• State TI Thinks Resolved
  • Locked Locked
  • Replies 3 replies
  • Answers 1 answer
  • Subscribers 63 subscribers
  • Views 117 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

TPS4H160-Q1: loss of GND resistor selection

Chuqiang Wen
Expert 5855 points
Part Number: TPS4H160-Q1
Other Parts Discussed in Thread: TPS4HC120-Q1,

Tool/software:

Hi team, 

I would like to check with you how to calculate the resistor(R839) and diode(D54) value for loss of GND protection circuit? 

Thanks!

Ethan Wen

  • 0
    TI__Mastermind 34935 points

    Ethan,

    The resistor is fine- but the diode is going to risk triggering the loss of ground protection. The forward voltage drop needs to be <=0.6:

    ... and it looks like the forward voltage drop of this diode is significantly  higher:

    You can estimate about 10mA of current maximum. 

    Would the TPS4HC120-Q1 be an option here? It will be a significantly better price perspective and would not have the 0.6V diode restriction.

    Best Regards,
    Tim 

  • 0 in reply to Timothy Logan
    TI__Expert 5855 points

    Hi Tim,

    Thanks for your reply. 

    Further question, how to calculate this resistor value? How customer can judge if this value works well? 

    Timothy Logan said:
    It will be a significantly better price perspective and would not have the 0.6V diode restriction.

    Why TPS4HC120-Q1 do not 0.6V restriction? 

    Regards,

    Ethan Wen

  • 0 in reply to Chuqiang Wen
    TI__Mastermind 34935 points

    Ethan,

    It has to do with the way loss-of-ground is detected in the TPS4H160-Q1. For that device, there is a comparator that will compare a digital low signal with the potential on the IC ground. If it is greater than 0.6V (plus some margin), loss of ground protection is triggered. For TPS4HC120-Q1, the loss-of-ground mechanism was changed so the device does not have this limitation.

    The resistor is there for loss-of-supply + inductive turn-off scenarios. If the resistance is too high in this case, the current will not have a path to go. I misread your schematic- you should have 1k. With a 10k resistance it's possible that during a loss of supply + inductive turn-off scenario, the resistance would be too high-impedance and the current would rather go through the MCU's ground (or something else shared on the ground plane).

    Best Regards,
    Tim