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Thevenin equivalent for strain gauge bridge

  1. Jan 16, 2012 #1
    A strain-gauge bridge is made up of four resistive elements, each element having an unstrained
    resistance equal to R, as shown in the diagram below. When a strain is applied to the bridge
    elements R2 and R4 increase in resistance by an amount ΔR, whereas R1 and R3 decrease by ΔR,
    where ΔR/R is proportional to the applied strain. Find the Thévenin equivalent of the circuit in its
    strained state looking into nodes x and y.


    I cant seem to cut the image so hopefully I can describe it. Top node is Vbridge, left and right nodes are x and y respectively and bottom node is ground.



    3. I have R2+ΔR//R1-ΔR= R^2-(ΔR)^2/2R=(approx)R/2. Same will apply for R3 and R4 to give R/2. Equivalent resistors are now in parallel to give R? Would this be correct?

    Thanks in advance
     
  2. jcsd
  3. Jan 16, 2012 #2
    Anyone?
     
  4. Jan 16, 2012 #3

    gneill

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    Staff: Mentor

    If you can't post a drawing, you should at least make clear where in the bridge the resistors are located. Also, a bit more detail concerning your solution attempt (what was the plan of attack?) would not be amiss.
     
  5. Jan 16, 2012 #4
    Ok, R1 top left of bridge, R2 top right, R3 bottom left, R4 bottom right. I've taken R1//R2 and R3//R4 and input the strain states for each resistor and simplified as shown in my original question.

    edit: I believe I actually should be taking R1 and R2 in series and R3 and R4 in series instead?
     
    Last edited: Jan 16, 2012
  6. Jan 16, 2012 #5

    gneill

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    Staff: Mentor

    Okay, so the circuit looks something like this:

    attachment.php?attachmentid=42766&stc=1&d=1326760038.jpg

    Your method for finding the Thevenin resistance doesn't look right. What's the usual procedure for finding that resistance?
     

    Attached Files:

  7. Jan 16, 2012 #6
    The diagram is similar to this,

    http://en.wikipedia.org/wiki/File:Diode_bridge_alt_1.svg

    Obviously just replacing the diodes with resistors. I know I need to suppress the voltage and current sources but I am unsure as to where to take it from there really. I am short on examples for this to refer to.
     
  8. Jan 16, 2012 #7

    gneill

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    Staff: Mentor

    I think you will find that my diagram is faithful to the described circuit. It is only slightly rearranged for clarity, but all the components and connections are the same.

    If the voltage source is suppressed it is equivalent to short-circuiting the top node (Vbridge) to the bottom node (ground). Think of the circuit being "folded" along a line passing through terminals x and y, the top rail being brought down to coincide with the bottom rail. Which resistances will be in parallel?
     
  9. Jan 16, 2012 #8
    So R1//R2 and R3//R4.
     
  10. Jan 16, 2012 #9

    gneill

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    Staff: Mentor

    Nope. Look again at the circuit. Remove the voltage source and fold it along the x-y line so that the top node coincides with the bottom node. What's in parallel? When two components are in parallel the ends of the two components share the same two nodes.
     
  11. Jan 16, 2012 #10
    Ok, so it should be R1//R4 and R2//R3. Sorry this is dragging on, struggle with the electronics side of the course.
     
  12. Jan 16, 2012 #11

    gneill

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    Staff: Mentor

    No again. R1 parallels R3, R2 parallels R4. Don't sweat it if it's not immediately obvious; intuition and understanding comes with practice.

    attachment.php?attachmentid=42768&stc=1&d=1326764885.jpg
     

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