Does increasing the current make a thermistor more accurate?

Question

For example, if the resistance/temperature ranges were fixed to 7500ohms (0C) and 400ohms (70C). At 30mA, the voltage would be 225V (7500ohm), and 12V at (400ohms). If the current was to 50mA, it would be 20V at 400ohms and 375V at 7500ohms. So the range between the volts at 30mA is smaller than at 50mA. Does this increase in a range increase the accuracy of a thermistor, if the analog signal is converted to a digital signal with a 10-bit ADC? If this doesn't increase accuracy, what will?

Answer 1

Running a thermistor with a typical resistance of 1k (somewhere in your range) with 30mA will mean it's developing 30V, or dissipating 1 watt. At 50mA, it will be 2.5 watts.

This means it will be waaaay hotter than the ambient you're trying to measure at either current, and much further from your ambient at the higher current. Very inaccurate.

You only need a few volts swing.

Answer 2

No, a greater range of absolute voltage does not imply anything about the accuracy of the measurement. The more important consideration is to maximize the range of the voltage that you're sending to the ADC, within its limits.

Thermistors are frequently used as one half of a voltage divider in order to create a voltage that an ADC can read. It can be shown that for a given ADC range, using a larger resistor and a higher source voltage uses more of the ADC range. Therefore the best configuration would be to use an infinite resistance, which means using a current source.

Let's say you want a range of 5V to match the input of your ADC. That means you can put no more than 5 V / 7500 Ω = 0.667 mA through the thermistor. At this current, you'll be dissipating about 3.3 mW in the thermistor, which should keep self-heating to a minimum.

At minimum resistance, you'll be putting 400 Ω × 0.667 mA = 0.2667 V into the ADC, which means that you're using almost 95% of the range of the ADC, which is about as good as you're going to get. And the self-heating drops to 0.18 mW.

I keep mentioning self-heating, because if you're using a thermistor for temperature measurement, you don't want self-heating to introduce any significant error. Obviously, using currents as large as what you're talking about would create massive problems in this area.

Answer 3

Increasing of thermistor current will increase it's own heat sourcing. This will affect on resulting temperature of sensing element, therefor, measurement accuracity will decrease in the same way. Thus, theoretically, current of thermistor should be keept as low, as possible. On other hand, decreasing of current will make circuit more sensitive to an external noise (collected from the long connection lines, acted like coreless transformer, for example).

Well, there is an idea to use greater voltage with greater pull-up resistor that, firstly, will decrease current deviation across measurement range, and, secondly, dramatically increase voltage drop range, making circuit more suitable for ADC's. In this case, You'll surely get increased resulting accuracity.

Answer 4

Increaseing the current will increase the voltage produced across the thermistor, this will reduce the effect of noise relative to the signal voltage giving greater accuracy.

However doing this also increases self-heating of the thermistor, which will make the reading higher than ambient, so if you should only apply the sensing current briefly to avoid excessive heating.