Sold/Expired Help installing different Luxeon into Q3

I was wondering if there is anyone on this site that could install a different bin Luxeon into my Q3? I was hoping to get a TWOK installed, but I lack the experience, and I don't want to screw it up. If anyone could help me, that would be great.

don't worry, it won't die. just unsolder the old star and clean the junk goo off the q-iii heatsink. then put your own thermal compound there and stick your new twok there.

I think the Q3 is just about the most upgrade friendly flashlight there is once you get the head off the body.

The only tools you will need is a plain old pencil style soldering iron of around 30 watts rating and a pair of fine tipped straight needle-nose pliers.

Insert the tips of the pliers into the two notches in the black plastic Lux star retaining ring and spin it up and out of the battery tube.
Make a note of which wire is + and which is - and go ahead and unsolder the wires from your original star.

Check the bin code on the back of the old star, it will likely be a S**K or a S**L binned star.
By now you've taken a good look at the aluminium ring the star rests on...that's your heatsink. It's not much to look at but it serves the purpose as long as your not going to overdrive the new star.

Remove the aluminium heatsink ring and apply a very thin coating of heat sink compound on the inside wall of the battery tube that comes into contact with the heatsink ring. The important thing to remember about heatsink compound is that less is more, if you know what I mean. It's not the type of stuff where if a little is good, then twice as much is twice as good. A little dab will do ya.

Now that the inside of the tube has heatsink compound on it, you can go ahead and put a very thin coat of heatsink compound on the outside wall of the heatsink ring.
Carefully slip the heatsink ring back into the battery tube. Spinning the ring while slipping the ring into place will help ensure you've got a very thin even layer of heatsink compound between the battery tube and the heatsink ring.

By applying the heatsink compound in this fashion you've just maximized the amount of heat the original heatsink can transfer to the body of the flashlight. Apply heatsink compound to the top edge of the now back in place heatsink ring and drop your new star in place making sure the wires are sticking up above the star and near the proper pads for them.

Solder the + and - wires to the correct pads and screw the black retaining ring (by putting the needle-nose plier tips into the small notches) down until it makes firm contact with the star. It's crucial for the ring to push the star against the heatsink so make sure the ring is firmly in place.

That's it. Your done!

On a side note, you might want to consider replacing the original metal reflector of your Q3 with a 20mm IMS reflector. The IMS reflector drops right in and gives a slightly brighter and tighter hotspot. They cost about 5 or 6 dollars. Replacing the original Q3 plastic lens with a much more transparent and scratch resistant mineral crystal lens is another popular easy Q3 upgrade as well.

Been there and done this many times.
Once you get it apart without tearing anything up, the upgrade is pretty much the reverse of disassembly.
Be sure to mark the (-) wire with a black marker during disassembly, to help with re-assembly.

If still using a 3 volt nonrechargeable battery, a TWOH star would be the better bet if you can find one.

If using a 3.6 volt rechargeable, the TWOK you were eyeing is a very bright choice.
A Photonfanatic TWOM star could also work fine using the stock Q3 boost circuit, and stay cooler longer.

Ok, I was actually hoping that I could find someone to do it for me (for a price obviously) and find a place to buy a TWOK. Any help?

where abouts are you from?
And after a few posts about bins are you still stuck on the TWOK star?

Rogue_monkey said:

where abouts are you from?
And after a few posts about bins are you still stuck on the TWOK star?

Click to expand...

Upstate NY. Well, I'm not really stuck on anything, but I have heard alot of good things about the TWOK when installed into a Q3. If there is anything better, I definetly want that.

First of all, in general you will get more light output with an LED having a higher luminous flux bin. Thus, a bin number closer to the end of the alphabet would produce more light at a given current level, but these LEDs are rare and highly sought after. You will also have more difficulty finding the color bin you desire in a higher output LED because everyone wants them!
"Better" might depend on where the Vf (last of the four digits in Luxeon codes) for your particular TXOK falls. It also depends on the current level at which you would like to operate the LED.
If you intend to use a 3.6V rechargeable, and if the Vf for your particular "K" voltage binned LED is higher, then the circuit will kick up the voltage to control to 750mA (some QIIIs run at a different current). In this mode the circuit will convert about 17% of the energy passing through it into heat (83% efficiency, which is generous for the stock QIII circuit). So, in the case that Vf for your "K" bin is 3.75V, then the LED consumes 2.81W at 750mA and the circuit dissipates about an additional 0.58W for a total of 3.39W. If the Vf for the LED is lower than 3.6V, then the circuit should go into bypass mode and direct drive off of the battery voltage. This may result in overdriving your LED, which is not recommended with the stock QIII heatsink (I made a custom copper one instead).
I chose to purchase a "J" bin for Vf and ensure that the circuit would not be needed (I removed it too to make room for more copper) by direct driving with a resistor to set the current. For example, assuming 3.6V from the battery and a drive current of 750mA and you have a Vf at the lower end of the "J" bin = 3.27V, then you will need a resistor of 0.44 Ohms which will dissipate 0.25W as heat. (Use LED Pro 2.12, courtesy of Jtice, which can be found on this site to simplify these calculations) In this case the LED running at 750mA consumes 2.45W for a total of 2.70W being drawn from the battery. Thus, direct drive can conserve battery power compared to situations where the regulation circuit is needed. Less energy would need to be dissipated through a resistor for Vf closer to V battery, but for an LED with a higher Vf a higher voltage would be required to achieve a certain current, meaning that the LED has a higher "resistance" (an LED is non-ohmic though) and so more heat will be dissipated through the LED as I squared times R.
So, you also need to consider that the amount of heat that will be generated by the LED is influenced by the Vf bin. The rate at which the battery is drained can also be affected by Vf bin, depending upon whether you use a higher Vf with a regulating circuit or not. Since in direct drive the current at the LED is equal to the current through the resistor, the total heat generated by the circuit is not affected by choice in Vf in this case, but where would you rather have that extra heat? If you need a hint consult the light output vs junction temperature curves from Lumileds. This is why overdriving should not be done without good heat sinking: you might run the LED at a higher power level but not create any additional light while shortening it's life!