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I would like to make multiple strings of sensor having about 10 thermistors per string, each wired individually, but terminating at different lengths so that they're evenly spaced along each string. This is to verify that 10 cu. yards of unknown materials having high thermal resistance reach a yet-to-be-determined target-temperature within the 65°F to 200°F range for a specified period of time. The target is currently 150°F but may change.

To reduce self-heating effects, I chose this 100K NTC thermistor:
NTC Thermistor 100K 4250K Bead Murata NXRT15WF104FA1B040 Digi-Key 490-7169-ND

But I want to know if a 1Meg NTC would be better, and if there are any other factors that I really need to consider. Why not choose the 1Meg for the least self heating effect?

So then, what are the tradeoffs for NTC Thermistors regarding 10K vs 100K vs 1Meg?

I have noticed in the desktop power supplies that I work with, that the 10K NTC seem to be a de-facto standard. A really good, thorough answer to this general question might really help others as well. Thanks ahead of time.

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    \$\begingroup\$ You can easily filter noise with a cap on high R values as the rate of change is slower than noise \$\endgroup\$
    – D.A.S.
    Commented Jul 11, 2021 at 6:23
  • \$\begingroup\$ Thank you, Tony. But is this all I have to consider? \$\endgroup\$ Commented Jul 15, 2021 at 16:27
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    \$\begingroup\$ Sensitivity to current , input bias current, , amplification required , accuracy, range , power savings and filtering are all considered. \$\endgroup\$
    – D.A.S.
    Commented Jul 15, 2021 at 17:43

3 Answers 3

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Why not choose the 1Meg for the least self heating effect?

Noise is a big reason, the thermal noise for a resistor is:

$$V_n = \sqrt{4k_BT\Delta f R}$$

so all things being the same the Vn for a 1MegΩ resistor will be 10 times higher than a 10kΩ resistor (R would be 100x more). Also the difference between 1MegΩ and 100kΩ would be 3 times. In addtiion, usually these resistors are placed in a wheatstone bridge, with a resistor in the same order of magnitude so that resistor noise will also increase (to keep the voltage range of the bridge the same) and the two noise sources will add with the sum of the squares, which will (in most cases) be up to two times the noise.

There are tradeoffs, increasing resistance means less self heating, but more noise. It is up to the designer to find out which is best for their application.

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    \$\begingroup\$ How much effect do you reckon Johnson-Nyquist noise has on the temperature reading, say with a 1Hz bandwidth? \$\endgroup\$ Commented Jul 15, 2021 at 16:35
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    \$\begingroup\$ @sphero Temperature noise actually is a large source (if not the largest source ) of noise in the thermistor circuits I design. \$\endgroup\$
    – Voltage Spike
    Commented Jul 15, 2021 at 17:05
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    \$\begingroup\$ @micro 1Hz is the circuit bandwidth, determined by filtering and it's the Delta f term in the equation \$\endgroup\$
    – Voltage Spike
    Commented Jul 15, 2021 at 17:06
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    \$\begingroup\$ @VoltageSpike So if a typical NTC thermistor has 1V across it, and therefore changes about 40mV/°C and we have 105nV RMS thermal noise (1M || 1M at 130°C) over a 1Hz BW, that's not very many micro kelvins of noise. It will be lost in the 1/f noise of most op-amps. \$\endgroup\$ Commented Jul 15, 2021 at 17:31
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    \$\begingroup\$ @SpehroPefhany and that is exactly what I'm doing, is measuring uK, and I also use the lowest noise opamps I can buy \$\endgroup\$
    – Voltage Spike
    Commented Jul 15, 2021 at 17:38
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It depends a lot on actual situation. The self-heating may or may not be significant. Remember it's the power dissipation at the temperature(s) of interest that matters, and the thermal resistance to whatever medium they are in, and the required accuracy. That's at least 3 to 5 variables when you count sensor tolerance and the circuitry involved.

Unless power is at a premium (eg. battery) lower resistance has more noise immunity and immunity to electrical leakage (for example if there is moisture).

Sometimes thermocouples are used in this type of application, not for their stellar accuracy (it's actually tricky electrically and mechanically to get good and stable readings, and cheap solutions lead to a lot of problems) but because they are mechanically quite robust and are very low impedance.

So higher design complexity and cost is traded off against superior reliability in the field.

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  • \$\begingroup\$ Thanks for mentioning the leakage -- hadn't thought about that, and thank you for answering! \$\endgroup\$ Commented Jul 15, 2021 at 16:30
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Thermal Sensitivity to constant current , input bias current, , amplification required ,input noise current and voltage, accuracy, range , power , cost savings and filtering are all considered when choosing the solution to meet your design specs and the tradeoffs for selection.

The electrical filter time constant ought to match the mechanical thermal time constant for optimal response time at low noise unless you can tolerate a slower response.

Start with your “must have” design specs then add “nice to haves”

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  • \$\begingroup\$ I must have an accurate answer (0.1 deg F, a self-challenge). I think that I can tolerate a slower response for a more accurate answer... Once every 1-5 minutes for each of either 60 or 120 sensors sounds okay to me. The process is in hours (>2). \$\endgroup\$ Commented Jul 15, 2021 at 19:05
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    \$\begingroup\$ i got these inexpensive digital thermometers with 3 remote sensors and one in the master and all give the same temperature with 0.1’C resolution. So repeatability is great but accuracy is only as good as the calibration you rely on. \$\endgroup\$
    – D.A.S.
    Commented Jul 15, 2021 at 21:08
  • \$\begingroup\$ I can get distilled h2o and boil it for one point, and freeze it for the other point, and get my calibration, right? (Taking atmospheric pressure into account, of course, and altitude, I think). \$\endgroup\$ Commented Jul 15, 2021 at 21:14
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    \$\begingroup\$ 0.1'F is harder to achieve than 0.1'C You should convert ;) \$\endgroup\$
    – D.A.S.
    Commented Jul 15, 2021 at 23:35
  • \$\begingroup\$ You're right -- i should! ;) \$\endgroup\$ Commented Jul 17, 2021 at 12:47

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