Yet another Wheatstone bridge with thermistor

1. Dec 20, 2012

Femme_physics

1. The problem statement, all variables and given/known data

So I got a thermistor here and am told that in the range 0-80 celsius the thermistor's resistance is given by Rt = R0 - kT ...whereas R0 = 100 ohms which is the thermistor resistance at 0 celsius. k = 0.5 kohms/celsius. T is of course temperature. The op-amps are ideal and their supply voltages are 15 volts

Question: The measured temperature is 80 deg celsius. What re the voltages in V1 an V2.

http://img854.imageshack.us/img854/80/circuit1.jpg [Broken]

3. The attempt at a solution

According to my logic, V1 has to be 12 volts

Since V+ = V- = V1

http://img837.imageshack.us/img837/8282/loopex.jpg [Broken]

Last edited by a moderator: May 6, 2017
2. Dec 20, 2012

Staff: Mentor

Hi FP.

Can you check the values you've posted for Ro and k for the thermistor? 100 Ω seems a bit low for a starting resistance, and .5 kΩ/C seems awfully large if Ro is so small -- the thermistor would be showing negative resistance after about 0.2 °C. This is not physically possible. I think your Ro should be something like 100 kΩ.

Next, you should recall the "rules of thumb" for op-amps. How much current does an op-amp input draw?

3. Dec 20, 2012

Emilyjoint

I agree with Gneill. Your equation indicates that over the range 0 - 80 the resistance of the thermistor decreases by 500ohms/degree. I also think that your starting resistance is probably 100kohms, not 100ohms

4. Dec 20, 2012

Femme_physics

You're right, it's 100k, I miswrote.

The op-amp draw 0 current, because of an infinite internal resistance! :) which is why I allowed myself to make a loop there. Perhaps I've forgotten elementary electronics already. I think I should just use the Wheatstone Bridge equation to solve it though

http://img577.imageshack.us/img577/672/solutttt.jpg [Broken]

Last edited by a moderator: May 6, 2017
5. Dec 20, 2012

Staff: Mentor

You should determine V1 and V2 separately. After all, they are independent of each other, and determined by separate voltage dividers. They won't be equal in value except for a very particular value of RT.

Last edited: Dec 20, 2012
6. Dec 21, 2012

aralbrec

That's right, which means it's like they aren't there :) They sample the voltage but don't affect the circuit in any way.

7. Jan 22, 2013

Femme_physics

Last edited by a moderator: May 6, 2017
8. Jan 22, 2013

Staff: Mentor

For your analysis you've offset the input voltages by subtracting 6V from each, yet you've left the output unshifted at 12V. That's going to produce incorrect results.

Rather than trying to force the circuit into a form for which you've memorized the gain formula, you can directly analyze the circuit as given without too much trouble.

Notice that you have the potentials at either end of the 1 kΩ resistor. So what is I? You've also got the potentials at either end of R2...

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9. Jan 22, 2013

Femme_physics

ah...easier than I thought.
I1 = 1.5/1000
VR2 = 12-7.5 =4.5V
R2 = VR2 / I1

I forget my basic electronics sometimes, I didn't think it would come in handy in control systems! Who'd thunk it.

Much obliged gneill, sorry for the long delay :)

10. Jan 22, 2013

Staff: Mentor

That'll do it
Devilishly pervasive this electronic stuff
No worries. Always happy to help.