Urgent help with R-L-C series resonant circuit!

Discussion in 'General Electronics Chat' started by Tristanorr, Jan 27, 2011.

  1. Tristanorr

    Thread Starter New Member

    Jan 27, 2011
    Sorry I posted this in the wrong part earlier today, was in a rush at uni.

    After proforming a test at college, I am left to write a report about it. I am now stuck and need help explaining when when you double R-L-C in turn in the circuit you have the same, larger or smaller bandwidth.

    Here are my results, i cant show you the circuit as I dont have multisim8 on this PC.

    RESULTS Resonant frequency Hz Amplitude at resonance
    dB Bandwidth Hz
    f2 - f1
    R = 80 ohms
    L = 40 mH
    C = 400 nF 1.254K -38.191 321
    Resistance doubled
    R = 160 ohms
    L = 40 mH
    C = 400 nF 1.254K -44.142 645.761
    Inductance doubled
    R = 80 ohms
    L= 80mH
    C = 400nF 887.692 -38.212 161.742
    Capacitance doubled
    R = 80 ohms
    L = 40 mH
    C = 800nF 887.692 -38.18 324.985

    So as you can see, I can tell where the bandwidth is the same or increased / decreased, but I don't understand why if you double the inducatance you get a lower Bandwidth, and if you double the capaistance you get the same... I believe with the resistance doubled you have a much larger bandwidth as you have much less gain (if you plot gain vs. frequncy).

    Any help tonight would be very helpful.
  2. edgetrigger


    Dec 19, 2010
    The ratio of center frequency(resonant frequency) to its bandwidth is called as q factor

    Q = fc /(f2-f1) or f2-f1 = fc/Q --- eq1

    Q is also = √L/ (R√C) ----- eq2

    with increase in L, Q increases square root times L as per eq2 and as per eq1 bandwidth will decrease.

    If C is increased Q will reduce quare root times C as per eq2 so bandwidth will increase.
  3. Tristanorr

    Thread Starter New Member

    Jan 27, 2011
    Thank you very nuch I hope that should explain it enough :)