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See this picture (from Wikipedia):

enter image description here

The electret microphone consists of a capacitor with a static charge and a JFET. I wonder how the electrical signal relates to the sound signal.

I understand that with only a capacitor capsule the sound, with a change in distance between the two plates and capacitance, changes accordingly.

But what happens when we add a JFET to the system? Why does it require 2 V to operate?

In the final circuit we have a voltage signal but for the microphone itself, does it convert sound into current or resistance?

I want to understand how power supply ripple or noise affects the SNR of the microphone circuit.

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  • \$\begingroup\$ ti.com/lit/ug/tidu765/tidu765.pdf, section 2, Theory of Operation, under equation (1) '...the microphone capacitor acts as an ac-coupled voltage source.' Next page, see Figure 5, and right above equation (2) you read 'Op amp U1 acts as a transimpedance amplifier, and attempts to hold its inverting input at a constant voltage (VB) by varying its output.' The mic capacitor acts as an ac-coupled voltage source, JFET CS amp generates a current \$I_{ac}\$ (it flows through C3 in Fig 5). JFET CS amp requires an external power to operate. \$\endgroup\$
    – V.V.T
    Commented Nov 29, 2021 at 10:04
  • \$\begingroup\$ “does it convert sound into current or resistance” - I tend to say voltage/E-field. Once produced by changing H-fields/current in a speaker. \$\endgroup\$
    – RemyHx
    Commented Oct 5, 2022 at 20:07

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The JFET requires 2V in order to a) have energy for creating a lower output impedance b) be biased in an operating region where current changes more or less linearly depend on voltage changes on the JFET gate (by multiplying with its transmittance rating). The microphone capsule converts pressure changes into capacitance changes; and with a constant (electret foil provided) charge on that capacitor, this results into small charge changes on the JFET gate that, due to the gate capacitance, create gate voltage changes.

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