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disign of 1st HPF for audio transformer

Discussion in 'Audio' started by DuctDuck, May 3, 2013.

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  1. DuctDuck

    DuctDuck

    52
    0
    Jan 26, 2013
    Good...day...Gentlemen, Ladies.

    I have encountered a troubling aspect in my audio project. A balanced line's filter and the reflected impedance from the amplifier (it's the reflected Z throwing me off).

    I would like to use a 1st order highpass filter (fc=30Hz)
    Ctotal=110nF
    Rtotal=48.2kOhm

    [​IMG]

    C1=C2=220nF and Rtotal is RIN to amplifier and all.

    I am used to seeing R parallel to L but if R is suitable, can't I omitt R for Rreflect?

    Thanks in advance,
     
  2. john monks

    john monks

    693
    2
    Mar 9, 2012
    Your schematic is unclear.
    What is connected to the right side of R1?
    What is connected to the bottom the the winding on the right?
    Are you assuming an ideal transformer?

    If the transformer is ideal and the turns ratio is 1:1 then whatever is on the right is reflected to the left.
     
  3. DuctDuck

    DuctDuck

    52
    0
    Jan 26, 2013
    Thanks for input, J.Monks!

    I can try again with a schematic and description for you...
    [​IMG]

    This is the high pass filter's parameters, for fc=30Hz. But I have an isolation audio transformer at the audio signal generator (before the cables to the small amp; using its own power supply).

    I would like to ask you is about Rreflect. Can I use my amplifier's Rin as the filter's resistance? What can I do for a balanced high pass filter on the primary?

    [​IMG]
     
  4. john monks

    john monks

    693
    2
    Mar 9, 2012
    The way you have your new schematic drawn is that the capacitors are seeing R1(48.2kOhm + Rreflect). And you two 220 nF capacitors connected in series.
    Your 48.2kOhm is reflected right back through the transformer.
    Bottom line is that you can now place your total capacitance C1 + C2 (110 nF) and the series (Rreflect + 48.2kOhm) into your formula to find out what your turnover or transition frequency is.
    I would love to simply give you the formula but I got in trouble doing peoples homework for them so give me what you come up with and we'll take it from there.

    I am assuming that the transformer is ideal.
     
  5. DuctDuck

    DuctDuck

    52
    0
    Jan 26, 2013
    Thanks, J.Monks! It's good you don't want to get into trouble; neither do I.

    If you can crunch an fc-transition point from the present parameters, please try so we may verify. But I would rather take up some of your time to confirm my present reasoning and progress further.

    [​IMG]
    On the primary side, where the filter is, the reflected impedance loads the filter's resistor (it's a balanced line so everything is distributed equally; for sensing).

    Then Rreflect is parallel to Rfilter and this looks more like a filter I am familiar with!

    Additionally, can you write an S-parmeter transfer function. I am used to Tau developed transfers and am suffering for it.

    [​IMG]

    ...the -j operator is made positive (j squared or j/j), for the second line blah blah blah, apply j-omega-tau product, for the third line to develope a frequency domain blah blah blah.

    It's limitting me to deriving 1st order filters; s-parameter is versatile, right?
     
  6. john monks

    john monks

    693
    2
    Mar 9, 2012
    So far your logic sounds good.
     
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