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MCP3551 and PT100: Understanding Measurement Accuracy, Resolution & AN1154 Application

FastProject 6321 8
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  • #1 13255873
    FastProject
    Level 28  
    Hello.
    Regarding the application AN1154 I would like to use Microchip Ratiometric measurement of PT100 temperature using the MCP3551 transducer.

    I read the description from this documentation and I can not understand from where and how the resolution and measurement accuracy were calculated and the resistance measurement itself. Therefore, I do not know how to write a temperature calculation program, based on reading from ADC.

    So please, explain this passage in particular:
      MCP3551 and PT100: Understanding Measurement Accuracy, Resolution & AN1154 Application

    As far as I know the 1 µ V resolution, I do not know why Ra and Rb are 6.8 k Ω and from where value 111.6 μ V / ° C and value 0.008 ° C / LSb ...

    The author of the application probably uses some simple patterns and rounding, so please help in deciphering all these calculations.
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  • #2 13257831
    jarek_lnx
    Level 43  
    Quote:
    I do not know why Ra and Rb are 6.8 k Ω
    A trade-off between signal level and self-heating.

    Quote:
    and where from value 111.6 μ V / ° C
    The entire divisor is 6.9k Ω RTD sensitivity is 0.385 Ω / ° C reference voltage 2V so 0.385 Ω / ° C / 6900 Ω * 2V ≈ 111.6 µ V / ° C

    Quote:
    and value 0.008 ° C / LSb ...
    1µV/111.6µV/°C

    Watch out for thermoelectric voltages, these ones look good on paper, but each combination of two metals is a thermocouple that can generate tens of voltages under unfavorable conditions µ V.
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  • #3 13258329
    FastProject
    Level 28  
    And why is it 6.9k Ω instead of 6,8k & # 937 ;. Is 100 Ω with PT100, but why? After all, RA is everywhere as 6.8k Ω ...?

    Ok, so what does it receive as a result of ADC conversion, what represents the value of the Code and how to transform it into temperature? Lookout table every 1 ° C will be good? Fractional values count interpolation from 2 points?
  • Helpful post
    #4 13258582
    jarek_lnx
    Level 43  
    Quote:
    And why is it 6.9k Ω instead of 6,8k & # 937 ;. Is 100 Ω with PT100, but why? After all, RA is everywhere as 6.8k Ω ...?
    So the sum, count like any other resistive divider in the nominative will be the sum.
    Quote:

    Ok, so what does it receive as a result of ADC conversion, what represents the value of the Code and how to transform it into temperature? Lookout table every 1 ° C will be good? Fractional values count interpolation from 2 points?

    Have you ever had an analogue ohmmeter? here you have the same, only the "pointer" is digital, in the analog equivalent there was a nonlinear scale here too, you linearize by "equation 1", you get resistances from the resistance you will calculate the temperature.
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  • #5 13283302
    FastProject
    Level 28  
    Quote:

    Quote:
    and where from value 111.6 μ V / ° C
    The entire divisor is 6.9k Ω RTD sensitivity is 0.385 Ω / ° C reference voltage 2V so 0.385 Ω / ° C / 6900 Ω * 2V ≈ 111.6 µ V / ° C


    I am wondering this value of resolution 111.6 µ V / ° C, because yes
    R min 18.52 Ω For -200 ° C
    R max 390,48 Ω For + 850 ° C

    With this at 290 μ A, the voltage at PT100 is:
    Urtd min 0.005368V For -200 ° C
    Urtd max 0.13183V For + 850 ° C

    The range of samples from -200 to 850 ° C is 1050 samples.
    111.6 µ V / ° C * 1050 = 0.11717V

    Where does this discrepancy between 0.113183V and 0.11717V come from (discrepancy of almost 4mV)?
  • #6 13339813
    FastProject
    Level 28  
    I think that it is best to deduce t from the formula for resistance Pt100 and substitute the calculated resistances Rrdt from equation 1 for this formula.

    Just damn how to draw t from this formula: R = R0 * (1 + At + Bt ^ 2)
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  • Helpful post
    #7 13342464
    jarek_lnx
    Level 43  
    As you noted in post # 5, the non-linearity of the sensor does not allow for such a simplification in such a wide temperature range.

    You have to solve the quadratic equation that you mentioned, it is difficult, because of some useful tips for implementation, I recommend to look at the application note Analog Devices AN-709.
  • #8 13342603
    FastProject
    Level 28  
    Thank you guy jarek_lnx, pattern with DIRECT MATHEMATICAL METHOD AN-709 (which defakto lay on my disk) solves my problem.

    Now since this equation:
      MCP3551 and PT100: Understanding Measurement Accuracy, Resolution & AN1154 Application
    I will measure and face Rrtd, then from this:

      MCP3551 and PT100: Understanding Measurement Accuracy, Resolution & AN1154 Application
     I will calculate the temperature.

    The calculations will take a while, but the system will control the heater so the calculation speed is not needed. As for me, this is the most convenient method, because you do not need a data table.

    And for the sake of curiosity, I was transformed into a model procedure by: http://www.wolframalpha.com/
      MCP3551 and PT100: Understanding Measurement Accuracy, Resolution & AN1154 Application
  • #9 13796248
    luzik
    Level 11  
    what would you have to do with the system to narrow the measurement to 10-60st? (losing in range, gaining resolution?
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Topic summary

The discussion revolves around the application of the MCP3551 ADC for ratiometric measurement of PT100 temperature sensors, specifically addressing the calculation of measurement accuracy and resolution as outlined in application note AN1154. Users seek clarification on the derivation of key values such as the 111.6 µV/°C sensitivity and the choice of resistor values (6.8 kΩ vs. 6.9 kΩ). The conversation highlights the importance of understanding the non-linear characteristics of PT100 sensors and the need for precise calculations to convert ADC readings into temperature values. Various methods for calculating temperature from resistance values are discussed, including the use of quadratic equations and the recommendation to refer to Analog Devices' AN-709 for further guidance.
Summary generated by the language model.
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