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I am looking to make a simple circuit where i want to measure a freezer's temperature at -80 C. The problem is that most NTC thermistors i see have a low range of about -40 C. Is there a trick in designing the voltage divider to capture only low temperatures. I would accept an error of +-10C.

I would like to use a cheap thermistor if possible

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  • \$\begingroup\$ define low? what is your simple circuit? \$\endgroup\$
    – Nick B
    Sep 20, 2017 at 12:30
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    \$\begingroup\$ Why NTC? Why not choose a suitable sensor for your need? \$\endgroup\$
    – winny
    Sep 20, 2017 at 12:30
  • \$\begingroup\$ Diodes make good temperature sensors in this range. (there are probably other choices... platinum resistance for one.) \$\endgroup\$ Sep 20, 2017 at 13:42
  • \$\begingroup\$ @winny do you have a sensor in mind that is not too expensive? <5$ \$\endgroup\$ Sep 20, 2017 at 13:55
  • \$\begingroup\$ Search internet/DigiKey. Questions for specific product recommendations are off-topic here. \$\endgroup\$
    – winny
    Sep 20, 2017 at 15:21

1 Answer 1

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No, there's no electrical way to make a sensor measure values outside its operating range.

However, heat flow is in many ways analogous to electrical flow, so it is at least theoretically possible to create a "temperature divider" using a pair of thermal resistances. For example, you could have an insulated metal rod that has one end at the freezer temperature and the other end at ambient temperature. (Obviously, this would require drilling a hole through the wall of the freezer.) Use two thermistors, one at the ambient end and the other part way along the rod. This would allow you to extrapolate the temperature of the freezer.

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  • \$\begingroup\$ +1 though you may want to add such a system introduces a thermal time lag and geometry and linear insulation so the heat only travels along the length of the rod is important. \$\endgroup\$
    – Trevor_G
    Sep 20, 2017 at 14:02
  • \$\begingroup\$ @Trevor: As I said, "theoretically possible" -- I never meant to imply that this would be better than picking a suitable sensor in the first place. \$\endgroup\$
    – Dave Tweed
    Sep 20, 2017 at 14:07
  • \$\begingroup\$ :) yup, just adding to an already good answer \$\endgroup\$
    – Trevor_G
    Sep 20, 2017 at 14:12

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