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Thermistor theory and practice

Discussion in 'Electronic Basics' started by Rex, Mar 21, 2005.

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  1. Rex

    Rex Guest

    I want to use a thermistor to measure room temperature. I never did this
    before so I have been reading about thermistors. I have been trying to
    reconcile a table of R vs T to the values computed by formula and B
    value. They don't match. I am wondering if anyone here can explain why.

    I want to build a PIC-based thing to record room temperature at
    intervals over a period of days. I'll use a serial port on the PIC to
    gather data to a PC to save it. Relative is more important than absolute
    accuracy for this, but I wanted to try to get reasonably calibrated for
    C or F.

    I checked my junk drawer and found a 10K thermistor that I bought at
    Radio Shack some years ago. After some web searching, I finally figured
    out that it is a Semitec 103AT thermistor. I found a web page that has a
    datasheet (unfortunately, only in a couple pages of their 800K catalog
    pdf). At 25 C it has an R of 10K and its B value is 3435K between 25 and
    85 C.

    I did some general reading on the use and theory of thermistos via the

    I found this equation for R at a T value...

    R = Ro exp( B (1/T - 1/To))

    For my 103AT
    Ro = 10K = 10000
    To = 25C = 298K
    B = 3435K

    The data sheet says that B was determined by R values at 25C and 85C. If
    I run the equation for T = 85C = 358K I get an R value that doesn't
    match what is in the T vs R table for 85C.

    If I calculate B from the table values for resistance at 25 and 85C I
    get a B of 3477. At different T values in the table I get different B
    values and none that I tried are 3435.

    Is this normal? I thought the equation and table data would be closer.
    I'm trying to figure out a way to generate a conversion table in the PIC
    code but I'm not sure how to start. Do I have to do my own regression on
    data from the table? This is looking more complicated than I expected.

    If anyone wants to see the table data, here is a link to an Excel sheet
    for the 103AT from the manufacturer.

    Other pages on that site have the datasheet but only as part of the
    catalog, so I didn't post that link.
  2. Bob Masta

    Bob Masta Guest


    Instead of a thermistor, you can use the forward drop across an
    ordinary diode junction. Unlike a thermistor this is inherently
    linear. You basically push a small constant current through it
    and measure the drop. A DMM ohmmeter does this. You will
    need to calibrate it using an ice bath and boiling water.

    There are fancier variants using dual junctions that don't
    require calibration, or you can buy a temperature sensing
    chip that does it all.

    Best regards,

    Bob Masta

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
  3. Rich Grise

    Rich Grise Guest

    Why mess with nonlinear stuff when there are chips that give a
    linear readout? A really quick google search on "temperature sensor
    IC" turns up a couple of chips - the DS1620 looks like a likely
    suspect. :) Or the LM35, if you can still get them. It's analog
    output, the DS1620 is 9-bits serial digital.

    Good Luck!
  4. Rex

    Rex Guest

    Thanks for the reply. I'm aware of the Dallas chips. There are several
    reasons why I want to use the thermisor.

    - I already have all the parts I need here.

    - I have more thermistors coming from eBay for future projects.

    - The termistor only needs two wires to remote, while the dallas chip
    needs 4.

    - The thermistor is smaller so should respond faster.

    - The thermistor is cheaper.

    - I want to understand how to do it.

    I just went back and reviewed my original question. Looks like I
    transposed numbers in my calculation. The R value derived from B value
    and equation is slightly different compared to the value in the table,
    but not enough to worry about. The B value I gave from my reverse
    calculation was wrong and therefore farther off.

    Nevermind. Sorry.

    Now I need to get back to figuring out how to implement the code after I
    do some linearization by putting the thermistor in series with a
  5. Bob Masta

    Bob Masta Guest

    The diode junction is the most linear sensor around,
    down to near absolute zero and up to near-melting of the
    junction. It is the heart of the chips you mention. The
    advantage of the chips is that they use a dual-junction design
    and a calibrated current source so you don't need to calibrate them.

    However, most of us have plenty of diodes or transistors on hand.
    They are also pretty small and fast-responding. It's true that you
    may be able to find a smaller thermistor than the diodes you have
    on hand. I haven't looked to see how small a diode I could find.

    Best regards,

    Bob Masta

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
  6. Rex

    Rex Guest

    I have considered that. I think I would use a diode connected smt
    transistor. What do you recommend for the current source, a FET?
  7. Bob Masta

    Bob Masta Guest

    The current source controls the stability of the
    readings. I don't know the relative drift vs temperature
    of various self-biased FETs versus bipolars with various Zeners
    or whatever as references. Of course, you can buy pretty
    decent references for a couple of bucks if you need really
    tight stability.

    I'm always interested in "junk box" solutions, so I'd love to
    hear what others have come up with for stable references,
    either current or voltage, that don't involve just buying a
    reference chip.

    Best regards,

    Bob Masta

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
  8. Rex

    Rex Guest

    So I was playing with MS Excel to plot the raw data from the web page
    for the 10K termistor. I made a circuit of +5V through the termistor
    then to a 10K resistor to ground. Then I did the math in Excel for the
    voltage across the 10K resisitor. Over -10 to +50 deg C it seems nicely
    linear with one breakpoint.

    Here is the Excel xls file with the plot from that data.

    I'll add an op amp to offset 0 deg to 0 V and multiply the output to 5V
    at 50C.

    Actually I have 47K thermistors coming which should have very low self
    heating at 5V, so I think I'll wait for them to arrive.

    Looks like just adding the series resistor makes them positive temp co
    and also nicely linear with one breakpoint.

    Thought it might help someone to post this info after asking my original


    Here's one of the best links I found while researching the subject...

    Found another link describing how to plot with Excel and find linear fit
    over a range. Here's one of those pages...
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