I have an issue with my headset. I can’t stand not being able to hear myself speak. I’d like to build a passive device to loopback the mic to the headphones that would be plugged in-line between the headset and device, with a potentiometer to adjust the mic volume to headphones.

Would it be as simple as this?


Are additional fixed resistors required? I assume it would be a good idea to put something in series with the pot for when it is turned up fully, but I’m not sure what size..100ohm?

The headset speaker impedance is 32ohms each and the microphone impedance is 2200 ohms. I’m not sure of the impedances of the device, it can be assumed to be a typical laptop computer or smartphone.

As for power ratings, I assume the typical voltage at max volume would be less than 1V for the headphone jack on these devices. So, would a standard 0.06W dual gang logarithmic pot would be adequate for this application?

Appreciate the help on this.


Thanks for everyone’s feedback. I’ve added an amplifier with variable gain for the mic input to the best of my understanding, but I’m not sure about splitting/mixing the signal. Will I be able to do this passively without any more amplifiers? Would anyone suggest different values for resistors/capacitors? Any other errors?



EDIT 2: New question

I’m looking to create a circuit to add sidetone to a headset. Looking to add it to both right and left channels, with a volume adjustment. The circuit would be connected in-line between the headset and the device to which it is connected (laptop, smartphone, etc). The device would be powered by 2AA batteries.

The speakers on the headset are 32ohm and the microphone is 2200 ohm.



Would this even work?

Could the number of amplifiers be reduced?

Could the number of pots be reduced? Note that there would probably be 3 off 2-gang pots, as there is no need for balancing between right and left. There is also no need for master volume control, but I’m worried I won’t be able to select the proper resistor value.

Are the resistors/pots/capacitors sizes critical? If so, any suggestions?

Is it even feasible to run this off 2AA batteries?

Is there any risk of damaging the headset/device?

Amplifier data sheet for reference: https://www.njr.com/semicon/PDF/NJM2113_E.pdf


Since the amount of gain on preamp is unknown, would it make sense to add a potentiometer? If so, what min max values would you suggest?

Regarding your final statement on ground wiring, are you suggesting arranging the ground wires in a manner such as the below diagram:



When the circuit is powered (3V on) the mic output to headset is loud and clear, but the mic output to the device is very muffled and quiet. When the circuit is not powered (3v off) the output to the device returns to normal (and nothing can be heard in headset as the speaker amps have no power, of course). Connected to oscilloscope to demonstrate. Mic is picking up a loud tone from a tone generator. Cursor shows where circuit is switched OFF.

Yellow - M+

Green - Preamp Term 4

Red - Preamp Term 5

Blue - Input of Volume Pot


When switched off, preamp output goes to zero, and M+ signal goes to normal.

Any suggestions for retaining signal quality at M+ when circuit is powered ON?

To answer previous question on mic voltages (headset connected directly to device):

Vrms = 1.54V

Vp-p quiet = 44-56 mV

Vp-p talking = ~80 mV


Looking to clarify circuit:


Note I am using 100k POT for R1

  • 2
    \$\begingroup\$ I believe what you are suggesting will turn you mic into a speaker. \$\endgroup\$
    – HandyHowie
    Apr 8, 2019 at 12:43
  • 5
    \$\begingroup\$ The signal from your mic is too small to directly drive the headset speakers. You'll need to add an amplifier or do the loopback function in software (assuming there isn't too much latency). \$\endgroup\$
    – brhans
    Apr 8, 2019 at 12:47
  • 1
    \$\begingroup\$ Most likely whatever audio the computer is playing is fed back to computer mic input, and by the look of impedances involved, it will be louder than you talking to the mic. \$\endgroup\$
    – Justme
    Apr 8, 2019 at 12:56
  • 3
    \$\begingroup\$ The technical term is "sidetone". If you search for this term, you'll find all kinds of ways to achieve your goal -- some not requiring any hardware at all! \$\endgroup\$
    – Dave Tweed
    Apr 8, 2019 at 13:09
  • 1
    \$\begingroup\$ What are your headphones plugged into? A PC? \$\endgroup\$
    – Ian Bland
    May 13, 2019 at 1:42

2 Answers 2


Obviously you cannot feed the mic signal back with the available settings in "the device", so you need a mixer - one which accepts your mic signal as one mono input and the headphone signals as one stereo input. The mixer should be able to supply your mic with the common 1...3VDC, if the mic is of the usual electret type.

The stereo input should be able to handle headphone signal level, which is pretty much more than normal audio device line signals.

The mixer should be able to produce the needed output power for your headphones and split the mic signal to where it's actually used.

I bet someone has already invented this device and it's for sale somewhere. Unfortunately I do not know, where

If I needed it, I would build it - one battery, one dual audio amp IC, one potentiometer, few caps and few resistors and proper case+connectors, maybe also a battery switch if it cannot be a part of a connector.

ADD due questioner's new attempt:

Split = branch a wire as you have done, only be sure that the possible mic supply voltage DC from "the device" doesn't get into your own circuit. 1uF cap in your circuit does the separation, no need to add parts. Remove 10kOhm resistor between the mic wire and your own battery, the mic gets its DC from "the device". Find how to change your amp to be a summing amp which gets another signal from the headphone output of "the device". Let your headphones get signal only from your amp. Now you have very inefficient passive mixer, it's well possible you get too low volume to the headphones.

You need 2 amps if you want listen as stereo. If you accept mono listening, one amp is enough.

Your amp IC of choice NJM2113 is designed to make a bridged speaker amp. That cannot be utilized because from your own schematic I can see you seem to have only single ended (=L,R,GND) headphone wiring. But the output power can still be enough because you get over 1Vpp.

NJM2113 has low maximum voltage gain. I inserted a mic preamp. It's here:

enter image description here

Hopefully NJM2113 doesn't generate too much noise and stays stable with low current loads and the other bridge output as unloaded. How much gain you really need, that can only be guessed. Increasing R1 increase gain, decreasing R1 makes gain smaller. C1 should be reduced if there's still too much bass umble. If you want more bass, increase C1. I have assumed your mic system (=headset + "the device" together) look outwards as 2kOhm signal source and you get 5mV signal.

The mixer (only one amp is drawn, the +3V supply is common to all amps):

enter image description here

The mixer outputs headphone speaker voltage without amplification. Mic preamp output voltage is amplified to 2500%. At the output you see the common zobel circuit to ensure some resistive loading. Inductive headphones alone would easily make the circuit unstable. There's 100uF coupling caps for the headphones. Anything less will kill the bass. The battery has 100uF bypassing. I hope it's enough to prevent signal feedback via battery lines.

Hopefully you can make the wiring so that output's current doesn't go through the GND wires which feed input signal to the mic preamp IC. If not, do not expect stable function.

EDIT: This circuit has a trivial error. The preamp input is in current mode, it pulls the mic output signal to knees and the original audio device gets nothing altough in the headphones the signal is loud. It's fixed in the discussion. The idea is to change the preamp circuit to non-inverting which has high Z input. The questioner drew the fixed image for his Edit 5: https://i.stack.imgur.com/DZl1x.png

  • 1
    \$\begingroup\$ @Pat Edit2,3,4 ....with better circuits are not clutter. Creating new question when the old was still incompletely solved is clutter. I know how to reduce the amps to 2. You get it as soon as you have removed the new branch and inserted the improved, but too complex version here. If someone other happens to answer something more useful than me, you can move the acceptance to him without obstacles or ask the moderator to do it. \$\endgroup\$
    – user136077
    Apr 8, 2019 at 18:37
  • 1
    \$\begingroup\$ @Pat NJM2113 isn't a best device for this purpose. It's bridged speaker amp with unnecessary complexity and quite low max. voltage gain. Do you already have a bunch of them (= they must be used, if possible)? \$\endgroup\$
    – user136077
    Apr 8, 2019 at 19:05
  • 1
    \$\begingroup\$ Ok. Key question: Do you know how many volts you can get out of your mic as peak to peak voltage? NJM2113 amplifies voltage maximally about 100..140 times. You need a mic preamp if you haven't at least 20mVpp output from the mic, which is much. I guess without screaming you get only few millivolts. \$\endgroup\$
    – user136077
    Apr 8, 2019 at 19:57
  • 1
    \$\begingroup\$ Check which causes the noise. Is it born in the mic or preamp or does it come fron "the device"? Disconnect C1 from the mic line and connect instead a 2,7 kOhm resistor between the free end of C1 and GND. Then remove the new resistor, insert the device only and finally the mic (it needs DC power from the device, I guess, but confirm) \$\endgroup\$
    – user136077
    Apr 23, 2019 at 6:46
  • 1
    \$\begingroup\$ @Pat It's my error. The preamp input works in low-impedance (=current) mode, It loads the mic signal to low voltage. You may hear it in your headphones, but the "device" gets much less than before. The simplest solution is to add a couple of kOhm resistor in series with preamp's C1. I'll try to find a better way to use the given IC. \$\endgroup\$
    – user136077
    May 13, 2019 at 0:55

The headset speaker impedance is 32ohms each and the microphone impedance is 2200 ohms.

Most of your answer is in that statement. The mic and speakers will act as a potential divider of \$ \frac {32}{2200+32} \$ giving about 1.5% of the mic output (which was already a small signal) to the 32 Ω speakers. It won't work.

What you are seaking is called "sidetone" and was used in ordinary old-fashioned telephones so you might gain some insight by searching for telephone sidetone.

  • \$\begingroup\$ Yes, that’s it! Thanks. Found some helpful info but not quite a working solution. I updated the original question, appreciate any new feedback you might have. \$\endgroup\$
    – Pat
    Apr 8, 2019 at 15:58

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