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Thermistor theory and practice

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.

Background: 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 web.

Question: 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.

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

Reply to
Rex
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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! Rich

Reply to
Rich Grise

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 dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis

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Reply to
Bob Masta

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 resistor.

Reply to
Rex

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 dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis

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Reply to
Bob Masta

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

Reply to
Rex

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 dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis

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Reply to
Bob Masta

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. ftp://ftp.sonic.net/pub/users/rexa/Thermistor/103AT1_plt2.xls

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 question.

P.S.

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

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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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Reply to
Rex

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