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Author Topic: Decoding thermistor characteristics table  (Read 2341 times)
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So as my first real Arduino project, I'm building a custom multi-temp monitor for smoking meats.  I picked up some cheap thermistor meat probes from Thermoworks and believe I tracked down the OEM to a company called Amwei.  Anyways, I'm trying to understand the information about the themistors- especially the B-value and resistance.  Some brief information about the probe is listed here (search for 'meat probe'): http://www.amwei.com/views.asp?hw_id=61

And the characteristics table is here: http://www.amwei.com/news.asp?news_id=79

I've done some tests, and I get the widest range of voltages when using a 10K resistor with it.  Does that seem reasonable? Is the beta value really 4300k?

Anyways, I'm looking into using the Thermistor4 library: http://www.arduino.cc/playground/ComponentLib/Thermistor4.  Looks like I'll have to grab my scientific calculator to figure this out!  Any help or advice would be appreciated.  Thanks!
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Thermistors are odd devices and its not a trivial task to get a precise temp reading with them. I've used them in the past for trip point applications. Anyways, you will need to use the Steinhart-Hart equation to calculate the temp.
Have a look at this site: http://www.facstaff.bucknell.edu/mastascu/elessonshtml/Sensors/TempR.html    I would use a thermocouple.
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Thermocouples require low level amplification and junction compensation circuitry ($$), plus they are not all the impressive in accuracy.

Here are some info on reading thermistors from the arduino playground. http://arduino.cc/playground/ComponentLib/Thermistor2

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Granted TC are not as cheap as a NTC but, you don't need a $20 AD582 to used a TC either. Just a little thinking out of the box smiley. But, that's another post.
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Thanks for the replies everyone.  I'm still learning about the Steinhart-Hart equation and how to apply it with a thermistor, but it seems quite doable.  The main issue I'm having is trying to understand/decode the information that Amwei has posted about the thermistor.  Basically, the data seems to be there, but I'm not sure what I'm looking at.  Any help would be greatly appreciated.

Thanks,
Aaron
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Actually, looking at the tables, am I confused or is that not for a NTC?  Looks like the resistance is *decreasing* as temperature increases.  Seems backwards.  Or am I just really confused?
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Actually, looking at the tables, am I confused or is that not for a NTC?  Looks like the resistance is *decreasing* as temperature increases.  Seems backwards.  Or am I just really confused?
Seems right to me.

Like an LDR, the resistance gets less with more light, with a thermistor, resistance gets less with more heat.
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Thermistor are available with either negative temperature coefficient or positive temperature coefficient factors. The NTC is a negative variety. How the thermistor is wired in the voltage divider (either top of bottom element) also effects the direction of voltage change Vs temperature change.

Lefty
« Last Edit: January 30, 2011, 12:53:14 pm by retrolefty » Logged

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Yep, you're right.   I just measured the resistance of the thermistor using my multimeter, and it's definitely increasing as the temp decreases (ice water vs room temp).

So if V = IR, I'm guess current must be the same regardless of temperature, so V is proportional to R.  As R increases, so does the voltage.  Hence, as temp drops, the resistance increases which increases the voltage towards the max, which in this case is the input: 5V.  That' what I had wrong- I thought V & R were inversely proportional.

So, I guess the last thing to figure out is which column to use in the tables.  The information about the thermistor says the beta value (100/200 - what does that mean?) is 4300k.  The 2nd to last column has a label of 4300 at the bottom (labeled b value @ 25/50C).  I guess that must be it?  So at 0C it should read 352.4 * 100Kohm (.3524M Ohm)?  

Doesn't seem to match that or any column though.  My multimeter (+/-(%2 + 4) accuracy) is showing .75M @ 0C and .28M at 18C.  Or am I confused again?

Thanks!
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