trying to learn more about battery choices

[olympus mons]

I'm having difficulty finding more information about the different size, mAh, and V battery advantages and disadvantages relating to lasers. I'm not very educated in electronics.
Is the main differences simply about power output?
For example what is the reasoning some laser use a single 18650 vs. 2 18350 or 16340 batteries?
When you use 2 3.7V batteries does the combined voltage double?
Do higher mAh produce a brighter beam or is it only a longer use time?
When building a laser what are some reasons you would chose which power cell? How do you know which would be the best power cell for the build?

 

[Alaskan]

I am just learning some of the particulars about these batteries myself, but here is what I've found:

1 each 18650 3.7 volt (some are 3.6). Peak voltage at 4.2 volts after freshly charged.
1 each 18350 3.7 volts. Peak voltage at 4.2 volts after freshly charged.

Every time you stack batteries, you get double the voltage, put them in parallel the voltage is the same but the current capacity is double.

Stack two 18650's for double the voltage, or up to 8.4 volts.
Stack two 18350's for double the voltage, or up to 8.4 volts.

One 18650 is roughly the same length as two 18350's (depending if flat top or not). The advantage of stacking two 18350's is you get twice the voltage in the same amount of space or length of a single 18650. For example, one of my hosts requires two 18350 batteries so there is enough voltage to run my 900+ mw output 405nm DTR laser module, if I put a single 18650 in the holder which fits fine, there isn't enough voltage at 4.2 volts to run it, it won't light up at all.

More MaH only means longer run time, not brighter with the exception that the battery will stay at its peak voltage longer, this may cause your laser to be brighter, if the regulator needs the peak 4.2 (or higher) voltage of a freshly charged battery(s) for full current or power output. However, most designs do not need your battery to be at their peak freshly charged voltage for full output, not if properly designed, in my opinion. The more maH rating, the longer it takes (for a given load) to deplete the battery or for its voltage to turn downward to 3.7 volts and lower.

Someone correct me on this if I'm wrong, but I think 16340 batteries are a different diameter than 18650 or 18350 batteries. Someone else will need to chime in with more on batteries, this is the limit of my knowledge for these types right now except the batteries behave differently, depending upon their construction and chemistry, some are made for long life at moderate loads, others for high loads with less overall mAh or total capacity. Then you have variations of any of these parameters manufacturer to manufacturer, some outright lie about the real maH capacity of their batteries.

 

[ChaosLord]

The mAh is cell capacity, how much energy it can output before needing recharging. In general, the higher the number, the longer it can go. The only thing this affects is run time.

The number of batteries depends mostly on driver. If not a boost driver, it almost always will need 2+ batteries to have enough voltage to fully run. So yeah, two batteries in series for ~8.4V on a full charge.

If you wanted to build a smaller, lighter laser, you might use a boost drive for a single batt build. If you want a longer runtime, use multiple batteries with a buck driver.

 

[olympus mons]

That's great info thank you both. So with drivers are they rated with min/max Voltage requirements? Would a person every use 2 batteries and a boost driver?

 

[Alaskan]

I have a boost driver that only needs one 4.2 volt battery, I have a buck that needs two.

Edited: to remove the next sentence questioning whether a boost can use more than one battery or not, I wasn't completely sure if there were instances where two could be used and have edited so as not to confuse the issue. From the following post by ChaosLord (next post below) this is never the case, a boost uses just one battery, this sounds right to me, even a green laser diode where a boost is commonly used only needs 7 to 8 volts, two 4.2 volt batteries would be 8.4 volts.

 

[ChaosLord]

Yup, drivers have a min/max input voltage on their data sheets.

No, never use more than one battery with a boost drive. That's a good way to fry it. They generally can't take much more than 6V. They "boost" the input voltage up to power diodes with higher voltage drops. This eliminates the need for a second batteries voltage, rendering it unnecessary, even if the driver could take the voltage.