모뎀과 asterisk

asterisk로 pc에 pbx를 구현하려고 하다가 fxo 카드 대신에 굴러다니는 모뎀을 사용할 수는 없을까 하여 조사해보았다. 역시 내가 하는 생각은 이미 다른 사람들도 많이 하는 생각들;;. 해외 웹에는 관련 질문들이 많았다. 그러나 대부분의 thread에서는 이런 질문에 대해 NO라고 하더라..-_-;;

하지만 digium에서 파는 fxo카드인 x100p가 근본적으로는 intel V92 modem이라고 한다. 몇가지 수정만 하면 다른 v92 full-duplex 모뎀들도 사용 가능할 듯 한데....

 

One important point when you use an Intel V92 modem card that has not been purchased as the X100P card from Digium, but otherwise look identical to it, is that the electronic vendorID read from the card will be different. There are two ways to get around this: 1. If you are compiling Asterisk from source, edit the zaptel/wcfxo.c file as follows: Existing code: static struct pci_device_id wcfxo_pci_tbl[] __devinitdata = { { 0xe159, 0x0001, 0x8085, PCI_ANY_ID, 0, 0, (unsigned long) &wcx101p }, { 0x1057, 0x5608, PCI_ANY_ID, PCI_ANY_ID, 0, 0, (unsigned long) &wcx100$ }; Now change the wcfxo_pci_tbl[] in zaptel/wsfxo.c to: static struct pci_device_id wcfxo_pci_tbl[] __devinitdata = { { 0xe159, 0x0001, 0x8085, PCI_ANY_ID, 0, 0, (unsigned long) &wcx101p }, { 0xe159, 0x0001, 0x8086, PCI_ANY_ID, 0, 0, (unsigned long) &wcx101p }, { 0x1057, 0x5608, PCI_ANY_ID, PCI_ANY_ID, 0, 0, (unsigned long) &wcx100$ }; The added line in the middle will allow the wcfxo driver to work with off the shelf Intel v92 Winmodem cards as well as with the Digium X100P cards. 2. The other alternative is to make the same modification to the card that Digium did when they turned them into "genuine" X100P cards. Simply carefully remove R13 and R19 with a soldering iron. R13 & R19 are pull down resistors that affects the vendorID number that is read from the card. With these two resistors gone, you now have a card that will appear as a genuine Digium X100P card to the Asterisk software. Cards purchased from DigitNetworks will already have been modified this way, but then they charge quite a bit more than the US $5 - $15 that one can pick up these Intel Winmodem PCI cards for elsewhere. It is considerably cheaper when you modify the wcfxo driver source code, and/or remove the 2 resistors yourself.