HELP: internal electret microphone circuitry

I want to understand how an electret microphone works.
I need to understand if the schematic below is correct.
Possible values for resistors are RL = 2KOhm, RG = 1GOhm; is it correct?

Try to tell me if below I write correct things!
The resistor RG works for let current enter in the JFET gate, since the microphone (capacitor C) is only polarized but actually cannot give any current in DC.
As the value of C vary, the voltage at the gate of the JFET vary and the JFET converts this voltage variation in current variation on the resistor RL. Then, we read the voltage value on the resistor RL (thru pin1 and pin2) and bring the sound signal to output through some decoupling capacitor. Normally, the resistor RL is working at a certain current defined by the voltage source attached to pin1 and by a resistor (R1) that can be put between the pin1 and the voltage source (Vcc), and another resistor (R2) between the pin2 and ground. [Obviously I am referring to differential mode bias.] If any sound is present, then pin1-pin2 = 0 and it means silence. Otherwise you can read the sound value as pin1-pin2.
The sum R1+R2 must be as closer as possible to have the voltage on RL centered between the values of ground and the value Vcc of the voltage source, in order that the voltage signal can go higher or lower of the same quantity with respect to that center value.


Thank you!!

N.

No one can help me!?
Really I would appreciate...
Thx.

McShamrock said:

I want to understand how an electret microphone works.
I need to understand if the schematic below is correct.
Possible values for resistors are RL = 2KOhm, RG = 1GOhm; is it correct?

Try to tell me if below I write correct things!
The resistor RG works for let current enter in the JFET gate, since the microphone (capacitor C) is only polarized but actually cannot give any current in DC.
As the value of C vary, the voltage at the gate of the JFET vary and the JFET converts this voltage variation in current variation on the resistor RL. Then, we

Thank you!!

N.

Click to expand...

I think the value for RL could be correct, the value for RG seems a bit overkill.
Since the electret mike is shown as 'C', you possible need another cap (0.1uf or so) between the gate and RG.

I don't think that most microphone capsules have RL inside, the fet source is connected to gnd (the electret's case). RL is external and connected between the drain and Vcc.

There is a lot of information about the electret microphone, start at https://en.wikipedia.org/wiki/Electret_microphone and then try Google.

About RG, actually it is not needed since in the JFET gate the current can be also zero and only a voltage is required to make the gate voltage lower than source. In this case the voltage is indeed the mere microphone capsule (C) since it has a fixed Q (Q=Vc*C) and then a range of possible values of Vc. If it used a BJT instead of a JFET, then RG is needed since a current should be sent in the gate.


About RL, maybe it is not present. I only know that datasheets say a value for RL, then I have 2 options:
1) RL is inside the JFET and you must put a value of external bias resistor equal to RL, in order to have pin1-pin2 = Vbias/2. This can work fine, but the voltage between drain and source must be really small and negligible. In this way, the operating dc voltage is almost pin-pin2 = Vbias/2, and by connecting to pin1 and pin2 two decoupling capacitors I can extract the ac components, which are the really small variations of the drain-source voltage. This ac voltage must go to an amplifier since very small.
2)RL is not a physical resistor but a value given to put an external bias resistor equal to that in order to let the JFET working in the expected and corrent operating point. Then, from pin1-pin2 directly the drain-source voltage can be seen and the output is taken from there.

Actually, I think the more realistic is option 1) but can anyone confirm that?
I would really appreciate your help, please...


Thank you,
N.