2
\$\begingroup\$

Thermistor / Voltmeter

Assume that I had a circuit as such above. I was wondering, if I increased the temperature of the surroundings, what would a graph of the temperature/PD graph look like?

I know that Resistance decreases with temperature, as more electrons 'escape' from their atoms, so more charge carriers are available, but would the graph show a general decrease in PD as temperature increases?

Thanks.

\$\endgroup\$
  • 3
    \$\begingroup\$ Would Electrical Engineering be a better home for this question? \$\endgroup\$ – Qmechanic Apr 30 '16 at 21:40
  • 2
    \$\begingroup\$ The temperature curves of thermistors are technology dependent and can be found in the data sheets. Once you have that, you simply apply a resistor divider formula. \$\endgroup\$ – CuriousOne Apr 30 '16 at 21:42
1
\$\begingroup\$

There are a few formulas to model the behavior of an NTC. The most common (and simlest) one is the b-Formula:

$$R(T)=R_0\cdot \exp\left(b\cdot \left(\frac{1}{T}-\frac{1}{T_0})\right)\right)$$

T is given in Kelvin, and b is a parameter of the NTC type, typically in the order of 3000-5000. \$R_0\$ is the resistance at \$T_0=298.15K(=25°C)\$.

Putting this into the formula of a voltage divider, one gets this plots:

enter image description here

For the first curves, the normal resistor \$R_N\$has a value equal to \$R_0\$, for the two, it has a value of \$200R_0\$ or \$(1/200)R_0\$

So there are three conclusions:

  • The higher the b-value, the higher the sensivity (slope), but the smaller the temperature range.
  • The highest sensitivity is at the temperature T where \$R(T)=R_N\$, and it decreases with temperature difference from T.
  • Choosing an other value for \$R_N\$ allows to shift the usable range, but the sensitivy / range will change, too.
\$\endgroup\$
  • \$\begingroup\$ Nice answer. I recommend putting correct units on b (it is a temperature), as I find it adds a level of understanding. \$\endgroup\$ – Scott Seidman Aug 3 '16 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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