SHIN

Hi. jermstur.

I'm still a amateur, too.

If you use 5 of 1N5405 and 0.2ohm at 1.2A like this:
flexdrive(+)--1N5404--1N5404--1N5404--1N5404--1N5404--0.2ohm--flexdrive(-)

1. the minimum wattage of resistor should be 0.288W.
[ P = I * V = I * (I * R) = 1.2 * 1.2 * 0.2 = 0.288 W ]

2. total voltage drop
= Vdrop of 1N5404 + Vdrop of reistor[ V = I * R ]
= 0.8 * 5 + 1.2 * 0.2
= 4.24V

3. the current of your test load should be 5 times of the voltage measured at your DMM.
V = I * R = I * 0.2
I = 5 * V

It is difficult to simulate exact Vdrop because there is some variation between laser diodes, and Vdrop of 1Nxxxx is 0.7-0.9V fixed step, not a spectrum.

I tested with 1ohm. It's simple V=I.

And remember one thing. At near 1A, 5-pin chip of flexdrive should be heatsinked.

I wish this may help your 445nm build.
Good luck.

SHIN

Hey Shin, I have been reading some of your post on test loads for 445nm diodes. I have been doing nothing but reading posts for the past week... and I love it
I have a question that I know you can answer. I'm going to build a test load for a 445 a140 diode, pushed by a flexdrive v5 @ 1.2A. I've gathered from reading that 1N5404 diodes are the way to go for 1.2A, and that they have a voltage drop of .8v if given between 1.1-1.2A. I also read drlava recommends 0.2 ohm resistors for higher amps. the 445 diode has a vdrop of 4.1-4.2. so if I used 5x 1n5404 in series that should give me 4 volts. I would need .2 more volts correct, and I can get this from the .2 ohm resistor? How can I find out the vdrop of the 0.2 ohm resistor given 1.2 amps? thank you!

I live in South Korea too,
Where do you live? Seoul?