Hacker News new | comments | show | ask | jobs | submit login

For some reason they seem to be very interested in the peak current.

I guess the higher the current the better you vaporize the material?




Actually the most important figure is not just the peak current amp limit, but the actual continuous discharge amperage limit. While batteries are capable of higher peak discharge rates in very short bursts, you really do not want to push this and instead should stay below the continuous discharge amperage limit.

The higher this limit generally the "safer" the battery performs in high drain operations like actually vaping the ecig, and this means that not only can you go to higher power levels, but that lower ones will also generally not heat the battery as much, and reduce the risk of thermal runaway or venting.

Take a variable power mod using power regulation set at 50w (a relatively common power setting for enthusiast/higher end vapers but that can be a daily driver), with an atomizer at 0.18ohms resistance and a Sony VTC5 18650 as the battery with 2600mAh capacity and a sustained safe current amperage limit of 30A.

To get 50w you will need to apply 3.00V / 0.18ohms or 16.67 amps. If you for instance choose a poor battery re-wrap, like this Cylaid 3500mAh advertised at 10amp continuous discharge, which is actually just a "re-wrapped" battery which can only sustain 8 amps, you would be in dangerous territory where the battery would likely get much hotter than a quality one and have higher potential for thermal runaway/venting.

https://www.e-cigarette-forum.com/forum/threads/cylaid-10a-3...

The most interesting thing you'll notice is an inverse relationship with capacity in mAh and continuous discharge in amperage. You can either have a higher capacity battery with lower continuous discharge current, or a lower capacity battery with higher continuous discharge current, not both.


On a regulated device, the coil resistance is irrelevant to battery amp draw. 50W will consume the same amount of current regardless of what it is being used for. The right way to figure it is 50W/3.7V = 13.5A. Add a bit more on that to account for regulator inefficiency. It also rises as the battery discharges down to cutoff voltage, which is usually 3.0 or 3.2V.

This holds true even if the coil is 0.5Ω




Guidelines | FAQ | Support | API | Security | Lists | Bookmarklet | Legal | Apply to YC | Contact

Search: