[SI-LIST] Re: Drivers on a 50Ohm line

Andrew,

My conflict was with an issue regarding a fundamental concept, so I must
have made an incorrect fundamental assumption.  And I think you're on to
it near the end of the message regarding definition of drive strengths
of a driver.

My fundamental assumption so far has been that a driver designated as
"Max 25mA" can really drive 25mA *Maximum*.  The caveat I dropped was
that 25mA is the maximum the driver sends IF it is to send an output
within a given tolerance.

Therefore, in with an example of a 5V "25mA" driver, if the driver is
sending a clock to a 50Ohm, series terminated (I'm assuming Rint of
driver + Rterm =3D Zo) line, then for 2*Propogation delay, the driver =
WILL
send 5V/100 =3D 50mA, even though the driver spec is "25mA".  This will
have the effect of making the output of the driver less than the rated
voltage given in the spec (like 0.9*Vcc).

But we are ok, since what we are really concerned about is that the
voltage at the node between the external terminator and the line >=3D
Vcc/2 (in order for a successful voltage doubling to take place at the
end).  This will happen as long as Rint+Rterm <=3D Zo (ideally equal =
to).

However, for any end terminated scheme, we would still have an issue
since the driver now needs to drive the rail-to-rail voltage.  But
beyond the rated current spec, the rail may no longer be guaranteed to
go beyond some fraction of the final voltages, which might indeed be
unacceptable.

Thanks
-Nitin

-----Original Message-----
From: Ingraham, Andrew [mailto:a.ingraham@xxxxxxxx]=20
Sent: Thursday, July 29, 2004 6:43 AM
To: Bhagwath, Nitin; si-list@xxxxxxxxxxxxx
Subject: Re: [SI-LIST] Drivers on a 50Ohm line


Nitin,

When a driver drives a single trace with Zo =3D 50 ohms, what it means =
is
that a signal with a 2.5V edge on that trace, will see an initial
delta-I of 2.5/50 =3D 50mA every time it switches.  That's all it means.

There are lots of wrinkles to this analysis, some of which have been
mentioned already.

For delta-I to be 50mA, it might be 0mA and 50mA, or -25mA and +25mA,
etc.

If you put a driver in the middle of a 50 ohm trace, in order for that
driver to provide a full 2.5V (rail-to-rail) signal, the delta-I would
be 100mA every time it switches.

This delta-I pulse you've calculated lasts only for one round-trip trace
delay.  After that, the current requirement is dictated more by the
termination (if any) on the far end of the trace, eventually not by the
trace impedance.  If the ends are not terminated, then the delta-I pulse
is only required for a few nanoseconds or less, every time the driver
switches.

If the 50 ohm traces are very short, with delays less than the driver's
risetime, then all this analysis is moot because the reflection comes
back before the driver has switched fully.  Unless the far end is
terminated with a DC load, you can't calculate the driver's output
current this way, it's going to be a lot less.

Getting a full 2.5V rail-to-rail initial edge from a 2.5V clock driver,
requires a driver with zero source impedance, which doesn't exist!
Every 2.5V driver would impress an initial delta-V edge of something
less than 2.5V, maybe a little less, maybe a lot.

A special case of that is a series terminated driver that matches the
line ... whether that series termination is in the driver's own internal
impedance, or using an external resistor.  A 2.5V driver would switch
only 1.25V edges onto the line, and delta-I =3D 1.25/50 =3D 25mA.  If =
the
far end of the trace is unterminated, one round-trip later the voltage
swing at the driver's end of the line doubles to 2.5V, and the delta-I
pulse ends.  The far end of the trace would see just a full amplitude
2.5V signal.  This is basic transmission line stuff.

The combination of a parallel terminated clock line, with a driver where
you want the amplitude to be (nearly) rail-to-rail, is the most
demanding regarding output currents; and for that you would need to
select a clock driver whose output impedance is very low and whose
output current capability is very high.  More general purpose drivers
would suffice when you don't need to drive rail-to-rail signals on
parallel terminated lines.

When a driver specifies a drive strength of something like 30mA, it does
not necessarily mean that is all it can supply.  Depending on how the
vendor gave that number, it might mean any number of things.  It might
be a static spec, meaning they don't want you connecting the driver to a
load that requires more than 30mA continuously, but short pulses that
are larger are OK.  Or it might mean that the voltage drop is less than
some X when the current is 30mA, but you could actually get considerably
more than 30mA. How much more can vary a lot from one IC vendor to
another.  (I've seen some ICs that could supply many times the output
current spec, even at reasonable output voltage amplitudes.)  There
isn't a consistent definition for output current capability when taken
on its own, without all the qualifications that go along with it.

Regards,
Andy


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