15 Nov
2014
15 Nov
'14
7:49 p.m.
Howdy,
On Sat, Nov 15, 2014 at 9:10 AM, Tothwolf <tothwolf at concentric.net> wrote:
No joke, you got it right. That's the standard scheme for IDS (I said DIP,
which is the wrong term I think) headers.. across-and-back, as you wrote
it.
Standard numbering scheme? Surely you jest? ;)
Do you mean something like this?
2 4 6 8 10
. . . . .
. . . . .
1 3 5 7 9
It sounds like you've just about got it figured out though. If you have
the datasheet for the 1458 (aren't there 1459s on the board too?) then you
can follow the signals through the level converters back to the UART, and
its datasheet should then tell you exactly which signals are which.
Don't see any 1459s on there, just the pair of 1458s up near the IDS I/O
headers
But yes, you have a point.. I could probably have continued, tracing back
from the 1458 inputs, through the various logic gates,
to the 1014 / 1015
UARTs. But I stopped where I did as it was already pretty
tedious and I
figured I had enough info for you in-the-know types to figure it out at a
glance.. ;-)
That, and I am confused by a couple of my findings, particularly the Pin 4
- Pin 6 relationship. Why are these pins essentially common to each other,
and to the XNOR inputs? One pin has a 47R in series to the XNOR gate, the
other has a 150R in series to that same gate. It's as if the line with the
150R could be asserted 'high' by some line and then the line with the 47R
could re-assert it 'low' - but not all the way to zero, more like to 1/3 of
voltage. See, that makes no sense to me,.
Ditto for the Pin 1 / Pin 7 relationship. Both of these pins are
essentially outputs from the same op-amp output - one is basically a
direct-out from the op-amp #2, the other is via the collector of a PNP
(3906) transistor - the base of which is also tied to op-amp output #2.
This is the stuff that's really confusing me.. is the 3906 acting as an
inverter? If so, why?
To be clear, J1 (10-pin I/O header) has two pins (4 & 6) both tied to the
selfsame +input+ point, and two pins (1 & 7) tied to the selfsame +output+
point. J2 is identical, but it uses a different op-amp and a different XNOR
input. Otherwise, J1 and J2 are carbon-copies of each other, electronically
speaking.
It makes sense that pin 10 would be an extra ground too, since it is quite
possible they originally used a 9-conductor ribbon cable and wouldn't have
connected that pin. You may also find than the signals come out on the
correct pins of a DE9 connected to a 10-pin IDS connector with some
9-conductor ribbon cable.
Or to a 25-pin version of the same. Sure wish I had one! But I tossed /
gave all that away years ago, I think.. =/