1. Mar 2, 2005

### Noj

I have a massive problem! I am currently doing a practical coursework on thermistors. I am using a potential divider circuit to measure the current against temperature. I put the thermistor in a beaker of water for the experiment with a thermometer to measure the temperature, the problem being that the water records current between the crocodile clips which hold the thermistor even without the thermistor attached. (With them merely in the beaker together) How can I arrange the circuit so that I can measure the thermistor current again temperature without the water carrying an unneeded current and resistance. PLEASE HELP because the final day of the experiment is tomorrow. Greatly appreciated,

Noj

2. Mar 2, 2005

### Noj

Somebody!

Please someone reply, any help would be brilliant or I will fail my coursework.

3. Mar 2, 2005

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4. Mar 2, 2005

### Noj

yes it is an NTC thermistor and is a bead. Xanthym I believe you know Padman and you were invaluable to him so if you could help me in any way I would be enormously grateful. Thanks

Noj

5. Mar 2, 2005

### xanthym

Your best approach is to avoid actually submerging the thermister and its contacts into water. Here are 3 alternatives:

1) For this method, you'll need an ordinary Styrofoam Cup and a Paper Clip. Unfold the Paper Clip and punch a tiny hole 1/2 way up the cup. Then insert the small thermistor bead into the hole so it just contacts the inner surface and seals itself in the hole. Tape the leads to the outside of the cup for more stability. Fill the cup with water. Although there might be some leakage, this method should allow contacting the water without submerging the leads. For more stability, use TWO (2) cups, one inside the other. Punch a hole through both cups and set up like before. The added thickness should hold the thermistor more securely and also provide more insulation. The 1 cup method is shown below:
Code (Text):

|         |
|---------|
|         | Styrofoam
-------[COLOR=Red][B]O[/B][/COLOR]  Water  |  Cup
|         |
|_________|

(Use 2 cups for more stability and insulation.)
2) Can you make a leak-proof tin-foil container (probably in the shape of a tube) in which to place the thermistor?? (The contacts should be insulated so they don't touch the metal tin foil!!) You could then submerge the tin-foil container with thermistor inside into water while keeping contacts dry. The tin-foil should conduct heat fairly well and attain the water's temperature. However, there will probably be some measurement errors.

3) If nothing else works, you might consider placing the thermistor on the OUTSIDE of the beaker in contact with the glass wall surface (half way up the beaker's height). Cover it tightly with insulating material (like a piece of styrofoam cup), and tape the insulating material tightly to the beaker so it holds the thermistor in place. The beaker glass wall will attain approximately the same temperature of the water; however, it will not be exact. You can also try placing the thermometer in a similar set up so both thermistor and thermometer are measuring the same environment. This method will definitely introduce some measurement errors, but at least the contacts will not be in water.

Good Luck!!

Last edited: Mar 2, 2005
6. Mar 3, 2005

### Noj

Thank you so much X! I will attempt the first method as it seems to be the most accurate but the other two are far better than I could have imagined. I was simply thinking getting a large container and filling with water and then putting the beaker inside the container but this is unlikely to be possible. Thank you very much,

Noj

7. Mar 4, 2005

### xanthym

Without seeing your experiment and the actual thermistor, it's difficult to provide exact guidance on data techniques. However, if the thermistor is the NTC type, and you're using a standard linear voltage divider circuit, the general relationship between measured Voltage "V" (in volts) and measured Temperature "T" (in degKelvin) would likely be approximately similar to:

$$:(1): \ \ \ \ V \ = \ V_{s} \frac {R} {R + R_{0}e^{C(T - T_{0})} }$$

where Vs is the Voltage Source (in volts) used in the divider circuit, R the Resistance (in ohms) across which you are measuring V in the divider circuit, R0 the calibration resistance (in ohms) provided by the manufacturer, T0 the Calibration Temperature (in degKelvin) provided by the manufacturer, and C the Thermister Coefficient value provided by the manufacturer.

For {Vs= 10 volts}, {R = 1000 ohms}, {R0 = 1000 ohms}, {T0 = 300 degKelvin}, and a NEGATIVE Thermistor Coefficient {C=(-0.05)}, the graph of Voltage "V" (vertical axis, in volts) versus Temperature "T" (horizontal axis, in degKelvin) would be similar to:
http://img191.exs.cx/img191/6674/thermistorexp4zk.jpg

The above graph illustrates how you could display your data. On standard Linear-Linear scales, graph your measured Voltages (vertical axis, in volts) versus measured Temperatures (horizontal axis, in degKelvin). The graph should look like that above.

Now notice the "Linear Region" in the graph's center that's fairly straight like a line. That region is the best operating region for your thermistor because the relationship between "V" and "T" is almost a straight line. By drawing a "Best-Fit Line" thru that center region and determining the straight line's equation, you'll have important information about your thermistor's operating characteristics. (Use standard algebra to determine the line's equation from 2 points read from the line's endpoints.) More information on NTC thermistors can be found in the following document. Good Luck!!
http://www.thermometrics.com/assets/images/ntcnotes.pdf

Incidentally, methods to determine "Best-Fit Curves" for measured data are called "Regression" techniques. The following site describes and demonstrates some methods which you might find useful in the future. (Scroll all the way down the page for the interactive portion.)
http://www.arachnoid.com/polysolve/

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Last edited: Mar 4, 2005
8. Mar 4, 2005

### Noj

Your a saviour!! Thank you very much, have a good day..

9. Mar 13, 2005

### Noj

Your information was very useful X but I could only test between a range of about 30-90 degrees and so could not get to the range of values supplied by the graph you gave. What can I derive from my small portion of the graph? If you could get back to me as soon as possible that would be brilliant, Thanks again