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I would like to replace the bulb based tail/brake light and rear indicators/flashers on my bike with one integrated LED unit (flashers in brake light).

These are pretty expensive, and since I'm on a budget I'm looking at building one myself.

I've found some seemingly suitable 12v LEDs. I was hoping to use these to save me bothering with limiting resistors. However, one thing that bothers me is the max voltage rating of the LEDs and the rather unpredictable supply voltage. I've not yet measured it on my bike, but usually you would expect about 13v with the engine running. The absolute max voltage rating of the LEDs I'm looking at is 14v - do you think this is too close? Would the LEDs stand a higher voltage for short periods? (e.g. poss slightly higher V when at high engine revs when overtaking)

How would you protect against this? Would one resistor for the whole set of LEDs suffice? (one resistor supplying multiple parallel LEDs). Any suggestions on what kind of resistor to use? Perhaps it would just be better to use standard votage LEDs if I am forced to use resistors?

Anyway - any advice appreciated!

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If the voltage from the battery is fluctuating - you could look at using a voltage regulator to keep it stable. You can get fixed value regulators (no resistors or other components required!) or variable ones (2 resistors are used to set them, but they're both quite easy to use). The one I've used, LM317T adjustable voltage regulator, requires the input voltage to be around 2v more than the output voltage (in order to stay stable) and they have a maximum input to output differential of 40v. In other words - you want to have LEDs that are rated at a lower voltage than your supply - then use the voltage regulator to reduce the supply and provide the correct voltage for the LEDs.

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  • \$\begingroup\$ Yep, I'd recommend a regulator... \$\endgroup\$ – Earlz Apr 21 '10 at 0:30
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There will probably be some fast high-voltage transients on the supply, which will kill any LEDs. Transient suppressors would be advisable.

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  • 1
    \$\begingroup\$ In any automotive application you get nice spikes as other elements of the electrical system turn on and off (like the headlights). Just spreading knowledge I learned elsewhere on this site :) \$\endgroup\$ – jnylen Nov 1 '10 at 5:23
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Using a linear voltage regulator like an LM317T is a decent idea, but it means you're guaranteed to throw away 2 V, which is a little disappointing. You might instead try a low-dropout regulator like the LT3085-- with that, you'd only lose 0.275 V.

But in both cases, you'll be limited by the relatively low current limit of the regulator (1.5 A for the LM317T, only 500 mA for the LT3085). For a bike light, either would work fine, but I think you could build a more robust circuit without a regulator.

Here's what I'd suggest.

First, cap the voltage with a 13 V zener diode between power and ground. A zener diode is a diode that acts mostly like a normal diode-- a one-way valve. However, in the no-flow direction, it blocks current until you hit a certain threshold voltage. Once you hit that voltage, it opens up with very low resistance until you drop below the threshold again. It's great for capping voltage spikes. Be sure to get one that can handle the power you'll me pumping through it. Maybe start with a 5 W diode and then buy a fatter one if that burns?

Second, you need to limit the current through the LEDs. (If you did this part right, you might not even need the zener, but zeners are cheap.) LEDs burn out because they get too hot, and prolonged high current is what makes them hot. They can handle very high currents for short periods. Whatever LEDs you buy will have a current rating, like 20 mA at 4 V. There are no LEDs with voltages over roughly 5 V. The higher voltage LEDs you see advertised are either multiple LEDs in series or LEDs packaged with a current-limiting resistor. You can do better with discrete LEDs.

What you should do is put LEDs in series until you get close to your supply voltage-- maybe 3 of them, so you're trying to supply 3 * 4 V = 12 V, 20 mA. Then, take the remaining voltage between the cap (13 V) and the LED voltage (12 V) and pick a resistor that allows the right amount of current. In this case, 13 - 12 is 1 V, and you want 20 mA, so that's 1/0.020 = 50 ohms. I'd build one string like this and test it. Once that's working, build a few more. You can tweak the resistor value to make the LEDs a little brighter or dimmer (but don't burn them out).

Putting LEDs in parallel without giving each series string its own resistor is risky. LEDs are nonlinear elements-- raise the voltage a little near a certain threshold and the current changes dramatically. This means that they tend to not share current evenly. You might be able to pull it off if the LEDs are all from the same manufacturing run, but resistors are almost free; I wouldn't think it would be worth the savings.

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First of all, run the LEDs in parallel to share the current load, and also if one dies only the dead LED will not come on; if you have them in series if one dies they all switch off, which makes finding the dead one a PITA.

Secondly, Jim's suggestion of using an LM317T is a good one, don't forget to use a nice smoothing capacitor on it.

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I think the magic is to put a lot of led:s in series, that what you spread the voltage over a handful of led:s.

14V/7led:s = 2V over each led.

But my gut feeling tells me that you need to do more than this....

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