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BMS short circuit interruption capability
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Biduleohm:
Increasing the inductance can only make the problem worse as it'll stock more energy, also it's not very practical nor cheap given the current it should handle (300 A continuous...).


--- Quote from: Siwastaja on March 25, 2020, 04:24:59 pm ---Use freewheeling diodes to discharge the inductance into DC link capacitance; large part of capacitance should be of higher ESR to prevent ringing.  A crowbar circuit of some kind may be needed, but it can be slow because the capacitance limits the voltage rise speed. Though, just using enough electrolytic capacitors (calculate for maximum voltage swing when all the inductance-stored energy discharges into the capacitors) may be cheaper.

Freewheeling diodes need to be rated for single pulse only, hence shouldn't be expensive or too big.

--- End quote ---

I thought about using capacitors but I'm worried about resonance because I don't have a predetermined inductance as it's dependent on the wire size, length, etc... I took the pretty defavorable case of 2 m of wire but it can be anywhere from 10 cm to a few meters.

I can't use freewheeling diodes, the switch is bidirectional.


--- Quote from: Siwastaja on March 25, 2020, 04:24:59 pm ---The faster you can make the OCP, the less inductance you need. You may want to measure the current rise rate of a shorted battery; it may have enough inductance already. Inductance limits the current rise speed, while resistance limits the maximum current itself. Using enough inductance removes the short-circuit current (defined by the total system R) from the equation.

--- End quote ---

It's already as fast as I could make it. And I should already have received the battery but the corona virus decided to change that...

The problem isn't the short circuit current, the mosfets can handle it, it's the inductive voltage spike.

What I need to know is how much energy is stored in a 2.4 µH inductor fed by 64 V @ max 10 kA for a duration of 5 µs.
mzzj:

--- Quote from: Biduleohm on March 25, 2020, 04:57:32 pm ---
What I need to know is how much energy is stored in a 2.4 µH inductor fed by 64 V @ max 10 kA for a duration of 5 µs.

--- End quote ---
OCP set to 420 A would give you 550 A  after 5 us short and 0,35 J
Biduleohm:
Thanks a lot ;)

What formula did you use? As I want to be able to do the calculation for other values.
mzzj:

--- Quote from: Biduleohm on March 25, 2020, 10:45:28 pm ---Thanks a lot ;)

What formula did you use? As I want to be able to do the calculation for other values.

--- End quote ---
dI=dt*V/L
dI = 5us*64v/2.4uH = 133A
Siwastaja:
2.4uH of inductance stores way too much energy at 10000A, but the idea is that you won't need to ever put 10000A through that 2.4uH! Note the current has a second exponent in the energy stored, so you need to really prevent the current from reaching the resistance-limited short circuit current; act well before that happens.

Electrolytic caps are cheap and small, it's trivial to store the energy stored in the inductance in the capacitors, unless someone adds a MASSIVE ferrite core in the system, but this wouldn't be expected.

E = 0.5 * L * I^2 = 0.5 * 2.4uH * (550A)^2 = 0.36J
E = 0.5 * C * U^2, solve for C:
C = E/(0.5*U^2). Allow for 20V voltage rise on the capacitor voltage,
C = 0.36J/(0.5*(20V)^2) = 1800uF

You can clamp the voltage to a capacitor bank with a diode even if the switch is bidirectional. The advantage is you may be able to find fast diodes better than extremely fast TVS diodes (I don't know if I'm correct here; you may able to find a directly suitable TVS, as well, in which case the diode-capacitor solution does not have such advantage), and you can discharge the capacitors as slowly as you want.

Note, 2.4uH, whatever this is coming from, is already quite a lot. Using https://www.eeweb.com/tools/loop-inductance , this would be a circular loop 75cm in diameter. In a properly designed battery pack and wiring, the current and the return current run closer to each other.

The idea is, do not design your active efuse circuit around the short circuit current of the battery, because it sees this current only if you have failed the design somehow, i.e., it won't be able to detect the overcurrent in time, and switch off the transistors in time - in which case it doesn't matter, it will likely blow up anyway. Instead, design your efuse circuit to prevent the current from ever rising much over the expected maximum operating current. Then, protect (the wire insulation and the battery pack, and other things thermal) against a design failure using a traditional passive fuse, and for that, make sure it handles the 10kA current.
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