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Data problem for acquisition

  1. Dec 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Question is determine the maximum amount by which the Analogue to digital converter output can differ from the analogue input when a 8-bit Analogue to digital converter has a full scale input of 2.55 V. It has a specified error of 0.1% full - Scale.

    2. Relevant equations

    % resolution = step size/ full scale *100%
    3. The attempt at a solution

    I just did 0.1% * 2.55 to get the specified error which was 2.5*10^-3. Then I calculated the resolution which was (2.55/2^8-1)*100% = 0.01. So for each resolution the error was 2.5*10^-3. Thus, maximum amount by which it could differ was 0.01*2.5*10^-3 = 2.5*10^-5. Is this correct at all? Please help.....!
     
  2. jcsd
  3. Dec 26, 2014 #2

    berkeman

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    Staff: Mentor

    You correctly calculated 0.1% of full scale as 2.55mV. And the resolution is 10mV per step as well. Calculating the overall accuracy is a bit trickier, depending on how you want to define it.

    As long as the output of the ADC is monotonic (no missing codes, etc.), then the overall error would be how early/late the output code is versus a swept input voltage. Does your textbook show you a figure of how the overall error is calculated versus the step size of the ADC?
     
  4. Dec 26, 2014 #3
    Nope no diagrams on textbook. This question is all I was given. How do I calculate the maximum amount? Do I just add up the 2.55 mV and 10 mV?
     
  5. Dec 26, 2014 #4

    berkeman

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    Have a look at this tutorial from Maxim. Especially check out the ADC Gain Error and INL definitions to see if they help...

    http://www.maximintegrated.com/en/app-notes/index.mvp/id/641

    :-)
     
  6. Dec 26, 2014 #5
    They don't unfortunately....Are you sure its really so complicated? We didn't study those things in the lectures....aren't there some easy formulas to solve this question?
     
  7. Dec 30, 2014 #6

    gneill

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    Hi JI567,

    While it's not clear to me what the specifications of accuracy and resolution are meant to imply in the context of your course, I was thinking that it might be worthwhile considering what happens when an actual value lands somewhere in a 10 mV step range of the converter but has an error of +/- 2.5 mV associated with it. How far away from the the true value might the ADC's reading be if it has to round up or down to a step? A sketch might help.
     
  8. Dec 31, 2014 #7
    if you are talking about the +/- range then shouldn't it just be 2*2.5 mV?
     
  9. Dec 31, 2014 #8

    gneill

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    What's your reasoning? As I mentioned, a sketch of the situation may be helpful. Look for the worst case situation.
     
  10. Dec 31, 2014 #9
    a sketch of what though? Like what on the y axis and what on the x axis?
     
  11. Dec 31, 2014 #10

    gneill

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    Sketch one step of the ADC's "stairstep" transfer function (a 10 mV jump between output values). Imagine that a measured value falls somewhere in that range so that the when the ADC does the conversion it must round either up or down to the nearest 10 mV level. But the input value has an error of +/- 2.5 mV associated with it, so for a given actual value it could fall anywhere within a spread of values on the sketch. Find the worst case scenario for the difference between the actual value and the "measured" value.
     
  12. Dec 31, 2014 #11
    A measured value falls within a range of 10 mV? I mean how am I supposed to sketch it? I don't have any other points or values. Sorry I haven't been taught stairstep transfer function in lectures. But anyways why don't we just write 10mv - 2.5 mv as the measured value for the worst case scenario then? As for the actual value I don't know what it is as its not given in the question.
     
  13. Dec 31, 2014 #12

    gneill

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    The resolution of the ADC is 10 mV --- that's the value that you calculated. It can only return values in 10 mV increments. Any input value applied to the ADC will fall within some 10 mV step, and will be rounded to the nearest 10 mV value. So an input of 1.322 V would be rounded to the 1.23 level, while an input of 1.326 would round to the 1.33 level.
    Picture a staircase with a slope of 45 degrees. Each step has a height of 10 mV. Each level corresponds to one of the 28 possible outputs of the ADC.
    You don't need a particular actual value. You propose some value that will fall somewhere in the ADC's input range. A particular input value isn't important. It's the fact that any input value will be rounded to the nearest 10 mV level.

    So yes, 10 mV - 2.5 mV sounds like a reasonable worst-case scenario:

    Fig1.gif
     
  14. Dec 31, 2014 #13
    Alright that's some new information....so what's the answer to the question? 7.5 mV?
     
  15. Dec 31, 2014 #14

    gneill

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    By Forum rules I can't give you an answer to that... Homework Helpers are not allowed to provide solutions directly. We can provide hints, corrections, and background information to help you arrive at a solution yourself. I can state that I don't find any fault with your reasoning :)
     
  16. Dec 31, 2014 #15
    Why shouldn't the answer be 12.5 mV? I mean isn't that the maximum value obtained? I thought that's what the question was asking for.
     
  17. Dec 31, 2014 #16

    gneill

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    Can you sketch a situation where an input would lead to a 12.5 mV difference? Remember, the ADC will round to the nearest 10 mV level.
     
  18. Dec 31, 2014 #17
    so you mean because the resolution of the ADC is 10 mV the difference can't be higher than 10mV but lower than it?
     
  19. Dec 31, 2014 #18

    gneill

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    Yes, since the possible error in the input value is less than half a step size. There's no input value that would put its error range more than a half-step away, which is what would be required to jump a whole 10 mV away:

    Fig2.gif

    Slide the actual value and its error range up and down on the stairstep and see what values the extremes of the error range would round to.
     
  20. Dec 31, 2014 #19
    well lets say at 1 V. For the upper part will it be 1.0025 V? which will be rounded to 1.003? and for the down part it will be 0.9975 so will be rounded to 0.998?
     
  21. Dec 31, 2014 #20

    gneill

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    10 mV is 0.01 V. So you get two decimal digits in any rounded result. Both of those values would round to 1.00 V, the nearest step.

    Fig3.gif
     
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