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Discussion in 'Electronic Basics' started by jason, Apr 8, 2005.

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  1. jason

    jason Guest

    Hello All
    Say I have a resistor R, inductor L and capacitor C connected together
    in series

    The easiest way to find the total impedance is just plug the formula
    a+ bj

    Z= square root (a^2 + b^2 )

    without drawing any phasor diagram right?

    For example
    R + Xc + XL = R + jwL - (j /wC)
    = R + j (wL - (1/wC)
    = a + j b

    so Z total is square root( a^2 + b^2 )

    Am I right?

    Can I always plug in equation square root( a^2 + b^2 ) without drawing
    a phasor diagram?
    If cannot what is the reason and any example to explain it?

    Kindly enlighthen

    Thank you

  2. Lord Garth

    Lord Garth Guest


    yep, the resistance is the horizontal axis,
    the reactance is the vertical axis
    the impedance is the hypotenuse.

    Sounds just like the Pythagorean theorem, doesn't it....
  3. jason

    jason Guest

    Yes that's right Garth ,.. Thank you

    Sometimes in some cases, we compare the reactance of a capacitor with
    resistance of a resistance like the below case

    1/wC << Rs

    So my question is , what does it mean by this comparison?
    reactance < resistance

    What does it tell us?

    Should not we compare resistance with resistance
    while comparing reactance with reactance
    Both of them have the unit ohm?

    What is the implication of using such a comparison , what does it tell

    Thank you
  4. Andrew Holme

    Andrew Holme Guest

    When using AC-coupling between stages, the coupling capacitors can form
    unwanted high-pass filters, unless the reactance is negligible compared
    to the source resistance.

    Consider the RC high-pass filter:


    Transfer function = sCR / (1+sCR)

    At high frequencies Vout/Vin = 1 and phase = 0;
    At the break-point, phase is 45 degrees and output is 3dB down;
    Below the break-frequency, output falls at -20dB per decade.

    The break frequency is w = 1/RC

    So, Vout/Vin = 1 when 1/wC << R
  5. Andrew Holme

    Andrew Holme Guest

    Perhaps I should say: compared to load resistance.

    Another time you might want 1/wC << Rs would be when selecting
    decoupling capacitors. Here, it is the source resistance we are
    concerned with, and we are forming a low-pass filter with a VERY low
    break frequency.
  6. Lord Garth

    Lord Garth Guest

    I was trying to recall the original question.....

    You had a series RCL I think that all you are seeing is that
    the inductive reactance and the capacitive reactance are nearly
    equal but opposite therefore they cancel. The circuit appears to
    be very much like a resistance only.

    If 1/wC << Rs the resultant phasor is very close to Rs so it seems
    to not be very significant in a first approximation.

    I guess I'm not seeing why wL couldn't also be << Rs resulting in
    a phasor in quadrant 1 rather than Q4

    In either case, the circuit would be dominated by Rs, the power factor
    would be close to 1 therefore the current and the voltage would be
    (pretty much) in phase. The crest factor would be close to minimum.
    Rs would have to handle near maximum power. It sounds like reasonable
    matching for maximum power transfer.
  7. jason

    jason Guest

    Hi Andrew and Garth
    Thanks a lot for the help.
    Always gald to see positive guides from you all

    Well my case is the Rs at source node of a nmos, and a tap capacitor is
    connect to the top end of Rs.
    So it is using 1/wc <<Rs

    I was wondering why
    Thats all
    Other than the above explanation, is there anything can be added to my

    Thank you so much
    I have learnt so much from you all

  8. jason

    jason Guest

    by the way, mine is to create an oscillating circuit using nmosfet, tap
    capacitor, inductor as feedback.
    Kindly enlighthen if there is anything else to be added
    Thank you so much Garth and Andrew
  9. Lord Garth

    Lord Garth Guest

    Is Rs actually Rds? (the channel resistance)

    In any case, recall that you must get the proper amount of phase
    shift or there will be no oscillator.

    Good luck!

    BTW, why are you doing this?
  10. jason

    jason Guest

    Rs is the total ac resistance of cource and also resistance connecting
    source node and ground.
    I do for my assignment
  11. jason

    jason Guest

    I got my circuit oscillating but do not know how to do hand calculation
    to prove it
    .. so wish to know the reason for 1/wc<<Rs
  12. Lord Garth

    Lord Garth Guest

    Since the Xl & Xc cancel, you get the required 180 degree phase shift.
    Given the above, Z is Rs so the current through the node is high and
    the drop across the node is low. This should prove the circuit is
  13. jason

    jason Guest

    I am sorry for my lacking in basic knowledge, why when XL and Xc cancel
    , we can get 180 degree phase shift?

    Also the drop across the node is low meaning the current to Rs is low?

    AM I right?
    Thank you Garth
  14. Lord Garth

    Lord Garth Guest

    Do you know the memory aid for current and voltage in reactive circuits?
    The mnemonic is ELI the ICE man. ELI = in an L circuit the voltage is
    ahead of the current (in a perfect world, by 90 degrees). ICE = in a C
    circuit the current is ahead of the voltage ( by 90 degrees).

    So there is your 180 degree shift. The definition of resonance is where
    Xl = Xc a condition your circuit meets.

    The drop across the node is low so the current through Rs is high. Think
    of replacing the combined reactance with a resistor of value Z. Z would
    be a low value so I is high for a series LC. The story is different if LC
    were a parallel tank circuit. In that case the current through the tank
    be low so the voltage across the tank would be high.

    I hope that helps!
  15. jason

    jason Guest

    Hi Garth

    Thank you
    My understand is most current will be in the tank circuit no matter it
    is series or parallel LC tank.
    That means I am wrong?

    Say for a nmos transistor, where its source node is connected to a tap
    capacitor as a feedback. Then with the tap capacitors, a inductor in
    series with a capacitor is connected in parallel to the earlier
    mentioned tap capacitors.
    They form a clapp LC osccilator with feeedback to the source node.
    For the nmos, the source node is connected to a resistor named Rs.

    So this case is a parallel LC tank, is it?
    If yes, the condition 1/wc<< Rs still hold . This simply means the
    current must flow to the LC tank circuit for the oscillating operation
    when the XC=XL.
    So voltage drop across Rs is small.
    Is it right for my understanding?

    So when it is series LC tank circuit or only a 2 capacitor with
    inductor in parallel case, what will this be different?

    One thing I wish to enquire, that is, what actually it helps in
    osccilator when XC=XL, both impedance are the same, so what will it
    help in this case?

    I wish you could help me in understanding this.
    Thank you Garth

    I apologise for the inconveniences caused

    Thanks a lot

  16. Lord Garth

    Lord Garth Guest

    I think this is a point of view, whether you are looking at what happens
    inside the LC or what happens when you mentally replace that LC with
    a black box of impedance Z and consider the whole circuit.

    We have been discussing what happens inside the LC and then applying
    that to the what happens to the remainder of the circuit. Our point of
    view has been shifting and this can be confusing.

    In a series resonant LC, maximum current flows through LC and the rest of
    the oscillator circuit.

    In a parallel resonant LC, maximum current is contained within the LC tank.
    Which implies that minimum current flows through the rest of the oscillator

    You've got to mentally step back and replace the resonant LC with its
    resultant Z then look at the voltage across that Z
    No, its just a matter of whether your looking at the whole oscillator or
    just the
    LC portion alone.

    In either case, the losses are made up for by the active component or the
    system losses would dampen the oscillition and the circuit would ring to
    a stop. A mechanical equivalent would be a cymbal on a drum set, hit it
    once and it oscillates with a decreasing amplitude until it stops. Add an
    active component (a drummer that adds more energy to the system) and
    oscillation will continue until the input ceases, then the decay proceeds.
  17. jason

    jason Guest

    Hi garth

    Thank you
    That is really a helpful explanation
    I learnt about parallel LC tank only. I wonder what physics that lies
    behind the series LC tank to make the max current to flow to the rest
    of oscillator circuit without keeping most current in the LC tank(like
    in paralle)

    By the wayI tried to email you but it failed.
    Is your mail box full?

    Thank you so much for the help
    Will write again if there is any arising doubts :0

    rgds and thanks
  18. Lord Garth

    Lord Garth Guest

  19. Lord Garth

    Lord Garth Guest

    The email is fake, the name and email derive from an early Star Trek
    What's up? Feel free to ask here, I don't mind!

    My generic excite email got a netsky infected email the other day, so much
    their "spam" filters.

    Caps / inductors and op-amps can be fun. I built a high gain amp while in
    connected it to a geophone and jammed it into a knot hole in the wooden
    Scoping the output, I was able to detect a person walking up to the
    building at night when the campus was quiet.
  20. jason

    jason Guest

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