Below is the block diagram of the mentioned driver (used to drive dc light loads). My understanding is this driver is using an op-amp based voltage to current converter where one of its inputs i.e. A is VCC i.e. 12V and the other input B is equal to the source terminal (C) of the load's N-MOSFET.

The o/p current of this op-amp is (VB - VA)/ Rinput; don't know why the input resistance between B and the op-amp is not shown in the driver datasheet.

During Normal Operation
When the logic turns on the gate for both MOSFETs through pin 1, both the points B and A are almost equal to 12V and the o/p current is zero.

During Failure
When the logic turns on the the gate for both MOSFETs through pin 1 but there is a short to ground on the driver's output, then point B will be at 0V but A continues to be at VCC i.e. 12V. In this scenario, the o/p current = (0-12V)/ Rinput and hence its is a non-zero current. The microcontroller will know short to ground through this non-zero op-amp output current.

Is this analysis of the driver correct?

Q2. Why the i/p A is connected to VCC and not ground?

 

Input A is the reference current input and due the variable output resistanse through the FET in opamp output it generates current proportional to load current (Load_current / K). STM has some literature with explanations for high side load switces, go to the product page and see this Application Note

 

Input A is the reference current input and due the variable output resistanse through the FET in opamp output it generates current proportional to load current (Load_current / K). STM has some literature with explanations for high side load switces, go to the product page and see this Application Note

Thanks for sharing these 2 documents. I appreciate it if you could answer the following questions pertaining to the microcontroller I/Os protection section ?

1. What the ground protection network means?
2. Does negative transient means reverse polarity i.e. VCC receiving 0V and ground receiving +12V?
3. Why would the microcontroller pins latch if the control pin gets pulled negative?

 

Many MCUs have internal diodes on every input pin for protecting it from overvoltage. Upper diode is clamping positive overvoltage to Vdd(Vcc) and lover diode is clamping negative (below the ground) overvoltage to GND(Vss), leaving about +/-0.7V overvoltage to pin. These diodes can't drive lot of current away, maybe maximum 20mA. The side effect of these diodes is ability to power up MCU core if main power is missing. Depending of MCUs input/output buffer design the buffer input/output elements can be permanently open until power removed. So STM is recommending connect this VND load switch to the ground via diode which prevents reverse connection to the power source. In automotive designs is also often used TVS diodes which prevents overvoltage spikes and power diodes to ground after fuses or polyfuses (where the fuse cannot be changed) for clamping reverse voltage and cutting it off from device.