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ISO7721: FIT data for ISO7721

Part Number: ISO7721

Hi,

We have designed in the ISO7721 as a replacement alternative for Silicon Labs isolators. However, we did not manage to find any FIT related data from the online tool

http://www.ti.com/quality/docs/estimator.tsp#resultstable

A search for ISO7721, ISO7721FDWR yields no results.

We require details of the endurance test including number of samples assessed, number of test hours, failure rate, test temperature, etc,.... (ie, the usual results obtained from the online tool)

Can this be provided as soon as possible ?

  • Hi Christopher,
    Thanks for asking your question on E2E!
    Our apologies, our online tool seems to still be down. I can get our quality engineer to manually pull the data. I'll respond back by the end of the day.
    Best regards,
    Dan
  • In reply to Dan Kisling:

    Hi Christopher,

    Thank you for your patience. See below: 

    DEVICE

    MTBF

    FIT RATE

    MTBF CONF LEVEL

    MTBF USAGE TEMP

    ACTIVATION ENERGY

    MTBF SAMPLE SIZE

    MTBF TEST TEMP

    MTBF NUM FAILS

    TEST DURATION

    ISO7721FDWR

    2.50E+09

    0.4

    60

    55

    0.7

    47098

    125

    0

    1000

    Best regards, 

    Dan

  • In reply to Dan Kisling:

    Thank you for your reply Dan, really appreciate it.

    One final confirmation in regards to your answer.

    I see that it is stated "MTBF SAMPLE SIZE" to be 47098. Does it mean that in total, for this device silicon, 47098 samples were assessed at a test temperature of 125 degC for 1000 hours and there were 0 failures ? (ie , total equivalent number of test hours is 47098000 hours )

  • In reply to Christopher Sam Soon:

    I tried to recalculate the FIT value provided and I cannot get the FIT data of 0.4 based on the information you listed.

    Based on a test sample size of 47098, with a test run of 1000 hours at 125 degC and 0 failures, the 60% confidence level FIT rate should be 0.25 FIT @ at a use temperature of 55deg. However, the number you listed is 0.4 and I do not understand how TI is getting 0.4.

    In the past, I have always recalculated the FIT numbers using the details of the test run results (provided by the online tool) and verified them against TI's FIT numbers and the numbers always matched ( within +/- 0.1 FIT error due to rounding) . This case does not seem to be the same.

    Did I miss something?

    Please advise

    Thank you!

    Christopher

    Note: The details of the test results is important to us as in our application, our use temperature is not always 55 degC and we always use FIT numbers with 90% confidence level. The data of the test results allows us to regenerate the FIT values for other use case temperatures and confidence levels
  • In reply to Christopher Sam Soon:

    Hi Christopher,

    Sorry for the confusion.
    You are correct that the calculated value using statistics gives a FIT of 0.2. Since ISO7721 has two dies, we have taken a more conservative approach by adding the FITs to get the worst case FIT of 0.4.

    In regards to your previous post, your understanding is correct: 47098 samples were assessed at a test temperature of 125 degC for 1000 hours and there were 0 failures.

    Let me know if this answers your question or if you have any additional questions.

    Best regards,
    Dan
  • In reply to Dan Kisling:

    Hi Dan,

    Would that not make the FIT 0.5 ? Because based on the data, there should be a FIT of 0.25 per die, and if like you mention, you cater for two dies, then  the combined FIT should be 0.5, is it not ? Or is there some common mode failures that I am not aware of here (ie, packaging, isolation barrier, etc..)

    Of course, 0.1 FIT is statistically insignificant, but I would like to have more insights on the thinking of TI regarding this peculiar case (ie, why 0.4 FIT instead of 0.5 FIT as per the test results)

    Best

    Christopher

  • In reply to Christopher Sam Soon:

    Hi Christopher,

    I talked this over with our quality engineer some more. We calculate a 0.2 FIT and double it for 0.4 FIT. He said that you can use the 0.5 FIT number if you like since your math leads to 0.25 instead of 0.2. Sorry I don't have a better answer than that for you.

    Best regards,

    Dan

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