The center LED (on MOST LEDs, and ALL Cree LEDs) pad isn’t ground. It isn’t anything. It can be connected to ANYTHING and it will not affect operation of the LED.
PPtk
Unless it’s a boost driver!
Well, Yes, obviously. It sounded like he was talking about a linear current-regulating driver, but yes, that is a very valid point.
PPtk
Thanks, good to know. I’m planning on direct bonding some LEDs to homemade copper stars (OL made a thread about this) if the 16mm sinkpad thing takes too long to come to fruition.
Yes I was referring specifically to linear regulators in this example. I'm looking at modifying a light to be as small as possible had the brain wave last night that for my specific needs, direct drive was perfect.
skyrider1 - out of curiosity what is this 16mm sinkpad thingy you are refering to?
I did exactly this with an xre in a Lumatic special. Ground off the positive pad on the underside if the led, reflowed both center and negative to the sink, and ran a positive wire from an isolated B+ pad to led plus.
If you direct bond the center pad to copper or sth else with no dielectric inbetween, you should not connect LED- and the center pad. Some drivers regulate the negative path, some the positive..
Might be more accurate to say "If you think that one day you might decide to use a rare driver that does things in a nonstandard way you should not do it, otherwise it's fine."
I would venture to say that the AMC7135 is the most commonly used driver on this forum. It regulates in the return path. This means you probably won’t want to connect LED- to the pill since the pill is directly connected (in most hosts) to battery-
So, “might be more accurate to say if you are using the single most popular driver, or one like it, you probably should not do it, otherwise, it’s fine”
Connecting the Center slug to “Whatever” isn’t a problem at all… Connecting LED- to the pill will be a problem for a low side regulated driver (AMC7135 and many others).
PPtk
I figured that just had to be wrong, no sensible driver would be designed that way, but I just tested it, and dammit, you're right. Jumping the LED - to the pill body completely bypasses the driver. 
I've been mucking around with building my own drivers recently. It wasn't until I started doing this that I realised there was actually a different way to hook up LEDs than what is commonly practiced with the AMC7135 chips. I never knew it was any different because ever since I got into flash lights (about 2 years ago) I barely used anything but 7135 based drivers! I actually think it was Pilot that pointed this out to me when I asked him about a different linear regulator not too long ago...so thanks haha.
Surprisingly, because of the way electronic circuits work, it’s very sensible. Switching the return path is easier and more efficient. An N-Channel Mosfet can be easily used in the return path, whereas a P-Channel must be used in the feed path (an N-Channel can be used, but this requires a bunch of high(er) voltage driver circuitry.) In generalized terms, N-Channel Fets are cheaper, have a lower RdsON, switch faster, and have a lower Gate-Charge requirement. There are many benefits to using an N-Channel FET, and this is why most drivers are low-side switching.
The AMC7135 is but one example. A huge portion of switch-mode buck drivers are low-side switching as well. Boost drivers are ‘usually’ high-side controlled, but even this isn’t 100% true.
PPtk
Another quick and silly question - if you use direct drive, but put a resistor in series on the +ve LED terminal to reduce the current flowing to the LED, would doing the same on the negative terminal have the same effect? Not 2 resistors, just one resistor, but in two different cases...one on LED+ and in the second case one on LED-.
Would the same result occur?
Yes, The result will be identical.
Awesome :) gives me some flexibility in design.
So because my brain never stops ticking over, what happens with triple emitter direct drive?
I'm assuming that each emitter will see 3V max, and thus each emitter will draw whatever current they normally draw at said voltage. Ignoring voltage sag, that means your current draw will triple. At 3V an XPG2 emitter will pull at most 600mA, but if I put 3 in parallel it will try and pull 1800mA. If I want to limit this current draw to a safer level (say 900mA) what do I need to do?
I've seen the resistance calculation written as R=((Vsource - Vforwardled)/Iled). Which in this case Vs=3V and Vf=2.9V (at 300mA, one third of the desired maximum current drain), therefore R=((3-2.9)/300)=0.33ohm.
Do I need 1 .33ohm resistor per LED, or one large 1ohm resistor on the battery input?
The resistor/s I use obviously also need to be capable of burning off 0.1x.3 Watts of power as heat.
Man, I'm all over the shop at the moment...
All of this assumes they are wired in parallel, but that is how most are wired. With 1 CR123 and 1 XP-G2 there will be a 600 ma draw, with 1 CR123 and 3 XP-G2s the draw will be approximately the same. It could end up being slightly higher due to the lower combined vf of the emitters, or it it could end up slightly lower due to the added connections dropping slightly more voltage.
So basically, no matter what combination of LEDs you use (assuming they are white LEDs) you will not need a resistor to drop current; it will always be in a safe range.
Thanks scaru. Maybe because it was late on Sunday evening for me, and now it's early Monday, but I'm struggling to get my head around this. Each emitter is seeing (for arguments sake) 3V. So should each emitter not pull 600mA (for a combined total of 1800mA)? Or will they operate at 2.9V but 200mA each?
It's possible I'm wrong here, but as I understand it this is because the battery can only supply 600ma without the voltage dropping below the vf of the LED.