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OPA348-Q1: Questions about FIT(Failure In TIme)

Intellectual 1430 points

Replies: 6

Views: 1821

Part Number: OPA348-Q1

Hi Team

my Customer has the following questions about FIT for OPA348-Q1/OPA2348-Q1/OPA4348-Q1.

may I expect your swift response ?

thanks.

[Q1] OPA348 is 1 CH, OPA2348 is 2 CH, OPA4348 is 4 CH.

        BTW,,, all ICs have same FIT = 1.6 as you see in below.

        all ICs have different channels, but why they show same FIT ?

        may I get your detail explanation on it ?

[Q2] our FIT rate calculation say FIT = 1.6 for OPA2348-Q1(2 Ch).

        populating OPA2348-Q1(2 Ch) on customer board level, but in fact, customer is using only one channel of the integrated two channels,

        and the unused another channel is left floating.  in this case, customer must use FIT = 0.8 for OPA2348-Q1(2 Ch) on his board level ?

        or customer must use FIT = 1.6 on his board level even if using one of the integrated two channels on board level.

         let me expect your detial reason explanation.

  • Paul,

    IC components are in general very reliable, therefore, it is difficult to accumulate enough test time to get statistically valid number on their Mean Time Between Failures (MTBF) and Failure in Time (FIT). For that reason, the manufacturer will test over a period of time a large number of wafer lots of single, dual and quads versions, as well as similar products fabricated on the same semiconductor process, to determine their overall failure rate. 

    There are standard formulas that convert the number of failures in a given test time to MTBF for a selected confidence level.  For a system of components, one method of predicting the MTBF is to add the failure rates of each component and then taking the reciprocal. For example, if one component has a failure rate of 100 FITs (100 failures per billion of hours), another 200 FITs and another 300 FITs, then the total failure rate is 600 FITs and the MTBF is 1/(600 failures per one billion hours) or 1.67 million hours.

    Therefore, all three versions of OPAx348 have a combined MTBF of 6.42E8 which results in FIT of 1/(6.42E8) or 1.56 failures per billion of hours of operation.

    Since FIT number refers to an individual channel, you should use FIT of 1.6 for your calculation.  Using both channels would double the combined time of operation resulting in combined MTBF of 6.48E8/2 - this in turn would lead to FIT of 1/(6.48E8/2) or 3.12 failures per billion of hours.

    BTW, you should NOT allow unused channel to float as this may lead to unpredictable results - read the blog below:

    Marek Lis, MGTS
    Sr Application Engineer
    Precision Analog - TI Tucson

  • In reply to Marek Lis:

    Hi Lis

    May I get your swift response again ?

    thanks a lot.

    [Q1] you means like below ?

            the FIT calculation on TI.com is based on 1 channel,

            so that OPA348(1 Ch)/OPA2348(2 Ch)/OPA4348(4 Ch) should have same FIT = 1.6,

            however, if customer uses OPA348(1 Ch), customer can use FIT = 1.6 on his board level,

            but if customer uses OPA2348(2 Ch), customer has to use FIT = 3.2( = 1.6 x 2) on his board level,

            and if customer uses OPA4348(4 Ch), customer has to use FIT = 6.4( = 1.6 x 4) on his board level ?

            hence, customer must use our All OP AMP device's FIT by calculating [TI.com showing FIT x Channel number] ?

    [Q2] how about our comparator devices ? their FIT on Ti.com is also based on 1 Ch ? 

            for comparator device, I also see same FIT value = 0.4 for TLV3701(1 Ch)/TLV3702(2 CH)/TLV3704(4 Ch) comparator devices as like OPA348/2348/4348.

            even if  TLV3701(1 Ch)/TLV3702(2 CH)/TLV3704(4 Ch) comparator devices has same FIT = 0.4,

            customer also has to use FIT = 0.4 for TLV3701(1 Ch), and customer also has to use FIT = 0.8 for TLV3702(2 Ch),

            customer also has to use FIT = 1.2 for TLV3704(4 Ch) ?

  • In reply to Paul Kim:

    Hi Lis

    I added one more question.

    May I get your swift response again ?

    thanks a lot.

    [Q1] you means like below ?

            the FIT calculation on TI.com is based on 1 channel,

            so that OPA348(1 Ch)/OPA2348(2 Ch)/OPA4348(4 Ch) should have same FIT = 1.6,

            however, if customer uses OPA348(1 Ch), customer can use FIT = 1.6 on his board level,

            but if customer uses OPA2348(2 Ch), customer has to use FIT = 3.2( = 1.6 x 2) on his board level,

            and if customer uses OPA4348(4 Ch), customer has to use FIT = 6.4( = 1.6 x 4) on his board level ?

            hence, customer must use our All OP AMP device's FIT by calculating [TI.com showing FIT x Channel number] ?

    [Q2] how about our comparator devices ? their FIT on Ti.com is also based on 1 Ch ? 

            for comparator device, I also see same FIT value = 0.4 for TLV3701(1 Ch)/TLV3702(2 CH)/TLV3704(4 Ch) comparator devices as like OPA348/2348/4348.

            even if  TLV3701(1 Ch)/TLV3702(2 CH)/TLV3704(4 Ch) comparator devices has same FIT = 0.4,

            customer also has to use FIT = 0.4 for TLV3701(1 Ch), and customer also has to use FIT = 0.8 for TLV3702(2 Ch),

            customer also has to use FIT = 1.2 for TLV3704(4 Ch) ?

    [Q3] for OPA2348(2 Ch), if user will not use one of two channels,

           user can use FIT = 1.6 for OPA2348(2 Ch), right ?

  • In reply to Paul Kim:

    Paul,

    That is correct - the FIT calculations are based on a single channel because even if one channel fails, it does not affect the other channels from continue to operate properly.  Therefore, I would not go as far as to change the FIT numbers the way you suggest - e.g. if you use only a single channel in a quad, the failure of any of the unused channels will not change the probability of failure of the channel you use; thus its FIT number should stay the same regardless whether you use it in a single, dual or quad version.  It is a completely different question in case you want to determine the overall probability of system failure containing multiple IC's - in such case, you may need to add the FIT numbers of all the IC's and/or used channels to determine total FIT number of your system.

     

    Marek Lis, MGTS
    Sr Application Engineer
    Precision Analog - TI Tucson

  • In reply to Marek Lis:

    Hi Lis

    I am confused....you mean whether I am correct or I have something wrong...

    if user board targets at ASIL-D on board level, the all populated components's summed FIT on critical safety path on board level should be below 10 FIT.

    the FIT calculation on TI.com is based on 1 channel for OPAMP IC, and Ti.com shows same FIT = 1.6 for OPA348(1-CH)/2348(2-Ch)/4348(4-Ch).

    hence, I think user has to (TI.com calculated FIT x Channels number) for board level FIT summing, because TI provide one channel based FIT.

    because TI.com calculates FIT based on one channel and OPA2348 has 2 channels inside, I think user has to sum each channel's FIT = 1.6 for board level FIT

    budgeting to meet ASIL target, and finally user has to use FIT = 3.2 for OPA2348 because it has 2 channels inside and Ti.com provides FIT = 1.6 for one of two

    channels. hence, I think the following should be correct for user to apply FIT for OPA348/2348/4348 on board level FIT summing.

    please let me get your clear response on the below each example question respectively to make the clear communication.

    thanks.

    [Q1] so if he uses OPA348, he has to use FIT = 1.6 for OPA348 because it has the only one channel and he will use the only one channel on board level.

            so if he uses OPA2348, he has to use FIT = 3.2( 1.6 x 2) for OPA2348 because it has two channels inside and he will use all 2 channels on board level.

            so if he uses OPA4348, he has to use FIT = 6.4 (1.6 x4) for OPA4348 because it has 4 channels inside and he will use all 4 channels on board level.

    [Q2] so if he uses OPA2348, he has to use FIT = 1.6 for OPA2348 because it has two channels inside and he will use only one channels on board level.

            so if he uses OPA4348, he has to use FIT = 1.6 for OPA4348 because it has 4 channels inside and he will use only one channels on board level.

    [Q3] so if he uses OPA4348, he has to use FIT = 3.2(1.6 x 2) for OPA4348 because it has 4 channels inside and he will use only two channels on board level.

     

  • In reply to Paul Kim:

    Paul,
    Your latest interpretation of the FIT number is correct. In case of OPAx348, you need to multiply FIT of 1.6 by the number of USED channels. Therefore, it would not matter whether you use three singles, one single and one dual, or three channels in a quad - for all three cases, the overall FIT number would be equal to 4.8 (3x1.6).

    Marek Lis, MGTS
    Sr Application Engineer
    Precision Analog - TI Tucson

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