Let's look at the datasheet of a typical electret condenser: https://www.mouser.com/datasheet/2/334/AOM-5024L-HD-R-1219369.pdf


What exactly do the resistor and capacitor do and how does one determine the capacitor value? Looking at the datasheet it looks like the max voltage of the capsule is 9V so adding a 2kOhm resistor limits the current to ~0.4 mA, which is the max current rating of the capsule. So, the resistor is limiting the current to the capsule. I don't understand why we have a capacitor and the physics of the capacitor in this design. Is my interpretation of the resistor correct?

Additionally, suppose we wire this for a 3.5mm TRS or TRRS jack. Presumably, the tip and ring on a TRS jack both provide a small positive voltage. How does the device separate out the voltage and the output signal from the mic and how is this wired? For a TRRS configuration assuming CTIA standard, I'm guessing the sleeve provides this voltage and also is the output?

Edit: From https://www.epanorama.net/circuits/microphone_powering.html

    +---------------------------- battery +ve (3 to 12 Volts)
   2k2 R1
    o---------- 10uF ------o----- output
    |+                     |
 CAPSULE                  10k R2
    |-                     |
    +----------------------o----- GND, and battery -ve

This is the basic electret microphone powering circuit which you can use as generic reference when receivign circuits which use electret microphones. The putput impedance is determined by R1 and R2. If you leave out R2 the output impedance is roughly the resistance of R2.

Finally, this 10kOhm resistor connected between the output and the ground, is confusing me, as well as the statement in quotes. What does the 10k resistor do and how does not including this leave the output impedance roughly half the resistance of R2 (now left out of the circuit)?

  • 2
    \$\begingroup\$ It is called an Electret Condenser Microphone, not Electric Condenser. An old condenser mic needs a 48V polarizing voltage and an impedance converter circuit but an electret mic has the 48V built into its electret material and it has a Jfet that is powered with the external resistor. The value of the output capacitor is calculated using the same formula used with any other audio coupling capacitor, 1 divided by (pi x lowest frequency x the load resistance). \$\endgroup\$
    – Audioguru
    Jan 23 at 1:09
  • \$\begingroup\$ There is a special he'll for those who draw stuff like this. It would be much more clear if the bias resistor would be vertical \$\endgroup\$
    – user76844
    Aug 13 at 11:11

The resistor and microphone unit together form a voltage divider. Yes, the resistor limits the current, but importantly for this application, it also drops a voltage proportional to the current that the microphone unit allows to pass through it. The capacitor blocks the DC that is supplied to the microphone unit. Only AC passes through the capacitor to the output.

Edit: The impedance of 2.2k referred to in the datasheet is the small signal impedance. It is the ratio of how much the voltage would change if there were to be a small change in the current, when the reason for that change in current is external to the microphone unit. That is,

$$Z = \frac{dv}{di}$$

It is NOT the voltage drop across the microphone unit divided by the current.

$$Z \ne \frac{v}{i}$$

At least not necessarily.

Thus, your guess that

so presumably, the voltage divider (using a 2.2kOhm pull up) is dropping the voltage in half at this frequency?

is mistaken.

The instantaneous voltage drop across the microphone unit depends upon the sound signal it is transducing. The average of these instantaneous voltages may or may not be \$Z \cdot I\$

The resistor R2 may be there to provide some impedance matching between the microphone unit and the amplifier. As you correctly surmise, the 10k resistor and 10uF capacitor form a high pass filter, with cutoff below the audible range. As implemented, that filter will completely block DC. But that can be achieved simply with a capacitor. So, the resistor is likely for impedance matching, as I mentioned.

  • \$\begingroup\$ Thanks that makes sense. Do we know what the impedance of the microphone unit is? The datasheet specifies that at 1kHz the impedance is 2.2kOhm, so presumably, the voltage divider (using a 2.2kOhm pull up) is dropping the voltage in half at this frequency? \$\endgroup\$
    – mu7z
    Jan 22 at 4:43
  • \$\begingroup\$ In addition, I modified my question to add details regarding a 10k resistor often seen. Clarification on that would be helpful. \$\endgroup\$
    – mu7z
    Jan 22 at 5:02
  • \$\begingroup\$ @mu7z: I think that the 10K resistor is there to ensure that the audio output will have zero DC bias. \$\endgroup\$ Jan 23 at 0:19
  • \$\begingroup\$ As far as I can see the 10k resistor and 10uF capacitor form a highpass at ~1Hz. I'm not sure how zero DC bias comes into the picture here? \$\endgroup\$
    – mu7z
    Jan 23 at 2:19
  • \$\begingroup\$ @mu7z Disregarding leakage conductance, DC is blocked by capacitors. \$\endgroup\$ Jan 24 at 4:50

The datasheet calls it a "GSM buzz-blocking capacitors"

Normally one is sufficient, but they must have found 2 parts work better in their layout.

Microwave has a nasty wave of inducing rectified carrier into audio systems. Cell phone pings and GSM are notorious for this. So this simply shunts these little signals above 10 MHz so the FET does not rectify it.

enter image description here

I plotted the FET drain R and shunt caps here. The intersection is where the signal "rolls off".(attenuates)


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