I understand the problems with parallel LEDs sharing a current source. However, I came acrosse this LED from Cree, which, in one configuration described in the datasheet, has its internal LEDs electrically in parallel: 9V Config

It seems like Cree knows what they're doing with LEDs, so there must be some reason this works. I am imagining some sort of extremely tight binning scheme for the individual chips before being put into the modules, and I supposed they're pretty well coupled thermally inside the package. My understanding is that they would have to bin not only based on forward voltage but also based on temperature coefficients, which seems infeasible(?)

If you look very carefully you will see the electrical connection putting the two strings of LEDs in parallel are outside the package, and they describe an alternate configuration where the strings are put in series. I also can imagine another setup where each string gets its own constant-current driver (or really beefy resistor), but that adds a lot of cost and complexity.

My question is, will using a Cree XLamp MHD-E series LED in a 9-V configuration degrade the performance or lifespan of the LED, and if not, how is this possible?

  • \$\begingroup\$ LED's are driven by current, not voltage. But your right in that this LED needs at least 9 volts to push any current through it. This also seems like and LED that needs to be epoxied to a metal plate so it does not burn up. \$\endgroup\$
    – user105652
    May 20, 2016 at 19:41
  • \$\begingroup\$ Could you please clarify your questions by putting then at the end of your observations and analysis. \$\endgroup\$
    – user105652
    May 20, 2016 at 19:47
  • \$\begingroup\$ I don't understand how I'm implying that LEDs are driven by voltage, or even what that would mean. My use of the term "9-V configuration" is borrowed from the datasheet. I don't see how I could avoid referring to forward voltage as it's central to the problem. I will try to edit the question to put the questions at the end, let me know if I get at what you wanted \$\endgroup\$
    – user371366
    May 20, 2016 at 20:56
  • \$\begingroup\$ To respond to your epoxy comment, it's meant to be soldered to a PCB, with thermal vias and a large copper plane / heat sink. The middle pad is not electrically connected to anything, meaning you could connect it to your ground plane even if you use low-side switching, or connect it to a floating poly and heatsink if some weird requirements need this. \$\endgroup\$
    – user371366
    May 20, 2016 at 21:00
  • 1
    \$\begingroup\$ Yes you're right, paralleling leds without some kind of current sharing can be problematic. Cree is most likely using the fact that (I assume) all those leds were next to each other on the wafer when they were processed so they probably have pretty well matched characteristics. Just FYI current sharing problems can be an issue even inside a single large led, one small spot gets a bit hotter than the rest of the die, its Vf goes down, it draws more current... you know the rest. But if anyone knows what they're doing, it's Cree, so they've probably got some trick we don't know about. \$\endgroup\$
    – Sam
    May 20, 2016 at 22:11

1 Answer 1


There are three points that may explain how it is possible:

  • The led dies that are grouped are matched. Either they do the matching during the manufacturing of the whole led group, or they know they will always match because they take them from the same wafer and they made some tests that show it is enough to guarantee it works.
  • The substrate guarantees enough uniformity in the dies temperatures. So that if a led sees a little more current than another, the excess temperature is spread over the other leds and the current difference does not get much worse.
  • The Vf/If relation allows this. If you look at the Forward Current vs Forward Voltage graphs in the datasheet, you'll see that over the interesting range, the relation is actually almost linear (although steep). If you extrapolate these graphs to get the Vf/If of a three-die string, you'll see that if you have a Vf mismatch of +/-0.25V (over the whole string), it will lead to a mismatch of +/-50mA (relating to a current of typically 200mA for a single string).

So, if Cree measured that, for example, +/-0.25V was the maximum Vf mismatch they could have, considering different dies on a single wafer, and maximum temperature mismatch between dies once manufactured, they just have to account for the fact that, in the worst case, one string may see 150mA and another 250mA. That is manageable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.