Converting 10bit ADC result to 8 bit


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  1. #1
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    Default Converting 10bit ADC result to 8 bit

    I need to write some ADC results in EEprom for calibrating a design. This wont be needed in the final product, its just to calibrate a resistor divider.

    I'm using a 10bit ADC result and need the result in 10 bits for elsewhere in the program. If I put the result into a duplicate VAR WORD and divide it by 4 , can I put that result in a new VAR BYTE and write the byte to EEprom?

    It *looks* like this works from what I'm reading back from EEprom but I just wanted a second opinion.

    Thanks!

  2. #2
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    Default Re: Converting 10bit ADC result to 8 bit

    shift the value 2 places to the right.... 10bits shift >> 2, 8 bits....
    Dave Purola,
    N8NTA
    EN82fn

  3. #3
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    Default Re: Converting 10bit ADC result to 8 bit

    If you simply shift out the least significant bits you lose any small variations between values
    that might not have occurred with a divide, but then when you multiply back to the original range
    you also drop resolution.

    You can calculate a new range like this:
    NewValue = (((OriginalValue - OriginalMin) * (NewMax - NewMin)) / (OriginalMax - OriginalMin)) + NewMin

  4. #4
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    Default Re: Converting 10bit ADC result to 8 bit

    Unless I am missing the obvious. Shifting right by two, or dividing by two (integer division) is identical. The only difference should be in speed. Shifts should always be faster, and divides may be slower depending on the compiler optimization.

  5. #5
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    Default Re: Converting 10bit ADC result to 8 bit

    Lanquer, That is correct...
    Dave Purola,
    N8NTA
    EN82fn

  6. #6
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    Default Re: Converting 10bit ADC result to 8 bit

    Ignore my whole post... don't know where my head was,
    you're going to drop the same resolution anyway!

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