# Thermistor theory and practice

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

1. ### RexGuest

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.
http://www.semitec.co.jp/RT/AT/103AT1_11.xls

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

2. ### Bob MastaGuest

<snip>

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
www.daqarta.com

3. ### Rich GriseGuest

Why mess with nonlinear stuff when there are chips that give a
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

4. ### RexGuest

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.

5. ### Bob MastaGuest

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
www.daqarta.com

6. ### RexGuest

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 MastaGuest

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
www.daqarta.com

8. ### RexGuest

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...
http://www.thermometrics.com/assets/images/ntcnotes.pdf

Found another link describing how to plot with Excel and find linear fit
over a range. Here's one of those pages...
http://phoenix.phys.clemson.edu/tutorials/excel/graph.html#5