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ADS5296A: What, precisely, are the PRBS training pattern definitions?

John Handwork
Intellectual 420 points
Part Number: ADS5296A

The data sheet for the ADS5296A doesn't elaborate on the PRBS sequences and I cannot find a common reference for these patterns with a quick TI search.  I'd love more detail.

Pseudorandom Bit Sequences are not standardized, though some standards call out specific PRBS patterns.  They are typically implemented with Linear Feedback Shift Registers with an XOR or XNOR from taps in the shift register to generate the new bit shifted in.  Only some combinations of taps deliver a Maximal Length Sequence (MLS) of 2ⁿ-1 bits for an n-bit PRBS.

Does the PRBS9 pattern use an XOR of the 9th and 5th bits to generate the new (0th) bit to shift in?

Does the PRBS23 XOR bits 23 and 18 to generate the new 0 bit?

XORs with a zero seed will stay 0.  XNORs with an all-one seed will stay all ones.  Both XOR and XNOR deliver MLS sequences for the proper taps.

What should I expect from the ADS5296A so I can get my alignment code up and running without a good scope capture of the actual data?

Sincere thanks for any insights.

  • 0
    TI__Intellectual 1590 points

    Hi John,

    Please find the relevant description below-

    The PRBS generator in ADS52J96 can give 9-bit or 23-bit LFSR Pseudo random pattern on the channel outputs that are controlled by the register 0x25. To enable the PRBS pattern PRBS_TP_EN bit in the register 0x25 needs to be set. The default is the 23-bit LFSR. To select the 9-bit LFSR, set the PRBS_MODE_2 bit. The seed value for the PRBS patterns can be chosen by enabling the PRBS_SEE_FROM_REG bit to 1 and the value written to the PRBS_SEED registers in 0x24 and 0x23.

    As you understood correctly, when 23-bit LFSR is used, XOR of bit 23 and bit 18 is used to generate the new LSB bit. When 9-bit LFSR is used, XOR of bit 9 and bit 5 is used to generate the new LSB bit. The polynomial equations for both PRBS modes are as follows:

    PRBS23 : x^23 + x^18 + 1

    PRBS9 : x^9 + x^5 + 1

    Kindly let us know for any further information required.

    Regards,

    Kalyan