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14 bits at the output means 16 bits -2 on the MSB side?

  1. Mar 25, 2012 #1


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    Just trying to check myself, as 14 bits doesn't seem to make sense...so I wanna see how does it make sense. Is what written at the title true?
  2. jcsd
  3. Mar 25, 2012 #2
    14 bits means just 14 bits. You dont have to fight war for that.
  4. Mar 25, 2012 #3

    jim hardy

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    I suspect it's device specific.

    My old system(ca 1973) gave 14 significant bits in its 16 bit word, two MSB's were hardwired together and used for sign .
  5. Mar 25, 2012 #4

    I like Serena

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    If you have to connect 14 bits of output to 16 bits of input, you would typically connect those 14 bits to the 14 least significant bits (lsb), leaving the 2 most significant bits (msb).

    So yes, what you wrote at the title is true.
  6. Mar 25, 2012 #5


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    Seems simple enough-- thank you.
  7. Mar 25, 2012 #6


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    This is an interface question, I guess and the solution will depend upon the specifics.. If you have only 14 bits and want to connect to a 16 bit input then where you connect will depend upon what the 16 bit input ' expects' to receive. If you want 16 1s to be correspond to the maximum value that your 14 bits correspond to then you would need to connect MSB to MSB and put up with the coarser resolution. If the maximum value your 14 bit signal corresponds to 1/4 of the 16 Bit range, then you would connect LSB to LSB and accept the 1/4 reduction in range but the same resolution.

    Or is your 'lack of belief' in 14 bits because you expected data to be in two 8bit bytes?
  8. Mar 25, 2012 #7
    Where to connect the 14 bits to 16bits depend on the range you want. Usually for maximum range, you map MSB to MSB so you just map the 14MSB to the MS 14 bits of the 16 bit system. Then you just ground the LS 2 bits. With this, you get the full range of the 16 bit system BUT with only 14 bit resolution.
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