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TMP20: How to calculate accuracy at a specific temperature?

Izak Liebenberg
Prodigy 10 points
Part Number: TMP20
Other Parts Discussed in Thread: TMP236

Hi,

We use the TMP20 in a temperature control application which is set to 30 °C.

The temp error at 30 °C is between 0.8 % and 1 %.

The max deviation (error) around 30 °C is 0.004 °C.

Accuracy vs supply is given as 0.05 °C/V which translates to 0.076 °C at 1.5154 V (30 °C).

What is the accuracy at this temperature? How is it calculated around 30 °C?

  • 0
    TI__Expert 7180 points

    Hi Izak,

    Deviation and power supply rejection are both sources of error that can affect the accuracy of the TMP20:

    • The maximum deviation parameter is a post-processing offset error. If you use the parabolic transfer function, this offset from linear approximation will not be present so it all comes down to how much processing time you are willing to spend in calculating temperature vs. how much accuracy you are willing to give up in your system.
    • Please note that the supply range of this device is from 1.8 V to 5.5 V. Power supply rejection refers to supply of the device, not the output voltage level.

    The graph shown from the datasheet comes from bench characterization of the device and it gives you an idea of typical values seen from a batch of devices. However, due to process variation, what you should expect in your system is the guaranteed accuracy spec of ±2.5°C over the full range.

    I would recommend looking at TMP236 for better accuracy and gain (resolutiion over temp range). 

    Best regards,

    Simon Rojas

  • 0 in reply to Simon Rojas Avila
    Prodigy 10 points

    Thanks for your quick response SImon.

    Ok, so if I use the parabolic transfer function to calculate the temperature, what accuracy can I expect? I struggle to see how one gets a ±2.5°C accuracy.

    The eq. 4 states that T equals the temperature in the middle of the range. Should T in Example 1 not be 75 °C instead of 35 °C?

    Also, working through the example eq. 5 doesn't give the correct offset.

    And the offsets given in eq. 3 and 6 doesn't match that in Table 1. The latter seems correct.

  • +1 in reply to Izak Liebenberg
    TI__Expert 7180 points

    Izak,

    Whether parabolic or linear approach, these are post processing techniques. The parabolic approach gives you the proper gain over the full range so you would not be adding any additional error to the guaranteed ±2.5°C spec. The ±2.5°C error is intrinsic to the part due to process variation in production. For this reason, all error calculations in your system should use the maximum error.

    In terms of the example, the middle point temperature T is 35°C. As 40+35=75 and 110-35=75--hopefully this makes sense. The offset is given by: b = (0.5520+1.4566+0.01177(110+35))/2 which yields b = 1.8576 V.

    Best regards,

    Simon Rojas