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PCM1690 thermal pad - connect to ground?

Andrew Ostler
Prodigy 60 points
Other Parts Discussed in Thread: PCM1690

Hi,

in the PCM1690 datasheet it's not clear if the thermal pad should be connected to ground, left open, or connected to something else. Can anyone let me know?

thanks.

  • 0
    TI__Expert 8505 points

    Hi Andrew,

     

    The part uses the PowerPad as mentioned in page 43 of the datasheet. It implements an exposed thermal pad which is designed to attached to the PCB directly and using the PCB as a heat sink with vias connecting from the pad to internal copper planes let it be ground, power, or signal depending on your board structure. If interested to explore further please take a look at this application note which discusses the PowerPad design:

     

    2388.powerpad.pdf

    Hope this answers your question.

     

    Regards,

     

    Brian Wang 

  • 0 in reply to Brian Wang
    TI__Guru**** 189401 points

    Hi, Andrew,

    Welcome to e2e, and thanks for your interest in our products!

    Let me expand a little bit on what Brian said.

    The exposed pad of the PowerPAD package is tied to the substrate of the silicon with sort of electrically-conductive die attach epoxy. For single-supply chips, the substrate usually wants to be held at ground. Hence, you should ground the PowerPADs in these cases.

    In cases with split supply, charge pump, etc, it may change. Sometimes those are tied to ground, sometimes they need to be floating, it all depends on the silicon technology used inside the package.

    For this case, PCM1690, you want to ground it. 

    -d2

  • 0 in reply to Don Dapkus
    Prodigy 60 points

    Thanks Don, that's the answer I was looking for.

    I had another question about this part. Page 38 of the datasheet shows recommended output filters. It shows the DAC's VOUT+ connected to the op-amp inverting input, and VOUT- connected to the non-inverting input. Is that intentional? The absolute phase of the output voltage relative to the digital codes is important to me.

  • 0 in reply to Andrew Ostler
    TI__Expert 8505 points

    Hi Andrew,

     

    The VOUT+ and VOUT- could be one way or the other where one will give you an Analog Output 180 degrees different from the other. With the current topology on page 38 you will get Analog Output = -R2/R1 * [(VOUT+) - (VOUT-)]. With the values chosen for bandwidth at 56kHz, it is designed so VOUT- will be in phase with Analog Output accross the audio band. Please let me know if you have further questions. Thanks!

     

    Regards,

     

    Brian Wang