Daniel, thanks for the confidence boost!

I couldn't find a seller that was willing to sell only 4, and I was too anctious to wait - so I bought 4 of the Cree Q5 XR-E and a 3023 DE 1000 buckpuck. I guess I will wait on the high CRI LED's until they become a little more common.

The lights came in yesterday!! Yay!!

Anvit2a, I would very much like to see how you have put together your lights. I like to make sure of what I'm doing before I put a hot solder gun next to some computer chips!!

I was studying the buckpuck pdf and was having problems getting my head around the point that as you INCREASE resistance to the {ctrl} pin the lights would get BRIGHTER...

That seemed counter intuitive and my confusion was definitely not helped by the completely innacurate (as far as I could tell) statement under figure 16: %IOUT = R1/50

I noticed the "Output current vs. control voltage" graph.

Referencing the top line for the 1000 mA buckpuck - If there is zero voltage going to the {ctrl} pin then it will be full bright, and if we tie the 5V {ref} pin to {ctrl} the output will be zero. That all falls in line with what Daniel said earlier in this thread and what he demonstrated in the video.

I finally realised that the voltage going to the {ctrl} does not directly control the output current - instead there is some hidden voodoo magic inside the buckpuck that uses that voltage to LIMIT the current. The more the voltage coming into the {ctrl} the more the current is limited. The specifications state the Control pin adjustment threshold is 1.65V, the Control pin shutoff threshold is 4.2V, and the

Control pin input impedance is 1000 ohms.All my calculations up till now didn't take into account the added resistance of the {ctrl} pin itself. Once I added that value into the equations, all the empirical evidence coincided with the numbers. I made a table if anyone cares - that

should work - showing the % of current output with different resistors:

Resistor

>=2000 ohms Full Bright

1750 ohms 85%

1500 ohms 80%

1250 ohms 75%

1000 ohms 65%

750 ohms 50%

600 ohms 45%

500 ohms 35%

400 ohms 25%

300 ohms 20%

<=200 ohms Off

If you are using the 1000mA buckpuck then the percentage is roughly equivilant to the output current.

Here are the equations I used:

I'll use a 500 ohm resistor for the example. Rctrl = 1000 ohms from specifications.

1) need to find the total resistance between {ref} and {ctrl}. Assumed resistors in series Rtot = R1 + R2... + Rctrl

Rtot = 500 + 1000 = 1500 ohms

2) need to find the current between {ref} and {ctrl} with a given resistor. Ohm's Law: V = I * R; rearranged I = V/R

I = 5V / 1500 ohms

I = 0.0033333334 amps

3) need to find the voltage drop over the resistor(s) - NOT including the {ctrl} impedance. Ohm's law again V = I * R

V = 0.0033333334 * 500

V = 1.6666667 V

That is the voltage DROP so we need to subtract that from the 5V {ref} puts out.

V into {ctrl} = 5 - 1.6666667 = 3.334

If you enter the "Output current vs. control voltage" graph at the bottom with 3.334 and go up until you intersect with the top line (1000 mA) you will get approximately .35 Amps output current (or 35% illumination)

I hope I didn't bore anyone and this helps someone else who may be having "issues". I sure hope I'm not the ONLY one who had issues with this.... lol

Cheers!