Re: Triton DMA

[Rogier Wolff]

Gabriel Paubert wrote:
> 
> 
> 
> On Sun, 30 Nov 1997, Rogier Wolff wrote:
> 
> > If you have a cable, and hoist one end of it from 0V to 5V in a few
> > ns, the other end will take a while before it notices that you did
> > that (speed of light at least). Worst case the other end will have a
> > completely different wave form, and you will have trouble deducing what
> > the other end did to generate this. However with a small trick, you
> > can make the wave form travel intact along the wire (the "transmission
> > line effect"): you have to make the conductor be surrounded by a
> > cylindrical grounded plane. Coax.  This effect is reasonably strong: A
> > grounded wire on both sides of the conductor in a flat cable is already
> > a reasonable approximation of the coax leading to the transmission
> > line effect.
> > 
> > The trouble starts when such a conductor abruptly ends. At that point,
> > the travelling wave form simply bounces back and starts propagating
> > back to where it came from. However with current performance
> > requirements, the source will be wanting to send the next bit along
> > the wire by then. This will lead to data corruption. To prevent this
> > the transmission line will have to be lead to believe that there is no
> > abrupt ending to the cable. This turns out to be relatively easy: a
> > simple resistor to a DC level will do. With SCSI termination, they
> > have one more trick: The DC level has been chosen such that without
> > any drivers the signal level will be somewhere around the switch point
> > of the then-common TTL chips. This requires someone to actively pull
> > it down to get a reliable "0", or to actively pull it up to make a
> > reliable "1". This distributes the burden of transmitting data over
> > both the pull-down and the pull-up output transistor, instead of only
> > requiring just one. (You also get faster circuits if you use both of
> > them.)
> 
> This is slightly wrong, TTL thresholds are defined as:
> (I don't use 0 or 1 since SCSI uses inverted signals, inactive (logical 0) 
> is actually the high level and active (logical 1) is low).
> - any voltage between -0.5 and 0.8 V is a low level
> - any voltage between 2.0 and 5.5 V is a high level
> - any other voltage is undefined or illegal. Voltages well below ground or 
> well above 5 V may kill internal protection diodes or "zener" the input
> transistors which permanently degrades their gain.

Actually:
  [II = +/- 2,0 mA maximum.]

     The II requirement shall be met with the input voltage varying
     between -7 V d.c. and +12 V d.c., with power on or off, and with
     the hysteresis equaling 35 mv, minimum.

So: No diodes allowed to +5 or GND. 

> The level of an undriven SCSI bus is actually close to 3 V 
> (active terminators often regulate it at 2.85 V). So it is a _high_ level.
> And SCSI bus drivers need to be able to sink a significant current
> to drive the bus down to below 0.5V, but do not need to source much to 
> drive the high level. The SCSI specification is IMHO a consequence of the 
> assymetry of TTL output circuitry (and CMOS chips are also similar: for the 
> same geometry, an N-channel will conduct more current than a P-channel).

I reread the electrical part of the SCSI spec and you're right. The
termination resistors are the other way around. The terminators indeed
drive the bus to a well-defined "high" (0 or inactive) level.

The spec however says that you're supposed to be able to source or
sink both 55ma. So the assymetry that TTL and CMOS drivers exhibit is
not allowed to be visible on the SCSI bus....

> OTOH, I doubt that IDE Ultra-DMA is terminated in the sense that the cable
> has at both ends a resistor equal to its characteristic impedance. 
> Because of the following reasons: 
> - bus drivers would need to be able to sink and/or source reliably much 
>  larger current. Old devices would probably stop working.
> - the cable length is limited to 50 cm, which is about 5 nS round-trip,
>  with a 60 ns cycle time, you don't need to be so strict with cable length
>  if there are no or few reflections on the cable. 
> - terminations consume power, and considerations of mobile computer battery
>  life weigh in quite a lot these days. 
> - many other problems, like how a device decides it needs to be terminated
>  when one is an Ultra and the other an old device. 

As far as I know, there is more like "half" a termination at the
motherboard side of the cable. This means that a transition takes a
round-trip (10ns) before it more or less settles. In the unterminated
case, there could be a signal still bouncing back-and-forth between
the two sides of the cable after 60 ns.

                               Roger.

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