Re: [SI-LIST] : Q: Plane-jumping return currents

Larry Smith (ldsmith@lisboa.eng.sun.com)
Wed, 22 Sep 1999 10:10:22 -0700 (PDT)

Eric - The situation you have described is completely true. Return
current associated with a trace via must jump from one reference plane
to another in stackup 2 and 3 below and in most of the stackups in use
in the computer industry today. This is not to much of a problem for a
single trace, but can be a major problem for a wide bus (perhaps 64
signals) with similar vias and all of the signals switching the same
way at the same time. It becomes an SSN noise problem that I call the
"power plane bounce" issue. This happens not only at a via but also at
a connector and at electronic packages containing drivers or
receivers.

You are also correct in saying that the return current must come
through a capacitance somewhere. The parallel plate capacitance
between the planes is generally sufficient for one signal for a short
time period (several nSec). But if there is a far end parallel
termination, DC current may have to go through this capacitance for a
long period of time. Discrete capacitors may be required.

How close does a capacitor have to be to be effective? First, look at
the time of flight of energy on the power planes: 6 inches/nSec in FR4
dielectric. The return current associated with the via creates a
radial disturbance that emanates out from the via at that velocity. It
will take at least 2 time of flights for the capacitor to be
effective: one time of flight for the disturbance to reach the
capacitor and another for charge to make it back to the via. For a
capacitor to be effective during a 1 nSec rise time, it must be within
3 inches of the via.

The loop inductance from a capacitor to the power planes is probably
more important than the capacitance. Suppose we have a 50 Ohm trace
and 1V/nSec edge. The current and return current for the trace is
1V/50Ohms = 20 mA. Most of that return current has to jump planes in
stackup 2 and all of the current must jump planes in stackup 3. The
inductance of a discrete capacitor is determined mostly by the pads and
vias to the power planes. If the power planes are reasonably near the
surface of the PCB and the distance between vias is short, the
inductance in the capacitor/power-plane loop is around 1 nH. The
voltage drop across the loop inductance is V=L*di/dt = 1nH*.02A/1nSec =
20mV, not bad. But if 10 vias are involved then we have 10X the
current and 200mV of drop across the capacitor loop inductance.

I have come to the conclusion that the power plane capacitance must
conduct the return current during the rise time and for the first
several nSec. After that, the time constant involving the loop
inductance of the capacitor will allow it to be effective. The amount
of capacitance required can be calculated by I=C*dv/dt.

The interplane capacitance is easily calculated from
C=e0*eR*Area/thickness. Calculate the capacitance and then calculate
the voltage drop from dv=I*dt/C. In this case, dt is the amount of
time that the interplane capacitance must support the return current.
For source terminated lines (open circuit far end) it will just be a
few nSec. For far end terminated lines, the return current goes on for
ever and discrete capacitors will be required. Eventually, the return
path involves the power supply, but it takes micro seconds before
it can respond.

An interesting side note, the power plane capacitance is not all
available immediately. As described above, only the capacitance within
a 3 inch radius is available in 1 nSec. If we wait long enough, the
radial disturbance on the power plane reaches the edges of the board
and bounces back. Power plane resonance's occur. Some other responses
to the list have already mentioned that the planes can appear to be
inductive at certain frequencies and at certain positions on the
board. The best way to look at the planes is by impedance rather than
capacitance calculations. But that is a long discussion and this note
has gotten too long already..

regards,
Larry Smith
Sun Microsystems

PS - Mike Jenkins has already mentioned that many of these issues
go away with differential signals (true!). But for those 'gluttons for
punishment' who are trying to push single ended signals as far as
they can, this stuff is real important.

> Hi,
>
> Consider a few different partial stackups each with a via:
>
>
> Stackup 1
> | |----------- trace B
> plane =========== | | ==========
> trace A ------------| |
>
>
> Stackup 2
>
> plane =========== | | ==========
> trace A ------------| |
> trace B | |-----------
> plane =========== | | ==========
>
>
> Stackup 3
> | |----------- trace B
> plane =========== | | ==========
> plane =========== | | ==========
> trace A ------------| |
>
>
> In stackup 1, the return currents for trace A and trace B just need to
> migrate to the other side of the the plane which is fairly easy and has a
> low impedance. However, this stackup is not preferred for PCB
> manufacturability reasons.
>
> In stackups 2 and 3, the return current for trace A moving to trace B has
> to jump planes. The only place that can occur is via a capacitance. Some
> capacitance is provided by the interplane capacitance which works better in
> stackup 3 than it does in stackup 2. Otherwise, a nearby bypass cap must
> be found. The farther away the cap is, the larger the inductance (and
> impedance) of the return current path.
>
> My system is running pretty fast (> 1 Gbps).
>
> My questions:
>
> 1. Is what I've described generally true?
>
> 2. How could one analyze how far away a "nearby" cap can be and not degrade
> the signal too much?
>
> 3. How does the value of the cap affect this? Clearly we want a low
> inductance package. Do I just go for the largest capacitance that fits in
> a low-inductance package?
>
> 4. How could one analyze if the interplane capacitance is sufficient for
> this purpose?
>
> -Eric
>
> --
>
> Eric Goodill Cisco Systems M/S SJ-N2
> mailto:ericg@cisco.com 170 W Tasman Dr
> voice: (408) 527-3460 San Jose CA 95134-1706
> fax: (408) 527-3460 (yes, the same)