On Sun, 17 Mar 2002 at 13:29:15, Dilwyn Jones wrote:
(ref: <004401c1cdbe$f5efd360$81065cc3@default>)
>Tony Firshman wrote:
>>This is all done after 1am, just having returned from Eindhoven, so
>>would welcome correction (even by myself when I re-awake later)
>>
>>DSR/DTR connections are not needed.
>>
>>QL ser1 is identical to ser2 other than RX/TX and CTS/DTR(RTS)
>swapped.
>
>
>=====================================================================
>QL QL QL CABLE QL PC PC PC
>ser1 ser2 COLOUR SIGNAL SIGNAL 25D 9D
>
>(modem?) (terminal?)
>=====================================================================
>3 (RxD o/p) 2 (TxD o/p) <- white TxD RxD <- 3 RxD 2
>2 (TxD i/p) 3 (RxD i/p) -> green RxD TxD -> 2 TxD 3
>5 (CTS) 4 (DTR=RTS) -> blue DTR(=RTS) CTS -> 5 CTS 8
>4 (DTR=RTS) 5 (CTS) <- red CTS RTS <- 4 RTS 7
>1 1 -- black GND GND -- 7 GND 5
>N/A N/A 6 DSR 6
>N/A N/A 20 DTR 4
>=====================================================================
I knew I should not have done it so late. Pinouts are right, but
directions are worng.
This should be OK (but I am not laying my life on - I am still tired
(8-)# )
=====================================================================
QL QL QL CABLE PC PC PC
ser1 ser2 COLOUR SIGNAL 25D 9D
=====================================================================
3 RxD 2 TxD white -> RxD 3 2
2 TxD 3 RxD green <- TxD 2 3
5 CTS 4 (DTR=RTS) blue -> CTS 5 8
4 (DTR=RTS) 5 (CTS) red <- RTS 4 7
1 1 black -- GND 7 5
N/A N/A DSR 6 6
N/A N/A DTR 20 4
>=====================================================================
This whole diagram is QL <-> PC only of course.
If you think of QL to 25D by pin number, then ser1 is 'straight through'
and ser2 is 'crossover'. That is always my starting point, and all the
rest can be derived from that.
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