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Electret microphone ?

Discussion in 'Electronic Basics' started by phil, Apr 29, 2004.

  1. phil

    phil Guest

    Hello all,

    This is actually a continuation of my previous post. The circuit in question
    is http://www.uoguelph.ca/~antoon/circ/fmt1.htm and the question is about
    electret microphones. I have pulled the specifications of an electret from
    maplin, as follows:

    Response: Omni-directional
    Frequency response: 50Hz to 13kHz
    Sensitivity: -60dB ±3dB (0dB = 1V/µbar at 1kHz, Vcc = 4·5V, Rl = 1k
    Ohms)
    Impedance: 1k Ohms maximum
    S/N ratio: >40dB
    Sound pressure level: 120dB maximum
    Power supply: 1·5V to 10Vdc
    Recommended voltage: 4·5V (optimum performance)
    Current drain: 0·4mA


    My question is what sort of ac voltage output range can I expect from the
    microphone and what does it mean 0db = 1V/ubar and do we (and if so , how)
    use it to find our output from the electret.

    Also, is it possile to remove the electret and the 1st stage amplifier, and
    connect the output from e.g. my computer sound card directly to C2
    and control the gain from the pc volume control ?

    Tia
     
  2. Bob Masta

    Bob Masta Guest

    SOME FORMULAS FOR WORKING WITH SOUND:

    1 Pascal (Pa) = 1 Newton/m²

    = 10 dyne/cm²

    = 10 µbar

    = 94 dB SPL

    Sound Power Level (dB SPL):

    SPL = 20 * log(P / P0)

    where P is the measured pressure and P0 is
    a reference pressure in the same system of
    units:

    P0 = 20 µPa (or µNewton/m²)

    = 0.0002 µbar (or dyne/cm²)

    OK, so your mic sensitivity is -60 dB re 1V/ubar. That means
    that if you get an output that is 60 dB below 1 volt (ie 1 mV)
    then the sound pressure is 1 ubar. Since 10 ubar = 94 dB SPL, then
    1 ubar = 74 dB SPL.

    So if you feed it 74 dB SPL, expect about 1 mV.
    For 94 dB SPL, expect about 10 mV.

    I'm not sure what you are asking here. The PC can provide
    a controlled amount of amplification of the mic signal. Why
    do you want to take the mic apart? If you try to take it
    apart, for whatever reason, you will probably destroy its
    usefullness as a mic. This is a mechanical system that
    is sensitive to tiny vibrations, so it's not terribly robust
    inside.

    Hope this helps!




    Bob Masta
    dqatechATdaqartaDOTcom

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
    www.daqarta.com
     
  3. phil

    phil Guest


    When you say -60 dB re 1V/ubar, what does "re" stand for

    What I meant was not to actually dismantle the microphone, simply to
    remove all components up to C2, and feed the signal into C2 from
    my PC sound card. This is just to get a prototype running a bit faster,
    and I can control the gain of the audio from the PC volume control,
    which sounds rather convenient ?

    Yes, very much so, thanks!!
     
  4. Ian Bell

    Ian Bell Guest

    phil wrote:

    snip
    relative to or with respect to

    Ian
     
  5. Soeren

    Soeren Guest

    Hi Phil,

    You would need a potential divider like this:

    PC-soundcard O---[33k]---+
    _|_
    | |
    | |<-----O to left side of C2
    |_| 1k
    PC |
    Gnd O--------------------+-------O to trnsmitter gnd.
     
  6. It appears to me to be a bit 'reverse engineered' right now. Aren't
    the ins and outs swapped?
    GG
     
  7. Soeren

    Soeren Guest

    Hi Glenn,

    Nope.
    The input from the soundcard should be as light a load as possible
    (within reason), while the output must have a low impedance for driving
    the next stage (the x-mitter).
     
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