[SI-LIST] Re: PDS analysis?
- From: "Chris Cheng" <Chris.Cheng@xxxxxxxx>
- To: "steve weir" <weirsi@xxxxxxxxxx>
- Date: Mon, 31 Mar 2008 19:17:25 -0700
>> It is the tight coupling between the signal traces and its=20
>> power/ground return path that will sustain your GHz signal.
>No, we have two separate phenomena: The signal launch and the signal=20
>propagation. At the signal launch ( driver ) the signal can only be=20
>sustained when the driver power does not collapse. For the signal to=20
>get anywhere, the signal propagation path needs to be contiguous. A=20
>discontinuity through a junky power system will badly impair=20
>propagation. But even the most perfect propagation path cannot fix a=20
>starved driver.
>> The overall loop inductance between the signal and its ground return, =
>> when connected through a driver that is driving low, determines=20
>> whether you can sustain your I/O to go from high to low. The reverse=20
>> for I/O power from low to high. You can't use decoupling caps to=20
>> "decouple" your way out of this outside the package because you can't =
>> decouple between the signal and its reference plane !
>The inductance determines whether we can initiate the edge, not whether =
>we can sustain it.
Remember Chris Rule B
The signal consist of the signal trace AND its return path. While your =
driver may not be switching, the return path that is most likely shared =
between multiple drivers may not.=20
A classic example is your processor FSB or DDR1/2/3 during a write =
cycle. The DQS is switching between the DQ data packet. While your DQS =
may be quiet and not switching, the entire DQ bus can be switching and =
raise or drop your return path high or low or ring like a gong. Your DQS =
will not be in a steady state of voltage.
>> Power is delivered through the I/O signal leads between I/O power for =
>> low to high and I/O signal leads between I/O ground for high to low=20
>> transition.NOT between I/O power and ground (again assuming you don't =
>> have large crowbar current).
>No this is only partially true.=20
>
>The wave propagates away through whatever the transmission path is. We =
>can arrange the transmission path to be virtually entirely referenced =
to=20
>either power rail or something else entirely. For the driver to be =
able=20
>to launch a wavefront, it must be able to impress or remove charge at=20
>the launch. If we switch from high to low and are VCC referenced, then =
>local capacitance connects the low side of the driver to the return =
side=20
>of the launch. If we switch from low to high and are VSS referenced,=20
>local capacitance connects the high side of the driver to the return=20
>side of the launch. If we reference to something else, we have to=20
>couple both VSS and VCC to whatever that something else is.=20
>
>As long as the driver has a sustained energy source / sink available it =
>can maintain the switched level. The wiring that supplies that energy=20
>can be made almost entirely independent of the signal propagation path. =
=20
>An example is the MOLEX Z axis powering scheme.
Agree and the decoupling will be done by either I/O decoupling cap on =
die or the plane capacitance of the reference plane that sandwich the =
signal combined with the proper via that switch together with the signal =
when it entires or leave a reference plane.
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