• State Resolved
  • Locked Locked
  • Replies 4 replies
  • Subscribers 29 subscribers
  • Views 236 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

TAS2562: 1% output power under 5V PVDD

Will Wang47
Expert 7100 points
Part Number: TAS2562

Hi team,

Customer plan to use TAS2562. 

They plan to connect 5V to both VABT and PVDD. This should be OK for their application, right?

They checked the following figure, only get PVDD4.5V waveform.

They would check about the output power of 5V. Could you help check the 1% THD output power under PVDD =5V?

Thanks and Best Regards,

Will

  • 0
    TI__Guru** 104725 points

    Hi Will,

    If they're planning on using the same supply for both VBAT and PVDD, perhaps it's better to just use the boost in passthrough mode.
    External PVDD setup is most useful if PVDD voltage is larger than VBAT.

    I'll test for 5V and share some plots, please allow a couple of days to get further results.

    Best regards,
    -Ivan Salazar
    Applications Engineer

  • 0 in reply to Ivan Salazar
    TI__Expert 7100 points

    Hi Ivan,

    Yes, they will use pass through mode. They could remove the inductor to save cost, right?

    For the 5V, please help on test it.

    Thanks and Best Regards,

    Will

  • 0 in reply to Will Wang47
    TI__Guru** 104725 points

    Hi Will,

    Yes, removing the inductor is possible, I'll test and share further results Monday next week.

    Best regards,
    -Ivan Salazar
    Applications Engineer

  • +1 in reply to Ivan Salazar
    TI__Guru** 104725 points

    Hi Will,

    Here's the test capture for 5V direct PVDD:

    You can get ~1.1W before distortion starts increasing due to saturation.
    One thing to note is that you must connect 5V supply directly to PVDD, if you rely on VBAT connection through internal circuitry the output power will be limited to <1W.

    Best regards,
    -Ivan Salazar
    Applications Engineer