• State
  • Locked Locked
  • Replies 3 replies
  • Subscribers 29 subscribers
  • Views 495 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.

tas5630b: switchable channels, one or two channel operation

Bernhard Waechter
Intellectual 620 points
Other Parts Discussed in Thread: TAS5630B, TAS5630

Hello,

I'm actually designing a amplifier based on the tas5630b chip.

I need to switch between one and two channels in btl mode. Most of the time I work just with one channel (A+B), so I need to shutdown the second (C+D) one to keep the switching losses low.

Is there a way to disable the second channel by a µC without having a failure on ready pin?

I thought about several possibilities:

- disabling the bootstrap fets with a fet, controlled by a µC

- shutting down PVDD with relay and µC

- shutting down GVDD with a analog switch and µC

I guess the best way is to shut down GVDD using a analog switch, controlled by µC. But I don't know if ready still stays high if I do so?

Can you please help me with this?

Thanks a lot.

Best regards

Bernhard Waechter

  • 0
    TI__Intellectual 2360 points

    Hi, Bernhard

    You can shutdown output C+D by

    1) disabling boostrap by fets and MCU

    2) disabling input C+D from audio source

    But TAS5630 internal switching is still working, which means it'll has idle loss.

     

    About shutting down PVDD and GVDD, it can't fulfill your requirement because GVDD_A, GVDD_B, GVDD_C and GVDD_D are tied internally. PVDD_A, PVDD_B, PVDD_C and PVDD_D are also tied internally.

     

    Hope it's helpful for you. Thanks.

     

    Regards,

    Will

  • 0 in reply to Will Liu
    Intellectual 620 points
    Hi Will,

    first of all thanks for your reply. That helps me a bit.
    I Need to reduce switching losses of the unused channel. It's a high power battery device where I usually just need one channel. If I could reduce the losses through the inductors of the second channel this will save me about 2Watts.

    Are you sure there will be switching if I disable (short circuit) the bootsrap fets? The high side fet will not be able to open so there is no voltage across the inductors.

    And stays ready high if I disable the bootstrap of the second channel or will it report a global error and stay in shutdown?

    Thanks for your help.
  • 0 in reply to Bernhard Waechter
    TI__Intellectual 2360 points

    Hi, Bernhard

     

    Sorry for my misunderstood. It doesn't work if you short circuit. But it'll be dangerous, there will be a path from PVDD to GND as below picture.

    Thanks for your patient.

     

    Regards,

    Will