Constant current laser driver (buck converter)

Thread Starter

MrCooI

Joined Sep 10, 2019
1
Hello, it's my first post here, so I hope it's the right topic.

I want to build compact constant current laser/LED drivers for different diodes in 1.8-5A range. I managed to do it using linear regulators or operational amplifiers but both method was very inefficient and they didn't really fit in the laser/flashlight case, so now I want to build a driver using a step down buck regulator. I found this product:
https://images.bonanzastatic.com/afu/images/a3ee/bffa/bd89_7115368102/__57.jpg
It seems small and very easy to build, but I have no idea what regulator ic is used on this board because the labels are removed. So, my question is, does anybody know a constant current buck regulator ic which is easy to use and can be used for this purpose?

Thanks in advance!
 

Delta prime

Joined Nov 15, 2019
52
This is going to sound silly but just scroll down I believe it'll get you some information..... This place is great... I hope it helps you
 

Bordodynov

Joined May 20, 2015
2,455
I've been developing a 10A laser driver. But after two samples were made, it turned out to be a problem. The laser current was regulated up to 5 Amperes, but when trying to increase the current at the level of 6 A, the driver went sharply to a current of about 5 A. The used chip is very complicated and I did not understand, and in one day I developed a more classical scheme and a month later received the manufactured driver with a current of up to 14 A. You can use my scheme:2019-12-12_13-42-08.png
 

kubeek

Joined Sep 20, 2005
5,650
I've been developing a 10A laser driver. But after two samples were made, it turned out to be a problem. The laser current was regulated up to 5 Amperes, but when trying to increase the current at the level of 6 A, the driver went sharply to a current of about 5 A.
The current-mode control of that chip doesn´t really help with regulating output current, it is more related to the peak inductor current. Best option would be to put a current shunt between ground and the diode, and feed this signal to the feedback pin (most likely amplified, because you don´t want to dissipate 1.25V*10A just in the current sense).
 
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