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0 pointsssl-31y ago0 comments

> I'm not sure why I would need/want to push 10A to each battery?

Because the soldering iron in question dissipates 100W, and 100W is a substantial amount of power?

If a soldering iron is drawing 100W from a 2S pair of 18650 batteries, then:

The cells are supplying ~13.5A of current through their contacts.

To get below 5A, we'd need a six 18650s in a 6S configuration. Certainly doable (a relatively inexpensive [for a packaged battery] 24V Kobalt power tool battery has 6S 18650s), but beginning to be chonky.

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xxs1y ago· 1 in thread

just pointing that even in 6S, it'd be over 5A at the lower end of charge (I'd assume 3V to be the cut off).

I have not checked the schematics but I'd expect a PWM controlled resistive heater and Joule counting, i.e. constant power.

ssl-3OP1y ago

Ah.

You're right, of course.

But I guess I'm assuming that 100W is an ideal figure based on fully-charged cells of ~3.7v instead of a constant 100W, since that's the normal (if somewhat carpetbagging-esque) way to describe things in marketing world. This assumption points to 6s as staying below 5a.

At 3V per cell and 100W output, it takes 7s to stay below 5A. (7S little weird, and even chonkier, but not that weird.)

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