How Current Return Paths Affect Signal Integrity

When a signal is passed through a trace, it is safe toexpect the majority of the signal will be contained in the trace. But when asignal returns from its load, there are many paths it can take. No matter whatthe return path is, from a non-ground grid to a ground grid or even a groundplane, the return current will take the lowest impedance path. When designingwith a ground plane, the signal is expected to flow under the signal trace as amirror current (See Figure 1), but it is probably returning at lowerfrequencies as shown in Figure 2. Why is this?

Affects of Mutual Inductance

When the current is flowing inopposite directions, which the intended signal and return current are generallydoing, the total inductance looking into two wires is given by Lt = L1+L2 - 2M(Ott Eq. 10 -10)1. Lt is the total inductance looking into the two wires. L1and L2 are the inductance of the intended and return paths in the wires and Mis the mutual inductance between the two wires. This analogy of using two wiresis an approximation and really only works when using non-grounded wires (not atrace and a ground plane), but it helps to illustrate what is happening on thecircuit board. L1 and L2 are often nearly the same value if the two paths arewires and are identical. To minimize Lt, when L1 and L2 are almost the same value,the value for M should be equal to L1 or L2. If this were to happen, Lt wouldbe very small. The values of L1 and L2 come very close to each other in acoaxial cable where the two wires are positioned one inside the other, and thevalue of M is
approximately L1 or L2 , making the total inductance very low.

Considering the various circuit elements in this example,note that the dc resistance of the U path (the yellow path in Figure 1) is 1.5Ohms and the inductance is 10.7 nH looking into the source and ground. Thisinductance is low because of the mutual inductance between the trace and theground plane.

When looking at the path the U trace uses across theground plane as its return path from load to source, this total dc path isstill 1.5 Ohms because of the small amount of dc resistance added from theground plane of approximately 1 m Ohm. There is an inductance associated with this new path of 491nH due to the significant areaadded between the U and the short path from load to source.

Before moving on to the affects of different frequencieson this example, the following points highlight the main aspects of thiscircuit:

Different Frequencies

At 1 KHz our path is basically dc around the U trace of1.5 Ohms and across the very low dc resistance of the ground plane of 1 m Ohm.Figure 2 illustrates how most of the current is flowing in the trace and acrossfrom the load to the source.

At 1 MHz our path is basically around the U trace of 1.5Ohms and our return will be under the trace with its low inductance of 10.7 nH,giving a return impedance of .07 Ohms. The current will now prefer this pathcompared to returning through the large loop, as it did at 1 KHz, since ourlarger loop impedance would now be just a little over 3 Ohms because of its 491nH (See Figure 1). Here you can see most of the current is returning asintended.

Remember, the current will always flow through the leastimpendence path. As a result, on a circuit board with a ground plane, anunsuspecting designer might be surprised to learn the return current at lowerfrequencies could take a totally unexpected path (See Figure 2). Once theengineer realizes this, he or she must also recognize that these currents canaffect other currents also flowing through these unintended paths.

In addition, intended signals can also be distorted andinterfered with by those other intended currents. (Author's note: I have seeninterference with an analog signal at 180 Hz, where the interfering signal was4.5 times greater than the intended signal.)

Layer-to-Layer Transfer

Another area where the quality of the signal might becompromised is when a signal moving across a circuit board must transfer fromone layer to another using a via (see left image in Figure 3), where the returncurrent is on the bottom of the right plane and needs to be on the top of theleft plane.

In this case, the engineer must be concerned with whathappens to the return current if the mirror current on one layer is not commonwith the mirror current on the layer the signal is now flowing above or below.This must be known to know how it will transfer from one plane to another.

If you don't makethis transfer happen the way you want it to, the return current will find itsown way. It might be through a decoupling cap far from the via or it mighttransfer as displacement current using only the layer-to-layer capacitance ofthe planes as seen on the left image in Figure 3. When this happens, thecurrent is spread over a large area, greatly increasing the possibility ofcross contamination with other signal currents creating a loss of signalintegrity. A better way would be to add decoupling capacitors (the closer thebetter) near the via that the signal is passing through, as shown in the rightimage of Figure 3. This contains the current to the vicinity of the signal via.

Donald L. Sweeney, senior EMC engineer and president ofD.L.S. Electronic Systems Inc. has been teaching for more than 30 years at theUniversity of Wisconsin as well as at independent EMC design seminars.

References:
1. Ott, Henry; Electromagnetic CompatibilityEngineering 2009, Wiley.
2. All graphics usedby permission of Dr. Bruce Archambeault.