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

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  1. jason

    jason Guest

    Thanks Garth

    My doubt is
    " 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 parallel)

    "

    Jason
     
  2. Lord Garth

    Lord Garth Guest

    I'm trying to figure out how to explain it...

    I think you see that in your series resonant LC the Xl and Xc cancel
    leaving only the R term as the impedance of the LC combination.

    There is no added R to the series resonant LC so we are left with
    the ESR of the cap and the resistance of the coil wire. Both of these
    are really low so Z is consequentially low. Since the series resonant LC
    does not feed back to itself until the loop through the active component
    is complete, the current will not (can not) circulate within itself, as it
    is
    not yet a complete circuit.

    This is bit like asking why glass is transparent. The best answer seems
    to be that there is no reason it shouldn't be.
     
  3. jason

    jason Guest

    Hi Garth

    Thank you so much
    I learnt so much from you and other kind people up there :)
    So that means for parallel LC , there is a existing loop within itself.
    Therefore, it will keep the current in the parallel LC tank rather than
    let it flow to the active circuit?
    :)
    Thank you


    Jason
     
  4. jason

    jason Guest

    Hi Garth

    Thank you so much
    I learnt so much from you and other kind people up there :)
    So that means for parallel LC , there is a existing loop within itself.
    Therefore, it will keep the current in the parallel LC tank rather than
    let it flow to the active circuit?

    For parallel LC tank, the gain of the active circuit must be greater
    than 1. Then the initial loop gain must be greater than
    1(theoritically, in practical is at least 2)

    For series LC tank, what is the condition, do you know?

    Thank you


    Jason
     
  5. jason

    jason Guest

    Hi Garth

    Thank you so much
    I learnt so much from you and other kind people up there :)
    So that means for parallel LC , there is a existing loop within itself.
    Therefore, it will keep the current in the parallel LC tank rather than
    let it flow to the active circuit?

    For parallel LC tank, the gain of the active circuit must be greater
    than 1. Then the initial loop gain must be greater than
    1(theoritically, in practical is at least 2)

    For series LC tank, what is the condition, do you know?

    Thank you


    Jason
     
  6. Lord Garth

    Lord Garth Guest

    Yes, and it will decay over time due to losses within the components.
    Sorry, I don't know this one but I would guess that a gain of 2 is
    a practical minimum here as well.
     
  7. jason

    jason Guest

    Thank you Garth
    It is my honour to have learnt so much from you
    :)
    Thanks a lot and wish you good luck

    rgds and thanks
    Jason
     
  8. Lord Garth

    Lord Garth Guest

    You're welcome Jason, I hope you continue to study and in your own
    time, help others as well...

    Take care,

    Jon
     
  9. jason

    jason Guest

    Sure Garth(or Jon). I will do my best as a learner and help people who
    is in need
    Thank you so much
    :)

    Jason
     
  10. Lord Garth

    Lord Garth Guest

    Great, keep up with your studies!
    You are welcome
     
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