How does the discharge MOS of a BMS break down?
Why is the voltage higher in a BMS system than in a system?
Before the MOS tube is shut down, a large current flows through the MOS tube to power the load, which is almost the same voltage across the load as the battery at each end.
Because the current does not change much, the parasitic inductors L1 and L2 are like wires.
When the MOS tube is turned off, the current of the inductor cannot change suddenly, so a loop is required to maintain the current, and the inductor becomes a small power supply.
The energy previously stored in the inductor requires the release of an electric current, but the current creates a voltage on the inductor.
In other words, this circuit (including the battery) has three power supplies connected in series and applied to the MOS tube.
Therefore, this total voltage must be higher than the battery voltage.
How much taller is it?
The L2 inductance is generally related to the positive and negative poles of the battery and the leads of the BMS system, and the parameters are generally relatively small; L1 inductors are different. In the case of over-current protection and short-circuit protection, the current is relatively large when the current is turned off, and the voltage drop will be large enough to penetrate the MOS transistor.
How to solve it?
You can make the MOS tube close slower, so that the di/dt will be much smaller and the voltage on L1 will be lower.
