Re: Splitting serial cables ?

In article <v%vDf.2975$Iw6.196541@xxxxxxxxxxxxxxxxxxxxx>, "Neil Rieck" <n.rieck@xxxxxxxxxxxx> writes:
>
> <briggs@xxxxxxxxxxxxxxxxx> wrote in message
> news:oCgSttFou3DR@xxxxxxxxxxxxxxxxxxxxxxxxxxx
>> In article <43DD3DBC.6E32BF1E@xxxxxxxxxxxx>, JF Mezei
>> <jfmezei.spamnot@xxxxxxxxxxxx> writes:
> [...snip...]
>>
>> Transmit data is on one of those wires. You have two stations both
>> wired into transmit data. Both of them will be holding that lead
>> in the space condition (or mark -- I can never remember) when idle.
>> This is not tri-state logic where the lead floats when idle.
>>
>> If you've ever put a breakout box on an RS232 connector, you'll have
>> noticed that TD and RD are always live.
>>
> [...snip...]
>>
>> You can have one source sending data to multiple receivers. That's
>> no problem since the receivers aren't trying to coerce the signal line.
>>
>> You can't have multiple sources sending data to a single receiver.
>> That's a problem since multiple senders are each trying to coerce
>> the signal line differently.
>>
> Umm, if you read my original post you'll notice that I said 2 diodes were
> required to isolate the two transmit lines from each other. Only one phase
> of the signal (negative if memory serves) is required to send data from
> transmitter to receiver. Without the diodes one transmitter would be
> positive while the other is negative and you would have a problem.

I posted before seeing the reports that diodes were a possible solution
that had actually been deployed successfully.

What I recall of RS232 in action is that transmit and receive data
use positive and negative voltages to signal mark and space. Zero voltage
is never used.

Checking online I see that the signalling range is technically -3 to -25V
denoting a logic 1 (mark) and +3 to +25V denoting a logic 0 (space).

Actual implementations are more likely to keep the signals down to
8 to 14 volts according to the first reference I found.

No matter. A diode can still work in any case. It just needs to be
capable of dealing with 50 volts end to end without breaking down.

And it needs to leak enough current so that even when both transmitters
are idle and, consequently, both diodes are blocking current flow
that enough signal leaks through to keep the receiver seeing the
mark condition.

One poster mentioned a pull-up resister tied to DSR, presumably to
address with this leakage requirement.

Mind you, I'm speaking from theory, not from practice. Looking at
LEDs on a breakout box is about as far as I've ever actually mucked
about with the relevant hardware.
.

Relevant Pages

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