IBUFG and BUFG +xilinx

Discussion in 'VHDL' started by williams, Apr 4, 2005.

  1. williams

    williams Guest

    Hello Guys,
    I had a doubt about the IBUFG and BUFG in xilinx.
    1.I have connected clock from oscillator to CLKG IO of the Xilinx. In
    this case is it required to instantiate the IBUFG inside my code
    also?.
    2. The DCM output is already BUFG i think and so is it required to
    BUFG again in my code?

    Thanks and regards
    Williams
     
    williams, Apr 4, 2005
    #1
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  2. williams wrote:
    > Hello Guys,
    > I had a doubt about the IBUFG and BUFG in xilinx.
    > 1.I have connected clock from oscillator to CLKG IO of the Xilinx. In
    > this case is it required to instantiate the IBUFG inside my code
    > also?.


    Howdy Williams,

    I believe the tools will usually do it for you - the exception that
    comes to mind is differential clocks.

    > 2. The DCM output is already BUFG i think and so is it required to
    > BUFG again in my code?


    BUFG is another name for a global buffer. Although most of the time
    you'd want it on a global buffer, the output of a DCM is not "already
    BUFG", nor would you always want it to be. Even if it does automaticly
    insert a BUFG in some situations, I rarely trust that it will every
    time.

    Have fun,

    Marc
     
    Marc Randolph, Apr 4, 2005
    #2
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  3. Williams,

    By default, an IBUFG & BUFG are instantiated, but if you specify something
    like:

    wire dcmclkout;
    // synthesis attribute clock_signal of dcmclkout is true
    // synthesis attribute clock_buffer of dcmclkout is none

    the synthesis tool will not issue a clock buffer placement.

    i usually instantiate IBUFG & BUFG, for safety.
    hope this helps.

    Vladislav

    "williams" <> wrote in message
    news:...
    > Hello Guys,
    > I had a doubt about the IBUFG and BUFG in xilinx.
    > 1.I have connected clock from oscillator to CLKG IO of the Xilinx. In
    > this case is it required to instantiate the IBUFG inside my code
    > also?.
    > 2. The DCM output is already BUFG i think and so is it required to
    > BUFG again in my code?
    >
    > Thanks and regards
    > Williams
     
    Vladislav Muravin, Apr 5, 2005
    #3
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