Improve thermal extraction on xre

[saabluster]

Improve thermal extraction on xre(updated W/pics pg 2)

I have an idea that I thought I would put before everyone here and see if this might be viable. I thought about increasing the current to the xre beyond 2 amps and I think to do so will take even better heat extraction. On my "DEFT" I'm running it at 1.76A with it soldered directly to a copper plate and it does fantastic. Absolutely no problems with heat. But I would like to punch it up to around 2.5A on the new R2 I've got coming.

The question is this. If I sand down the ceramic base that the xre is built on to about half the thickness(or more) it currently is, will this improve the thermal resistance of the package? If it does one would assume more power could be sent into the LED knowing the extra heat could be extracted quick enough.

I would think with it being so thin it would be very easy to break. So to counter this I would solder my leads on first then add epoxy on the top side of the ceramic to give it strength. Then I would proceed with the sanding. What do you think?

 

[VanIsleDSM]

Well.. the metal pad on the bottom is good for distributing heat.. there are a bunch of little holes in the PCB filled with metal that connects the pad on the bottom to the LED junction.. taking away that pad may not be good, but a thinner PCB may pay off more than losing the pad will hurt you.. I guess you'll just have to try.

I solder wires onto the top, so always sand down the electrical contacts on the bottom so they don't short on the heatsink.. but I've never sanded down the heat pad.

 

[frenzee]

XR-E's output pretty much levels off by the time it reaches 2.0A. But if have an extra emitter you don't mind destroyuing, I guess it might be worth the try.

 

[jtr1962]

XR-E's output pretty much levels off by the time it reaches 2.0A. But if have an extra emitter you don't mind destroyuing, I guess it might be worth the try.

Not to mention that the limit of the bond wires is roughly 2.4A. In other words, it doesn't matter how well you heat sink it. If the bond wires fuse past a certain current, that's it. For the XR-E 2 amps is about the limit of safe operating current. Much beyond that, and chances are good the emitter will open circuit.

 

[saabluster]

Well.. the metal pad on the bottom is good for distributing heat.. there are a bunch of little holes in the PCB filled with metal that connects the pad on the bottom to the LED junction..

Not exactly. That metal on the bottom is not thick enough to distribute or spread any heat. It just allows for solderablity. Also the "holes" do not go thru the ceramic. Newbie disproved that idea some time ago.

 

[saabluster]

XR-E's output pretty much levels off by the time it reaches 2.0A. But if have an extra emitter you don't mind destroyuing, I guess it might be worth the try.

That was with a P3. The arc has changed somewhat with the latest LEDs. Also I certainly don't mind sacrificing a few leds.

 

[saabluster]

Not to mention that the limit of the bond wires is roughly 2.4A. In other words, it doesn't matter how well you heat sink it. If the bond wires fuse past a certain current, that's it. For the XR-E 2 amps is about the limit of safe operating current. Much beyond that, and chances are good the emitter will open circuit.

Just the man I was hoping to hear from. I was wondering what the limitations were for the bond wires. Thank you. Do you know this from personal experience? I'm sure cree doesn't list this as they wouldn't dream of anyone taking their leds anywhere near these currents. Here is an example of someone running it at 2.5A with no problems.

The question still stands. Would a thinner substrate have less thermal resistance? The benefit would be cooler die temps which would give more lumens at the same drive level along with better lumen maintenance.

 

[jtr1962]

Just the man I was hoping to hear from. I was wondering what the limitations were for the bond wires. Thank you. Do you know this from personal experience? I'm sure cree doesn't list this as they wouldn't dream of anyone taking their leds anywhere near these currents.

I vaguely remember a few posts from Newbie, or at least I think it was him, mentioning that the XR-Es open circuited at over 2.4A. Of course, this figure isn't exact. Some may do better than others. You could try it if you don't mind potentially losing a few parts. However, the difference between 2 amps and 2.5 amps, even on a perfect heat sink, would likely be under 10%. The main limit here is the thermal impedance between the die and package. I think it's 7°C/W. 2.5 amps means a power dissipation of roughly 9 to 10 watts. That means the die temperature is ~70°C above the package temperature, or probably in excess of 100°C. Not a good situation at all. If you try it, let us know the results, and if you get any additional light at all.

Regarding a thinner substrate, yes, it would have less thermal impedance. Assuming that the substrate is alumina ceramic (thermal conductivity about 1/7 that of copper) and 1mm thick, thermal impedance is on the order of 0.5°C/W. Cutting the thickness in half would reduce this to 0.25°C/W. At the power levels we're talking about that's a reduction in junction temperature of not much more than 2°C. This wouldn't make any measureable difference at all in light output. Even reducing thickness to zero wouldn't help much.

 

[saabluster]

The main limit here is the thermal impedance between the die and package. I think it's 7°C/W.

Its actually 8°C/W. Thank you VERY much for your input. I think I'll try it out just to have some cold hard facts to post.

 

[SteveDavis]

Another issue is that even if the heat pad is not thick enough to spread heat, it is still a part of the heat extraction mechanism. The ceramic is bonded to the heat pad very well, and then the heat pad is soldered to the PCB, giving a good thermal junction. By removing the heat pad, you roughen up the surface, yielding air gaps between your heat sink and the die. Air is a fantastic insulator.

If you could sand the ceramic smooth enough, and redeposit metal to better extract heat from the ceramic, then soldered the heat sink to the heat pad as you do now, then maybe you'd improve thermal performance.

 

[evan9162]

Assuming that the substrate is alumina ceramic (thermal conductivity about 1/7 that of copper) and 1mm thick, thermal impedance is on the order of 0.5°C/W. Cutting the thickness in half would reduce this to 0.25°C/W. At the power levels we're talking about that's a reduction in junction temperature of not much more than 2°C. This wouldn't make any measureable difference at all in light output. Even reducing thickness to zero wouldn't help much.

This is the crux of the issue. The majority of the thermal resistance is not in the ceramic substrate, it's in the junction->substrate interface. The junction is the photon generating region of the LED, sandiwched between the layers of semiconductors that make up the LED. The semiconductors aren't great thermal conductors, at least not as good as the remaining parts of the package.

The Seoul P4 uses the same LED dice as the XR-E. Instead of a ceramic base, they use a silver coated copper slug. Yet this only reduces thermal resistance to 6.9C/W. Again, the limitation is in the LED die its self, and not the package. Only Cree can do anything to significantly improve the thermal performance. Unfortunately, we cannot know for certian the best possible performance of the chips in the XR-E or P4 (the cree EZBright 1000), since Cree does not document the thermal resistance of the bare LED die in their datasheets.

We can see the same thing with the TFFC K2 and "classic" K2. The older K2 had a thermal resistance of 9C/W, but the TFFC K2 is 5.5C/W. The package didn't change a bit, but the die structure is completely different. For one, the sapphire substrate is gone, which was one additional layer between the junction and thermal interface for the die. The use of thin films probably helps in this as well.

 

[Illum]

you could flood the cree emitter with thermal epoxy until its even with the die's metal ring and divert a second thermal path to the copper heatsink that way

very impressive light by the way :wave:

 

[saabluster]

I wonder how Osram got the new Diamond Dragon to 2.5 K/W. Thats VERY low. Can you imagine if the cree was this good.

 

[saabluster]

you could flood the cree emitter with thermal epoxy until its even with the die's metal ring and divert a second thermal path to the copper heatsink that way

very impressive light by the way :wave:

Thanks! I was thinking of doing just that. It probably will not make very much difference though as the vast majority of heat is going out the bottom. But when you're running it on the ragged edge every little bit helps.

 

[Illum]

well, my original thought would be that even with thru-hole vias under the PCB, the majority of the heat will still accumulate on top of the PCB. Since the PCB material doesn't dissipate heat too well, I figured thermal epoxy can divert the heat elsewhere and correct to the copper heatsink more effectively.

I drive my X-RE's at 300ma for homemade fixed lighting purposes...I didn't need to worry about a overly large heatsink but I still potted the whole emitter assembly under thermal epoxy on a 5" x 1.5"x 1/8" aluminum bar, showing only the ring and dome. it runs all day in lukewarm temperates with no observable ill effects

 

[evan9162]

I wonder how Osram got the new Diamond Dragon to 2.5 K/W. Thats VERY low. Can you imagine if the cree was this good.

From the article:

The LED is based on a 2 mm2 chip manufactured in Thin-GaN technology with chip-level phosphor coating.

There's twice the area to transfer heat, which would result in the junction->substrate thermal resistance being cut in half. That's the biggest part of the reduction in thermal resistance. Better materials in the die also help.

This is much like when the Luxeon V came out over 5 years ago. At the time, all other power LEDs had junction->package thermal resistance ratings in the 15C/W range. The Luxeon V was 8C/W, using the same dies and same package - but since there were 4 dies, the junction->substrate thermal resistance (talking about the set of 4 dies as a single LED) was 1/4.

Increasing die size is one way to significantly reduce thermal resistance. Of course, you end up with a larger source volume and run into other constraints because of it (current spreading, thermal hot spots, semiconductor defects, etc). These seem to keep LED makers from persuing large die soltuions too much - we've had 1mmx1mmm dies as the "standard" for power LEDs since they came around.

 

[2xTrinity]

7°C/W. 2.5 amps means a power dissipation of roughly 9 to 10 watts. That means the die temperature is ~70°C above the package temperature, or probably in excess of 100°C. Not a good situation at all. If you try it, let us know the results, and if you get any additional light at all.

Keep in mind that the LED will be radiating about 25% of its power as visible light, which won't contribute to heating the package, which would mean closer to a 55C° difference.

 

[SteveDavis]

evan: The Diamond dragon die has 4x the area of the XR-E die, not 2. 2mm x 2mm / 1mm x 1mm = 4.

 

[evan9162]

evan: The Diamond dragon die has 4x the area of the XR-E die, not 2. 2mm x 2mm / 1mm x 1mm = 4.

If that's true, then the article is poorly written. It gives the impression that the total area is 2mm^2 (the exact quote is "The LED is based on a 2 mm2 chip manufactured in Thin-GaN technology with chip-level phosphor coating"). That to me says that the die size has a total area of 2mm^2, not that it's a 2mmx2mm square (4mm^2 of area).

If the diamond dragon is indeed 4mm^2 of area, then all of the thermal resistance improvements are due to the size increase (that would probably be a 9C/W device if it were 1mm^2)

 

[jtr1962]

Keep in mind that the LED will be radiating about 25% of its power as visible light, which won't contribute to heating the package, which would mean closer to a 55C° difference.

That would be correct running in the neighborhood of 350 mA where efficiencies are indeed 25% to 30%, depending upon bin. At 2.5A amp efficiency would be in the neighborhood of 45 lm/W. This means only about 13-14% of the power comes out as light. Assuming 10 watts in, we have 8.6 watts going to heating. Thermal impedance is 8°C/W, so the thermal rise is close to 70°C. You could probably add another 5 to 10 degrees for the thermal interface between the emitter and heatsink, so that puts you at least 75°C over ambient. Assuming the bond wires are up to it, 2.5A is definitely right on the edge with these parts.