Hi
I want to connect the handheld PC(with internal modem) and the desktop
computer, is the resistor + 9v batt method reliable, what baud rate could I
use ?? I want to connect them via modem cause I don4t have the serial cable
:-(
There is no problem if I invert the tip and ring wires, right?
Thanks
Javier
PD Are there any trafos in your schematic, Wagner ? I4m not very good
reading ascii4s.

-----Mensaje original-----
De: Wagner Lipnharski <wagnerl@EARTHLINK.NET>
Para: PICLIST@MITVMA.MIT.EDU <PICLIST@MITVMA.MIT.EDU>
Fecha: Sabado 2 de Octubre de 1999 16:17 p.m.
Asunto: Re: [MIGUEL][OT] TWO MODEMS


>Michael Rigby-Jones wrote:
>[snip]
>> >
>> >                      | < Connect to negative terminal of PS/Battery
>> > ---------+           |     +-------------+
>> > modem A  |Red -------+-----| modem B     |
>> >          |Green -------+---|             |
>> > ---------+             |   +-------------+
>> >                        | < Connect to positive terminal of PS/Battery
>> >
>> > 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.
>> >
>> >
>> This surely can not be correct, the presence of a nice low impedance
battery
>> or PSU would effectively short the lines for any signal, making comms
>> impossible.  I thought the the supply was effectively in SERIES with the
>> line?
>>
>
>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
>
>
>Wagner Lipnharski