I am trying to understand how an inductive load can pose a threat to its high side driver chip when the load is turned off. When the driver is ON, the load is connected to 12V supply through the driver's MOSFET switch. When the MOSFET switch is turned off, the electromagnetic field collapses which causes a high potential across the inductive load but with the opposite polarity as shown in scenario 2 below. This causes 112V appears at VCC which can potentially damage the chip. Note- I picked 100V as a random high vtg which generates across the inductive load when it is turned off.
Next question- How the MOSFET switch "T2" is working to protect VCC from the high voltage generated by turning off the inductive load?
Note- T1 and T2 are withing the high side driver IC with T1 to connect the load to 12V, whereas T2 to protect VCC from high voltage spikes.
Next question- How the MOSFET switch "T2" is working to protect VCC from the high voltage generated by turning off the inductive load?
Note- T1 and T2 are withing the high side driver IC with T1 to connect the load to 12V, whereas T2 to protect VCC from high voltage spikes.