7
\$\begingroup\$

How can I connect an arduino to a phone line such that, when a call rings, the arduino will 'answer' and be able to respond with (touch) tones and, if possible, 'hear' touch tones as well for 2-way communication? Is it possible to only answer on double rings?

I saw this page on a similar topic, but that does not describe answering with an arduino.

\$\endgroup\$
  • \$\begingroup\$ Take a look at this, it's from the comments in the link you provided and seems like it is able to answer calls. \$\endgroup\$ – Garrett Fogerlie Jan 6 '13 at 3:23
6
\$\begingroup\$

The page you link to references the Cermetek CH1817-LM Direct Access Arrangement Module. It provides a simple interrupt pin for Ring Detection, while handling some of the more complex POTS interfacing. They just didn't implement it.

The rest is simple. You just need a DTMF decoder and encoder.

Actually, the page gives you everything you need, just need to adjust the code for your own use.

Alternatively, you could use some other chips. TI's TCM1520A RING DETECTOR for ring detection. Or take apart a Answering Machine or older isa/pci 56k modems. The older, the bigger the ICs and better documented.

\$\endgroup\$
  • 1
    \$\begingroup\$ Why take apart an old modem? Why not just interface directly with it? Google the AT command set, modems can do a lot more than download pornography frustratingly slowly. \$\endgroup\$ – John U Jul 15 '14 at 9:02
0
\$\begingroup\$

There's an Arduino shield called MICO that can help you with this project. This shield connects to an Arduino and to a cell phone via the audio jack. The shield detects the phone is ringing and can answer the call.

Upon answering the call it presents the user with a voice prompt (audio files are stored in the onboard microSD card). Check it out at HelloMICO.com

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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