Can I assume that the signal from an electret microphone capsule is inverted due to common source configuration of jfet? enter image description here

I would like to have non-inverted signal on my sound's card line input. Then I pick up inverting preamplifier project. What about microphone input. Does it have inverting op-amp?

  • \$\begingroup\$ Why do you care whether or not the signal is inverted? It would only matter if you were trying to match another signal source. \$\endgroup\$
    – Transistor
    Sep 11, 2021 at 20:20
  • 1
    \$\begingroup\$ Polarity matters. See below. (Experience: preamp for karaoke gear). \$\endgroup\$ Sep 11, 2021 at 20:32
  • \$\begingroup\$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. \$\endgroup\$
    – Community Bot
    Sep 11, 2021 at 20:40
  • \$\begingroup\$ The FET is self-biased, and the sense capacitor has an electret (permanent charge). So, the question is unanswerable until you tell us the permanent charge sign, since the capacitance change (but not polarity) is pressure-proportional. At a guess, the sign of the output voltage is the 'standard' polarity by design. \$\endgroup\$
    – Whit3rd
    Sep 11, 2021 at 23:38
  • \$\begingroup\$ @Whit3rd the electret charge polarity doesn't matter, as the electret assembly is two capacitors in series which would model as (cap - voltage - cap.) Meanwhile, the overall change in capacitance follows the sound pressure, and the voltage will be inverted. See below. \$\endgroup\$ Sep 13, 2021 at 19:19

4 Answers 4


As others have stated, in general, in many situations not only does not matter as absolute phase is impossible to guarantee with audio over more than a short distance.

However, if you you wish to combine the output with other microphones then it is necessary to ensure that they all have the same polarity signal with a pressure wavefront.

This would be used for example in beam-forming or to locate a source of sound. Most cell-phones use multiple microphones to aid in noise cancellation to reject background noise.

Especially at low frequencies multiple microphones can be coherent over several feet.

Although the FET amplifying stage does invert the signal it also depends upon how the elecctret microphone capsule is designed and connected. If the capsule gives out a negative going signal for a positive pressure change the output from the combination will be positive going.

The information you need might be in the datasheet.

  • \$\begingroup\$ Sorry, have to disagree. Polarity absolutely matters, even for ordinary sound. \$\endgroup\$ Sep 11, 2021 at 20:41
  • 2
    \$\begingroup\$ @hacktastical - I did not disagree with that, however, it is only in select scenarios where any difference can be perceived. \$\endgroup\$ Sep 11, 2021 at 20:43
  • \$\begingroup\$ No, pretty much any scenario will reveal a mic polarity issue. The scenarios where it doesn’t matter? I really can’t think of any. Fact is, all mics follow this convention, from the lowliest MEMS type to a telefunken U47. \$\endgroup\$ Sep 11, 2021 at 20:46
  • 3
    \$\begingroup\$ @hacktastical - where there is only a single mic it will be extremely difficult to tell the polarity. If you combine it with others (as most cell-phones do) then they need to be coherent. \$\endgroup\$ Sep 11, 2021 at 20:59
  • \$\begingroup\$ Thank for all your answear. I was wrongly convinced that absolute phase is important while recording on mono microphone. I found out, I have to bear this in mind while recording in stereo and some situations as @KevinWhite and hacktastical said. \$\endgroup\$
    – Dom.in
    Sep 11, 2021 at 21:08

I would like to have non-inverted signal on my sound's card line input.

You are asking for an impossibility. The wavelength at 1 kHz is 340 mm hence, if your microphone is 170 mm from your speaker and the sound is 1 kHz, the signal received will be inverted. In fact it will be inverted every repeated 340 mm from the initial half wavelength of 170 mm. It will be in-phase every 340 mm from the speaker. That's 1 kHz and, for a complex audio signal the shape of the microphone signal will not at all look like the original signal fed to the loudspeaker.

You appear to be on a quest of disappointment.

I would like to have non-inverted signal on my sound's card line input.

If you are using a microphone to pick up this signal then.... it's not going to happen. Be realistic and think about it.

Then of course you have to remember that as the air warms the wavelength changes. Wavelength calculator. Speed of sound changes with air pressure too so, wavelength will also change with weather conditions.

  • 1
    \$\begingroup\$ Typo: The wavelength at 1 kHz is 340 mm. \$\endgroup\$
    – Transistor
    Sep 11, 2021 at 20:41
  • \$\begingroup\$ I got your point. Anyway I was thinking about t0 in function amplitude=f(t). \$\endgroup\$
    – Dom.in
    Sep 11, 2021 at 20:42
  • \$\begingroup\$ Technically isnt this only true for a constant tone starting at minus infinity? There is a physical difference that is especially apparent for impulses (not necessarily detectable by humans), where the first part of the change in frequency will be either positive or negative regardless of the distance between the mic and the speaker (hence the interesting conversation in the comments above) \$\endgroup\$
    – BeB00
    Sep 12, 2021 at 0:56
  • \$\begingroup\$ @BeB00 who are you talking to? \$\endgroup\$
    – Andy aka
    Sep 12, 2021 at 9:10

It inverts.

Applying voltage (properly) to a JFET game makes it conduct less unlike most other devices available. With no gate-source voltage it conducts maximally. On an N-channel JFET, the gate should be more negative than the source. That means the gate is biased negative.

So with a positive signal relative to the bias, the gate-source voltage gets less negative and the JFET conducts more. That causes more current to flow through the resistor which causes a larger voltage drop at the output away from +V which makes the output more negative.

With a negative signal relative to the bias, the gate source voltage gets more negative and the JFET conduct less. That causes less current to flow through the resistor which causes voltage drop at the output that is more towards +V which makes the output more positive.

  • \$\begingroup\$ The standing bias in no way affects the fact that the stage inverts. A positive-going signal from the microphone capsule results in a negative-going signal at the output. For these applications, the no-signal situation does not have the most negative output and the small signal from the microphone can go positive or negative. When it goes positive the output will go more negative (ie closer to ground). \$\endgroup\$ Sep 11, 2021 at 20:37
  • \$\begingroup\$ @KevinWhite I wasn't referring to the bias. My phrasing could have been better though than "no signal". But I don't see how an input signal that is positive relative the bias results in a signal that is negative on the output since it should cause the JFET to conduct less which should make the output voltage more positive. \$\endgroup\$
    – DKNguyen
    Sep 11, 2021 at 20:43
  • \$\begingroup\$ @KevinWhite Oh wait, I forgot the gate on JFETs has to be reverse biased. Corrected. Thanks for pointing that out. \$\endgroup\$
    – DKNguyen
    Sep 11, 2021 at 20:54
  • \$\begingroup\$ the gate wil be slighty positively biased when the input goes positive but not enough to cause current to flow into the gate, there may in fact be a source resistor that is not shown that would add additional reverse bias.. \$\endgroup\$ Sep 11, 2021 at 20:57
  • 1
    \$\begingroup\$ @DKNguyen - No. When the gate is moved positive, even it is already at zero volts the drain current will increase. There is no discontinuity at zero volts. \$\endgroup\$ Sep 12, 2021 at 1:59

Microphones, if configured correctly, register a positive pressure wave as a positive voltage (transducer moves in.) Likewise, a positive speaker drive pushes a positive wave toward the audience (voice coil moves out.) More about that here: https://www.sweetwater.com/insync/polarity-does-it-really-matter/

With that out of the way, you’re good to go with the circuit as-is. Positive pressure wave is positive output. The details of whether the JFET inverts or not aren’t important (it does, kinda: the varying capacitance of the electret cell causes displacement current at the JFET gate.)

Note also that your sound card probably has a mic input that injects a bias (called phantom power) to power the electret capsule, so you only need two wires to the capsule. The load resistor and DC blocking cap that inject and isolate phantom power are already on the sound card. All the other discussion about sound polarity aside, you do need to ensure the correct polarity for phantom power, otherwise the mic won’t work at all.

So, what actually happens at the JFET gate?

The electret structure consists of a permanently-charged capacitor in series with a variable capacitor. The moving part can be either the electret or the opposing plate. Works the same either way: the assembly has a fixed charge \$Q = CV\$. It would be modeled as a fixed voltage source (the pre-charged electret) in series with the variable capacitor.

The capsule JFET, with nothing connected to the gate, will self-bias with the gate close to ground via the miniscule gate leakage. This puts the JFET in the conducting region, as the pinch-off (threshold) for a depletion-mode n-JFET is below the source / drain voltage (pinch-off will be about -1V for the popular 2SK170 device), but still below the forward diode conduction threshold of 700mV or so.

With the electret connected between JFET source and gate, the electret assembly will also bias to about 0V across it via JFET gate leakage. It would be the same as just connecting a capacitor from drain to gate.

Now, here's the magic. When a positive pressure sound wave hits the diaphragm, it will push the electret assembly's opposing capacitor plates together. You may recall that the closer the plates are to each other, the higher the capacitance. This can be expressed more formally as follows:

  • \$ Q = CV = {\epsilon A \over d}V \$

where \$C\$ is the capacitance, \$V\$ the bias voltage, \$A\$ the plate area, \$\epsilon\$ is the dielectric constant, and - most relevant here - \$d\$ the separation of the plates.

What happens? Plates closer together, capacitance gets bigger, voltage decreases to maintain the constant charge. How?

See that equation above? The distance \$d\$ is in the denominator. Smaller \$d\$, bigger capacitance. Meanwhile, the electret's two terminals connected between the JFET source and gate are essentially open circuit. To maintain a constant charge Q across both the charged electret and the varying C, the voltage must change. So, the V becomes smaller.

The JFET gate-source voltage will see its voltage decrease with a positive-pressure impulse. Likewise, a negative pressure will pull the plates apart, and you will see an increase in gate voltage.

If you were to probe the gate with a special high-impedance probe you would see an inverted waveform vs. the instantaneous sound pressure impinging on the diaphragm. The JFET will invert this, producing a waveform that follows the sound pulse: positive pressure is more positive, negative pressure is less positive.

So the n-JFET buffered electret mic polarity is same as a dynamic mic: diaphragm moving in, positive-going voltage.

  • \$\begingroup\$ But of course an omnidirectional microphone doesn’t know where the audience is, it just measures changes in pressure at the diaphragm. \$\endgroup\$
    – Frog
    Sep 11, 2021 at 20:33
  • \$\begingroup\$ Regardless, positive pressure is positive voltage. \$\endgroup\$ Sep 11, 2021 at 20:36
  • 2
    \$\begingroup\$ Yes. However this has little meaning in a real-world situation as Andy aka has pointed out, since the phase will change with distance from the source, reversing every few cm depending on frequency. \$\endgroup\$
    – Frog
    Sep 11, 2021 at 20:42
  • \$\begingroup\$ also your last paragraph is wrong: more positive voltage on the gate of an N-JFET turns it more on ( like all N type transistors) \$\endgroup\$
    – tobalt
    Sep 11, 2021 at 20:49
  • \$\begingroup\$ @tobalt, you’re right, so I set that aside for now. It’s not really relevant. No one has really explained exactly what the transducer does (its capacitance varies, so displacement currents vary too). I’ll come back to that later. \$\endgroup\$ Sep 12, 2021 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.