On Tue, 8 May 2001 at 10:16:05, you wrote:
(ref: <H0000b5f107f99fa.0989313362.ln4p1327.ldn.swissbank.com@MHS>)
>> It is all down to handshaking and how fast the QL can accept and
>process
>> the data. If the QL invoked the handshake line, and DTR was connected
>> to RTS .....
>
>Just to throw in my two pen'orth (by the way hello, I'm a new Q40
>user), how exactly is RTS/CTS handshaking implemented on the [Linux] PC
>side?
Exactly the same as a PC - it is a std ISA serial card.
> I have a hardware book that shows several complicated ways of
>connecting two PCs via serial ports, but taking the following scheme as
>an example:
>
> PC 'A' PC 'B'
> TD ----------> RD
> RD <---------- TD
> RTS ----------> Input X
> CTS <---------- Output Y
>Input X <---------- RTS
>Output Y ----------> CTS
>
>Then if, for example, PC A wants to transmit to PC B, A would signal
>its intention on its RTS output driving an input pin on PC B. PC B then
>uses an output pin driving PC A's CTS to control the flow of data from
>PC A. The question is, when the Linux PC is the receiver,
Doesn't matter which is set as receiver - the data and handshake
direction is always the same.
>which pin
>does it expect the sender's RTS to come in on and which output pin does
>it use to drive the sender's CTS?
Simple - you need a crossover lead.
TX -> RX
RX <- TX
RTS -> CTS
CTS <- RTS
and that is it.
No complications with DSR and DTR - forget them.
A standard Laplink serial cable works fine.
Goodness knows why your hardware manual complicates matters by adding
'input x' etc - just confuses. What do they mean I wonder?
This applies to QL ser2 as well ('DTR' is actually RTS on the QL).
QL ser2 is identical to Q40 port. ser1 includes a crossover internally.
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