Solving Resistance to Increase Thermistor Temperature Reading

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In summary, the conversation discusses the issue of manipulating a thermistor to read a higher temperature than it actually is. The solution involves adding a resistor in parallel with the thermistor. The specific calculations and values for the resistor are also mentioned. It is acknowledged that there may be potential problems with accuracy, but it is not a critical issue for the application. The conversation also expresses a desire to learn and problem solve.
  • #1
EclipseAlex
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Homework Statement



Hi all this is my first post so bare with me! My background is that I am new to the industry of controls engineering but my problem is more fundamental than any related specifics of control engineering so please read on!
Basically. I have a thermistor. It reads 9700 Ohms at 10decC and 6500 Ohms at 20decC (NTC). For a reason that I won't go too far into (controller not letting me do what I want with it because the manufacturer never specced it for the application I am using it for) I want to make it think it's 10decC higher than it actually is. This clearly involves putting another resistor between the temp sensor (thermistor) and controller. But how, where, parralell, series, size etc.

I have given it a go. Everyday is a huge learning experience for me. I love to problem solve, to engineer, to become informed! I just don't have the formal education I know a lot of people on this site do so please share the knowledge!

I know there will be potential problems with linearity, accuracy isn't hugely important. It's not a critical application, it needs to work, not be spot on perfect.

Thanks in advance


Homework Equations



1/R = 1/Ra + 1/Rb etc

The Attempt at a Solution



6500^-1 - 9700^-1 = 1/R
 
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  • #2
So I suppose the question I'm asking is how do I proportionally remove resistance from a dynamic resistance(resistor). I thought putting in parrallel would work as the final value of resistance is based on the dynamic thermistor.
 
Last edited:
  • #3
If all you are doing is measuring the resistance then effectively when the thermistor is at 10C you want to record a resistance of 6500Ω
This means you want a resistor,R, in parallel with the thermistor such that
1/R + 1/9700 = 1/6500
1/R + 0.00010 = 0.000154
1/R = 0.000054 gives R = 18.5kΩ

With this in parallel with the thermistor when the temp is 20C the combined resistance will be
1/18.5kΩ + 1/6.5kΩ =1/R gives R = 4800

Hope this is on the right tracks for you
 

1. How does a thermistor work to measure temperature?

A thermistor works by changing its electrical resistance in response to changes in temperature. As the temperature increases, the resistance decreases, and vice versa.

2. What is resistance and how does it affect the accuracy of temperature readings?

Resistance is the measure of how difficult it is for electricity to flow through a material. In a thermistor, resistance is directly related to temperature. If there is resistance to the flow of electricity, it can affect the accuracy of the temperature readings.

3. How can I increase the thermistor temperature reading?

One way to increase the thermistor temperature reading is by reducing the resistance in the circuit. This can be done by using a lower resistance thermistor or adding a parallel resistor to the circuit to decrease the overall resistance.

4. What are some common causes of resistance in a thermistor circuit?

Some common causes of resistance in a thermistor circuit include poor connections, improper placement of the thermistor, and external factors such as humidity or vibrations affecting the circuit.

5. How can I solve resistance in a thermistor circuit to get more accurate temperature readings?

To solve resistance in a thermistor circuit, you can try using a higher quality thermistor, ensuring proper placement and connections, and shielding the circuit from external factors. Additionally, using a low-pass filter can help reduce noise and improve the accuracy of temperature readings.

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