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Reactive impedance question

Discussion in 'Electronic Basics' started by Tim Williams, Oct 3, 2005.

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  1. Rich Grise

    Rich Grise Guest

    At resonance, Z = R. ;-) (these _are_ perfect components, right? ;-) )

  2. I believe you are thinking of all 3 components in series.
  3. Rich Grise

    Rich Grise Guest

    Yeah. Or in parallel. ;-)

    You know about the ohnosecond, right? ;-)

  4. I missed that.
  5. The Phantom

    The Phantom Guest

    Probably you don't have the expressions right. I haven't used the
    symbols C, L1, and L2 to represent their reactances, but rather have
    used them to represent the values of those components.

    The expression for input Z is:

    Sqrt[(4*f^2*Pi^2*((L1 + L2)^2*R^2 + 16*C^2*f^4*L1^2*L2^2*Pi^4*R^2 +
    4*f^2*L1*L2*Pi^2*(L1*(L2 - 2*C*R^2) - 2*C*L2*R^2)))/
    (4*f^2*L2^2*Pi^2 + R^2 - 8*C*f^2*L2*Pi^2*R^2 +

    and the voltage you have called Vx is given by:

    Vx = Vz*Sqrt[(L2^2*R^2)/((L1 + L2)^2*R^2 +
    16*C^2*f^4*L1^2*L2^2*Pi^4*R^2 +
    4*f^2*L1*L2*Pi^2*(-2*C*L2*R^2 + L1*(L2 - 2*C*R^2)))]
  6. Tim Williams

    Tim Williams Guest

    I went over the Vx problem twice and got the same result, so there must be
    something wrong with my procedure...but it's rock solid as far as I know:
    loop current = vector sum of Ir + IL2 + Ic (where Ireactive = IL2 - Ic) and
    Vz = vector sum Vx + VL1, where VL1 = XL1 * I. Plug first into second,
    switch component currents with expressions in terms of Vx, isolate Vx
    outside of the radical and solve for Vx.
    Fair enough, but makes for a messier result. I can just as well set Xc = 1
    / 2piFC and so forth when I graph it.
    Um.. I'll look at that later... why does it keep growing?...
    So that,

    Vx = Vz*L2*R / sqrt of:
    R^2 L1^2 + R^2 L1 L2 + R^2 L2^2 + 16 pi^4 f^4 L1^2 L2^2 R^2 C^2 +
    4 pi^2 f^2 L1^2 L2^2 - 8 pi^2 f^2 L1^2 L2 R^2 C - 8 pi^2 f^2 L1 L2^2 R^2


  7. Tim Williams

    Tim Williams Guest

    Well, that doesn't work the same as the original (if I copied it right...I
    copied it twice correctly...), but even so, it gets infinite values for the
    range zero to some value of F. The equation I came up with at least shows a
    resonance peak, but as I said, it doesn't go to infinity with R, at Fo.

  8. The Phantom

    The Phantom Guest

    The way to do it is to first write the voltage transfer ratio in terms of the
    complex variable s. Since it's just a voltage divider, that's pretty easy. The
    three components, L2, R, and C form the bottom leg of the divider and L1 forms
    the top leg. Let z1 be the impedance of the top leg (just L1), and let z2 be
    the impedance of the 3 components of the bottom leg in parallel. Then the
    voltage transfer ratio of the divider is given by z2/(z1+z2). Expanding, and
    using the complex variable s, we get:

    Vz = Vx * [(1/(1/(s L2) + 1/R + s C))/(s L1 +1/(1/(s L2) + 1/R + s C))]
    or Vz = Vx * CVTR (CVTR stands for complex voltage transfer ratio).

    Replace the variable s in the expression for CVTR with (2 Pi f j), so that
    CVTR = (1/(1/(2 Pi f j L2) + 1/R + 2 Pi f j C))/(2 Pi f j L1 +1/(1/(2 Pi f j L2)
    + 1/R + 2 Pi f j C))

    Let CONJ[CVTR] be the complex conjugate of CVTR; that is CONJ[CVTR] is just the
    same expression as CVTR, but with j (= SQRT[-1]) replaced by -j. Then form the
    product: (CVTR * CONJ[CVTR]) and take the square root, so that the (real
    magnitude) voltage transfer ratio you want is given as:

    Vz = Vx * SQRT[CVTR * CONJ[CVTR]]

    This is only the magnitude (or modulus) of the CVTR; the phase information has
    been lost with this procedure.
    You made a couple of errors in your algebra:
    should be ...+ 2 R^2 L1 L2 +...
    And this should be ...- 8 pi^2 f^2 L1 L2^2 R^2 C

    When I plot this with R=100, C=20uF, L1=50uH, L2=10uH, I get a peak at about
    12328 Hz, with a value at the peak of about 25.82

    I verify these expressions I've posted by plotting the simpler complex
    variable version on the same plot with the complicated real variable version,
    and seeing that the two plots are identical (actually, they're on top of each
  9. Tim Williams

    Tim Williams Guest

    We never covered s, and I don't remember how to handle conjugation of i/j.
    Ah, ok.
    Not a problem, I know phase runs from 0 degrees approaching zero to 180
    degrees approaching infinity, with 90 degrees at Fo.
    Ok. But how does that help me...

    As it turns out I can paste your equation into QBasic verbatim, making minor
    corrections (that I can't possibly screw up ;) and it works exactly
    perfectly! Well thanks for your help Phantom :)

  10. The Phantom

    The Phantom Guest

    In an earlier reply of yours to one of my postings, you said:

    "And you're certain that the above equation is the real derivation of the
    imaginary equation, which is also correct?"

    You wanted to know if I was certain, so I thought I'd tell you how I make sure
    the complicated real variable equation is correct. :)
    Good! Glad it's working out for you.
  11. Rich Grise

    Rich Grise Guest

    An ohnosecond is the fraction of time in which you realize you made a big
    mistake and you can't undo it. The first documented use of the term was
    by Elizabeth Crowe in her book The Electronic Traveler.

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