Yet another Wheatstone bridge with thermistor

[Femme_physics]

Homework Statement

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

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

 

[gneill]

Hi FP.

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

 

[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

 

[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

 

[gneill]

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 R_T.

 

[aralbrec]

The op-amp draw 0 current, because of an infinite internal resistance! :)

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.

 

[Femme_physics]

*bumps*

OK I figured out the V1 and V2 finally

http://img16.imageshack.us/img16/9590/br1pp.jpg

Problem is that I'm then asked how much is R2, at 80 celsius, provided V0=12V

I get R2 in minus. I guess it means it's not really an non-inverting amplifier but an inverting? Presuming my answer is even correct...

http://img829.imageshack.us/img829/4138/br2m.jpg

 

[gneill]

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...

 

[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 :)

 

[gneill]

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

That'll do it

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

Devilishly pervasive this electronic stuff

Much obliged gneill, sorry for the long delay :)

No worries. Always happy to help.