Help understanding electret microphones

Discussion in 'General Electronics Chat' started by scooter8, Jun 25, 2009.

  1. scooter8

    Thread Starter New Member

    Jun 25, 2009
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    I've been trying for a little over two weeks now to get an understanding of how an Electret microphone works. I've scoured web forums, manufacturer data sheets and have run several simulations inside of Multisim and still don't have a solid idea of how they work.

    I have the following questions...

    1.) What kind of signal does an Eletrect microphone produce? Specifically, is it AC centered at 0 (e.x. high peak at +1 V and low peak at -1 V ), or AC centered at somewhere else (with low peak at 0 and high peak somewhere +V).

    My current understanding is as follows. The microphone contains a N-channel JFET with its gate connected to the sound sensing element. Since its an N-channel type when the sound sensor "capicitor" is compressed it produces a negative bias on the gate this reducing current from source to drain. My reasoning says that if the microphone is not hearing any sound, the output would be a steady +V DC signal determined by the "biasing" resistor and the internal JFETs Ids spec. (See http://en.wikipedia.org/wiki/File:Electret_condenser_microphone_schematic.png ). When the microphone does hear sound, that DC signal is reduced in relation to the incomming sound pressure, thus producing and AC signal ranging from its "off" +V DC voltage down to a minimum of 0 volts. Am I correct or just dumb as a box of rocks?:confused:

    2.) Along the same lines, I've seen circuit drawings that have no apparent source of -V yet are connected to standard speakers. See http://www.elecfree.com/electronic/warbling-alarm-circuit-with-tone-generator/. Are these circuits driving the speakers with an AC signal that never goes below 0 volts? I don't see how that could work since the current is always moving in the same direction, or not at all (at 0 volts). The speaker cone should move out for any changeing +V signal but there would be nothing to pull it back in other than the natural elasticity of the speaker suround.

    That leaves 1 final question.
    3.) I've seen some proper audio amplifier circuit drawings that show an OPAMP that has +V, -V, and 0V connections. This makes sense to me as the +V input would be used to source the amplification on the incoming +V audio sine wave and the -V would amp the - side of the audio sine wave. My question is how do you get a -V reference from a battery power circuit? Say one that contains only 1 9volt battery?


    Thanks a ton in advance for all those who are helping my sanity :)
    Scott
     
  2. Audioguru

    New Member

    Dec 20, 2007
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    The 2-wire electret mic has a resistor feeding current to its fET so that its output is at a DC voltage. The signal causes its DC voltage to increase and decrease a little. The output signal level is small, tens of mV. An output coupling capacitor passes the signal and blocks DC.

    The horrible "audio amplifier" you show has DC current in its speaker which causes the cone to be positioned to one side. The signal causes the cone to move more to the one side and toward the center. Without current in a speaker its cone moves in one direction and in the other direction.

    A power amplifier can have a positive supply and a negative supply. Then its output can swing positive and swing negative.
    Or the opamp can have its output biased at half the single positive supply voltage with an output coupling capacitor. Its ouput swings more positive and less positive while the load swings positive and negative. The DC voltage across the coupling capacitor does not change.
     
  3. hgmjr

    Moderator

    Jan 28, 2005
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  4. scooter8

    Thread Starter New Member

    Jun 25, 2009
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    Awesome. Thanks for the reply.

    Am I correct in thinking this variable DC voltage output is standard across most styles of microphones minus dynamic mic's and thus some kind of circuity is required to de-offset the sine wave so that has equal +V and -V peaks?

    Glad to know that was a bad design I referenced. I thought so :)

    So there are two options for eventually driving a speaker. 1, using an opamp to amplify the signal. The opamp uses 1/2 the +V voltage as the common to the speaker and thus can amplify the signal so the sine wave is at maximum +1/2V to -1/2V. 2, using a packaged IC that uses a charge pump that charges a capcitor then switches it to the other direction to obtain full +V and -V sources limited only by how big that internal capacitor is expressed in watts.

    I have a another question about JFETs in general. Is it true that mesauring the Source to Drain resitance is not a valid indicator of how much current or voltage drop will occur when Gate-Source = 0V? Its related more to the physical arrangement of the N or P type channel material, the Source-Drain voltage and how it affects the depletion zone and can only be determined using the Ids parameter specificed on a specific JFETS spec sheet?


    Thanks a ton!!!!!!!!!
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    Only a simple coupling capacitor is needed to pass the signal but block the DC.

    Many circuits on the internet are designed by idiots who know nothing about electronics. Most are Instructables.

    No, no, no.
    An opamp does not have enough output current to drive a speaker. an opamp can provide only 20mA peak current. A power amp provides 100 times to 1000 times more current.

    Many power amps have a built-in bias circuit so that the output idles at half the supply voltage and is able to swing positively almost to the power supply voltage and almost to 0V. An output swing of only +1/2V and -1/2V p is nothing. Many power amps also have built-in negative feedback.

    No.
    An ordinary charge pump IC has a very low output current. It is useless for a power amplifier.
    A TDA1562 car radio amplifier IC has a high power charge pump that doubles the 13.6V supply voltage so that its output at clipping is 44W into 4 ohms.

    A JFET is turned on fully when its VGS is 0V. The spec has a broad range of current that will pass. It is called IDSS. It is from 2mA to 20mA for a 2N3819 JFET.
     
  6. scooter8

    Thread Starter New Member

    Jun 25, 2009
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    Thanks for your help Audioguru!!!! I can't tell you enough how much I appreciate you taking the time to answer my questions!

    I bought a USB scope a couple of days ago so I could learn about electret mics and their operations expirimently. A day later a fried let me borrow his Fluke scope as well. My USB scope doesn't seem to be precise enough but I did get some results with the fluke.

    Connected to ground and the output side of a standard electret mic circuit capacitor (see the wikipedia circuit I referenced in my first post), I measured -4.4mv min peak and 10.4mv max peak for a difference of 14.8mv while playing a 1khz tone into the mic. The fluke detected the frequency at 1khz while in AC scope mode at 10mv @ 5ms. This varies only a bit from the off state which was measured at -800uV min, 7mv max and 6.4mv P-P.

    I noticed both scopes seemed to pick up noise which was almost as strong as the signal when one scope lead was connected with sound off, or just plain disconnected. I also tried to scope the DC voltage change by connected the red lead onto the mic facing side of the electret biasing resistor. With that setup I couldn't detect any appreciable signal at all other than the static. Thats probably pretty normal for most scopes correct? i.e. picking up what looks like static when using such small mV scale?

    Can you explain the -4.4mv min reading I received when measuring from the output side of the circuit capacitor? Specifically how can it possibly be lower than ground or is that just the scope giving an inaccurate reading.

    Can I ask you for a HUGE favor? I'm still having trouble understanding exactly whats going on and the scope didn't help clear it up. Can you draw me an example pseudo ideal scope reading for the following: (assume 2volt supply power). Doesn't have to be accurate, just something to show the relative voltage waveform and most importantly where the peaks lie relative to 0 Volts for both the dc side of the biasing resistor and the coupled ac output.

    * Quite room, scope connected to ground and mic-side of the biasing resistor
    * 1khz audio played directly into the mic, scope connected to ground and mic-side of the biasing resistor
    * Quite room, scope connected to ground and output side of coupling capacitor
    * 1khz audio played directly into the mic, scope connected to ground and output side of coupling capacitor

    I've precreated two graph images in voltages I think will work to take the least amount of time for you as possible. You can add the waveform super easy by using the following two links.
    2v chart
    20mv chart

    Again, I really appreciate you help in teaching me.

    Thanks alot!
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    A microphone has a very low output level and has a fairly high impedance so its wiring picks up mains hum and other interference. Therfore shielded audio cables are used to block the hum and interference.
    A sine-wave sound fed to a microphone should produce a symmetrical sine-wave at its output. An electret mic witll have DC on its output with the signal going symmetrically positive and negative a little from the DC voltage.
     
  8. endolith

    Member

    Jun 21, 2010
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    Depends on where you're measuring.

    • An electret capsule by itself, with no FET buffer inside, will produce an AC voltage centered on 0 V (no DC offset). The output impedance is capacitive, so it requires an FET preamp after it to avoid loading problems.
    • A 2-terminal electret capsule with built-in FET will require an external resistor and power supply, and so will produce an AC voltage with a DC offset. The FET and the external resistor form a common-source amplifier. The output voltage of the FET will vary from 0 V up to the supply voltage. When silent, it will settle to some non-zero voltage in the middle.
    • If you look after the coupling capacitor, however, it will be AC with zero DC offset. Same exact signal, but shifted down so the average is 0 V instead of, say, 2.5 V, with the signal going positive and negative. Coupling capacitors remove DC offset.

    In a silent room, the diaphragm is flat, at average atmospheric pressure. When a sound goes past it, the atmospheric pressure ripples higher and lower than average atmospheric pressure, so the voltage on the capsule ripples higher or lower than its average value, so the current through the FET goes higher or lower than its average value. So the output voltage of the FET can decrease or increase from its average value. So if the supply is 2.5 V, for instance, silence might produce 1.25 V DC, then the high-pressure part of the wave might cause the FET output to drop to 0.1 V, while the low-pressure part of the wave might cause it to increase to 2.4 V. It's an AC wave summed with a 1.25 V DC offset.

    Yes, and don't do that. It will have a constant DC current flowing through the voice coil, pushing the cone out permanently and distorting the AC waveform. That's fine for a low-power noise-maker, but don't do it in anything where sound matters or the speaker voice coils can't handle the DC current without melting.

    With the electret mic circuit, though, you can adjust the supply voltage and the drain resistance. What I'd like to know is how to optimize these. If you increase the supply voltage, the output signal swing will increase, but does this improve SNR? I'm thinking it doesn't change the noise of the FET stage, since the capsule/FET noise gets amplified to a higher swing by the same amount as the signal. However, it allows the subsequent stages to run at a lower gain, so they contribute less noise? Could we go all the way up to the 20 V Absolute Maximum Drain to Source Voltage of the FET?

    And what should the drain resistor be? As the drain resistor of a common source amplifier increases, the gain increases (and in my tests, the distortion improves too). But it can't be increased too much or this stops happening. What's optimal for low distortion/best SNR? Can this be deduced from the datasheet parameters? How do we handle variation in gain from one FET to the next? This page says "design for around 80 percent of IDSS", but it's talking about biasing the gate voltage, which we have no control over in an electret. (This PDF says the gate self-biases to 0 V? So it's not a normal N-JFET that requires a negative bias at all times? The signal on the gate of the JFET goes positive and negative from 0, and as long as it doesn't exceed the PN junction of the gate that's ok?)
     
    Last edited: Sep 25, 2014
  9. ian field

    Distinguished Member

    Oct 27, 2012
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    Just a couple of points to pick at; 1A op-amps are not *THAT* uncommon, there are a few on the market that have even higher ratings and can comfortably drive a speaker.

    As for charge pump voltage multipliers; I believe one of the Philips audio amp chips has one on chip (external electrolytics obviously) - can't remember the class of operation, but it has a second pair of rails that are boosted, at low volume its exactly like any class A-B amplifier, if you turn the wick up; 2 more transistors come into play running from the boosted rails. The AC waveform to drive the charge pump comes from the basic class A-B section.
     
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