Help to find out the function of the resistor and diode in the open-drain switch

[washburn961]

Hi everyone,

I'm analyzing a low-side switch circuit that uses a BSP75NTA MOSFET from Diodes Inc to control a 24 V load, and I’m trying to understand the specific roles of two components: R1 (1k resistor) and D1 (1N5819 Schottky diode). Here’s the circuit schematic:



Circuit Overview:

The BSP75NTA is used as a low-side switch.
The gate is driven by an NPN BJT (2N3904), which is controlled by an input signal.
R1 is connected between the source and ground.
D1 is placed across R1.

My Observations and Questions:

I suspect R1 is there to ensure the source is not left floating.
For D1, I assume it blocks reverse currents, possibly protecting against transients. Could D1 be moved to another position, such as in series with the drain, to achieve the same functionality?
I don't understand why the people who designed this couldn't just connect the source directly to ground, since at the supply section of the board that contain this cirucit there is a giant unidirectinal TVS diode that could protect against negative transients.

Could anyone provide insights into the function of R1 and D1 in this configuration? Would removing or repositioning them make sense, or are they essential for proper operation and protection? Note this circuit is used in an automotive ECU.

Thanks in advance for your help!

Best,
Gabriel

Edit: had previously linked the wrong image.

[inse]

In automotive electronics there is often a requirement for reverse battery protection.
Haven‘t figured out why it’s in the source, maybe to avoid transient negative gate overvoltage in case of reversed battery.
I wonder why there is no protection against inductive kickback or is it a purely resistive load?
Edit: ah, it‘s a smart switch, so inductive clamping is integrated.