Laser Driver from Scratch

[Photonz]

Hello all, it has been awhile since I've been on the forum and now that I have some time for a little while I wanted to create a circuit from scratch instead of buying a pre-set driver. I'm assuming it is more complex than just a few resistors in series to limit the current for the diode so if anyone knew a good starting point that would be great. :thanks:

Edit: More info: I am working with a m140 diode and would like to run it at around 1.8A to 2A.

 

[petters]

There are tons of threads about drivers on the forum, from their basics to their complex ideas. This would be a good starting point: http://laserpointerforums.com/f67/how-laser-diode-drivers-work-explanatory-thread-71513.html

Use the search bar, there are open-source drivers you can take a look at to learn off of before attempting to design your own.

 

[DTR]

http://laserpointerforums.com/f67/laser-driver-can-done-rog8811-87507.html

 

[Photonz]

Thanks for the links. If I plan to run a m140 diode at around 1.8A and 4.5V, will this setup work?

I'm thinking of using the DDL/LM317 Constant Current Driver
I will have everything the same except supply 9V instead of 7.2V and have four 2.7ohm resistors in parallel to give me a resistance of .675ohms and that will give me a current of 1.852A. (1.25V/.675ohms = 1.852A)

 

[ChaosLord]

Yes, 9V in with that driver should run an m140 just fine.

 

[DTR]

The best advice I can give it to test the circuit with a proper test load prior to connecting it to the diode.

 

[Photonz]

Okay sounds good! Thanks for the help. I already have the resistors but do you guys know of the best place to buy the capacitor, diodes, LM317, and pot? Or should I just go to the local electronics store? (also, is the pot necessary?)

 

[Zeebit]

Thanks for the links. If I plan to run a m140 diode at around 1.8A and 4.5V, will this setup work?

The driver is constant current so you don't have to worry about the voltage.

I'm thinking of using the DDL/LM317 Constant Current Driver
I will have everything the same except supply 9V instead of 7.2V and have four 2.7ohm resistors in parallel to give me a resistance of .675ohms and that will give me a current of 1.852A. (1.25V/.675ohms = 1.852A)

Just keep in mind that it will be an inefficient setup. The driver is linear so it will turn excess voltage into heat. If you supply it with 9V and the diode is dropping 4.5V @ 1.8A, the driver will turn the excess voltage into heat, (9v-4.5v)*1.8A = 8.1W. That is a lot of heat.

Also, a 9v battery won't cut it. It cannot handle that much current draw. You're better off using Li-ion batteries.

 

[Photonz]

But doesn't the regulator use 3V and don't I need an additional 1.2V? Also is that a 10microfarad (10μF) or 10 millifarad (10mF) capacitor?