I have a 1976 Triumph Bonneville and recently replaced the garbage LED chips inside the eBay signal lights with brighter Cree LED chips. I bench tested the circuit shown below with a 12v bench supply and everything worked perfectly. I also tested it with the motorcycle battery with the same success. However, once I began installing the system and testing it on the motorbike as I went along, I've been burning out LED's (they only glow dimly when fried). I want to know what I am doing to be frying these things left right and center!

I will mention that the bike is positive ground. This means the LED lighting circuit cannot ground to the frame and must have it's own closed circuit directly to the battery. Occasionally when wiring I'll have the battery connected and one of these negative return wired will spark off the bike (potential culprit?). What do you guys think is killing the lights?

I've already come up with some potential solutions:

  1. Adding protection diodes in the reverse direction parallel to the led's (prevent reverse current)

  2. Make absolutely sure the wire attaching to the chips are insulated from the metal lamp housing.

  3. Add a tiny fuse to each leg of the circuit.

Here's the wiring diagram:signal lighting

Thank so much everyone!


  • Have you measured the voltage across and the current through the LEDs?
    – Ignacio Vazquez-Abrams
    Commented Jul 26, 2016 at 17:13
  • How about connecting the battery ground to the frame?. Quick Edit : Don't.
    – ammar.cma
    Commented Jul 26, 2016 at 17:15
  • What specific cree led did you use (link to data sheet and not some ebay site)? How many in parallel?
    – Andy aka
    Commented Jul 26, 2016 at 17:19
  • @Andyaka cree.com/LED-Components-and-Modules/Products/XLamp/… 2 per leg in parallel. Commented Jul 26, 2016 at 17:50
  • @kyle_engineer, how did you know?
    – dlu
    Commented Jul 26, 2016 at 18:01

4 Answers 4


Did you factor in the resistance supplied by the fuse? Fuses will add resistance (of some level or another) and smaller fuses (smaller conductors) will supply more resistance.

But, as noted in the comments, check the voltages at various points (before and after each element).

Also, where is the fuse on your diagram?


Occasionally when wiring I'll have the battery connected and one of these negative return wired will spark off the bike (potential culprit?).

Yes. LEDs are generally very sensitive to voltage/amperage flux and shorting while doing wiring could definitely do that.

I will mention that the bike is positive ground. This means the LED lighting circuit cannot ground to the frame and must have it's own closed circuit directly to the battery.

As a slightly less important point, why can't you draw the positive (+) from the frame and the negative (-) from the power circuits?

In addition, as dlu mentioned, you will be pulling more than 12VDC if you tap the battery directly, or are otherwise not going through a voltage regulator (VR)... So if you are doing a straight battery pull, you need a VR.

  • The resistance of the fuse should help, no? I'm thinking that it would increase the series resistance to the LEDs and reduce the likelihood of burning them out.
    – dlu
    Commented Jul 26, 2016 at 18:00
  • Not necessarily. I haven't done much work with LED in cars, but I have handled a TON for custom interior lighting applications. It's been my experience that LEDs only perform correctly under specific ranges, and that under powering them will also cause accelerated burn-out. Commented Jul 26, 2016 at 18:04
  • Interesting. Guess it would good to go looking and see if Cree has an application note.
    – dlu
    Commented Jul 26, 2016 at 18:06
  • From an LED physics point of view, underpowering can't shorten the life of the LED itself. However it coudl easily shorten the life of an LED module. When working with LEDs for automotive application, you shoudl calculate the resistance required based on the maximum charging voltage rounded up (so call it 15V). But even this doesn't protect against spikes, which maybe the issue.
    – Chris H
    Commented Jul 27, 2016 at 10:50
  • I think you're right about the voltage regulator. The problem I think is that I may have sparked while bench testing (killing 2 chips) and the other being that I ran the circuit with only one led connected. Speaking of which, if I did only power one led off the circuit, would I have a larger voltage being dropped across the led causing it to fail? Commented Jul 27, 2016 at 22:58

There are two things here that sound like possible culprits:

  1. When the bike is running the voltage seen by the LEDs will be the output of the alternator or generator, which will be around 14.2 V, depending on the LED you're using that might be enough to cause problems.

  2. The sparks. They are not a good thing. They imply that there was a sudden rush of current. Off hand I wouldn't expect much current to be drawn by the LEDs in their correct configuration, so I'm not sure if you would see a spark in normal circumstances (if the circuit had enough resistance to limit the current to the LED's spec).

It sounds like what you are trying to do it to upgrade the LEDs in an already working circuit. In that case it would be good to look at what you had and what you're doing with an eye to what is different. It would be good to understand why any changes were made – or at the very least to know what the changes are and be able to ask about them.

I'm curious about the positive ground. I can't see any theoretical problem with running a separate ground for the turn signals back to the battery, but it seems like it adds a complication and may lead to confusion down the road. Couldn't use modify the circuit to accommodate the positive return and do your switching on the negative side? It seems like that would avoid possible problems with the return path and the confusion of having two different wiring conventions on the bike.

Looking at the data sheet for the Cree XPG2 I see that reverse voltage spec is only 5 V. That makes me think that it would be pretty easy to fry one of these if you hook it up backwards even for a second. So be very careful of that, both as you work and in your circuit design.

  • If his 1976 has a similar alternator to the Lucas RM21 in my 1973 Norton he's probably not getting 14.2 Volts until 2 or 3000 rpm with the lights off and higher and higher with the lights on.
    – Eric
    Commented Sep 18, 2016 at 0:07

The calculations I have done show that you will only be driving 0.5 amp through the LEDs when the supply is at 14 volts, so you should be ok, since they can take 1.5 amps.

Touching the negative wire against the chassis will be shorting the battery, and so bypassing the LEDs, so they shouldn't be damaged.

The only thing I can think of is that you haven't got a sufficient heat sink. These LEDs rely on the surface that they are soldered to to keep them cool. Have they got sufficient cooling? If not they will quickly get hot and become damaged.

Possibly on the bench running off a battery at 12volts, the lower current will stop them getting hot enough to damage them.

  • The led chips are soldered to a small 20mm heat sink. I tried to design the circuit so that each would be dissipating about 1W. The other thing I realized is I turned the circuit on when there was only one LED connected. Because that single led would draw less current than 2 (normal turnsignal operation) or four (4 way on), I figured that you would have a much larger drop across the chip than the maximum 3-ish volts. I am only using resistors, not a voltage regulator- so when I change the number of led's connected, I change the forward voltage drop across them...right? Commented Jul 27, 2016 at 22:51

Could the diodes in the rectifier be leaking voltage in the reverse direction? I could see that causing your problem but the battery wouldn't charge very well. Do you have the original Lucas rectifier?

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