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Removing resistance

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    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 in to (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


    2. Relevant equations

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

    3. The attempt at a solution

    6500^-1 - 9700^-1 = 1/R
     
  2. jcsd
  3. Nov 22, 2011 #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: Nov 22, 2011
  4. Nov 22, 2011 #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
     
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