second inductor in a butterworth......what does it do?

Discussion in 'General Electronics Chat' started by ninjaman, Nov 22, 2013.

  1. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    hello

    a third order butterworth filter has two inductors in series and a capacitor in parallel, placed between the two inductors.
    I thought that the second inductor would help filter out any remaining high frequencies. I spoke to someone who has a degree in electronics, he said the second inductor was there to match the first inductor and capacitor to the output of the circuit.
    I didn't understand and he didn't have time to explain.
    so if anyone would be kind enough to share that information I would appreciate it.

    thank you

    simon
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,791
    The three components act in combination to produce the low pass characteristic response. All three components act to attenuate high frequencies. They also act in combination to produce a particular input impedance and a particular output impedance.

    With a simulator you can investigate what happens if you change the value of one component while keeping the others constant. The greater the change in value the greater the deviation you will get from a Butterworth response.

    The key point here is the interaction between the components.
     
  3. LvW

    Active Member

    Jun 13, 2013
    674
    100
    *At first, a third order pasive lowpass filter needs THREE frequency dependent parts.
    That is the reason for the 2nd inductor.
    * Please keep in mind, that there is no filter that can "filter out any remaining high frequencies". All unwanted frequency components can be attenuated only, but not removed.
    * The third order lowpass you have mentioned needs an additional resistor at the output. Otherwise, there is no current through L2 (that means: no effect).
     
  4. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    the resistor after the inductor, is that not the driver? I know the driver is not a "resistor", it has inductance and a series DC resistance. also the Lpad and zobel circuits before the speakers. do these not act as the resistor you speak of. do these components not draw current?

    the circuit that I have has the filter components and L pad for the tweeter and zobel for the woofer.


    in the below formula, what is the Vd that I have highlighted? im thinking it is voltage drop? or could it be voltage decibels.

    Rt = 1 / (1 / Rp + 1 / Z) = 1 / (1 / 12 + 1 / 6) = 1 / (0.0833 + 0.1667) = 4 Ohms
    [Vd] = (Rs / Rt) + 1 = (2 / 4) + 1 = 1.5
    dB = 20 log(Vd) = 20 log (1.5) = 3.52 dB


    any "help" would be appreciated

    thanks

    simon
     
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,392
    497
    I want to see some pics.
     
  6. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    rs is the resistor in series, rt is the total resistance (which I believe is the speaker and resistor in parallel.

    -----mmmm-------------mmmmmm-------
    |
    |
    -----
    -----
    |
    |
    this is the low pass filter, the high pass is the same shape with reverse components, a T network I think its called
     
  7. LvW

    Active Member

    Jun 13, 2013
    674
    100
    Ninjaman, reading and trying to combine your post#4 and post#6 does not clear up what you really want.
    Are you not able to show us a clear circuit diagram with a corresponding question?
    In post#4 you write an equation without any indication what the used symbols are related to. Based on such "information" you cannot expect any meaningful answers.
     
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