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Newbie questions on impedance bridging in audio circuits

Discussion in 'Audio' started by Rainboyo, Jun 21, 2012.

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  1. Rainboyo

    Rainboyo

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    Jun 21, 2012
    So I'm trying to understand this - my understanding so far is that in most modern audio circuits, input impedances are kept low and output impedances are kept high in order to maximise voltage transfer. Thinking about this has inspired questions in my mind, which I hope someone can help me with:

    Question one
    When reading things explaining impedance bridging, I often come across a circuit diagram that shows an output impedance in series with a source signal connected to an input impedance in parallel with input terminals. Why the difference? (best guess so far... something to do with Thevenin vs Norton reductions...?)

    Question two
    If we want a flat response, we want the frequency-dependent part of the impedance to not impact our desired frequency range (ie the audio spectrum). So is part of the answer to how this works something to do with increasing resistance at the input, in order to 'overpower' the reactive parts of the input impedance? (this doesn't feel quite right, as this isn't anything to do with the ratio between output and input impedances, but hopefully someone can tell me where I'm wrong!)

    I have more questions, but even the questions get a bit garbled in my head when I try to write them down, so I think I'd better leave it there for now!

    Many thanks!

    Mark
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    9,659
    2,019
    Nov 17, 2011
    That wouldn't work. A high output impedance means a high voltage drop within the source, thus minimizing the voltage available to the input.
    Useful combinations are:
    high output Z, low input Z -> current controlled input
    low output Z, high input Z -> voltage controlled input
    output Z = input Z -> optimum power transfer


    Q1: This scheme models the real voltage source by an ideal voltage source (0 Ohm resistance) + a series resistance (equivalent to the output resistance of the real source). The real input is modeled as an ideal input (infinite input resistance) + a parallel input resistance.
    Therefore the current flows from the voltage source throught the series resistor on to the parallel resistor. The voltage drop across the parallel resistor is the input signal to the following stage (no further input current into this ideal stage is assumed).

    Q2: This is not as simple as you describe. The frequency response of the circuit is a complex function of all components used. Dpending on the type of components and the exact circuit the frequency response can have vey different characteristics: from rather flat in the pass band (which very generally implies a rather slow decline in the stop band) to rather wavy in the pass band (which very generally implies a rather steep decline in the stop band).
    See http://en.wikipedia.org/wiki/Transfer_function for various transfer functions.

    Harald
     
  3. Rainboyo

    Rainboyo

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    0
    Jun 21, 2012
    Thank you Harald

    Whoops! I got the words input and output mixed up... I should have proofread that better!

    Right, I think I understand it a little better now...

    The reason an input impedance is shown as being in parallel in these diagrams, is because it is essentially two terminals, which are the beginning part of the destination circuit - it is automatically in parallel in relation to the source circuit, due to the way the two circuits are connected together. The input impedance is defined as the impedance seen by those input terminals looking into the destination circuit, so being in series with the terminals (from the point of view of the source circuit) would simply not make sense.

    And in terms of why the input impedance should be high (one of the questions I didn't ask!)... the input impedance is the reduction of the destination circuit, and if it is too low, it will 'load' the outputs of the source circuit. That is to say, the output impedance of the source circuit will start to become significant compared with the input impedance of the destination circuit, meaning that voltage will be lost at that first output stage.

    I understand what you are saying about transfer functions - yep that makes much more sense!

    Thank you for your help, and please correct me if what I have said above still doesn't sound right! :) I find it really difficult to talk about this stuff, because until these words have an intuitive meaning, they're just words, no matter how many times you see/hear them repeated!

    Mark
     
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