What's the easiest way to mount a bunch of surface mount LEDs into an array or spotlight?

There are a bunch of very interesting new papers on red light therapy for a variety of conditions. I want to experiment with it. The most beneficial spectrum is around 650nm. I can't seem to find any bulbs that come ready made with the ~650nm LEDs in them. If you know otherwise, please speak up.

So I figure I can get thirty or so of these Phillips "deep red" LEDs, mount them parallel, and connect to a suitable power supply:

Philips Lumileds LXM3-PD01

I'd like to avoid having to solder them on to stripboard or something like that. Partly because I've never worked with SMD and haven't soldered in many years. Can anyone think of an easy way to mount them?

There are a bunch of LED spotlights that come with white LEDs already installed:

LED Spot Lights

Is there a way to get the fixtures without the LEDs? Should it be easy to remove the white LEDs and refit the Phillips red LEDs? I don't have any experience with such things.

  • \$\begingroup\$ Peopel will sell them to you mounted on "Stars"- for extra money of course. Such as here shop.stevesleds.com/… People who sell mounted LEDs may be willing to custom mount some for you. Street lights often use arrays of LEDs and such MAY be suitable for mounting your LEDs. As others have noted, heat dissipation needs proper attention. \$\endgroup\$ – Russell McMahon Jul 18 '14 at 16:10
  • \$\begingroup\$ Those stars mounted on heatsinks with some paste seem like the way to go. \$\endgroup\$ – rjt_jr Jul 19 '14 at 15:03
  • \$\begingroup\$ The stars also need heatsinking - but it's a lot easier to connect them mechanically than the LEDs themselves. Luxeon publish some excellent documents on LED thermal issues. \$\endgroup\$ – Russell McMahon Jul 19 '14 at 21:36

Yea, you can mount them on a circuit board just like everything else. The Phillips LEDs you link are meant to be mounted on a circuit board that is well cooled (heatsink and/or fan) as these are high-power, specifically ~750mW for red.

There are similar LEDs called Luxeon Rebel which look strikingly similar to those phillips ones which you can buy breakout boards that don't have any of them mounted already. I believe they are sold by sparkfun.com

Since your application sounds like it is targeted at a specific area, the solution needs to be compact. You'll need to design a circuit board and get it fabricated. You'll need to be able to draw all the heat away from the board since you'll be keeping it on for 15 minutes. This means a wicked heatsink and a beastly fan which mean this won't be the most silent device.

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    \$\begingroup\$ Philips Lumileds is the manufacturer of the Luxeon brand. \$\endgroup\$ – placeholder Jul 18 '14 at 14:14
  • \$\begingroup\$ Okay, I thought like, "Hey, they might be the same thing buddy" Thanks! \$\endgroup\$ – Funkyguy Jul 18 '14 at 14:17
  • \$\begingroup\$ The idea here is to have thirty or so in a fairly tight array held very close to the skin for about fifteen minutes. You want to blast the tissue with 650nm light. Those breakout boards don't look practical for such an array. I think a household box fan blowing on the array might be good enough for cooling? But then I don't know that some stripboard is a good enough mounting considering the potential heat. \$\endgroup\$ – rjt_jr Jul 18 '14 at 14:20
  • \$\begingroup\$ You will not be using strip board for this. The application honestly requires a custom circuit board. Professionally fabricated, you can use OSHpark or a better place if you want. The biggest concern for this is power, you need to be able to supply enough current at 2.1v and then be able to evacuate all the heat produced. Its going to need a wicked heatsink and a killer fan since this sounds like a fairly compact application. And for 15 minutes, yea you need a beastly fan and heatsink. \$\endgroup\$ – Funkyguy Jul 18 '14 at 14:34
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    \$\begingroup\$ The reason LEDs run so much cooler than e.g. incandescent is not because losses to heat are small, but because the much higher relative efficiency means the ratio of light to heat is much higher. (And also because high temperature isn't a necessary part of the physics) \$\endgroup\$ – Ben Voigt Jul 18 '14 at 18:30

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