Fast-20 active negation current

Tak Asami asami at
Tue Jan 24 19:37:57 PST 1995

Now that we are beginning to have fun, here's another piece from Tak
on Fast-20 driver spec.

Let's assume we have an 84ohm cable, 1.5m in both directions (i.e., your
driver is placed in the middle of 3m cable). There is a series 2.5V, 110ohm
terminator (active) on each end of the cable.

For simplicity, actually, we can consider just half of it, i.e., one 1.5m
and one terminator on that end.

  |\             Zo = 84ohm              -------
--| >------------------------------------| 110 |---\
  |/                1.5m                 -------   |

My driver is now driving low at t=0, when it decided to shut off, or negate.
Let's ignore the pull up transitor also for now, it comes in later.
In the simplest case where the pull-down device shuts off immediately, with
no delay, the output voltage at the pin will go immediately to:

Vo = (Zo/Rt)Vt
   = (84/110) x 2.5
   = 1.9V

Wow, we are almost there to 2.0V.

Now, due to slew rate specification of 520mV/nsec (and of course no transistor
I know turns off in 0nsec), we turns off this pull down device slowly, so
it goes up to 2.0V in about 4nsec.

So, my question is, where does it need 60mA of current to get it to 2.0V?
Right now, I am not adding any pull up transient current to my model, and
my output goes to nearly 2.0V very quickly unless I try to slow down.

At my pin, of course.

Is there a serious oversimplification in my model (probably), or is this
to do with actually propagating the "edge" across the cable?

Also (now I'm on roll) for 1.5m of cabling, I am not expecting any
help from terminator until about 8nsec after the driver decided to negate.

1) There shouldn't be any terminator in the load model to simulate this 
2) Since there are two 86ohm cables (going left and going right) we are 
   driving, the load model for low to high transition should look like a 
   parallel combination of 15pF capacitor and 42ohm resistor, both going 
   to ground.
3) With this load model, I do see a need for pull-up current supplied by
   the driver to go over 2.0V level in less than 8nsec, but again, I don't
   see the need for 60mA (the driver's share is 22mA).

Maybe there was a document handed out at June Study Group meeting, but
I wasn't present.  If someone can respond to this message or can fax
me the document, I really, really appreciate it.

Tak Asami ===============================================================
Western Digital Corp.
I/O Product Engineering
asami at
(714) 932-7621 : Voice
(714) 932-6496 : Fax

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