If I understand correctly modern microphones integrated with headphones with 4-contact 3.5mm jacks for use with cell phones are condenser microphones, and usually contain a small, integrated amplifier - consisting of at least one FET and a few passives for bias isolation.

These are very low current, but require a small DC bias available from the microphone input to operate, and usually a resistor divider there requires and draws at least a small current.

Question: Is there a standard, or generally and widely accepted minimum current that a designer can expect any modern phone to make available at the 3.5mm 4-contact plug - or voltage plus maximum internal series resistance (within the phone)? I'm guessing it's going to be somewhere in the ballpark of 1 to 100 micro-amperes and between 1 and 2 volts, but I don't actually know this for sure.

Here is just a random example, the TPA6166A2 3.5-mm Jack Detect and Headset Interface IC was the first I found in a search, and it seems to be rated to supply a maximum of 1.2 mA when it detects a need for it and the internal bias resister is set to bypass. However it is no evidence by itself of a minimum standard.

enter image description here

enter image description here

There are plenty of examples, and other questions within this stackexchange, but so far I haven't found the actual published, or industry accepted minimum current, or minimum voltage plus maximum internal(phone) series resistance.

Hypothetical use example: This is not what I'm doing but it's a good working example for the question. "pico-LEDs" - small surface mount LEDs attain useable brightness at only a few milliamperes. A power harvesting-like circuit could collect enough charge from say 50 microamperes over one minute to "blink" the LED for 100 milliseconds.

enter image description here

above: screenshot from an example picoLED-eco surface mount LED.

Comments on personal experience are welcome, but for an answer I need a link to a real, relatively reliable specification from which design something with the reasonable expectation that it should work with any modern cellphone with the 3.5mm 4-contact plug and draw a small but useful amount of power from it (microwatts).


3 Answers 3


It's not a universally applicable standard, but Android publishes specifications for headset plug and jack.

These documents (as of June 14, 2017) specify that the jack's mic bias voltage shall be in the range “1.8V - 2.9V”, and that the microphone shall have a DC resistance of “1000 ohms or higher”.

While it is not stated outright, from assuming that the two devices are compatible, we can thus conclude that if presented with a load of 1 kΩ, the jack must apply at least 1.8 V across it, resulting in a current of at least

$$I = \frac VR = \frac{1.8\,\mathrm{V}}{1\,\mathrm{k\Omega}} = 1.8\,\mathrm{mA}$$

and at most 2.9 mA.

Note that to follow this specification you must design your device's power supply so that it presents an input impedance of at least 1 kΩ. In particular, if it is less than 100 Ω then it may be considered a short and detected as a 3-pole headphone plug with no mic.

  • \$\begingroup\$ Bingo! Thank you very much for this. I thought there had to be something, somewhere. \$\endgroup\$
    – uhoh
    Jun 14, 2017 at 16:34

(Not enough reputation to comment on @Kevin Reid's answer so I have to do so in another answer.) The Android docs are a useful source of info regarding this question. I interpret them differently though: I think the 1.8 V minimum bias voltage in the spec is an open-circuit value. It's hard to be certain of course, because no output impedance for this voltage is stated.

However, the text also talks of a "2.2V mic bias applied through 2.2 kOhm resistor" which one might infer to be the reference method of providing the bias supply, and would allow for no more than 1 mA to flow (into a dead short). Given that mics which need bias tend to draw tiny currents, this would make sense.

Finally, the TI chip referenced in the question defaults to 2.0 V for the mic bias voltage and 2.2 kOhm for the mic bias resistance. This would give just over 0.9 mA into a dead short.

  • \$\begingroup\$ +1 This seems like a robust stand-alone answer, nonetheless we'll see what we can do about raising your reputation, Thanks! \$\endgroup\$
    – uhoh
    May 21, 2020 at 21:57
  • 1
    \$\begingroup\$ Much appreciated @uhoh ;-) \$\endgroup\$
    – Neilski
    May 21, 2020 at 22:11

I doubt mobile phone companies use an outside standard if they can use an improved mic. They would have internal standards which are proprietary.

Then they have to choose compatible external devices.

To look at the most sensitive devices, consider new MEMs microphones

Impedance 4.5 kOhm
Voltage - Rated -
Voltage Range 0.9 ~ 1.3V
Current - Supply 17µA

This might be your worst case for the near future.

  • \$\begingroup\$ Thanks for that. So the idea is that an Xphone made by company X might possibly only supply enough current for the Xmic sold by company X, and the marketing department wouldn't care if customers go home and then find it won't work with any of their current mics? Usually these things have standards, but possibly not in this case. Usually any socket in a commercial product that is expected to supply power will have a standard for the minimum power - agreed-upon standardizations usually make vendors more comfortable and more successful - there are exceptions of course. \$\endgroup\$
    – uhoh
    Aug 29, 2016 at 1:49
  • \$\begingroup\$ I looked at some of those specs, found that TIA stands for the Telecommunications Industry Association. From there, I found that TIA-810-B or TIA-920.110-A might contain what I'm looking for, but are paywalled. Update: I found this chip supplies a mic bias of 1.2 mA! http://www.ti.com/lit/ds/symlink/tpa6166a2.pdf \$\endgroup\$
    – uhoh
    Aug 29, 2016 at 2:07
  • 1
    \$\begingroup\$ TIA apply to many quality standards for communication, not how to bias the mic. \$\endgroup\$ Aug 29, 2016 at 2:23
  • 1
    \$\begingroup\$ @uhoh It would probably be best to not accept an answer quite so immediately since the question may now receive more attention and you seem to have found a possibility of rather more current being available. \$\endgroup\$ Aug 29, 2016 at 2:31
  • \$\begingroup\$ @user2943160 I understand - I found that soon after I accepted. (at)Tony I am going to reverse the accept - I appreciate your help, you got me to dig deeper! \$\endgroup\$
    – uhoh
    Aug 29, 2016 at 2:38

Your Answer

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

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