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TPS274160: GND network for protection against turning off of inductive loads and surge

Bernd Harrer
Prodigy 20 points
Part Number: TPS274160

Tool/software:

Hello TI team,

concerning the recommended GND network to protect the TPS274160 I have.the following questions.

1) GND network to protect against turning off of inductive loads
If the outputs of the TPS274160 are continuously switched on, a turn-off of the loads (inclusive inductive loads) will only occur at power-off of the input voltage. In this case I think it should be sufficient to have only a diode at the GND pin (without an 1k ohm resistor in parallel). Or would you recommend the 1k ohm resistor also in this case? If so, why?

Would you recommend a diode (or a diode/1k ohm network) at the GND pin also at light inductive loads (for example at Lmax = 0.5mH and Imax = 0.3A)?

2) GND network for surge protection
If a bidirectional 33V TVS is used between VS and GND and no continuous reverse current can occur, is a diode (or a diode/1k ohm network) at the GND pin necessary to protect the device against surge?

Thank you.

Best regards,

Bernd

  • 0
    TI__Mastermind 34915 points

    Bernd,

    Welcome to E2E and excellent questions!

    For question #1, the resistor is there for the cases where there is a loss of supply condition on an inductive load. In normal conditions, with a valid supply, when the inductor on the output turns off it will go very negative and cause the integrated VDS clamp to engage limiting the output to VS - VCLAMP. The energy then is dissipated over the FET (VS -> FET -> LOAD -> GROUND). If the supply is lost, however, the device automatically shuts off the FET which would cause the inductive load to also turn off. In this case the clamp would still engage, but the current would go through the ground pin, across an internal ESD  clamp to VS, through the FET, and back through the load. The current going into the ground pin would without a resistor would potentially cause damage on the device.

    For trace inductances (normally just cabling) that are in the range of a few uH, the diode/resistor network could be removed. In this case it would still be recommended to have the series resistors with the MCU digital pins as the negative voltage could cause current to go through the high-side switch to the digital IOs of the MCU and cause damage.

    For question number 2- you should use a unidirectional TVS. Otherwise for negative surge events it would be our ESD diode that is conducting on the input. But you are correct in this case you could get rid of the ground network. 

    Best Regards,
    Tim

  • 0 in reply to Timothy Logan
    Prodigy 20 points

    Hello Tim,

    thank you very much for your quick and detailed response!

    Nevertheless, I still have a query and additional points on it.

    For question #1:

    You said, the reason for the resistor is to avoid damage of the device by reverse currents at a loss of supply with inductive loads. But with only a diode, no current at all could go into the GND pin! So its not clear for me, why the resistor is recommended in general.

    In table 9-1 of the data sheet it is said that this resistor is to stabilize GND potential during turn-off of inductive load. So I have thought, that if turning off of inductive loads can only occur at a loss of supply (power off of the application) an instable GND would be no problem.

    For question #2:

    We have a modular system with a separate power supply module. In this power supply module there is a bidirectional TVS between the VS an GND potentials. That means, I cannot change this bidirectional TVS!

    In this case a diode at the GND pin would avoid reverse currents into the GND pin.

    And with no separate voltage sources at the inputs (there is a galvanic isolation between the TPS274160 and the MCU, input high voltages I will generate from the 24V supply at VS) and no low resistances between the inputs and module GND, currents into the inputs will be limited to uncritical values.

    The nSTx outputs will go to an optocoupler. So no current can go into these pins.

    Can you confirm to this, or do you have an other suggestion or remarks on it?

     

    Thanks again and best regards

    Bernd

  • 0 in reply to Bernd Harrer
    TI__Mastermind 34915 points

    Bernd,

    See my comments below:

    If you just have a diode, you are right in the sense that no current will go into the high-side switch, however the current from the charged inductor has to go somewhere. With just a diode, the current would find another path- possibly through your microcontroller, and cause damage somewhere in the system. 

    Do you have any strict timing requirements for the turn-off of the inductive load? If you put a free-wheeling diode in parallel with the inductive load, then the output would be clamped to negative the forward voltage drop of the diode. This would eliminate the need to change the TVS diode on the input. 

    We do have a pretty good presentation attached below on the different ways of discharging the inductive load and also an application note for loss of supply conditions below:

    Driving inductive loads using power switches.pptx

    https://www.ti.com/lit/pdf/slvaes9

    Best Regards,
    Tim