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Thermistor's relationship

  1. Jan 15, 2007 #1
    Apologies for not using the provided template, the help I require is not spoon-fed, more a method of directing me in the right direction hopefully.

    For my Physics coursework, I chose the rather simplistic thermistor experiment rather than being bold and going for a slightly more adventurous type of rotary potentiometer experiment. Anyway, I understand the basic principles of the thermistor, like it's negative coefficient of temperature etc, but I'm having trouble reaching the top grade boundary. I've noticed that dividing the graphs into segments (such as 10 degrees centigrade) causes the relationship to become linear and not exponential, yet I cannot explain why. I've worked out it's equation of ΔR = kΔT, I just don't understand why it happens (if indeed there is any worthwhile Physics behind it).

    Another problem is my explanation of why the thermistor differs to a metal with an increase in temperature. I've got the whole collision of charge carriers in metals causes resistance to increase, whilst in a thermistor it just releases more charge carriers, but am struggling to delve into it in more detail.

    I'm not expecting spoon-fed answers, but any help that can be provided will be much appreciated.
     
  2. jcsd
  3. Jan 15, 2007 #2

    berkeman

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    Staff: Mentor

    I'm not much help, but have you read the level of detail in this introductory wikipedia.org page?

    http://en.wikipedia.org/wiki/Thermistor

    Are there specific things on that page that are confusing you?
     
  4. Jan 15, 2007 #3
    Everything bar Steinhart I have included it appears (never thought to check Wiki admittedly). However, Steinhart is already out of the question as I don't know the specifications of the thermistor required (as it was an old one not found in the catalogue). My solution would be to find the equation of the exponential curve, but that's too mathematical for someone like me who doesn't do Maths!

    The problem I have is that by doing a thermistor experiment, I automatically forfeited 2 marks as it wasn't interesting enough. I am now struggling to get those extra marks to boost me into an A. So far, my entire scientific theory is based on more thermal energy=an increase in charge carriers so therefore, an increase in conductance. This has resulted in quite a short explanation and diagram, which apparently needs more detail and expanding (quite how, I am not sure).

    Also, for my analysis, I found that when using smaller temperature ranges, the relationship between temperature and resistance is linear and not exponential, but I can't understand why that's the case. As far as the Wiki page is concerned though, I understand it all.
     
  5. Jan 15, 2007 #4

    berkeman

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    Staff: Mentor

    Well, again, I won't be of a lot of help, but I can offer a couple thoughts:

    -- If learning some of the math of exponentials will help your grade significantly, it's probably worth a couple hours of reading/study. Exponentials are a fairly simple area of math.

    -- You mention the increase in conductivity due to more free charge carriers at higher temperatures, but that's only half the story, right? Remember that the normal collision-related reduction in conductivity is going on still, so the increase in the number of free carriers has to contribute more conduction capability over and above that lost to increased collisions. Is that accounted for in your model?

    -- You might look at other devices that use this thermistor effect. Like, how is a thermopile related to thermistors?
     
  6. Jan 16, 2007 #5
    Seems so obvious now you've mentioned it, can't believe I missed it out. It makes relatively little difference in the actual conductance model, though of course, it adds a little more substance science wise. Cheers.

    Thanks for the other points too, the reading up on Maths bit, it's something I have considered, but as lazy as it may sound, I'm hoping to compensate for it in other areas.

    Any other points will still be appreciated, but I reckon it's shaping up quite nicely at the moment.
     
    Last edited: Jan 16, 2007
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