How to implement short-circuit protection for MOS and IGBT devices under high voltage? Let me explain.

Firstly, in the pin arrangement, designate the pin with short-circuit protection lock as the short-circuit signal detection input; the second pin as the drive ground; and the last pin as the drive output.

In case of a short circuit in the circuit, a large current flows through the MOSFET (Metal-Oxide-Semiconductor Field-Effect Transistor, abbreviated as MOS), causing a voltage drop across the S-terminal resistor, which triggers the transistor to switch from cutoff to conduction (the extent of conduction depends on the MOS transconductance). As a result, there is a voltage drop across the drive resistor, and the MOS enters the amplification region. At this point, high voltage will not cause strong current discharge through the MOS, preventing the chip from local overheating. The circuit can detect the short circuit, close, and lock the drive voltage output in a very short time, so that the MOS only bears the constant current power loss during the action time, greatly improving safety.

However, the circuit presented here is for understanding the principle. Specific parameters for actual use should depend on the actual working conditions.

The value of the resistor R17 is determined based on the IDM (Pulsed Drain Current margin) of the selected MOS, and the transconductance is not a major factor. The transistor does not work in a switching state; it adjusts the MOS drive voltage based on the transient current value of the MOS to ensure that the IDM does not exceed the safe working range when a short circuit occurs. At this point, the drive has been turned off and locked in 1μs or even shorter, so there is no excessive power loss. The faster the transistor reacts, the better. In fact, this circuit has evolved from a current negative feedback circuit. The circuit is as follows: