Abuse kills......
"What kills Luxeons?"
This one's simple, too much current.
Does it one of several ways. Too many amps times volts (watts) for too long a time without enough heatsink cooks 'em. Higher currents will eventually cause failure, no mater what sort of heatsink. Think of that thin bond wire you can see if you squint hard and look sideways through the lens as a bitty fuse.
So, how much current are we talking about? Voltage bins J and K (at 350 mA) are 3.27 to 3.51 and 3.51 to 3.75, let's say 3.5 volts. This is to say a 'direct drive' of 3.5 volts will give us (full rated) current of 350 mA, the maximum the maker intends. Typically the one ohm internal resistance accounts for .35 Volts ( .35 Amps times one ohm) of that total, meaning 3.15 Volts 'across the diode itself'. Now we can add the internal resistance of our three D cells (.136 ohms each from <
http://www.duracell.com/oem/Primary/Alkaline/mn1300.asp> ), for a total of 1.408. Assuming 1.5 Volt cells, the total drop across this resistance is three times that less the 3.15 Volts, 1.35 Volts. This 1.35 Volts divided by the 1.408 ohms (since it 'appears across it') gives us a current of nearly an amp (.96). Nearly three times the 'absolute maximum' number specified by the maker, the point he guarantees the bond wire not to fail for instance. Over 3 Watts.
Interesting things can happen with heat. For instance the -2 mV per degree C number. A 100 degree rise (not itself fatal) will cause our 3.15 Volts to drop to under 3 Volts (2.95). This will further raise the current as the voltage 'left across the resistance' goes up. It's now 1.55 divided by 1.408 for well over an Amp (1.1 in fact), which of course heats the diode still further......
This is a slug temperature of 58 degrees (18 degrees per Watt times 3.245 Watts) lower for a heatsink that's 'warm not hot'. Yet the junction temperature is at the "absolute maximum" of 120 (assumeing a 20 degree C start). This assumes perfect slug to heatsink bond, of course not possible, our real world results will be even worse.
There's also the matter of saturation. There are only so many electon/hole pairs there, and the electrons can only move so fast, try to force more through and the voltage drop skyrockets (and with it heat).
Just the extra .3 ohms from using C cells probably kept you alive then? It represents what, a 20% reduction?
You're basically fooling around with the 'direct drive' condition CMG *didn't* use in the Reactor III. This was with AAA cells with even higher (.181 ohm) internal resistance. You really should consider adding some resistance. Using an additional two ohms or so (like they did) will hold your 4.5 Volt current to (1.35 divided by 3.408) about 400 mA. Full rated output (plus a bit, at least at the start) and *increasing your run time* more than proportionally.
I know there are lots of folks out there hotrodding stars into this range, but then again they're also having some "accidents" as well.
At least that's how I see it.
Cheers.
Doug Owen