Luxeon voltage question

This is my first post so please bear with me.

I have read allot and just finished making a small task lite.
I used a Luxeon Star O with a 9volt dc wall wart feeding
a LM317 for the driver



It works great and I am getting a regulated current of just under 350ma.

Now my question is about the voltage. At 1.25 volts is this not under the voltage required to light up a Luxeon.
The light output is great but I need to understand the relationship to voltage and current before I make up a new driver for a LuxIII.

Thank you , and this is a great forum.

As shown in your circuit, the 317 will regulate the voltage across R1 at 1.25V. If you use 3.6ohm for R1, the current through it will be ~350mA. And because current is equal in all parts of a series circuit, a Luxeon (or two, or three, etc.) will also have ~350mA flow through them.

This current regulation requires only that the supply voltage be greater than voltage drops of each part in the circuit. R1 will always be 1.25V, each Lux will be 3.2V-3.7V (depending on bin), and the 317 itself needs about 2V to maintain regulation. When the supply voltage is higher than needed, the excess voltage appears across the 317 and is dissipated as heat. The amount of dissipation is calculated as:
P(power in watts) = V(volts) x I(amps).
In your example with one Lux and 9 Volts supplied, the voltage across the 317 is probably 4V, times the 350mA, equals 1.4 Watts. The 317 could probably use a heat sink...

I hope this helps a little. What was the question again?/ubbthreads/images/graemlins/twakfl.gif

Crux

AFAIK, the 1.25v is used by the regulator to determine what current it's being "programmed" for output.

Based from Ohm's law:
I = E/R

or

output current = 1.25v/r1

@ ~350ma or .35amp out
r1 = 1.25/.35
r1 = 3.57ohm

Thank you both.
Think I have got it