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I've been trying for a few days to understand how an electret microphone with two pins works.

What I think I understand is:
The microphone has inside its capsule a capacitor that changes its capacitance from the vibrations reaching it (i.e. sound).

The changes of the capacitance make the current flow. When the capacitance grows, the capacitor will start storing more power (electrons), and when it shrinks, it will release power.

This change will have the effect that the voltage across the capacitor will be positive/negative in respect to the ground zero during charging/discharging.

Then the JFET's base is connected to the positive side of the capacitor to use the voltage change.

My question is, is my understanding correct? And additionally, how much will be the peak-to-peak voltage change across the capacitor?

I read that it's +-VCC/resistance, imagine a VCC of 3.3 V, but from where to get the resistance?

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    \$\begingroup\$ verbessern, There are at least 3 kinds of electrets, today. Maybe 4 or 5, even. I can't keep up, anymore. There are bare electrets, without any FET at all. These used to be what I had to use because, at the time, no one was adding a FET inside. Probably can still get them or at least find them in older equipment. Soon, almost all electrets had FETs inside them. This helped a lot and it quickly caught on. Soon, that was about all you could get. And they are still available. Then came various 'additions' which were slightly more complicated. Now it is an IC. Read datasheets! \$\endgroup\$
    – jonk
    Nov 22, 2022 at 21:03

4 Answers 4

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The microphone element is a capacitor that is pre-charged. The charge is fixed and cannot flow. The acoustic vibration changes the capacitance. From $$q=Cv\tag{1}$$ the capacitor’s voltage changes in response. The voltage is then amplified by the FET.

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  • \$\begingroup\$ Thank you, I understand. One of the plates is connected to GND (say). Will then the other plate have voltage changes +-X relative to the GND? With other words, if the other pole is left unconnected, will it still be +-X to GND? How can I know X, or it has to be just experimentally/approximately found per microphone? \$\endgroup\$
    – verbessern
    Nov 23, 2022 at 14:23
  • \$\begingroup\$ @verbessern: The FET is an n channel jfet so the end of the the capacitor connected to the gate should be negative. Without more info from the manufacturer the gate voltage must be tailored to the FET. If there is no FET built in then the information would be in the specification sheet. \$\endgroup\$
    – RussellH
    Nov 23, 2022 at 14:34
  • \$\begingroup\$ There is a FET, but on the schema, one side of the capacitor is connected to the gate, the other to the GND. I'm trying to understand how much is the voltage between the FET's gate and the capacitor. That voltage is then biased and amplified, but how can I know how much voltage I will get after all of this? It seems not possible to know exactly, except if the actual values of q/C are given, and the exact values of the FET (that I do not have) \$\endgroup\$
    – verbessern
    Nov 23, 2022 at 14:58
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There are different technologies that implement a mechanical audio vibrations to electrical signals transducer.

One can be, as you have mentioned, a capacitor which has one of its plates loosely bound to the transducer. This setup changes its capacitance as the distance between the plates changes under the audio's mechanical vibration. This technology can be implemented in very small form factors (such as MEMS).

Another option is a piezo- electric transducer, which is able to convert mechanical vibrations to an electrical signal. This transducer is made out of a thin film that vibrates when placed inside vibrating air and directly produces an electric varying charge. This charge is proportional to the audio signal. This technology can be quite cheap and is widely available.

The rest of the chain deals with all or a subset of the following:

  • Amplification: using amplifiers like op amps or transistors to convert small signal (usually ranging between \$10^{-6}\$ to \$10^{-2}\$ volts p-p) to a strong one.
  • Filtration: lowering unwanted signals that the sensor had picked up.
  • Conversion - from analog to digital using ADCs.
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    \$\begingroup\$ This question is specifically about electrets, so talking about other microphone technology seems a bit irrelevant. \$\endgroup\$
    – Hearth
    Nov 23, 2022 at 16:17
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A "condenser" mic has a thin diaphragm that is suspended with a 48VDC externally powered plate near it. When sound causes the diaphragm to move towards and away from the plate then the capacitance voltage divided between the distance between the diaphragm and the plate caused by the changing distance changes the output voltage. An electret mic has the 48VDC permanently stored on some electret material and a Jfet is used to convert the extremely high impedance of the capacitance voltage divider to a much lower impedance.

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The 2-terminal electret microphone has an inbuilt FET. The resistor and output coupling capacitor are required to be connected externally.

enter image description here

The 3-terminal electret microphone has an inbuilt FET and resistor. Only the output coupling capacitor is required to be connected externally.

enter image description here

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