• qjkxbmwvz@startrek.website
      251·
      9 hours ago

      I’m not sure though — the power output and the charging input are both regulated and (almost certainly) current limited. So I think (not positive…) that you’re basically dissipating your power in the inefficiency the charging and output circuits, with this power coming from the battery.

      The inefficiency should (I think…) just be the round-trip inefficiency of the charging/discharging of your power bank — this should be way, way less than the short-circuit power dissipation.

      The simplest toy model is to take a battery and try to charge itself. So you put jumpers on the + terminal and you connect those to the + terminal, and same for - (charging is + to +, NOT + to -). But this is silly because you’ve just attached a loop of wire to your terminals, which is equivalent to doing nothing. With charging circuits in between things get much more complicated, but I’m not sure if it goes full catastrophic short…

      • OsrsNeedsF2P
        6·
        7 hours ago

        I think you’re right and I was just memeing, but I’m curious how the battery percentage went up

          • BarbecueCowboy@lemmy.worldEnglish
            5·
            4 hours ago

            You’re probably mostly correct. Some of them do literally count that, but (to my knowledge) most measure voltage as a battery with lower charge usually outputs less and vice versa.

        • Sterile_Technique@lemmy.worldEnglish
          6·
          7 hours ago

          My guess is it didn’t, and the numbers were pulled out the OP’s ass.

          Otherwise, idk how power banks monitor their percentage of charge, but being that it’s a percentage, if you fuck up the capacity, the same amount of energy will take up a higher percentage of that capacity. /shrug

  • Sibbo@sopuli.xyz
    641·
    13 hours ago

    Does burning down your own house count as science? Give this guy a medal!

    • qjkxbmwvz@startrek.website
      31·
      9 hours ago

      Is that true though? As in, is it really that dangerous? It seems that you’ll dissipate power equal to the inefficiency times the nominal charging power, so something like 5V x 2A x inefficiency (inefficiency being 1-efficiency), which will probably be of order a watt.

      I can use my car battery to charge itself without any issues — I just plug the red terminal to itself, and same with the black, which is to say, a battery is always connected in a way that “charges itself.”

      I think the key is that the battery probably isn’t really playing a big role in OOP’s setup — electricity doesn’t “go through the battery,” it just goes from the charging input to the power output circuits, with the additional power (due to inefficiency) being provided by the battery.