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Re: Expansion board decoupling specs
> >Not a good idea. The decoupling requirement doesn't have anything
> >to do with power, it has to do with signal integrity.
> Quite true - if we talk about the board with the slots. I thought
> Craig was talking about a card which will be plugged in a PCI slot,
You're correct; he was talking about a card which will be plugged
into a PCI slot.
> where decoupling unused power pins does not make much sense because
> the bus will not be decoupled if the card is out of its slot.
True. If there is no card plugged into a slot, then the power pins
of that slot will not be decoupled. However, if there is no card plugged
into a slot, there is also no current flowing through the signal pins
into that connector either. Therefore, that there is no current that
needs a low-impedance return path.
Suppose I have a driver that is driving a net from one board to a
receiver on another board and that net goes through a connector.
|\ current XX |\
| \ --> XX | \
|D >--------------XX------------|R >
| / XX | /
|/ XX |/
Where D is the driver, R is the receiver, and C is the connector.
If the line is high, the driver is sourcing current into the line and
the receiver is sinking it and current flows in the direction indicated.
However, current always flows in loops, so there needs to be a return
current from the right board back to the left board which equals the
current flowing in the wire. This return current flows on the ground
There are two components to the current: DC and AC. During steady state
the current flow is DC and the return current flows on the DC ground plane.
During switching, there is some AC current component and the return
current for that flows on the AC ground plane. An AC ground is any plane
of constant potential: Ground, +5V, +3.3V, +12V, and -12V. So, during
switching, the AC return current will want to take the path of least
AC resistance back to the driver and this could be on one of the power
planes even if that plane has no DC current draw.
The AC return current will want to stay as physically close to the driving
current as possible. This keeps the area in the current loop as small as
possible. By decoupling all of the power planes, you give the AC return
current the ability to use the closest plane for return. If you don't
decouple all of the power planes close to the connector, you are forcing the
AC return current to go out of it's way to find a return path. This exta
bit of loop causes the return current to (a) be out-of-phase with the driving
current and (b) increases the area of the current loop. Both of these
will increase the noise on the signal and since this is all happening during
switching time, it shows up as degraded signal quality.
Again, if there is no card plugged into a slot, then there is no driving
current which requires a return current and therefore you don't need the
> I guess I just have misunderstood the term 'board'.
If you don't understand why something is in a spec, don't just ignore it.
You ought to ask questions to find out why it is there and understand the
I'm not an analog/EMI guy, so some of this might be oversimplification
and I'm sure that there are other people out there who could explain
it better, but there is a good reason why the PCI connector is pinned
out such that every signal is next to either a power or a ground pin.
And from pins 34-38, there is a one-to-one ratio of AC ground to control
signal because signal quality on control signals is (arguably) slightly
more important that on data signals, who only have a one-to-two ratio of
ground to signal in their sections of the connector.
> Dimiter Popoff
> Transgalactic Instruments, Gourko Str. 25 b, 1000 Sofia, Bulgaria
> Email: firstname.lastname@example.org, email@example.com
> Phone: 00359/2992/3340, 00359/2/9805997, Fax: 00359/2/9540384
Note: I speak for myself, not for Auspex.