Thank you EasyB, you reminded me that I failed to include calculations. I’ll do that now.
According to The Driver, XHP35 HI has thermal pad of 4.11 mm²
I’ll use this number as I suppose it comes from polarity indicator. It’s not a big deal…
I wonder why did you use 50 µm? I have seen no data on what would be joint thickness and used to use the smallest recommended stencil-produced 70 µm. At this point I’ll use both numbers, I have no idea where lies the truth.
Let’s calculate temperature increase on the solder joint when driving XHP35 to its peak (note: peak with SnPb solder; a better joint should move the peak slightly).
With 50 µm bond thickness the temp difference would be:
49W / 4.11 mm² / 50W/mK * 50 µm = 11.9 K difference.
With 70 µm bond thickness that would be 16.7 K difference.
That’s assuming there are no voids in the joint. This is unrealistic. Let’s use 10% voiding and assume that thermal resistance scales linearly with voids (it’s probably worse than that). So the real differences would be 13.1 and 18.4 K respectively.
According to PCT, dropping temperature by 10 K (150°C->140°C) improves flux by 2.5% at maximum current that they specify. Let’s assume that improvements that we make are linear to this gain.
With 50 µm bond we can gain:
table(table#posts).
|Material|ΔT between LED and PCB|ΔT improvement over SnPb|output improvement|
|Sn63Pb37|13.1|0|0%|
|Indium|7.6|5.5|1.4%|
|Conductonaut|9|4.1|1%|
|H9890-6A|4.7|8.4|2.1%|
With 70 µm bond we can gain:
table(table#posts).
|Material|ΔT between LED and PCB|ΔT improvement over SnPb|output improvement|
|Sn63Pb37|18.4|0|0%|
|Indium|10.7|7.7|1.9%|
|Conductonaut|12.6|5.8|1.5%|
|H9890-6A|6.6|11.8|3%|
Theoretically, a 10 µm bond of 9890-6A would bring improvement of 3.1% to 50 µm SnPb and 4.4% to 70 µm SnPb.