Here it is written:

Some references use the negative temperature coefficient (NTC) a to describe the sensitivity of a thermistor alpha = -B/T².

and here it is

The B constant expresses a degree of thermistor sensitivity (change rate of its resistance) to temperature changes.

Which is common now?


2 Answers 2


I don't use either one. Both are approximations that vary with the temperature being measured.

I use this equation (or a variation thereof), from the eFunda page,

enter image description here

put this into an Excel worksheet or Mlab code, and compute the sensitivity for any temperature I desire.

Note that you also should consider how you're going to measure the resistance of the thermistor (current source, voltage divider, etc). In the end, you probably care about the sensitivity of the thermistor + measuring circuit, in mV per deg C, or something similar.

Added Example Below

This is an example. Thermistor is a Betatherm 10K3CG3 (10 Kohms @ 25 C), with a parallel resistor excited with a 250 uA current source. Curve is the sensitivity of the circuit.

enter image description here

  • \$\begingroup\$ What kind of thermistor that? \$\endgroup\$
    – Ben
    Jun 10, 2020 at 13:11
  • \$\begingroup\$ @Ben - It's a typical NTC thermistor. Manufacturer is Betatherm. P/N is 10K3CG3, as stated in my answer. \$\endgroup\$
    – SteveSh
    Jun 11, 2020 at 0:19

Sensitivity is the derivative of the sensor's characteristic curve (equation) evaluated in certain point. If you want to know your sensor's or your system's sensitivity you have to find it's equation and derive it. The post you've referenced shows a simple approximation to NTC's curve given by

$$R_t = R_0e^{B(\frac{1}{T}-\frac{1}{T_0})}$$ Thus sensitivity is $$S = \frac{dR_t}{dT}$$ evaluated in a given temperature point. As you can see, a NTC have a non linear curve, so S will vary depending on temperature.

note The Steinhart-Hart equation: $$\frac{1}{T}=A+Bln(R_t)+Cln^3(R_t)$$ Fits better to NTC curve, so it can be used in a wider temperature range and it has better accuracy


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.