You’re probably right, but I’m trying to find a balance between Useful Throw and those unfortunate ugly color rings Cree LEDs tend to produce. The reflectors I’ve tried (even OP) don’t blend the colors at all, and may even exacerbate the “Cree Rainbow”.
Plus, parabolic reflectors, being designed for a point source of light like an incandescent filament; are a compromise with used with a planar source like an LED (even the XP-G — the answer to the OP question). I wouldn’t be surprised to find a properly-constructed TIR could actually ~equal Throw from a reflector, while solving the color problem at the same time. One can but hope.
PS: The 15° and 60° optics show that Full Yellow die image, just so you’ll know…
The yellow die image is referring to the color of the phosphor when the LED is off.
This makes me think of a question: from Dr. Jones thread on LED optics, I learned that the luminous intensity (cd) of a flashlight is equal to the luminance if the LED multiplied by the effective area (I=LA). The effective area is the area of the reflector that the die image seems to fill. So for TIR optics with equal areas but different beam widths, how can they produce different cd numbers, with the same LED and current? My hypothesis is that the wider beam TIRs are either defocused so that the effective area is less than the actual area, or they are diffusely textured, like an OP reflector, also decreasing the effective area. Any thoughts on this?
I am interested in the khatod PL136915 for the S2. I can imagine it actually beating a smooth reflector. It might be able to get rid of the center “dead” hole of a reflector, and might beat the refection/transmission efficiency of the reflector/lens combination.
Thanks. I was afraid I was going to have to post a picture.
Of course, I can’t tell if it properly fills the optic, like it does when the C8 reflector is fully seated, but the angle at which the yellow appears to pull away from the edge “feels” like the angle it’s supposed to be.
