Overdrive a Luxeon LED?

Hi all, sorry if this has been answered elsewhere. Here's the question:
At what point is a Luxeon LED overdriven too far? Example: A VV0TW bin which would have Vf of 6.39 - 6.87 voltage drop, if driven with 3 123's, would have 9v applied. Is this dangerous for the LED? Or does it just make it brighter? Should I be using a DB750, or some other means of current regulation?
I'm a newbie to modding, and I want to do it right. Thanks in advance for all the help.

Greg

The stress on the LED is not defined by the voltage. You can run an LED on 1000V, if you want to, as long as you adjust the voltage drop across the LED. Important for the lifetime/brightness of the LED is the CURRENT. If a LED is over/underdriven is defined by the current going through it, not the voltage (the current given in the datasheet of the LED defines the point between over- and underdrive). Given a certain voltage, the current is defined by the resistance of your circuit (Ohm's Law). So in a direct drive circuit, you have the LED, a resistor (for voltage adjustment and current limiting) and the INTERNAL RESISTANCE of your battery. Lithium batteries have generally a lower resistance than alkalines, would therefore generate a higher current with the same voltage. So there's a good possibility that you overdrive/fry the LED with your setup, if you don't use a regulation circuit.

Now, I don't know the internal resistances of different battery types, so I don't know if your setup will or will not work. I would advice you to post your question over on the flashlight electronics board. Some people around here are much more knowledgable than I am.

Welcome to CPF. You can get a good indication of how much current will flow from a look at the battery datasheets (energizer has good ones for 123 type cells). Very roughly, at 1 amp, the cell voltage will drop rapidly to approx 2 - 2.5 volts (x3 cells) putting you in the range of the LED Vf.

If you take your

(battery pack voltage) - (LED Vf) = "extra voltage".

This "extra voltage" must either be dissipated as heat (in a resistor or similar) or you can use a number of constant current power supplies. (examples, CPF'r georges80, LED Dynamics buckpuck, and some from Wayne Y at the "Sandwich Shoppe" (see the banners section)

I have used almost all of these, and they all work fine. Resistors are less efficient, but simple, you just buy a resistor that has the resistance needed, and enough wattage to handle the power. (quick example - since V = IR, just take your 2 Volts = 1 amp x R, so R = 2 ohms.

2 ohms x 1 amp = 2 watts.

(Even a non EE like me can do this)

For a Lux V, a good current target range is 700ma - 1 amp.

Just remember, 5 watts from the Lux V and 2 watts from the resistor - the heat has to be removed or the LED will dim.

I think my math is more or less right, but wait a few days and someone else will improve on the information.

There are people who "direct drive" their LEDs to obtain maximum power, etc. I personally have killed some this way, so I tend not to do this anymore.

Direct drive with 3 x CR123 to a LuxV is fine... as long as it is heatsinked properly. DD'ing with fresh CR123s will put ~1.8A to the LED initially with a lot of heat generated. Get the heat away and the LED will live. Have a dodgy thermal path and pooooooof! Take a look at the graph in the MR-X link below. It shows the current drop over time when using 3 x CR123s and a U5U LED.