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 =...
Well, here's the deal, when I was checking the threshold current the first time, I was using a smaller heatsink than a big block of aluminum.
As the device gets hotter, the current (voltage) required for lasing increases. Each diode has it own idea about what the threshold is. A row of six...
An update on NUBM34 lasing thresholds:
The diode lasing current thresholds were tested, one row at a time, to minimize heating. (and mistakes)
For heat sinking purposes, the module is clamped to a 14 pound aluminum block.
For each row, current was increased until lasing was observed, the number...