How is the discharge MOS of BMS broken down?

Why is there a higher voltage in the BMS system than the system?

Before the MOS tube is turned off, a large current flows through the MOS tube to power the load, and the voltage across the load is almost the same as the voltage across the battery.

 Since the current does not change much, the parasitic inductances L1 and L2 act like wires.

After the MOS tube is turned off, the current in the inductor cannot change suddenly, so a loop is needed to maintain the current. At this time, the inductor becomes a small power supply.

 The energy previously stored in the inductor requires the flow of current to be released, but the flow of current will generate a voltage across the inductor.

 That is to say, this circuit includes a battery, and three power sources are connected in series and applied to the MOS tube.

Therefore, this total voltage must be higher than the battery voltage.

How much higher is it?

The L2 inductor is generally related to the positive and negative poles of the battery and the leads of the BMS system, and its parameters are generally small; the L1 inductor is different. During overcurrent protection and short-circuit protection, the current is relatively large when it is turned off, and the voltage drop will be very large, enough to break down the MOS tube.

How to solve it?

The MOS tube can be turned off more slowly, so that the di/dt will be much smaller and the voltage on L1 will also be reduced.

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