Relationship between emitter size and lumen output for throwing?

For any given reflector size, how would the lumen output of a larger emitter need to scale with its size in order to retain the same throw/candela rating of a smaller emitter?

E.G. (Assuming emitters can be powered to maximum output) If a 50mm SMO reflector had a XHP35 emitter (3.45mm on each side, 1483 max lumen), and then had a XHP70 emitter swapped in (7mm, 4022 max lumen) I suspect that the XHP35 would still throw further.

Is there a way to calculate how many lumens the XHP70 would need to output before it matched the the throw of the smaller emitter? If my assumption is wrong, how many lumens would the smaller emitter need to match the throw of the larger.

Basically, you want to divide total lumens by the size of the die in square millimeters. That will give you the number you need to compare across the different emitters.


Also, it’s the actual light emitting surface you’re interested in, not the package area. For example, the XHP35 die area is right around 2.5x2.5mm

Yeah. Thanks EasyB. I was thinking about that too, but forgot to mention it.

You’re measuring the side of the emitter, which is wrong, you need to measure the size of the die.

I’m not sure about the XHP35, so let’s assume that it is 2.5mm x 2.5mm.
1483 / 2.5^2 = 137.28lm per mm^2

With the XHP70 it is four 2x2mm dies so total 16mm^2
4022 / 16 = 251.375 lm/mm^2

The XHP35 however can go much brighter than 1483 lumens, and the XHP70 can go much brighter than 4k lumens, so instead of looking up numbers on the cree datasheets you should look for tests people have done and pick what current and voltage you would use.

I just wanted to toss a note in, that this is pretty good for approximations for domeless emitters. And for my purposes I haven’t bothered with anything further.
However, if you’re looking for anything with domes, you have to account for how the dome on domed emitters modifies the light emitting area. (I basically researched enough to know that this is a thing, but I haven’t bothered to learn how to do it - however, if you’d like, a good place to start is probably this thread , which also contains/links to other resources on the subject).

Also, a correction: The above math is incorrect (likely a typo). 1483/(2.5^2) = 1483/6.25 = 237.28

Let’s compare the XHP35 Hi and XHP70.2 to show the problem with our calculations/data:
XHP35 Hi has been pushed to 2000+ lumens in flashlights. 2000/(2.5^2) = 320 lm/mm^2
XHP70.2 has been pushed to about 7000 lumens, for 437.5 lm/mm^2.
But the BLF GT out-throws the GT70 significantly, so something is wrong here.