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Can you answer some of my thoughts!


New member
Sep 1, 2013
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Im going to describe my build here, and just know that I am all about safety, so no worries there! Hoping to get ~2W

Host: Saik SA-305
Diode/Module: M140 445nm
Lens: 405-G-2
Driver: X-Drive @ 1.8A mounted to large pill in SA-305
Batteries: 2x AW IMR 14500 (AA size to fit stock AA battery holder, safe @ up to 4A draw) in series. The stock holder holds 3 batteries, so I will probably use some sort of "dummy battery" rig to fill the last slot. There will be no benefit to adding the 3rd 14500 right? All it would do is bump the voltage from ~7.4 (which is enough) to ~11.1 (which would just just be expelled as more heat from the driver). It would still be 700mAh, therefor being discharged in ~30 min with a 1.8A draw. This is where I am stumped. I need to look for better alternatives? Something that I can fab into the SA-305 easily with more mAh. Or is this just a problem I have to deal with in a high powered/host setup? High power and short run time as far as batteries go. Or i guess I could just find another host that fits some 18650's.


Another thing I can't wrap my brain around is how to figure out how many volts the drive is going to use to send the specified current (in my case 1.8A) to the diode. I know the voltage figures itself out when you set the current, but to my understanding, you want the battery voltage > driver output voltage. So i guess I just dont understand where you get this info from? Can you not figure it out with math? Is it something that actually testing it with a multimeter gets? Or is it included in some data sheet?

Would it be as easy at this for example...

I cant find a datasheet for m140 but take this into account...

In the case of the ML501P73-02 Mitsu 500mW 638nm...

datasheet says the output is 500mW when given 650mA which in the case of this laser = 2.2V
datasheet says the MAX is 800mA but who knows, theoretically speaking someone tested this diode and said its perfectly stable when given 1400mA. So is it safe to say at 1400mA of power, the amount of voltage needed will be about 4.7V????
Last edited:


Well-known member
Sep 12, 2007
The voltage of a diode depends on the type of diode, the temperature of the diode, and the drive current of the diode. The graph of current versus voltage is not linear, but logarithmic. Doubling the current will only give a few hundred mV increase in forward voltage.

For diode lasers, a very general rule of thumb is IR is 2V, red is 2-3V, blue is 4-5V, and violet is 5-6V.