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AMC1305M25: what's DOUT frequency?

Yuan Tan63
Expert 3280 points

Part Number: AMC1305M25

Hi,

When CLKIN=20MHz, is DOUT fixed to 10MHz since data change at CLKIN falling edge?

But how to understand the picture as below?  when analog input increase, It seems not only the "1" density but also frequency changes.

customer test our AMC1305 with 0mV input. it shown as below:

the "1" density is 50%, no problem, but DOUT frequency is 10MHz or 5MHz. I assume that it's due to 01 01 01 01 or 01 10 01 10, its always 10MHz, but when it shown as 01 10 01 10, it looks like 5MHz. right? is it normal? thank you!

Best regards,

Yuan Tan

  • 0
    TI__Mastermind 31655 points

    Hi Yuan,

    DOUT is clocked at the same frequency as the CLKIN. One DOUT bit is one MCLK cycle wide. Since there may be adjacent 1s or 0s this will look like a longer period/lower frequency at certain points, but the data output rate is fixed at 20Mhz.

    As a note, I edited this response because I previously put the incorrect data output rate. The latest post is the correct one, with the data output rate at 20Mhz with a 20Mhz clock.


    Regards,

  • 0 in reply to Scott Cummins
    TI__Expert 3280 points

    Hi Scott,

    I am not agree the data output rate is 20Mhz with a 20MHz clock, it should be 10MHz since data change at CLKIN falling edge. 

    I can understand it’s a 50% 1 and 50% 0 with 10MHz but sometimes it looks like 5M, right?  (Since in one cycle should appear 50% 1 and 50% 0, so that the continuous 4 data should be 10 10 or 01 01 or 0110 or 10 01, the latter two seems like 5MHz)

    Want to confirm 2 questions:

    1 from datasheet, how to understand that? when analog input increase, It seems not only the "1" density but also frequency changes. 

    2 customer test other supplier's part, it also shows 10M and “5M” but their “5M” have periodicity as green wave as below. does this mean it is better? Thank you!

    thank you!

    Best regards,

    Yuan

  • 0 in reply to Yuan Tan63
    TI__Mastermind 31655 points

    Hi Yuan,

    You're right about the DOUT frequency, and I had it right the first time but not in my edit. Since it is changing only on the falling edge of the clock signal, the output will be 10Mhz.

    This is a fixed 10Mhz, and the frequency of the output does not change. If you have two 1s followed by two 0s followed by another 1, this will look like the frequency is at 5Mhz. As the density of 1s increases, the chance of seeing that pattern I just described is increased. This does not mean the frequency that the data is provided at is changing, only that the data itself is changing.

    The bitstream that contains the data will look like it has a varying frequency, but again, the frequency that the data is being output is never changing.

    Regards,