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Altera_Forum's avatar
Altera_Forum
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16 years ago

tx_clkout in alt2gx

I am working with alt2gx modul, and I can't understand the conection between the clocks, nad data rate.

If I have 8 bits channel width, and the input clock fo pll is 156MHz, and I want to work at data rate of 780Mbps, Why the tx_clkout frequency is 78MHz and not 97.5MHz (780/8)? Is it because the syncronization bits, the actual frequency of the channel becomes higher?

4 Replies

  • Altera_Forum's avatar
    Altera_Forum
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    The input clock rate is pretty configurable such that it is endependent of your serial or parallel data rate. What's the parallel data width coming in/out of the module? Are you doing anything that adds/loses data, like 8b10b?

  • Altera_Forum's avatar
    Altera_Forum
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    Yes I am using 8 to 10 bit. So the modul gives me the actual rate of the parallel channel that i need to use in order to get the data rate that I choose in the serial channel. Is this the reason why the serial data rate is devided by 10 (since the 8to 10 bit adds 2 more bites)?

    I will check the frequency without 8to10 bit, but I should get output frequency of data rate devided by 8, am I right?
  • Altera_Forum's avatar
    Altera_Forum
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    You put in 8 bits of data, 2 more bits are added by 8b10b block, and so 10 bits go through the serializer and out the device. So you do have a 10:1 ratio. (Same thing if being received.) High-speed I/O pretty much always lose some of their throughput for the encoding. (There's a 64/66b encoding, I can't think the name of off the top of my head, that might work better. I also can't remember if it's a hard block, but it would also depend on the device you're using.)

  • Altera_Forum's avatar
    Altera_Forum
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    Well ratio of 64/66 is not critical. I could not understand at the begining why is the data rate devided by 10 and not 8. You explained it. Thanks.