Keelan Lightfoot says:
The modems need to see a little line voltage (off hook voltage) to work correctly with out telephone service. This is usually from 3 to 9v. To achieve this, connect power supply that provides a clean DC output, or better yet, battery in parallel with the phone line between the two modems. The north american standard is a red wire (ring/B) for -, and a green wire (tip/A) for +.| < Connect to negative terminal of PS/Battery ---------+ | +-------------+ modem A |Red -------+-----| modem B | |Green -------+---| | ---------+ | +-------------+ | < Connect to positive terminal of PS/Battery [through resister]
The power supply MUST be well filtered & regulated. Any power supply hum on the telephone line will probably ruin any hopes of high speed data transmission.
Wagner Lipnharski [wagnerl@EARTHLINK.NET] says:
In real, both devices have a phone line transformer that connects to the phone line, and that's all. Of course, signals generated by one device is transfered to the other via the transformers, however, the signal's power is very low and it is not enough to make a clear "conversation".
What happens in real, is that if you supply a current to the phone line, it will flow through the secondary coil (impedance) of the transformers. When your modem inject a signal at the primary coil, it is changing the transformer coils impedance, so in real it is modulating a bit the current flowing at the secondary.
The other transformer is also affected by this current modulation, so the other device "hears" the signal sent by the first device, much better than without this "carrier modulated current".
You have two ways to do that:
a) You can insert a battery in series with the phone line, it means, one phone line wire will be cut and have a battery inserted closing the circuit. By this way, a closed loop current will be flowing in a circular path through this circuit. There is a problem; It creates echoes and spurious noises because the current is in loop.
b) Based on the ASCII design above, you can insert a "ballast" or "pull-up" resistor between the Green wire to the positive of the battery or power supply. This way, the current will be splitted and flow towards both transformers at the same time. The pull-up resistor ensures (Voltage) modulation from one transformer to another. The difference here is that *if* the pull-up resistor is in the middle of the phone line (not close to one transformer or another, as happens in our actual phone switching centrals), the current will flows in opposite directions from the supply point to the transformers, and it is less sensitive to echoes, reflections and EMI interference, and the customer will always know which side is grounded.- 12Vdc | R 270 to 500 Ohms Current R Modem #1 <--- | ---> Modem #2 ---. .-------------o--------------. .--- 3|( 3|( 3|( 600 Ohms 600 Ohms 3|( 3|( Phone Line 3|( ---' '-------------o--------------' '--- | Ground
If you want to simulate the phone line voltages and currents:- 48Vdc | R 2 kOhms Current R Modem #1 <--- | ---> Modem #2 | ---. .----RRR------o------RRR----. .--- 3|( 150 Ohms 150 Ohms 3|( 3|( 3|( 3|( Phone Line 3|( ---' '-------------o--------------' '--- 600 Ohms | 600 Ohms Ground
If you want to simulate the "Ringing Signal", use this circuit:- 48Vdc | o------>\ <----. | \ | Relay Contact, clicking | \ | at 20 Hz | o GND | | 2 kOhms R === R --- 10 to 30uF | | | | | / | o--o/ o---' | RING SW | Current | Modem #1 <--- | ---> Modem #2 | ---. .----RRR------o------RRR----. .--- 3|( 150 Ohms 150 Ohms 3|( 3|( 3|( 3|( Phone Line 3|( ---' '-------------o--------------' '--- 600 Ohms | 600 Ohms Ground
Ron Fial says:
The two wires of an analog telephone connection are called Tip and Ring. Tip is normally at ground potential. Ring is normally at minus 48 volts with respect to ground, not + 48 volts as the diagram shows. Negative battery supplies (with respect to ground) have always been used with phone systems, so that leakage currents to ground caused by moisture do not electroplate away the copper in the wires. Some modems, e.g. certain US Robotics models, do not work correctly if the polarity is reversed (they will not detect ring), but most modern modems use diode bridges so that polarity doesn't matter. The 2KOhm resistor shown should work fine, but it only provides about 1/2 the normal current that each modem normally expects to see-- (each phone wants to 'see' about 20 ma) -- this should work fine, tho 1K to 1.5K would be more accurate.
Wagner's ring circuit is very clever, do you see that when the relay contact switches, the phone 'sees' a change of 96 volts. But be sure you get the capacitor the right way, or you will have quite an explosion, and with tantalum caps, you will have flames.
The transformers shown are 'inside' the modem -- but the vast majority of modem circuits do not pass the loop current thru the transformer, because the transformer core size required to avoid saturation means you need a bigger more expensive transformer. They couple to the transformer thru a cap, and send the loop current thru a resistor or resistive bridge.
|file: /Techref/modem/wotele.htm, 6KB, , updated: 1999/10/8 10:08, local time: 2022/5/20 03:25,
|©2022 These pages are served without commercial sponsorship. (No popup ads, etc...).Bandwidth abuse increases hosting cost forcing sponsorship or shutdown. This server aggressively defends against automated copying for any reason including offline viewing, duplication, etc... Please respect this requirement and DO NOT RIP THIS SITE. Questions?|
<A HREF="http://www.piclist.com/techref/modem/wotele.htm"> modem wotele</A>
|Did you find what you needed?|
PICList 2022 contributors:
o List host: MIT, Site host massmind.org, Top posters @20220520
* Page Editors: James Newton, David Cary, and YOU!
* Roman Black of Black Robotics donates from sales of Linistep stepper controller kits.
* Ashley Roll of Digital Nemesis donates from sales of RCL-1 RS232 to TTL converters.
* Monthly Subscribers: Gregg Rew. on-going support is MOST appreciated!
* Contributors: Richard Seriani, Sr.
Welcome to www.piclist.com!