Date: Tue Jul 05 2011 - 09:33:34 PDT
OK, so here are the basics I was referring to:
Imagine a buffer driving a T-line which is not
terminated at the other end. Let's assume a
simple 5 V CMOS buffer. When the buffer begins
to switch high to low the voltages are at 5 V
on both ends of the T-line. When the buffer
turns off its pullup and turns on its pulldown,
The voltage at the output of the buffer (the
beginning of the T-line) will be whatever the
voltage division due to the pulldown drive
impedance and the T-line impedance yields. For
example, if the pulldown of the buffer and the
T-line are both 50 Ohms, the voltage will be
2.5 V. The end of the T-line remains at 5 V
because it will take some time for this change
to propagate down to the end of the line.
When the incident wave arrives at the end of the
T-line, the voltage will drop to 2.5 V at the end
also. If the T-line is unterminated, a full reflection
will occur, and the 2.5 V transition will double all
the way down to 0 V. (This usually happens in a
continuous slope, you don't see a stop at the 2.5 V
point). If the T-line is terminated with a matched
resistor (50 Ohms to Vcc), the transition at the
end will remain at 2.5 V since no reflections occur.
If there are no reflections, the waveform will remain
at the 2.5 V level as long as the buffer is driving
low on both ends of the T-line.
If the end of the T-line is not terminated, than the
reflection at the end of the T-line will propagate
back to the driver. This is called the reflected
wave. It takes the same amount of time for the
incident wave to go from the driver to the other end,
as it takes for the reflected wave to go back to the
driver. Once it arrives at the driver, the waveform
at the driver will also go from 2.5 V to 0 V and the
transition is complete, because the pulldown will hold
the waveform at 0 V.
Note that the waveform at the driver will have a ledge
at 2.5 V during the time the wave travels down the line
and back.
If the driver is weaker than the T-line's impedance,
say 70 Ohms vs. 50 Ohm, then the first ledge voltage
will not even go half way and the doubling at the end
will not reach the full (rail to rail) swing. In this
case the full swing will take several round trips, and
you will see multiple ledge voltages in the waveforms
on both ends of the T-line. This is what the waveforms
looked like in the original posting.
You can prove all this to yourself by putting together
a simple experiment. Use an ideal pulse source, a
series resistor (to represent the driver impedance)
and an ideal T-line, and watch the waveforms on both
ends of the line. That should help you to answer your
own questions...
Vpls pls 0 PULSE (5.0V 0.0V 0.1ns 1.0ps 1.0ps 25.0ns 50.0ns)
Rdrv pls drv R=50
Tline drv rcv Zo=50 Td=5ns
Arpad
========================================================
From: naftali refaeli [mailto:Naftali.Refaeli@ecitele.com]
Sent: Tuesday, July 05, 2011 11:03 AM
To: Muranyi, Arpad; 'ibis-users'
Subject: RE: [IBIS-Users] A question about the ibis simulation.
To clarify the question - is it a simple reflection that can be fixed by a simple Termination?
On case of "one on one" the termination is a simple series resistor nearby the Driver or a parallel one nearby the RCV - is that the case?
Thanks.
Naftali Refaeli
CAE Engineer
ITO
HP Enterprise Services
Telephone +972 3.9266973
Mobile +972 52.3542043
Email Naftali.Refaeli@ecitele.com<mailto:Naftali.Refaeli@ecitele.com>
From: owner-ibis-users@eda.org [mailto:owner-ibis-users@eda.org] On Behalf Of Muranyi, Arpad
Sent: Tuesday, July 05, 2011 6:52 PM
To: 'ibis-users'
Subject: RE: [IBIS-Users] A question about the ibis simulation.
Naftali,
I am not sure I understand what the "The topology is one on one" is
referring to in the original question. So I am not sure what
you mean by "set a very same length to both T branches" either. Please
clarify your question.
Thanks,
Arpad
==============================================================
From: naftali refaeli [mailto:Naftali.Refaeli@ecitele.com]
Sent: Tuesday, July 05, 2011 10:49 AM
To: Muranyi, Arpad; 'ibis-users'
Subject: RE: [IBIS-Users] A question about the ibis simulation.
Do you mean that If he will set a very same length to both T branches - then the phenomenon will disappear?
Naftali Refaeli
CAE Engineer
ITO
HP Enterprise Services
Telephone +972 3.9266973
Mobile +972 52.3542043
Email Naftali.Refaeli@ecitele.com<mailto:Naftali.Refaeli@ecitele.com>
From: owner-ibis-users@eda.org [mailto:owner-ibis-users@eda.org] On Behalf Of Muranyi, Arpad
Sent: Tuesday, July 05, 2011 6:45 PM
To: 'ibis-users'
Subject: RE: [IBIS-Users] A question about the ibis simulation.
Grace,
It looks like that the first ledge is the voltage that the
driver can drive with the load imposed on it by the T-line.
Once the reflection comes back to the driver the voltage
steps up to the full swing levels. This is basic T-line
theory.
Arpad
============================================================
From: owner-ibis-users@eda.org [mailto:owner-ibis-users@eda.org] On Behalf Of Zhanggezi
Sent: Monday, July 04, 2011 8:54 PM
To: 'ibis-users'
Subject: [IBIS-Users] A question about the ibis simulation.
Hi All,
The topology is one on one. The simulation result is below. I am curious about the simulation circled
by red pink. What's the cause? Could someone give me any hint?
[cid:image001.png@01CC3B04.575AACD0]
Thank you.
Grace
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