I have this LED.

And I have this driver.

I'm trying to use it to drive the LED however when I hooked up the one LED to the driver it started smoking. I measured a voltage of 48V while it was working but before it started smoking.

I've been looking over the data sheet to see why, if the driver is rated for 10-43V I would be seeing 48V. I noticed that the open circuit voltages for the driver is 48.

Can anyone help me come up with a way to drive this LED using this power supply?

  • \$\begingroup\$ That package is a multi-chip array, so the # of LEDs in series shouldn't matter. For a constant current source, you don't need any resistors in series. Do you have the 500 or 700mA recom constant current source model? Did you heatsink the LED? How long did it run before it started smoking? I am guessing you overdrove it and the package got too hot and started smoking. \$\endgroup\$
    – user16461
    Nov 27, 2012 at 22:56

2 Answers 2


According to the datasheet for your LED driver, it has a 3 LED minimum load. The 10-43V rating of the driver equates to 3-12 LEDs wired in series. Driving a single LED will probably cause it to lose regulation with the smoky result you witnessed.

Get two more LEDs (3, preferably - I wouldn't re-use one that had produced smoke) and the driver should work well for you.


What smoked - driver or LED?

Device is rated for 3 LEDs minimum.
Exactly what happens depends on their circuitry but fewer LEDs may cause more power dissipation in driver.

Add a resistor in series with LED if using one LED.

LEDs drop about 3.5V.
If using 500 mA version Rmin = V/I = 2 x 3.5 / 0.5 = 14 R.
Try say 22R.
Rpower ~= I^2R = 0.5^2 x 22 = 5.5 Watt
Use 10 Watt 22R resistor.

For 0.7A version use say 15R resistor.

  • \$\begingroup\$ The LED smoked. I'm looking to avoid using resistors as much as possible. Do you think using two LEDs in series will solve this? Efficientcy is our main concern hence not wanting resistors. +1 for the math though. \$\endgroup\$
    – Mike
    Sep 7, 2012 at 2:45

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