Re: Vias and decoupling (was Re[x]: impedance across ground plane)

[email protected]
Mon, 26 Feb 1996 09:44:41 -0500

On Feb 25, 2:21pm, Andy Ingraham wrote:
> Subject: Vias and decoupling (was Re[x]: impedance across ground plane)
> > Consider a board stackup that has signals referenced against both ground and
> > power planes. from an AC point of view the power planes are as good as ground.
> > If the signal is driven in reference to that plane (usually not the case) it
> > will work fine. If the signal is ground referenced, and uses the power plane
> > for impedance control, it will work fine if the plane is decoupled to the
> > driver's reference electrically close to the driver and to the receiver.
> When you talk about which plane the "signal is driven in reference to",
> I think what you really need to think about is which plane the switching
> current comes from. The reference voltage is important at receivers,
> but not so much at drivers.
> For CMOS and TTL, the current comes from the power voltage plane 50% of
> the time, every time the driver switches high.
> Forget voltages. Think currents.

I agree. 3-D modeling can help. A feel for where current may want to flow can
be inferred by determining an impedance matrix in neighborhood of each large
chip, termination bank, or geometric anomaly. Include decoupling caps and
multiple plane PWB geometry. Its helpful to start at Vss, Vdd, Vcc, etc. on
the die. Don't forget die capacitance. Once you've done this you can determine
if the hypothesis that all planes are equal as transmission line returns was OK.
More important is now that you determine accuracy limitations about certain time
domain tools which use the concept of "ground". Guess what? Sometimes the one
ground concept is useful. I wonder how may definitions for ground we can come
up with? :-) Maybe that should be the topic of another thread.

> Let's say an output is low, and the line has settled so there is little
> current remaining. The driver switches high, by closing a "switch"
> between Vcc/Vdd and the output pin. Current goes out that pin and into
> the etch. Simultaneously, return current comes back on the reference
> plane(s) surrounding the etch. If those planes are power planes, then
> all is happy ... for this case of a low-to-high driven edge.

A good reason to keep the drive strength as low as possible.

> If one or both planes around the etch is/are ground, then you must get
> the current coming back on the ground plane(s), into a current in the
> Vcc/Vdd pins, to complete the circuit into the "switch" inside the
> driver. You do that by bypassing the planes near the driving chip.
> And the situation is reversed for high-to-low edges.
> > Now,
> > if the signal is routed via vias in a dense board to use multiple power and
> > ground planes as its reference, how do I assure or even model signal integrity?
> >
> > One suggestion is that at each via, a matching ground to voltage plane
> > decoupling cap be added. Any ideas?
> This really isn't any different than the situation at the driver, or at
> loads that consume switching current. In either case, you must make
> sure there is enough bypassing between your reference planes, to handle
> the switching current that must pass from plane to plane.
> Some of this bypassing is internal to the board, through the intrinsic
> capacitance between planes. But that isn't enough when an IC switches
> all its outputs, so we add capacitors at such "hot spots". A single via
> by itself has enough bypassing around it so you don't need to add a
> discrete capacitor. But a mass of vias, far removed from any ICs (with
> their bypass capacitors), might need some help.
> Even surface mount capacitors self-resonate at rather low frequencies.
> At the highest switching speeds, they look like inductors, so you depend
> on the intrinsic capacitance between reference planes in the board, for
> those first few nanoseconds.
> And yet, consider a via from one outer layer to the other outer layer
> (as a worst-case scenario), with a number of intervening reference
> planes. The return current of the transient edge, runs under the etch
> on the first layer until it reaches the via anti-pad; and then it seems
> to have nowhere to go! It spreads out around the via, finding an
> increasing capacitance to the next reference plane down; and then to
> the next plane; and so on, until it reaches the bottom reference plane,
> where it re-forms as the mirror current under the etch there. If you
> have picosecond edges, this is something to worry about. It's somewhat
> similar to running etch over a reference plane split.
> I've been told that east coast (USA) engineers equate vias with
> capacitance, whereas west coast engineers say it's inductance. I've
> wondered if it's the inductance of the via barrel itself, or of the
> somewhat longer path the return current must take.
> Regards,
> Andy
>-- End of excerpt from Andy Ingraham