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

Eric Goodill ([email protected])
Tue, 21 Sep 1999 10:56:46 -0700


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 Goodill Cisco Systems M/S SJ-N2 mailto:[email protected] 170 W Tasman Dr voice: (408) 527-3460 San Jose CA 95134-1706 fax: (408) 527-3460 (yes, the same)

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