for the 1w so the problem I run into with wiring up for 3 watt LEDs is the heat. I've got twice the power going through at 700mA so how do I compensate for that? Yeah, I can run 2 LEDs, but I don't know if a big resister in series is going to be enough to keep the temps down. I'm trying to find a different circuit or IC that can handle the current and keep the temps down. I know high end flashlights use pulse modulation but that's too advanced for me.

Alternatively, a company called Cree has come out with a new LED that blows away the 5w luxeon and it only pulls around 300mA. I might have to do some experimenting with those, but they're not cheap at $15 each.

The LM317 is rated for 1 watt without a heatsink and something like 3-4 watts with a heatsink. Use the formula R = 1.23 / I to calculate the resistor you need. R = 1.23 / 0.70 = 1.75 ohm resistor; find whatever is closest to that value. (The 0.70 is the 700 mA 3-watt LED.) Then use P = 1.23 x 1.23 / R to calculate the watts you'll have to dissipate. P = 1.23 x 1.23 / 1.75 = 0.86 watts. As long as you use a resistor with a rating greater than 1.5 watts (to be safe), you will be fine.

The above was used just to calculate the resistor value. Now you have to calculate the power dissipation of the rest of the system. Running two 3.6V LEDs; Vx = max forward voltage with alternator running - LED voltage drop - 1.23 or Vx = 15V - (3.6V + 3.6V) - 1.23 = 6.57 volts. Px = Vx x I or 6.57 x 0.70 = 4.6 watts. Yikes, you're right, I'm not sure how you safely dissipate nearly 5 watts of heat.

I'm sure there is an easy do-it-yourself solution to this problem, but this is certainly one answer to the problem:

This easy to use driver is ideal for powering 3 or 5 watt Luxeon LEDs in automotive applications using a 12VDC power source.

The LuxDriveÖ PowerPuck 700mA LED Drive Module is a true constant current driver designed to drive High-Power LEDs, or strings of 5mm LEDs, from 5-32VDC with efficiency and stability. The different types and numbers of LEDs that can be driven depend on driver current output and circuit configuration. (See the detailed Datasheet for more information).

The 2008 uses filtered switching technologies combined with proprietary drive and sense electronics to provide constant current to LED arrays. The PowerPuck's high efficiency and low cost makes it an excellent choice for automotive and marine applications, portable lighting and solar lighting.

The PowerPuck is an 0.5" x 1.5" dia. encapsulated puck, is resistant to harsh environments and moisture and is supplied with 6", 24AWG colored leads for easy wiring and an aluminium mounting bracket.

Low Voltage DC input power up to 32V
Ideal for Automotive/Marine use
Extremely small form factor* measuring just 1.5"dia x 0.5" tall
Supplied with 24 AWG 6" colored leads for simple connection
Output short circuit protection up to 15 seconds
Output open circuit protection
Designed to power 3 or 5 watt Luxeon LEDs
Will last as long as the LEDs it powers
Fully encapsulated construction
1.5" Diameter x 0.5" Thick

LINK. Price: $17.95

Link for the ones IcySyrup has

http://www.plasmaled.com/high_power_leds.htm