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Question about TPS79333DBVR

Paul Schwartz
Prodigy 10 points

We have an application that uses the  TPS79333DBVR to regulate the power to a CPU. This product comes in to flavors, one using the TPS79333DBVR to regulate the voltage and another that gets its power from and external source. The question I have is on the externally powered unit. 

In the externally powered unit the TPS79333DBVR is omitted and and Vcc is connected to the other power source. The problem this causes is that we have to stock two different versions of our product. If we leave the TPS79333DBVR on the board and not even power it up would allow us to have only one unit simplifying our inventory. 

The question is what harm could this do to an unpowered TPS79333DBVR? The voltage output and the extern power voltage are the same, but if this might cause a failure of the TPS79333DBVR we would have failures in the field. So how much external voltage can the Vout output take in the unpowered state and if it fails what is the failure modality? If it fails open it wouldn't affect us, but if it fails as a short that's a big problem. So how much overvoltage can the Vout take before it fails? 

Thanks

Paul

  • 0
    TI__Guru 55940 points

    Hi Paul,

    I believe your question is what would happen if you have 3.3V on the output of the TPS79333 and 3.3V on the input.  Is this correct?

    If this is correct, there would be no issue if the device is disabled as long as the input does not dip below the output.  If the input does go below the output, current will be conducted back through the device via a diode.  This current is not regulated by the ldo and would need to be externally regulated to prevent damage.  If the device is enabled, the device would be operating in dropout.  This will not harm the device.

    Very Respectfully,

    Ryan

  • 0 in reply to Ryan Eslinger
    Prodigy 70 points

    I have the same question.  I believe, like me, Paul wants to know what happens if the input is open/unpowered.

    My original question to TI:

    Want to use the TPS79333DBVR in my design. Some versions of the design will use a 5v input and the TPS79333DBVR will regulate this down to 3.3-no problem. I have variations of the design that will apply 3.3v directly to the circuit. My question is this: If I apply 3.3v to my circuit (and to the output of the UNPOWERED TPS79333DBVR) will I have any issues? Just want to make sure I won't have any long term problems with the TPS79333DBVR with a constant 3.3v applied to the output from another source. (worst case is the TPS79333DBVR becomes damaged and I have eccessive current draw/short)

  • 0 in reply to Rob Rice
    TI__Guru 55940 points

    Hi Rob,

    If Vout will go above Vin, you will need to take precautions to limit any reverse current back through the device even if the input is not powered.  The body diode will permit reverse current; however, will not limit the current.  Depending on where you "open" Vin and the size of your input capacitor, you may still need to take external precautions to limit reverse current.

    Very Respectfully,

    Ryan

  • 0 in reply to Ryan Eslinger
    Prodigy 70 points

    Thanks.  So the capacitance on the input would give a momentary path to ground for the body diode.  What is the current rating for that diode (did not find it in data sheet)?  Also, this implies that the failure mode is an open, correct?

  • 0 in reply to Rob Rice
    TI__Guru 55940 points

    Hi Rob,

    External current limiting will be required if you exceed the rated current of the device (in this case 200mA).  There are many factors that determine the failure mode.  The failure mode is not necessarily an open.

    Very Respectfully,

    Ryan