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V and I not in phase at resonance Frequency in RLC network?

Discussion in 'General Electronics' started by Diego Stutzer, Oct 14, 2003.

  1. Hi,
    Well, I'm really confused.
    I simulate a simple serial R-C-L-Network (all in series).

    As far as I know the total (input-)Impedance of the network is:
    Z = R + jwL - j/(wC) resp. the resonance frequency (where Zin=R) is
    1/sqrt(L*C).
    At resonance frequency, the Impedance should be real and therefore in
    my hummel opinion Voltage and Current schould be in phase.

    The funny thing is, when i build up such a network in Schematics
    (Cadence PSD 14.1/Orcad 9.2) and simulate it with the PSpice A/D
    Simulator, the current is displaced (relative to the voltage) about
    lambda/4 - obviously not in phase!?

    Can anyone tell my where I made a mistake?
    Or why this Problem is showing up?
    Thanks to anyone reading this and especially to those who post
    answers.
    D. Stutzer
     
  2. Don Pearce

    Don Pearce Guest

    The impedance should be R + jwL + 1/(jwC)

    d

    _____________________________

    http://www.pearce.uk.com
     
  3. John Fields

    John Fields Guest

     
  4. WOW!! Thank you soo much. The world is sound and the laws of physics apply
    again. I just confused w (omega) and f.
    Best Regards
    D. Stutzer
     
  5. Michael

    Michael Guest

    The impedance should be R + jwL + 1/(jwC)
    You sure?, how do the j parts cancel at resonance if they are both added?
     
  6. Tom Bruhns

    Tom Bruhns Guest

    Which voltage and which current? Clearly the voltage across the
    capacitor will always be pi/2 relative to the current in that
    capacitor, and the same (but opposite sign) for an inductor. Also,
    are you doing an AC or a transient analysis? If transient, are the
    transients settled, are you really on the resonance frequency, and are
    you simulating with sufficient time resolution? Oh, and I'm not quite
    sure what you mean by "lambda/4" phase shift. Is that degrees or
    radians, and what's lambda?

    Cheers,
    Tom
     
  7. Dale Parfitt

    Dale Parfitt Guest

    -j/wc= 1/jwc
     
  8. Reg Edwards

    Reg Edwards Guest

    Multiply top and bottom of 1/jwC by j (This does not change its value) and
    you get 1/jwC = minus j/wC.

    Back to school with your algebra.
     
  9. budgie

    budgie Guest

    His algebra looks perfectly fine to me. But as others have pointed
    out, he's left the 2pi out.
     
  10. Reg Edwards

    Reg Edwards Guest

    ------------------------------------------

    Yes. I apologise for my remark about school. I gained the incorrect
    impression from the previous replies.

    The w in wC stands for omega = 2*Pi*F, the angular frequency.
     
  11. Active8

    Active8 Guest

    now that were all done playing with j...

    Z = sqrt[R^2 + (jwL)^2 - (1/jwC)^2] = sqrt[R^2 + (jwL)^2 + (j/wC)^2]

    this is scary shit.

    mike
     
  12. Active8

    Active8 Guest

    now that were all done playing with j...

    don't forget

    Z = sqrt{R^2 + [(wL) - (1/wC)]^2]}

    and

    Z(s) = R + Ls + 1/Cs

    which is just plain easier to deal with 'til you need to journey back
    into time domain land. no need to leave it f(t) for this deal, though.

    all that j stuff... that was scary shit. so easy to make a mistake.
    swapping w and f is another good one. only works for f/f stuff.

    mike
     
  13. Especially so given the limited typography of this particular medium.
    I suspect few of us would have a problem if we could only view these
    formulae in a suitably appropriate typeface!!!
     
  14. R + jwL + 1/(jwC)
    = R + jwL -j/(wC)
    so at resonance wL=1/(wC)
    ie w=1/sqrt(LC)

    Chris
     
  15. John Fields

    John Fields Guest

    ---
    Since he calculated the resonant frequency of the circuit using

    f = 1/sqrt(L*C),

    his answer will always be a frequency 6.28 times higher than it should
    be, so the reactance of the inductor will be greater than the reactance
    of the capacitor, making the phase angle positive.

    "Lambda" is usually taken to mean wavelength, so "lambda/4" would mean a
    quarter wavelength; in this context, 90°, the approximate phase
    difference between his simulated voltage and current.
     
  16. Active8

    Active8 Guest

    what pre tell, is a typeface that would make it hard? wing-dings?

    i admit, even in HTML with arial or fixed-pitch, where you have sub and
    super tags and greek letter codes, you can't do much. i tried a few free
    math notation tools for HTML and wasn't happy. plus you need a plugin
    for most if not all. W^3C has a standard. i'd like to see more math
    capabiliies in browsers. we could attach formulae without violating the
    no bianaries rule. of course microshaft and standards will probably
    never be seen in the same line of code.

    i still don't see why text based math legibility is font dependant. i
    see i could have eliminated some clarifying parens by

    - -
    | 2 1 2 |
    sqrt | R + ----- | or a 1/2 power instead of sqrt
    | jwC |
    - -
    created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de

    does that also foul up?

    brs,
    mike
     
  17. Fred Abse

    Fred Abse Guest


    Improbability level now zero. Normality restored. Anything you still can't
    cope with is your problem.
    (Douglas Adams)
     
  18. Michael

    Michael Guest

    I get a bit scared of all the j's too, I prefer polar. Usually you
    want a magnitude and angle in the end result anyway.
     
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