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I have this 9Ah 42v battery pack protected by a BMS wrapped in a shrink which I would rather not disassemble (so I don't currently have access to BMS' circuit). It has two wires for charging, and two for load. Both showed 42v.

I then accidentally shorted the ones for charging for a moment (my probe melted instantly). After this incident it showed 37v. But the load wires still had 42v.

Now BMS's job is to protect battery from such occurrences. If it had a fuse, I assume it would blow thus breaking the circuit and showing 0 volts. Which must mean it doesn't have a fuse.

The battery must be okay since it still has its voltage. And its voltage reaches the charging wires in some way. But how can it change from full 42v to mere 37v? Did it have a resistor which momentarily burned thus increasing its resistance and therefore voltage drop on itself?

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    \$\begingroup\$ A battery can have nonzero voltage and still be damaged. \$\endgroup\$ – Hearth May 27 at 18:03
  • \$\begingroup\$ My guess (just a guess) is that a fuse blew. There may be another current path other than the fuse, and this current path may have high resistance. Thus the abnormal voltage. There are various methods for determining the input resistance without removing the shrink-wrap. You can google "measuring input resistance" or some such if you are interested. Ultimately, I don't think your question can be answered as a true failure analysis will require opening the pack. This is why I am commenting instead of answering your question. Even though it is a reasonable question, I think it should be closed. \$\endgroup\$ – mkeith May 27 at 19:14
  • \$\begingroup\$ @Hearth Yup. The same way a malfunctioning bomb can still explode. \$\endgroup\$ – DKNguyen May 27 at 20:01
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Some BMS units can do many things well.
Protect a battery from:
Over Voltage (OVP)
Under Voltage (UVP)
Over current (OCP)
Over Temp (OTP)
Under Temp (UTP)
Battery States SoC/DoD, SoP, SOH (? maybe)
Ground leakage (?)
Fuse current protection (?)

But input transient overvoltage from V=LdI/dt from a short circuit and inductive cables, may be an overlooked fault or not if it had MOV's.

Conclusion
Don't assume a BMS will protect from this kind of fault.

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If you have 42v going into the BMS, and 37v coming out, then either the BMS is also getting hot, or you don't have a load connected. My guess is that when you connect a load, the voltage will drop to a very low value.

If it's the other way, you and have a load connected, that 5v drop times the current being drawn equals the power "lost" inside the BMS, which becomes heat, which is why I said it would get hot.

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