my colleague and me have hit a brick wall - and are hoping somebody can help us out here :)

Following scenario: We record some data using a computer with an NI card and some signal transducers. We need to record audio from a mic of an headset synchronously. First thought: Build a patch cabel for the headset. Measuring the mic signal to GND - no problem. However, the signal is offset by about 2.65V. My guess this is the necessary voltage for an electret microfone.

To get rid of the offset we then thought to measure the voltage between supply and mic signal. However the difference is 0 - a look with an oscilloscope confirmed that. Can anybody help explain, why supply as well as mic signal on a headset carry the signal information?


1 Answer 1


An electret microphone on a headset is typically wired up thus:

Electret Mic

Measuring the signal between V+ and Output (or Ground and Output) should give the sound waveform from the microphone. The resistor from V+ forms a voltage divider with the on resistance of the MOSFET within the microphone capsule, with the current through the MOSFET, and thereby the voltage at the Output, varying with sound.

If measuring between the supply line before the bias resistor, and the Output in the above schematic, shows no waveform, the probes are probably loading the microphone too much. Electret microphones have very high output impedance, so a lower impedance probe would attenuate the signal to nothing.

Also, the voltages involved are small, in millivolts, so it would be useful to set the oscilloscope to a more sensitive range.

  • \$\begingroup\$ Thanks for the explanation! We measured the voltage through a patch cable, thus after the resistor. However we cannot see differences. I'll try again with higher sensitivity and see what that gives \$\endgroup\$
    – JoeyD
    Jun 20, 2013 at 6:22
  • \$\begingroup\$ So I checked the signals, see the image of the oscilloscope. Yellow and Blue are "output" and "V+", Red is "output" - "V+"... ![bla]<img src="i.imgur.com/ITRttAt.png" title="Hosted by imgur.com" /> \$\endgroup\$
    – JoeyD
    Jun 20, 2013 at 7:43
  • \$\begingroup\$ So blue seems to be the way to go. \$\endgroup\$ Jun 20, 2013 at 10:25
  • \$\begingroup\$ Okay, so then I gather there is no simple way to get the signal offset free? \$\endgroup\$
    – JoeyD
    Jun 20, 2013 at 11:00
  • 1
    \$\begingroup\$ @JoeyD Please see the capacitor in the schematic in the answer. The "Output" connection is offset-free, as the DC is blocked by the capacitor. \$\endgroup\$ Jun 20, 2013 at 11:24

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