LM317 Question

[MaxMouseDLL]

Hello All,

I'm currently in the middle of my first laser build (Link), and am following the LM317 driver build here, with a 300-400mW red laser diode/extracted from LPC-826 (link).

The LM317 build I'm following suggests two 10ohm resistors in parallel, which produces 250ma of drive current, however the laser I've chosen suggests a drive current of 400-500mA.

So, my thoughts are to half the values of the resistors, thus doubling the current (2x10ohms Parallel = 5ohm resistance = 250ma - 2x5ohms Parallel = 2.5ohm resistance = 500ma), however (for some reason) I can't seem to find any normal 5ohm resistors. could I just connect two 10ohm resistors in parallel and then connect those two sets in parallel, like this (Forgive my horrible paint drawing):

Could someone please check my reasoning and math? it's been a very long time since I was in college.

Reference voltage of the LM317
(always 1.25v)

1.25 / 2.5ohm = .5A

.5A X 1000 = 500ma

 

[rhd]

You're basically just connecting 4x 10 ohm resistors in parallel. This produces 2.5 ohm, and 500mA from a DDL.

 

[MaxMouseDLL]

You're basically just connecting 4x 10 ohm resistors in parallel. This produces 2.5 ohm, and 500mA from a DDL.

So my math is correct... thank you!

I had to check since I've already bought the 10ohm resistors and wanted a higher current than the 250mA which is produced in the above linked driver.

Edit: Had a look at a few of your builds in your sig, awesome... especially the 1.5A old as dinosaurs host, my host is going to be a ABS plastic project box lol... well, it is my first build.