R value for CR123 to get 150ma

I have a simple LED light that requires IF of 150ma. I'm driving it with a single 3v CR123. Can someone tell me how to get the resistor value?
THanks!
A

I will give you an example of how to figure it out.

From the Duracell web site OEM data sheet area
- at 150ma, the voltage will be pretty steady at 2.8 - 2.9 V

If you are using a RED LED, it has a Vf less than this (usually around 2.5 V), so you need to add enough resistance to effectively dissipate this extra 0.5 V

Since

Volts = Current x resistance
0.5 V = 0.150 A x R

R = approx 3 ohms (for RED)

If you are going to use a white LED, the Vf is approx 3.5 V, so in theory, one cell of 123 is not enough voltage to drive the LED at all.

In practice, the LED will turn on at approx 3 V. (I have done this with 2 x 1.5 V D cells) If you are not too picky about being exactly at 150ma, you probably can get the LED to light up with just a direct connection and no purposeful resistance.

Hopefully, I did the math right.

Welcome to CPF.

edit - additional info - I assume this is for a small light project you are building. You might want to consider using an "boost" circuit, which will raise the voltage to the proper level, and manage the current, which is the real goal here.

A fairly inexpensive way is to but a Dorcy LED light which is running off of 3 V, and use their circuit. If you search for the CPF members Milkyspit and Doug S, they have posted on using this setup.

You can use a small app to do the calcs, seethis thread
http://www.candlepowerforums.com/ubbthreads/favlinker.php?Cat=&Entry=43305&F_Board=UBB14&Thread=833564&partnumber=&postmarker=

You can either use mine or scroll down to the link for LED_Pro; mine requires .NET 1.1.

The best way to do this is to get a potentiometer and experimentaly find the resistance you want.