I am doing a project that involves using a landline phone to work as the speaker and microphone for Skype calls. A representation of the final design is given below.

Overview of System

I have a landline phone in which I will remove all the internal circuitry. In this phone I want to build and add my own circuit. Only the headset of the phone will be used.

The mic and speaker of the headset, connected to an RJ9 connector will be first converted to a 3.5 mm mono speaker and mic connection. This will be connected to the PC using a miniature USB sound card.

When the user wants to use skype, he will dial from the PC and use the headset to converse. This much I can do. I also want the output from Skype to be routed to an external speaker (built into the phone case) if the user presses the Push-to-Talk button, thus enabling him to listen on to the conversation without talking.

Here is where I am stuck.

  1. When the user puts the telephone headset down while on a call, I want the mic to be muted and the audio from the PC routed to the internal speaker in the telephone. To achieve this, I will have to switch the audio source from the PC from the headset to the external speaker. I am not able to decide how to accomplish this given that the amplification levels of the headset and speaker will have to be very different. Can anyone help me with selection of a set of IC's that will allow me to do the same.

  2. I want the mic to be turned off when the telephone headset is put down while on a call in skype.

I will be using a microcontroller board for managing the IO's (mostly an Arduino Nano).


2 Answers 2


You should be able to utilize the phones internal button used to activate the hang up action to disconnect the "sense" function of the USB audio device. ie; tell the computer that you have disconnected the headphones and it should send the audio through the original default device, the desktop speakers.

You would want to like you said adjust the volume on the phone so that the levels would stay even and comfortable, I would suggest installing a 10k logarithmic potentiometer as a voltage divider to simply cut down the volume going to the handset speaker.

If you are going to be using a button for push to talk you should not have a problem with putting the phone down, it will already be muted when you release the PTT.

There should not be any need for an Arduino here, maybe a couple op-amps if your mic level is to low.

Sorry I dont have more time to point you to some components.

  • \$\begingroup\$ The speaker shown in the diagram are not the PC speakers, that is a speaker in the phone unit. \$\endgroup\$ Commented Jun 1, 2016 at 17:07

Josh explained almost everything. For the amplification part, you could use the LM386.

It is fairly easy to use, requires very few parts and no symmetric power source. It can be noisy and lower the quality of the sound, but I don't think you aim to reach studio quality.The example schematics given in the datasheet are a good starting point.

To adjust the amplification level, just use a logarithmic potentiometer, as Josh advised it.

  • \$\begingroup\$ I can use the LM386 for amplification. But do I need to amplify signals from mic before sending to the pc? Also, how do I switch the output from the Pc to the telephone headset and the ext speaker \$\endgroup\$ Commented Jun 2, 2016 at 1:42
  • \$\begingroup\$ My main doubt is what level of amplification is ok for the speaker and the headset. I thought i could switch the speaker output from the pc, send it directly to the headset, and send it to the speaker with amplification. \$\endgroup\$ Commented Jun 2, 2016 at 1:44
  • \$\begingroup\$ Yes, this option is good and simple. I think that the amplification levels given by the sound card on your computer meets those required by the headset speaker and mic. Just plug directly those two to the jack I/O of your sound card. (those are only advices, you still want to try before putting into application) \$\endgroup\$
    – Albits
    Commented Jun 2, 2016 at 5:21

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.