The customer is doing "power on under short circuit condition" test.
they are using LD pin to wake up the AFE. (VWAKEONLD)
They designed a start circuit. When they push the button, the B+ voltage will be connected to LD pin.
But every time the customer push the button, there will always a voltage pulse in the DSG FET Gate, and the Vgs will be larger than the FET threshold voltage.
And the FET will be turned on for nearly 40ms. In the short circuit condition, the MOSFET and fuse will be burnt.
(The SW is connected to ground when the button is pushed. i SW is connected to MCU)
(The magnitude of DSG could be 1/5) (Probe GND is B-)
They have tried the following tests:
1. disconnect LD pin with P+, the DSG will have the same issues. In this situation, there is circuit connection between LD with DSG outside the BQ76942.
(In this waveform, Probe GND is B+. We could see the GS voltage turn on the FET)
The BQ76942 is not OTPed. The customer used MCU to change the register. But in that time, the MCU is not powered on yet. LDO 3.3V need 20ms to prepare.
Do you know any potential reason that could explain why DSG pin voltage follows the LD voltage when BQ76942 power on?
Which could lead to turning on FET and burnt the FUSE in short.
It is leaked internally?
