Re: What does "electronically isolated" mean when referring to LEDs and heatsinking?
The answer that iamerror was looking for is, does he have to worry about shorting his LED. In the case of white Luxeons, the answer is no.
But this is still not true. THere are many circumstances that even a single luxeon can fail if the slug is not isolated.
If the slug gets connected to the power supply's positive node, then it will forward bias the ESD diode on the negative terminal, causing a high current short through the slug, through the ESD diode, and through the negative terminal, burning out the negative bond wire and destroying the LED.
If the slug is connected to negative and the LED is floating with respect to negative because of the power supply topology, you could bypass the negative return path of the power supply in favor of the built-in ESD diode, causing the LED to be dim (at best), not light, or again, destroy one of its bond wires, rendering it useless.
Finally, if the slug is connected to negative, and the power supply input is reversed (i.e. by inserting batteries backwards), it will forward bias one of the two ESD diodes, resulting in the first failure mode described above. While it takes ~2.5V to forward bias the luxeon (the first point it starts producing light), it only takes ~0.7V to forward bias the ESD diodes - so even a small voltage applied in this way can lead to damage.
Since all white Luxeons are binned at at least 350mA, what is the reason for this bizzarre connection vis a vis zener diode?
The ESD diodes have nothing to do with the current rating of the LED. They are there to shunt ESD charges away from the junction to somewhere safe. The ESD diodes will conduct before the LED will, dissipating the static discharge energy instead of the LED taking it. It is entirely possible that electrically connecting one of the power terminals to the slug actually reduces the effectiveness of this setup, causing the device to be more succeptible to an ESD discharge, increasing the likelihood of damage.