**Thermistor (Negative Temperature Coefficient Type, NTC)**

- From the I-V curve, the ratio V/I decreases for increasing current. Resistance decreases with increasing current.

**Reason:** The thermistor is a resistance element made of semiconductor material. Increased potential difference across the thermistor results in increased current which in turn causes the temperature to rise. As the temperature rise, the lattice ion’s vibration increases and reduces the drift velocity of the charged particles. However, the number of freed electrons and holes due to the temperature increase is more significant than the reduction in drift velocity. Hence, resistance of NTC thermistor decreases with increase in temperature.

Another type of thermistor, positive temperature coefficient type (PTC), exists as well. PTC type thermistors’ resistance increases as temperature increases.

Metals have positive temperature coefficients.

**Important:**

**Gradient of I-V curve is not equal to the resistance of the component. Resistance is ratio of potential difference to current. $R = \frac{V}{I}$ and NOT $\frac{dV}{dI}$**

Hi

This website was very informative for me

I would be thankful if you answer me a question

Is i-v graph different from v-i graph?

Yes. You will need to inverse the i-v graph to get the v-i graph.

Hi, Very useful website, thanks. Can you tell me what volatge ramge would be necessary to get the graph above? I tries between 0 – 6V but the graph is quite linear. Any help would greatly appreciated. Thanks

Hi! If your thermistor can handle the current, try 0 to 12V. What you are aiming to do is to increase the temperature of the thermistor by around ~2 to 3 degrees Celsius via the heating effect of current. If you are still getting a linear graph, try to “cheat” a little by using a hair dryer to blow on to the thermistor. (Only if you can convince yourself that the IV graph of a thermistor is what as shown above due to the heating effect)