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'[MIGUEL][OT] TWO MODEMS'
1999\10\01@125245 by - KITS EDUCACIONAIS NACIONAIS

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How can i program two modems for when one HANGS UP, an automatic
connection be estabilish?

       I want to connected two modems directly one with other, in a
NODIALTONE line.

       Today, i have 2 modems:

       A                               B

For a connection, i have to type ATA on B modem, and ATX3 on A modem.
The problem is that the machine that i will connect on modem B only wait
for a CONNECTION status, nothing more, the machine can't send a ATA
command to MODEM B.

       Thanks again!

       Miguel Wisintainer
       REXLAB - Real Experiment with 8051
       REXPIC - Real Experiment with PIC (now 24hrs day)

1999\10\02@074404 by Michael Rigby-Jones

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       <snip>
{Quote hidden}

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?

Mike Rigby-Jones

1999\10\02@091227 by paulb

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Michael Rigby-Jones wrote:

>>                      | < Connect to negative terminal of PS/Battery
>> ---------+           |     +-------------+
>> modem A  |Red -------+-----| modem B     |
>>          |Green -------+---|             |
>> ---------+             |   +-------------+
>>                        | < Connect to positive terminal of PS/Battery

> 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?

 Well PICked!  I let that one pass when I saw it, but now that you
mention it, the answer is that it should have an inductor or "virtual
inductor" in series with the power supply.  "Real" transmission bridges
use a balanced inductor.


 The virtual inductor circuit was referenced a couple of weeks back at
http://209.1.238.246/1st_pages/A1208.htm (it's Q1 in the circuit) and
has some desirable characteristics over a real inductor including - no
inductive "kick".
--
 Cheers,
       Paul B.

1999\10\02@151612 by Wagner Lipnharski

picon face
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

1999\10\02@175400 by Keelan Lightfoot

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Next time I will keep my mouth shut :)

- Keelan, Purveyor of misinformation.

1999\10\04@013359 by Javier

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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 <spam_OUTwagnerlTakeThisOuTspamEARTHLINK.NET>
Para: .....PICLISTKILLspamspam@spam@MITVMA.MIT.EDU <PICLISTspamKILLspamMITVMA.MIT.EDU>
Fecha: Sabado 2 de Octubre de 1999 16:17 p.m.
Asunto: Re: [MIGUEL][OT] TWO MODEMS


{Quote hidden}

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
{Quote hidden}

1999\10\04@023040 by Dennis Plunkett

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At 23:10 2/10/99 +1000, you wrote:
{Quote hidden}

Yeh, Paul I seen this too, and that stuff on TIP and RING.

As for that Q1 thingo....  . The 12V zen will clamp the line voltage and
shunt the audio a bit, as for Q1, it is confuigured in a slow start
configuration that also provides a bit of dynamic current limit control the
intention is not to seperate the audio, but is a side advantage (51 ohms is
not enough to ensure that levels of +3dBr are acheived with QI fully on,
but should be OK for most users)


Dennis

1999\10\04@025403 by Ron Fial

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A note about polarity and phone connections:

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 res
pect 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 leak
age 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 s
hould work fine, but it only provides about 1/2 the normal current that each mod
em normally expects to see-- (each phone wants to 'see' about 20 ma) -- this sho
uld 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 swi
tches, 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 transfor
mer core size required to avoid saturation means you need a bigger more expensiv
e transformer.  They couple to the transformer thru a cap, and send the loop cur
rent thru a resistor or resistive bridge.

Wagner, you have a great way with ASCII-based circuit diagrams!
  Regards,
    Ron Fial
--------------------------------------------------------------

{Quote hidden}

1999\10\04@071747 by paulb

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Dennis Plunkett wrote:

> As for that Q1 thingo....  . The 12V zen will clamp the line voltage
> and shunt the audio a bit,

 Quite a lot, I'd think, just *asking* for distortion, quite
undesirable for a modem.  I find it a trifle strange to put it there.

> as for Q1, it is configured in a slow start configuration that also
> provides a bit of dynamic current limit control the intention is not
> to seperate the audio, but is a side advantage

 I would *hope* it's intended to separate the audio, because there's a
22 µF capacitor on the other side of the transistor.  If it was just a
slow switch on (and why on earth would you *want* a slow switch on?),
you wouldn't really need the 51 ohm resistor at all.

> (51 ohms is not enough to ensure that levels of +3dBr are acheived
> with QI fully on, but should be OK for most users)

 The 2k7 ohm resistor prevents saturation, so the transistor is
definitely in constant-current mode as far as AC is concerned.  As such,
the AC component doesn't see the 51 ohm resistor, only the 2k7.  Again,
saturation is *most* undersirable as it generates distortion.
--
 Cheers,
       Paul B.

1999\10\04@095922 by Wagner Lipnharski

picon face
Ron and Javier,
The polarity is really -48V and Ground, even that in a bench simulator
it doesn't matter  :)
You are right about the phone line transformer, in real, actualy they
are smaller in size and most of them requires a 30-100 Ohms resistor in
series to reduce a bit power over the coils.

There are some solid state phone line transformers, using proportional
LEDs and PhotoTransistors. I believe I saw something at CPClare.

I already used two resistors and two capacitors instead the transformer,
it works pretty nice, but there are some problems related to voltage
saturation.


--------------.     0.47uF 250V    Answer Relay Sw
             |                        /
             |---o------||-----o-----o/  ^----> To Phone Line
             |   |             |
             |   R             R
MODEM Circuit |   R   600 Ohms  R
             |   |             |
             |---o------||-----o--------------> To Phone
Line               |
--------------'
                   0.47uF 250V


Javier wrote:
> 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
:-(

Your serial port UART can easily goes up to 115kbps in direct attachment
using a null modem cable (just cross wires between two computers), while
your portable PC internal modem (can goes from 2400 to 28800, if it is
old, I guess it is 2400 max 14400bps).

To hookup your handheld to talk to your PC it will give you some
problems, first because you will not have the ring signal at the fake
phone line, you will need to know how to make both modems handshake and
start the connection.

I would go after a null modem cable, if you have a free serial port in
each computer.
If you have two female DB9 connectors (or DB25 if your computer uses it
as a serial port), it is also easy to build this cable, in real it is
just few wires, nothing special.

Wagner

1999\10\04@180828 by Dennis Plunkett

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At 21:17 4/10/99 +1000, you wrote:
>Dennis Plunkett wrote:
>
>> As for that Q1 thingo....  . The 12V zen will clamp the line voltage
>> and shunt the audio a bit,
>
>  Quite a lot, I'd think, just *asking* for distortion, quite
>undesirable for a modem.  I find it a trifle strange to put it there.
>

Yes I agree too, nominaly this is there as protection for the CCT and the
cct will nominaly hold the line voltage to under the zen limits.



>> as for Q1, it is configured in a slow start configuration that also
>> provides a bit of dynamic current limit control the intention is not
>> to seperate the audio, but is a side advantage
>
>  I would *hope* it's intended to separate the audio, because there's a
>22 µF capacitor on the other side of the transistor.  If it was just a
>slow switch on (and why on earth would you *want* a slow switch on?),
>you wouldn't really need the 51 ohm resistor at all.
>

Take a look at TS002, it sets out the maximum current that can be drawn
from a line at any one time and also sets out the current/voltage grpahs
that *have* to be meet in order to pass. So slow switch on in a potentialy
high current part of a cct is desireable. However, for the fist 300mS you
can almost do whatever you want, but in the case of this cct the 12V zen is
too high, and should limit the line voltage to 8 to 9V, this then lets the
designer draw as much currrent as possible during this period.


>> (51 ohms is not enough to ensure that levels of +3dBr are acheived
>> with QI fully on, but should be OK for most users)
>
>  The 2k7 ohm resistor prevents saturation, so the transistor is
>definitely in constant-current mode as far as AC is concerned.  As such,
>the AC component doesn't see the 51 ohm resistor, only the 2k7.  Again,
>saturation is *most* undersirable as it generates distortion.

Huh? The 2k7 only limits maximum base current, I don't see how this
prevents saturation, as a dynamic current change on the emitter will shift
this point around, with only the cap controlling the dynamic current
response. It should then be possible to drag the emitter more than one Vbe
down and thus place the transistor into staturation, all be it for short
periods of time. Also if saturation was not a problem then the 51 ohm
resistor would not be required. If I recall this is not there to limit
current into the cct, but as a discharge path for the cap. But then again I
am more than likely incorrect as usual.

Oh yes, add a PIC to control the ring cadence detection and off hook
control <G>



Dennis




>--
>  Cheers,
>        Paul B.
>
>

1999\10\06@043722 by paulb

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Hello Dennis.

>>  Quite a lot, I'd think, just *asking* for distortion, quite
>> undesirable for a modem.  I find it a trifle strange to put it there.
> Yes I agree too, nominaly this is there as protection for the CCT and
> the cct will nominaly hold the line voltage to under the zen limits.

 Ah!  That's a better explanation.  I'd have thought 18V though,
obviously must match the switcher rating.

> Take a look at TS002, it sets out the maximum current that can be
> drawn from a line at any one time and also sets out the
> current/voltage grpahs that *have* to be meet in order to pass.

 Do I gather that's Telecom Standard 002?  Is that web-able?

> So slow switch on in a potentialy high current part of a cct is
> desireable.

 I don't believe a word of it.  I remember what a rotary *dial* does.

> However, for the fist 300mS you can almost do whatever you want, but
> in the case of this cct the 12V zen is too high, and should limit the
> line voltage to 8 to 9V, this then lets the designer draw as much
> currrent as possible during this period.

 You have me thoroughly bamboozled.

> Huh? The 2k7 only limits maximum base current, I don't see how this
> prevents saturation, as a dynamic current change on the emitter will
> shift this point around, with only the cap controlling the dynamic
> current response.

 The more I look at this circuit, the more I conclude it's a virtual
choke.  DC-wise, the 2k7 defines an operating point for the transistor;
it is "reflected" by the transistor as 2k7 divided by the beta, if we
say 100, then the arrangement appears as 27 ohms and VCB becomes 1.25 V
at 50 mA.  For the overall circuit add VBE of 0.6V and 2.5V across the
51 ohm resistor.

> It should then be possible to drag the emitter more than one Vbe down
> and thus place the transistor into staturation, all be it for short
> periods of time.  Also if saturation was not a problem then the 51 ohm
> resistor would not be required. If I recall this is not there to limit
> current into the cct, but as a discharge path for the cap.

 AC-wise, the capacitor locks the voltage drop in so for a given
current and the 51 ohm resistor creates a very constant-current drive,
standard way of doing so in op-amps.  Now, this certainly *could*
saturate if driven with enough audio, but it is shunted by the voice
(modem) circuit in parallel with the line so its function is to
"reflect" the voice current from one to the other.

 Let's look then not at current, but at voltage, because the
relationship of one to the other is defined by the impedance at this
point.  It will only saturate if the VCB (not VBE) voltage drop is lost
which I calculate above as 1.25 V at 50 mA on the line, that's 1.25 V
peak in one excursion.  I believe most audio is less than 2.5 V p-p on a
phone line at 600 ohm?
--
 Cheers,
       Paul B.

1999\10\08@100846 by eplus1

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Listen,
often if no one opens their mouth and says the wrong thing (or slightly
wrong) the "experts" will not be interested enough to correct them with the
truth. The best way to get some real help is to start f**ing it up and then
watch people say "no,no,no, its like this."

Reminds me of the advice of a biology Prof. on how to run field trips:
1. make sure you include someone stupid and thick skinned enough to ask all
the dumb questions that all your other students want to ask but are
embarrassed to.
2. walk far enough ahead of your group to step on any specimens you don't
recognize.
3. make some obvious mistakes so your students can build confidence in
questioning you.

James Newton .....jamesnewtonKILLspamspam.....geocities.com phone:1-619-652-0593)
webmaster http://techref.homepage.com NOW OPEN TO NON-MEMBERS!
Members: Add your own private/public comments/pages (TANSTAAFL web hosting)




-----Original Message-----
From: pic microcontroller discussion list
[EraseMEPICLISTspam_OUTspamTakeThisOuTMITVMA.MIT.EDU]On Behalf Of Keelan Lightfoot
Sent: Saturday, October 02, 1999 3:55 PM
To: PICLISTspamspam_OUTMITVMA.MIT.EDU
Subject: Re: [MIGUEL][OT] TWO MODEMS


Next time I will keep my mouth shut :)

- Keelan, Purveyor of misinformation.

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