Using AST 4port or clone on uPort (long)

[John Plocher]

In article <705 at tukki.jyu.fi> makela at tukki.jyu.fi (Otto J. Makela) writes:
>the motherboard COM2 and used IRQ3, all worked ok.  The big question now
>is, how does one define "different cards" ?

The answer requires a bit of understanding as to how interrupts work on
the PC/AT/386 bus.


First, lets look at the circuit that IBM used to drive the BUS level
IRQ lines:

      (+5v) -------+
                   |
	      [Transistor]
             /     |
  <control> <	   +-------------> IRQ
             \     |
	      [Transistor]
                   |
      (gnd) -------+

The "control" is used to make sure that only one of the transistors
(switches) is on at a time. i.e.,

      (+5v) -------+
                   |
             	   +-------------> IRQ

      (gnd) -------+

    -or-

      (+5v) -------+

  		   +-------------> IRQ
                   |
      (gnd) -------+


This works well when only ONE circuit is used on a single line.
When you try to use more than one, though, you get:

      (+5v) -------+-------+
                   |       |
             	   +-------+-----> IRQ

      (gnd) -------+-------+

(Both in the same state - no problem)

      (+5v) -------+-------+
                   |
             	   +-------+-----> IRQ
                           |
      (gnd) -------+-------+

(each in a different state - a big mess.  Note the "short" between
the +5v line and gnd? - the IRQ is at an indeterminate state - it even
could be smoking :-)


Normally, each CARD will have ONE of these circuits on it, so the
directions say "one card per IRQ".  Cards that have several
ports on them usually have one driver circuit per IRQ line,
and the on-board electronics takes care of gating the "I want to
generate an interrupt" signals from the ports together before 
the driver circuit sees them.  This means that you could have
an 8 port card with 4 ports using IRQ3; the BUS level IRQ signal
is generated by only ONE IRQ driver circuit.


  -John Plocher