this post was submitted on 10 Jan 2025
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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.