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by tarancato 3542 days ago
>if the energy is allowed to be released too quickly

Wasn't this the problem? That the kernel was drawing too much voltage from the battery, or something like that I read.

5 comments

I'm no electronics engineer, but I don't think "drawing too much voltage" is a thing.

As current goes up, resistance has to go down, so there will be a voltage drop across the battery terminals.

That's my understanding, someone correct me if I'm wrong, or otherwise help clear this up?

Resistance is a simplification of characteristic curves, which themselves are simplifications. You can easily get a voltage drop across battery terminals, you just have to put the battery in series with another battery. This is dangerous. You could also hook up an active current sink. Again, dangerous. There is nothing preventing the internal battery resistance from causing the terminals to go negative except for the simple fact that most equipment we connect to batteries will stop working before that point is reached (unless you hook batteries in series—which is why batteries in series must be close to identical, so they run out at the same time).
I just want to point out that batteries are often connected in series in regular devices.

Many devices where there are two AA batteries (TV remotes, etc...) usually have the batteries wired in series. They need to do it to get the voltage high enough for some of the components.

Additionally, many batteries are actually made from multiple smaller batteries wired in series internally. For example, some 9V batteries are actually just 6 regular 1.5V batteries wired in series.

Of course, in these configurations, they generally discharge at a similar rate.

I believe the resistance is sort of constant (for most normal usage cases) for passive circuits and in solid state electronic it would be a function of how many parts of the chip are engaged.

This being said anything short-circuiting the loop, can create a new pathway with less resistance (think molten metal, sparks etc). But modern batteries can send a lot more current then, magnifying the effect.

There was a theory that they charge the battery to a too high voltage, creating internal shorts.
Lithium ion batteries can supply a lot of power, especially relative to what a cell phone uses. I would be very surprised if the batteries were failing because of too much power draw, rather than a combination of faulty manufacturing and overheating.
Little wording issue there if this is the case it probably drew too much current causing a massive voltage drop overheating and causing a runaway condition.
Something stuck in an infinite loop involving the antenna or other power-hungry module of the phone ?

Can this happen in theory ?