I have a question regarding VN5025AJ-E: High side driver IC. Datasheet link - https://www.st.com/resource/en/datasheet/vn5025aj-e.pdf enter image description here

In the figure you can see the application schematic. In the datasheet they have suggested to use a resistor at the ground pin for reverse battery protection (automotive circuits). I was just trying to analyze the path the current flows during normal operation and also during reverse battery condition.

Why are there two Zeners in parallel and also a resistor between the anodes of the Zener diodes? What will the output be under reverse battery?

  • \$\begingroup\$ That application schematics does not make any sense to me. I'd also would like to know an answer if one exists. \$\endgroup\$ – Maple Aug 4 '18 at 22:07

Automotive compatibility requires many repetitive impulse levels tests such as 5000 impulses at 50V.

This method of protection is essential to attenuate the over-voltage in 3stages.

1) an external MOV ( Dd ).
2) a series R current limited zener.
3) a direct clamp zener

Dgnd cathode protects reverse voltages to chassis Gnd below it.

  • \$\begingroup\$ Dgnd is one of the options for reverse battery protection as mentioned in the data sheet. I'm currently trying to use only Rgnd (without Dgnd)since that's an option too. Also I'm not concerned about the Did in thea above circuit because this IC used in the circuit that I'm trying to understand does not have a Did connected . \$\endgroup\$ – NIDHI Aug 5 '18 at 17:39
  • \$\begingroup\$ Just like ESD clamps in CMOS, there is a 2 stage diode to each rail with current limiting to the inner diode so that it only has to clamp the rise voltage on the outer diode and not the external pulse. Is that what you need to know? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 5 '18 at 18:29
  • \$\begingroup\$ Sorry ,could you elaborate? I'm not concerned about Did since that is only to be used in case of load dump voltage higher than Vcc and also since my application does not have inductive loads I'm not worried about Dgnd.i want to know why the two zeners on the inside of the IC are in parallel with a resistor in between them , and in reverse battery condition , what will the voltage at the output pin be ? \$\endgroup\$ – NIDHI Aug 7 '18 at 2:14
  • \$\begingroup\$ -0.7V Dgnd = open cct. in reverse battery \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 7 '18 at 14:28
  • \$\begingroup\$ Could you please elaborate about the current limiting to the inner diode you were talking about in the earlier comment . \$\endgroup\$ – NIDHI Aug 8 '18 at 1:12

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