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Thread: **SOLVED coefficents of resistance

  1. #1

    Thread Starter
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    **SOLVED coefficents of resistance

    I have used the stienhar-hart
    1/T = a + b(LN R) + c(Ln R)^3

    to get my coefficents
    0.000484976
    0.000311308
    0.000000490

    Now with this information how do I determine
    what the temperature is ?


    Brad

    The answer is simply plugging in the numbers where 5875 is the resistance of the range 0 to 65535

    1 / 0.000484976 + 0.000311308 * LN(5875) + 0.000000490 *
    (LN(5875)^3) - 273.15 == 12C

    Brad
    Last edited by oddprime; Mar 4th, 2002 at 09:36 AM.
    OddPrime
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  2. #2

    Thread Starter
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    I guess I should of added the range :-)
    0 to 65535

    and a brief explamation.
    I have hooked up (2) 10k ohm thermistors
    to the joystick port since it is an analog to
    digital converter with a range of 0 to 65535
    I can see the resistance being done by
    the thermistors with the X and Y position of a
    standard joystick driver. I have an application
    already that after I put in the coefficients will
    determine the temperature. My question is
    how it does it... I have the coeficcients as listed
    above I just dont know how it takes coefficeients
    and the range 0 to 65535 and determines the temperature.

    thanks for any help
    Brad
    OddPrime
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  3. #3
    Hyperactive Member
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    Are you asking for an explanation of the relationship between the resistance of a thermistor and its temperature?

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