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TPS4H160-Q1: Reverse-Current external protection question

Ronald Wang
Intellectual 390 points
Part Number: TPS4H160-Q1

Dear TI,

In D/S 8.3.7.4 Reverse-Current Protection.

I have the following questions.

1.Can I use method 1 only to achieve the protection function? (don't use method 2)   

2.Comparison Method 1 and Method 2. Is the loss of diodes smaller in Method B? 

3.Is using Diode=SK34 and Resistor=4.7K correct in Method 2?

Thank you

Best regards,

Ronald

  • 0
    TI__Mastermind 28200 points
    Hello,
    1- Method 1 is enough to protect the device and the laod as well against reverse polarity. The reverse current can flow through the FET body diode and the ESD diode from GND to VS. Diode from VBAT to VS blocks the current to flow. Device and load are protected in this method. Any type of load load is protected in with this method. The protection diode adds power losses Vf x load current
    2- Method 2 is protecting only the device. Reverse current flow in the ESD diode is limited by the GND network resistor. This method does not protect the load and the FET body diode. The body diode current is only limited by the load itself. For example if the load is electrolytic capacitor the current is not controlled and reverse polarity may damage the load and the FET. In method 2 the diode loss is insignificant because the forward current is few Milli-amps.
    3- GND Network diode forward voltage should be lower than 0.6V in any circumstances. SK34 is OK but you may use smaller diode because the forward current is only few milli-amps.

    Regards
  • 0 in reply to Mahmoud Harmouch
    Intellectual 390 points
    Dear Mahmoud Harmouch ,
    When method 1 and method 2 are used together.
    Can method 2 make method 1 diode losses even smaller?

    thank you for your help
    Regards
    Ronald
  • 0 in reply to Ronald Wang
    TI__Mastermind 28200 points
    Hello Ronald,
    The power losses in the diode is forward voltage times load current Vf x IOUT. If method 1 and 2 are used together, the power losses in the diode of method 1 will not change. The power losses in the diode (method 1) is independent of method 2
    Regards