I have not worked with this device for years, but have a recommendation. Acquire a lab power supply with independent voltage and current limits! Plan to work with a single row of diodes at a time initially. I DO NOT recommend putting the four rows in series until you know what to expect...
Well, that is right interesting. Two rows of diodes is easier to deal with than four.
Of course, the question still remains – what are we gonna do with it. We'll burn that bridge when I get to it, I guess.
By the way, while we cant use matrix lasers individually, I don't know that it will always...
Searching for a clue, I wanted to see the beam shape of an uncollimated laser.
A blue 7 Watt NUBM44 was on hand.
A piece of white paper was put in the beam path about an inch from the front of the 12mm diode mount.
The beam was less unattractive than I had expected, but the ends are truncated...
If starting from 12V... I might be inclined to do something quick and dirty, like a 5 volt regulator or just resistors. One gets no extra credit for making something efficient. Throw in a capacitor and resistor to deal with transients. Good enough is the enemy of better.
Laser tail lights...
I say NO to the specifically protected driver requirement, it is not so. I too, have never seen a higher voltage laser diode specific driver.
Actually, I have never used a laser diode driver. There are options which are less idiot-proof but are much more flexible.
A plain bench power supply...
That makes sense to me as well.
I am inclined to use modules rated for well above the current required.
Some suppliers exaggerate their capacity.
OK!! I have to take the gang lens off, in any case.
I don't know if I have a suitable dark glass bottle, but should be able to find something. I have...
Sigh, I always have trouble. ;) Yes I≠A, R≠Ω, E≠V
I knew this stuff 60 years ago, but don't use it often enough and my braincell had a parity error.
Back then, we had to use "O" and "I" because and had not been invented, yet.
Agreed, although there are fuses that are extra fast, it...
A fuse needs to be in series with each channel, like the resistors.
3A through 1Ω would give a 3V drop across the resistor. A x Ω = V
3A current through a 3 volt drop is 9 watts. V x A = watts
Current ( A = amps ) may be shown as I i.e. V x I = watts ( I don't remember why)
Also, A^2 / Ω =...