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DDC112: Page 7 of AB-136

Clifford Greenblatt
Prodigy 90 points
Part Number: DDC112

I am confused by a point made in Application Bulletin AB-136, “RETRIEVING DATA FROM THE DDC112”. At the top of the right-hand column on page 7, I read, “For TINT > t11, the first integration’s data is ready before the second integration completes. Data retrieval must be delayed until the second integration completes leaving less time available for retrieval.” I am having difficulty seeing how this situation corresponds with the Figure 10 on the same page. Maybe I am getting something wrong, so I will explain my reasoning below. 

In FIGURE 10, the first TINT must be integrating Side A in State 3 or State 5. The second TINT must be integrating Side B in State 4. If the second TINT is shorter than t11, then the DDC112 must go into State 1 and no integration occurs, as shown in FIGURE 10. This much makes perfect sense. However, if the second TINT is longer than t11, then my understanding is that the DDC112 will go into State 5 rather than State 1, so FIGURE 10 would not apply. So, I question the need to wait for the second TINT, if greater than t11, to complete before retrieving data integrated in the first TINT. Am I misunderstanding something?

  • 0
    TI__Genius 11550 points

    Hi Clifford,

    Thanks for the post. Please let me clarify this with the team and get back to you. 

    Thanks

    -TC

  • +1 in reply to TCT
    TI__Genius 11550 points

    Hi Clifford,

    Thanks for the patience.

    Clifford Greenblatt said:
    “For TINT > t11, the first integration’s data is ready before the second integration completes. Data retrieval must be delayed until the second integration completes leaving less time available for retrieval.”

    The primary intention for this statement is to avoid any digital noise coupling into the analog integration process prior to the completion of the second integration (Side-B). In this situation, we are still assuming the TINT > t11 but not beyond the Cont mode m/r/az cycle for the device to go into the continuous mode. Therefore, the time period t12 - (TINT - t11) is where the second integration (Side-B) is still ongoing, and data retrieval in this period should be avoided.

    Thanks

    -TC

  • +1 in reply to TCT
    Prodigy 90 points

    I think the answer resolves my issue. Before clicking on the resolved button, I will describe what I think it means, to ensure I properly understand it. It appears that the TINT > t11 issue is a concern only when TINT ends inside the r/az tail of m/r/az. I think the r/az tail is 582 clock cycles long. Because my design never allows TINT to end inside the r/az tail of m/r/az, I think I do not need to be at all concerned about the TINT > t11 issue.

  • +1 in reply to Clifford Greenblatt
    TI__Genius 11550 points

    Hi Clifford,

    Yes, that is correct. As long as the TINT (Side-B) does not extend into that region, you do not have to worry about where data retrieval occurs.

    Thanks

    -TC