Maximum value of ADCIN


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  1. #1
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    Fratello,

    The result should be in mVolts.
    Last edited by rsocor01; - 12th November 2010 at 16:36. Reason: Incorrect statement
    "No one is completely worthless. They can always serve as a bad example."

    Anonymous

  2. #2
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    You almost had it right Fratello.
    And the timing will be more accurate if you don't do the math on each loop.
    Code:
    vmax = 0
    for cnt=1 to 70       			
        adcin 3, v_cal
        vmax=vmax max v_cal
        pause 100
    next cnt
    v_cal=vmax * 5000            ' I use VDD as Vref
    v_cal=div32 1023
    DT

  3. #3
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    Thank You all for reply !
    Both variant (mine = 0 max v_cal and variant of Mr.Robert ) give me the same results : 24 .
    According of my math, I think this is the value of 24 mVolts. I am right ?
    Now I try the variant of Mr.Darrel (remember ? My guardian angel in PBP problems ...).
    I post the results.
    LE : Something strange...with this variant, I read only 5 mV ?!
    Maybe the schematic help...
    Attached Images Attached Images  
    Last edited by fratello; - 12th November 2010 at 16:16. Reason: schematic added

  4. #4
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    Oops, disregard my comment. Yes, you should get your result in millivolts.

    Robert
    "No one is completely worthless. They can always serve as a bad example."

    Anonymous

  5. #5
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    Tested and re-tested today both variant : Mr.Darrel's and Mr.Aratti.
    Still different results : 5mV vs. 24 mV. It's so strange for me...

  6. #6
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    fratello,
    I'm not sure what you've done in your code, but I've tested it here and it works.

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    Attached Files Attached Files
    DT

  7. #7
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    Wow, Thank You again Mr.Darrel !
    Maybe my code it's wrong ; I use
    Code:
    DEFINE OSC  4
    DEFINE ADC_BITS 10
    DEFINE ADC_CLOCK 3
    DEFINE ADC_SAMPLEUS 50
    instead
    Code:
    DEFINE ADC_BITS 10     ' ADCIN resolution  (Bits)
    DEFINE ADC_CLOCK 1     ' ADC clock source  (Fosc/8)
    DEFINE ADC_SAMPLEUS 11 ' ADC sampling time (uSec)
    This is the explanation, don't ?
    NO, I think this is NOT the explanation ! My math teacher it's verry, verry angry ...
    I learn more here : http://www.darreltaylor.com/DT_Analog/
    1 step=0.0048875. So, if I have Max_result 450, Volts= 450 * 0.0048875 =2.199 Volts; for 696 --> Volts= 696 * 0.0048875 =3.401 Volts...
    Stupid me, again
    Last edited by fratello; - 13th November 2010 at 16:54.

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