Thermistor Temperature- Ohms Law?

  • Thread starter ElBell
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Thermistor Temperature- Ohms Law??

Hi All!

I am trying to convert a current of 795mA to a temperature in degrees celcius but I dont understand how to do it.

I realise that Ohms law of V= A/ R is involved somehow but thats the extent of my knowledge.

PLEASE HELP!

I have also been told that in this case the voltage is constant....but I dont even know how this helps me!?

Thanks very much, any help would be much appreciated!

ellie
 

Answers and Replies

  • #2
vk6kro
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You would need a lot more information.

If you knew the voltage across the thermistor, you could calculate its resistance.

For example, if the voltage was 12 volts the resistance would be

R (in ohms) = Voltage (in volts) / Current (in amps)

795 mA is 0.795 amps, so

R = 12 V / .795 A = 15.09 ohms.

But this isn't much use unless you know what temperature this corresponds to. So you would need to calibrate the thermister first by measuring its resistance at different temperatures and maybe drawing a calibration graph. Then you could look up what temperature corresponded to 15.09 ohms.

There is another complication. If this thermistor had 12 volts across it, it would be dissipating 9.54 watts.
Power = voltage * current = 12 * 0.795 A = 9.54 watts
So it would probably be getting quite hot, just from its own heat.
 
  • #3
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I know that at 0deg celcius it is 120mA and at 100deg celcius it is 540mA. Are these the figures I would use to calibrate, and how would I go about doing this?

Thanks alot :)
 
  • #4
dlgoff
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  • #5
vk6kro
Science Advisor
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You could place the thermistor and an accurate thermometer in a non-conducting oil in a test tube. Connect the thermister to a digital multimeter on "ohms".

Gradually heat the oil while stirring it and taking resistance and temperature readings as you do this.

Note that you need resistance readings, not current readings. It is the resistance that changes, not the current, which also depends on the voltage.

Once you have a good set of data, graph it on graph paper or enter it into Excel and graph it there.

Excel will give you a curve of best fit formula which you could use to find unknown temperatures, or you could just look at the graph to get them.
 
  • #6


The last thermistor I used was nonlinear. We were able to approximate a linear curve for a specific temperature range to simplify the problem though. I'm not sure if that helps make this easier but could be considered.
 
  • #7
vk6kro
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Assuming this is homework, you could try this formula:

[PLAIN]http://dl.dropbox.com/u/4222062/thermistor%20formula.PNG [Broken]

Where t is the temperature in degrees C.

This assumes that the resistance of the thermistor drops by half every 46 degrees C. (The 46 figure was derived from the same formula, knowing the resistance at 100 degrees C was 22.2 ohms if 12 volts was placed across the thermistor to give 540 mA. Very messy, but that is how the problem is stated.)

For example, at temperature 0 deg C
R = 12 V / 0.120 A = 100 ohms
So at 80 deg C,
R = 100 / 2 ^ (80 / 46) = 29.95 ohms



You need to find a value of t so that R = 12 / 0.795 = 15.094 ohms

So 15.094 = 100 / 2 ^ (t / 46)

2 ^ (t / 46) = 100 / 15.095 = 6.624
(t /46) log 2 = log 6.624
find t.
 
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  • #8
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OK thanks so much!! This is what I now have:

T at 0 degrees:
R= 12/ 0.12 = 100 ohms

T at 100 degrees:
100/2^(100/46) = 22.2 ohms

R= 12/ 0.795= 15.094 ohms

So:

15.094= 100/2^(T/46)

2^(T/ 46)= 100/ 15.094= 6.625

(T/46)log 2= log 6.625

T/46 = log(6.625-2) ???

is this right? I dont know where to go from here to get T.
 
  • #9
vk6kro
Science Advisor
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(T/46)log 2= log 6.625

T/46 = log(6.625-2) ???

is this right? I dont know where to go from here to get T.


Look up log 6.625 and log 2 on your calculator.
Divide the first by the second and multiply by 46

I got about 125 degrees C
 

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