[SI-LIST] Re: A question about PCI bus termination.
- From: "Ingraham, Andrew" <Andrew.Ingraham@xxxxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Sat, 29 Dec 2001 15:44:25 -0500
Some would say that PCI uses no terminations, and that is partly true.
There are no provisions for parallel or RC terminations on the bus.
Effectively, PCI uses a combination of source (series) termination, and
diode termination, built into each PCI device (IC).
The source termination is there by sizing the drivers so that you get
"reflected wave" switching. This merely means that, ideally, the
driver's output impedance is roughly equal to the characteristic
impedance of the bus. When a driver switches, its output voltage
initially goes only half way to the other state, until the reflection
from the unterminated end of the bus returns to the driver and the
voltage there doubles to reach the full swing. In practice it is not
this ideal. For one thing, you have a very different situation when a
driver is at one end of the bus, versus in the middle of the bus (where
it sees one-half the characteristic impedance). The drivers are no
different whether the PCI bus has one other IC or several, or if it is
plugged into the first slot or the last. Also, IC drive strengths have
a fairly wide tolerance, so the concept of "reflected wave switching" is
only approximate.
To help matters, each PCI device incorporates clamp diodes, which help
control overshoot and absorb switching energy.
CompactPCI improves on the situation somewhat by adding small discrete
series resistors on each cPCI card, between the IC and the connector.
There might be nothing stopping you from adding your own terminations to
a PCI bus, as long as they are not on a PCI plug-in card. If you were
to add something like terminating resistors to the bus, you take your
chances, because nothing was designed and tested with this in mind.
Regards,
Andy
> ----------
> I want to use PCI bus in my design, but I don't know what kind of
> termination can be use in PCI bus with multi-load . Can someone help
> me? =20
>=20
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