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TMS320F2807x maximum QEP clock input from encoder

BillyO
Genius 17555 points

Hi team,

 

On the new F28075, can someone please confirm:

 

-The maximum QEP clock frequency input from encoder?

 

-Can all 3 QEP’s operate simultaneously without any pin overlapping?

 

 

Thanks in advance,

Billy

  • 0
    TI__Guru 61235 points

    Hello Billy,

    BillyO said:

    On the new F28057, can someone please confirm:

    I believe you mean the F28075.

    BillyO said:

    The maximum QEP clock frequency input from encoder?

    The QEP input timing requirements are provided in the device datasheet, SPRS902C, Table 5-59.

    BillyO said:

    Can all 3 QEP’s operate simultaneously without any pin overlapping?

    Yes.  Please see the pin muxing Table 4-1 in the datasheet SPRS902C, p.11.

    - David

  • 0 in reply to David M. Alter
    TI__Genius 17555 points
    Thanks David, good catch you’re right I meant F2807x and I've fixed the post. We saw that table with equations and looked at it for a while, but the question came up where the customer was wanting to know from us what the maximum supported QEP clock could be. I don’t have an easy answer for that so wanted to ask.

    Do you happen to have a known max frequency figure?

    Thanks,
    Billy
  • 0 in reply to BillyO
    TI__Guru 61235 points
    Billy,

    Per table 5-59, the maximum QEP period tw(QEP) is 2*tc(SYSCLK) (twice the system clock frequency). Maximum SYSCLK is 120 MHz from Table 5-12. Hence, maximum QEP frequency is 60 MHz.

    Basically, what the spec says is that the QEP edges must not occur faster than twice the system clock. This ensures that each transition on the QEP will get sampled on (caught by) a system clock edge.

    Note that the encoder may have trouble driving the signals at that speed. Most encoders do not operate that fast, much less the motor to which it is typically attached. Consider 1000 pulses per revolution of the quadrature encoder. 60MHz QEP edges on QEPA and QEPB signals of a quadrature encoder would mean (60 MHz)/(1000/4) = 240000 rps, which is 14.4M rpm. I don't think most motors would work too long at that speed.

    Regards,
    David
  • 0 in reply to David M. Alter
    TI__Genius 17555 points
    Thanks a lot for that David, that helped put it in perspective.

    Thanks,
    Billy