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Thermistor in Wheatstone's bridge

  1. Dec 14, 2012 #1

    Femme_physics

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    I'm told to presume that R1=R2=R3

    http://img716.imageshack.us/img716/7610/thermistor.jpg [Broken]

    And I have a chart for the value of the thermistor based on the temperature. At 20 degrees the thermistor is 2814 ohms

    My question is since they told me that R1=R2=R3, does it necessarily mean that R1=R2=R3=R(thermistor)

    ?
     
    Last edited by a moderator: May 6, 2017
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  3. Dec 14, 2012 #2

    gneill

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    I wouldn't think so, no. Besides, the thermistor will probably be changing value at some point in the question, or you'll need to find an R that yields a certain voltage swing for a given temperature range, or something along those lines.
     
  4. Dec 14, 2012 #3

    Femme_physics

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    I'm a bit confused then why my teacher decided that R1=R2=R3=R(thermistor) at 20 degrees. The question saying nothing of the sort. It says:

    There's a need to plan a heat transducer based on thermistor and wheatstone's bridge, and that we want in 20 degrees celsus for the bridge to output Vab = 0 V and in 90 degree celsius the bridge's output Vb = 1.0 V ...and I'm told to "find the resistance at the bridge arm's (wheras R1=R2=R3) and the excitation voltage E".

    And of course we have the chart FOR THE THERMISTOR:

    http://img534.imageshack.us/img534/1903/chartie.jpg [Broken]

    My teacher solution starts like this:

    R1=R2=R3=R(thermistor at 20 degrees) = 2814 ohms
     
    Last edited by a moderator: May 6, 2017
  5. Dec 14, 2012 #4

    gneill

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    More information is always better :smile: So there are some temperature-based constraints on the desired operation of the circuit, as I surmised. The initial condition, that Vab = 0 when the temperature is 20° certainly narrows your choice of value for R. Can you see why the professor's choice for R is the only practical possibility? (Hint: Vab is measured between two voltage dividers).
     
  6. Dec 14, 2012 #5

    Femme_physics

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    Last edited by a moderator: May 6, 2017
  7. Dec 14, 2012 #6

    gneill

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    An unpowered circuit that produces zero output under all conditions is hardly practical :smile: How will you manage to obtain Vb = 1.0 V when the temp goes to 90C?
     
  8. Dec 15, 2012 #7

    Femme_physics

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    Choose a correct thermistor?
     
  9. Dec 15, 2012 #8

    NascentOxygen

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    Good to see you still around, FP. :smile: You do realize the implication of the word "unpowered" in gneill's rhetorical question?

    Have you determined the equation relating Vab to R, Rt, and E?
     
  10. Dec 15, 2012 #9

    Femme_physics

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    Good to see you guys still around :)

    Is what I wrote above not the corrent equation?

    Unpowered means E = 0

    So therefor no current in the circuit and Vab must equal 0 as well
     
  11. Dec 15, 2012 #10

    NascentOxygen

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    My mistake. I glanced though looking for Vab= ... and didn't notice it renamed Vf.

    The bridge will not be much use without E being set to some value.

    So you have the equation for Vab, and it contains 2 unknowns. You are provided with 2 conditions to be met, so using these you can assign values to the unknowns. You have determined the value for R. Now determine E.
     
  12. Dec 15, 2012 #11

    gneill

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    G'day FP. You've written an appropriate equation to describe the bridge (although you've changed the names of some of the given parameters: Vin for E, Vf for Vab). However, it's not fair setting the voltage supply E to zero when you're looking for R; the circuit stops behaving as intended if it has no power supply and will not meet the other operating conditions. E should be assumed to be a fixed, non-zero value (value to be determined later).

    You will have noticed that the bridge is made up of two independent voltage dividers, and that both dividers split the same voltage (Vin or E). It's also been given that the numbered resistors will all have the same value. It's handy to remember that a voltage divider with two equal valued resistors always divides the potential in half.

    https://www.physicsforums.com/attachment.php?attachmentid=53963&stc=1&d=1355578834

    Knowing this allows you to immediately conclude that the potential at node b is 1/2 E, no matter what E eventually turns out to be.

    If Vab is to be zero, then the other voltage divider containing the thermistor must produce the same voltage division since it is powered from the same voltage source, E. The same principle of equal resistors dividing the potential in half compels us to choose R3 to be equal in value to the thermistor (under the specified conditions: temp is 20 C; Vab = 0).
     

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  13. Dec 15, 2012 #12

    Femme_physics

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    Yes, you two are right-- I changed the variables since I looked at an older example exercise and copied it.

    OK, I perfectly accept that, and even googled more info about Wheatstone to make sure I understand the formulas. But what do they mean by "excitation voltage"? What a weird question. Do they mean the voltage where Vab = 1 V, at 90 degrees, and the thermistor is therefor 208.5 ohms? (just guessing here)
     
  14. Dec 15, 2012 #13

    gneill

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    There may be some historical reason why they chose the term "excitation voltage" to describe the voltage that powers the Wheatstone bridge, but I must confess that I'm not familiar with the story. I suppose it's as good a term as any, since E does 'excite' the operation of the circuit, in a sort of Victorian-era interpretation of the word 'excite'.

    So, yes, you're looking for a suitable value for E which will produce Vab = 1 V when Rt = 208.5 Ω.
     
  15. Dec 15, 2012 #14

    Femme_physics

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    Great :) That clears it up! Thank you!

    They're asking me if there is a linear ratio...does that mean I need to start plugging in thermistor values and drawing them on a graph to see, or is there an easier shortcut?
     
  16. Dec 15, 2012 #15

    gneill

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    You're quite welcome.
    Can you expand on that a bit? Do they specify what ratio they're referring to?

    I suppose you could start by doing a plot of Vab versus Rt, letting Rt go linearly from 2814 to 208.5 Ohms (so no table lookups required). You might find the resulting curve interesting.

    You might also plot a handful of resistance values versus temperature from the table to see what that curve looks like. Is the thermistor a linear R vs temperature device?

    But I suspect that what they want to know is if there's a linear relationship between temperature and Vab, which will require plotting Vab versus temperature. So the table comes into play. I'd think that half a dozen points strategically chosen to make the curve obvious would suffice.
     
  17. Dec 15, 2012 #16

    Femme_physics

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    The question is " whether there is a linear relation between voltage Vab and the temperature at the range (20-90 degrees celcius) . Explain your answer."

    Isn't it enough to look at the thermistor chart to see that it's exponential and not linear?
     
  18. Dec 15, 2012 #17

    gneill

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    This is why I suggested plotting Vab versus Rt first; The bridge has its own 'response curve' to linearly varying Rt. The question then becomes, does this bridge response curve help to correct or exacerbate any non-linearity of the thermistor v. temperature curve?

    You could go by the linearity of the thermistor versus temperature curve alone IF the Wheatstone bridge produced linear response to a linear change in resistance. But it doesn't...
     
  19. Dec 19, 2012 #18

    Femme_physics

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    Hi Gneill, my teacher did this exercise yesterday in class :) Your feedback was very helpful in me participating in class, so thank you! Although he did decide it's exponential just based on the chart :P But whatever! I'm just glad I understand it.
     
  20. Dec 19, 2012 #19

    NascentOxygen

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    To demonstrate that it is exponential, he could plot Vab vs. temp on a sheet of semilog graph paper. If it really were exponential, that plot would be a straight line.

    This would be a good exercise to investigate using a spreadsheet.
     
  21. Dec 19, 2012 #20

    gneill

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    Hey, always glad to help if I can!

    Cheers!
     
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