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Yet another Wheatstone bridge with thermistor

  1. Dec 20, 2012 #1

    Femme_physics

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    Gold Member

    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. jcsd
  3. Dec 20, 2012 #2

    gneill

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    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?
     
  4. Dec 20, 2012 #3
    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
     
  5. Dec 20, 2012 #4

    Femme_physics

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    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
  6. Dec 20, 2012 #5

    gneill

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    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
  7. Dec 21, 2012 #6
    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.
     
  8. Jan 22, 2013 #7

    Femme_physics

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

    gneill

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    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.

    attachment.php?attachmentid=54922&stc=1&d=1358862041.gif

    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...
     

    Attached Files:

  10. Jan 22, 2013 #9

    Femme_physics

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    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 :)
     
  11. Jan 22, 2013 #10

    gneill

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