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Help with powering SF-AW210

hanzyfranzy

hanzyfranzy

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So I plan on making a handheld laser with this diode and the Guidesman kit from Mohrenberg:
http://laserpointerforums.com/f64/mohrenbergs-laser-emporium-58611.html

However, looking around I can't find a clear answer on how exactly to power it. I am pretty sure a single 18650 won't do the job, as it only hits 3.7 volts, and from what I understand these diodes need a higher voltage. I will also be using a rckstr driver, as I have some lying around, unless the build absolutely requires a driver that eats up less voltage...

Would like to avoid a 9V at all costs :yh:
 





I have the same diode in one of those hosts as well. I used a microboost set at 170ma powered by one 18650. If you want to use one of your rckstr drivers you'll need to use the original battery holder with 3x10440s. That should give you enough juice to power it but 10440s max out at ~500mA where 18650s go up to 2400-2800mA. I posted a thread about the build if you want any of the details.
 
I was afraid you would say that ;) . Guess I will just fork over $23 for a microboost so I can use just one 18650, as it will last a whole lot longer. Can I get a link to that build, by chance? :thanks:
 
remember befor you buy your driver. always think first of your battery options. your battery space generally dictates what driver you will use.

i have a dorcy led it take the 3 aaa holster. i have 2 of them. in one i rigged it to take 1- 18650. with a flex drive. in the other i use 3- 10440's with a rkcstr.

i hope this is helpful.
 
rhd said:
For sure?
Because I used 3x10440s in a guidesman, to power an LM317 driver. With the test load at least, I was supplying 900mA of current.
Click to expand...

Li-Ion cells should be able to deliver 2C (that is, discharging entirely in 1/2 an hour). Double its rated capacity, knock off the hours, and that's the current it can safely deliver. That'd suggest you can indeed get an amp or so out of a decent 10440 cell. I'd not run it for long though, and that's presuming the cell is actually 500mAh.

Reading the post you were quoting, it looks like they might be confused between the battery's capacity and maximum current.

Capacity = current * time the battery can deliver said current
Max current = the maximum rate of discharge a battery can manage safely without overheating or becoming damaged
 
Last edited:
Makes sense - that's more on par with my real world experience :)
 
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