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Sensor - Calculate the Capacitance

  1. Jun 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Airbag Sensor
    To trigger the airbag, accelerometers are used.
    A simple sensor is composed of three parallel plates of the surface A, which are connected to a condenser system.
    The mass m of the middle plate can move between the two fixed outer
    Moving plates. She is stopped by two identical k, isolated springs of spring constants. The springs are not stressed when the plate is in the center.


    To what maximum distance x, the middle plate
    is deflected, when the vehicle is braked suddenly from movement at a constant acceleration a.

    Determine how large the additional capacitance C must be when the airbag to trigger a voltage Uc = 0.15 V and this needs to happen at a speed of 10 ms ^ -2.
    To use the values ​​d = 1.0 cm, A = 250 cm2, m = 50 g, k = 5.0 × 102 N m ^ -1 and U = 12 V.

    2. The attempt at a solution

    x(t)=x0e−δt cos(ωt+φ)

    x(t)= F(t)/D

    C2 -C1 ≈∆d/d


    Is it right what I have done?
    What should I do now? How can I calculate the additional capacitance C ?
     
  2. jcsd
  3. Jun 30, 2013 #2
    It would be very nice if somebody could help me.
    To make to clearer for you how the sensor looks like I have got a picture of it.
     

    Attached Files:

  4. Jun 30, 2013 #3

    gneill

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    Staff: Mentor

    What is the reasoning behind your solution attempt? What does it mean to set a difference in capacitances to a pure number (ratio of distances)?

    Can you list the things you need to accomplish in order to solve the problem?
     
  5. Jun 30, 2013 #4
    differential equation (Accelerating force generated):
    m*d^{2}/dt^{2}+k*dx/dt+Dx = F(t) = m*a(t)

    damped harmonic oscillation:
    x(t)=xo*e−δt cos(ωt*φ)

    when the force or acceleration slower is than the time constant (T=2π/w):
    x(t)= F(t)/D

    through F=m*a:
    F=D*x
    a= D/m * x

    Is it right?
    How can I calculate the capacitance?
     
  6. Jun 30, 2013 #5

    gneill

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    You won't need the differential equation for the spring-mass system; Use conservation of energy to find the maximum displacement after the constant acceleration is suddenly applied.

    For the capacitance, what is the capacitance given plate area and separation of plates?

    For the electrical part of the problem, start simple and assume that the sensor consists of two capacitors, Ca and Cb. Find an expression for the potential across the unknown capacitor Cx in terms of Ca and Cb. Later you can plug in expressions or values for Ca and Cb.
     
  7. Jun 30, 2013 #6
    F(x)=-kx
    U(x)=∫(0-x) F(x) dx = 0,5k*x^2
    E=T+U=const.

    Q=C*U
    W=0,5C*U^2

    Is it right? How should I go on?
     
  8. Jun 30, 2013 #7

    gneill

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    Sure, the total energy is constant. So write an expression for the current mechanical system. Find the maximum displacement given the applied acceleration.

    Hint: The applied acceleration is equivalent to a gravitational field... imagine a vertical spring with a mass on the end held there at the spring's relaxed position. Then the mass is released. The "gravitational" acceleration acting is the applied acceleration a rather than g. What's the lowest position the mass will reach relative to its initial position? The concepts of conservation of potential and kinetic energy apply.
    You haven't written the expression for the capacitance of a capacitor given the plate area and separation. Once you do, write expressions for both variable capacitors given a displacement from the rest position, Δx.

    You have to solve the circuit for the potential across the unknown capacitor. Start with assigning variables Ca and Cb to the two variable capacitors; you can worry about their actual values later.
     
    Last edited: Jun 30, 2013
  9. Jun 30, 2013 #8
    W=Fg*∆x=m*g*∆x
    ∆x=W/m*g

    C=ԑ*A/d
    ∆C=Cb-Ca=ԑA/d±∆d - ԑA/d = ±ԑA*∆d/d*(d±∆d) = ±Ca*∆d/d±∆d
     
  10. Jun 30, 2013 #9
    Is this right?
    How should I go on?
     
  11. Jun 30, 2013 #10

    gneill

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    I don't see the spring constant in there. I see g rather than a. g does not come into play in this problem, I only drew a comparison with a spring and mass in a gravitational field as an analogous example. Use a for acceleration. What is W in terms of the given values (spring constant, mass)?
    The formula for capacitance is right. But rather than ∆C, work out the capacitance values of the individual sections. The sensor has three plates, which translates into two capacitors in series (Ca, Cb) The unknown capacitor Cx is connected where the two sensor capacitors join.

    attachment.php?attachmentid=59987&stc=1&d=1372612999.gif
    You still need to complete the expression for the displacement distance Δd in terms of the spring constant, the mass, and the acceleration. Leaving a new variable, W, in the expression is not useful.

    You still need to analyze the circuit to determine the voltage across the unknown capacitor Vx in terms of the two varying capacitors Ca and Cb.
     

    Attached Files:

  12. Jun 30, 2013 #11
    ∆d=k/m*a


    Ca=ԑA/d-∆d
    Cb=ԑA/d+∆d
    Ua=Uo/2 * ∆d/d

    Is it right? And then?
     
  13. Jun 30, 2013 #12

    gneill

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    What units does your expression yield? Check your derivation. Also use parentheses when order of operations is ambiguous. Is your expression (k/m)*a, or k/(m*a)?
    I have no idea what Ua or Uo are. What's Vx in terms of Ca, Cb, Cx?

    Surely you can work out some strategy to move forward if you have Δd and expressions for the capacitors and the voltage Vx? What conditions have to be met?
     
    Last edited: Jun 30, 2013
  14. Jun 30, 2013 #13
    ∆d=k/(m*a)
    k[N/m]; m[g]; a[m/s^2]

    Is it right? I am not sure with the units.

    Vx=Vo/2 * ∆d/d
     
  15. Jun 30, 2013 #14

    gneill

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    Multiply out and reduce the units for your expression. You've got the right units for the individual variables (keep in mind that the Newton is a composite unit itself -- kg*m*s-2).
    You'll have to explain how you arrived at the above. What's Vo?
     
  16. Jun 30, 2013 #15
    [kg*m/s^2] : [m] : [kg * m/s^2] = [kg] : [m] : [kg] =m

    I think the formula is right.?

    Uo is an alternating voltage is applied to this circuit, the voltage drop V1
    to a resistor equal to Vo / 2
    Is in the left stitch V2=Vo * Zc2/(Zc1+Zc2) of the complex resistance Z=(iwC)^-1 of the capacitors. This results in the output voltage according to some equivalence transformations Va=V1-V2=Vo/2 * (C2-C1)/(C2+C1) with the capacity Ca=ԑA/d-∆d and Cb=ԑA/d+∆d it follows after another equivalence transformation:
    Vx=Vo/2 * ∆d/d

    Is it right? How should I go on?
     
  17. Jun 30, 2013 #16

    gneill

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    kg/m/kg yields 1/m.

    The equation is inverted. Better check your derivation steps.
    Here's your circuit:
    attachment.php?attachmentid=59993&stc=1&d=1372623481.gif
    There's no alternating voltage source in the circuit, and no resistors... just two 12V DC supplies and three capacitors. You'll probably want to use the capacitor voltage division rule and superposition to analyze the circuit. Here's the same circuit rearranged slightly with the capacitors made obvious:
    attachment.php?attachmentid=59995&stc=1&d=1372623850.gif
     

    Attached Files:

  18. Jul 1, 2013 #17
    I have really no idea, what I have to do.
    I know: C=Ca+Cx+Cb
    V=V1=V2...

    Could you please help me?
     
  19. Jul 1, 2013 #18

    gneill

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    No, the capacitors are not in parallel; the voltage sources are separating their lower connections.
    I think you'll have to review your circuit analysis classwork. We can't just solve the problem for you; you must make a realistic attempt and we can then offer suggestions or point out errors.

    As I mentioned previously, you might want to look at superposition and the capacitor voltage divider rule.
     
  20. Jul 1, 2013 #19
    V=(Vc*Cc)/C
    Cc=(V*C)/Vc

    C=εo*εr*A/d
    d=k/(m*a)

    Cc=(V*εo*εr*(A/k/(m*a)))/Vc

    Is this right?
    What have I do wrong?
     
  21. Jul 1, 2013 #20

    gneill

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    What do the variables in the above pertain to? What does the first expression describe?
    The dielectric in the capacitor is assumed to be air, with a dielectric constant pretty much indistinguishable from 1.00, so no need to introduce an εr here. And as we discussed previously, your formula for Δd is inverted --- it yields 1/Δd. It doesn't yield the capacitor separation, it yields the CHANGE in its separation from its rest position. One capacitor's separation gets smaller by Δd, the other gets larger by Δd.
    We still need to see a correct circuit analysis for the 3 capacitors and 2 voltage supplies. If you were given only the circuit which I posted in post #10, what expression would you derive for the voltage Vx?
     
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