When a signal is passed through a trace, it is safe to
expect the majority of the signal will be contained in the trace. But when a
signal returns from its load, there are many paths it can take. No matter what
the return path is, from a non-ground grid to a ground grid or even a ground
plane, the return current will take the lowest impedance path. When designing
with a ground plane, the signal is expected to flow under the signal trace as a
mirror current (See Figure 1), but it is probably returning at lower
frequencies as shown in Figure 2. Why is this?
Affects of Mutual Inductance
When the current is flowing in
opposite directions, which the intended signal and return current are generally
doing, 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. L1
and L2 are the inductance of the intended and return paths in the wires and M
is the mutual inductance between the two wires. This analogy of using two wires
is an approximation and really only works when using non-grounded wires (not a
trace and a ground plane), but it helps to illustrate what is happening on the
circuit board. L1 and L2 are often nearly the same value if the two paths are
wires 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 would
be very small. The values of L1 and L2 come very close to each other in a
coaxial cable where the two wires are positioned one inside the other, and the
value 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 1.5
Ohms and the inductance is 10.7 nH looking into the source and ground. This
inductance is low because of the mutual inductance between the trace and the
When looking at the path the U trace uses across the
ground plane as its return path from load to source, this total dc path is
still 1.5 Ohms because of the small amount of dc resistance added from the
ground plane of approximately 1 m Ohm. There is an inductance associated with this new path of 491nH due to the significant area
added between the U and the short path from load to source.
Before moving on to the affects of different frequencies
on this example, the following points highlight the main aspects of this
1.5 Ohms in the U trace.
- 1 m Ohm across the left end on the ground plane.
- 10.7 nH looking into the source and returning via the ground plane under the
- 491 nH looking into the source and returning via the left end of the ground
At 1 KHz our path is basically dc around the U trace of
1.5 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 across
from the load to the source.
At 1 MHz our path is basically around the U trace of 1.5
Ohms 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 path
compared to returning through the large loop, as it did at 1 KHz, since our
larger loop impedance would now be just a little over 3 Ohms because of its 491
nH (See Figure 1). Here you can see most of the current is returning as
Remember, the current will always flow through the least
impendence path. As a result, on a circuit board with a ground plane, an
unsuspecting designer might be surprised to learn the return current at lower
frequencies could take a totally unexpected path (See Figure 2). Once the
engineer realizes this, he or she must also recognize that these currents can
affect other currents also flowing through these unintended paths.
In addition, intended signals can also be distorted and
interfered with by those other intended currents. (Author's note: I have seen
interference with an analog signal at 180 Hz, where the interfering signal was
4.5 times greater than the intended signal.)
Click here for larger image
Another area where the quality of the signal might be
compromised is when a signal moving across a circuit board must transfer from
one layer to another using a via (see left image in Figure 3), where the return
current is on the bottom of the right plane and needs to be on the top of the
In this case, the engineer must be concerned with what
happens to the return current if the mirror current on one layer is not common
with 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 make
this transfer happen the way you want it to, the return current will find its
own way. It might be through a decoupling cap far from the via or it might
transfer as displacement current using only the layer-to-layer capacitance of
the planes as seen on the left image in Figure 3. When this happens, the
current is spread over a large area, greatly increasing the possibility of
cross contamination with other signal currents creating a loss of signal
integrity. A better way would be to add decoupling capacitors (the closer the
better) near the via that the signal is passing through, as shown in the right
image of Figure 3. This contains the current to the vicinity of the signal via.Donald L. Sweeney, senior EMC engineer and president of
D.L.S. Electronic Systems Inc. has been teaching for more than 30 years at the
University of Wisconsin as well as at independent EMC design seminars.
References:1. Ott, Henry; Electromagnetic Compatibility
Engineering 2009, Wiley.2. All graphics used
by permission of Dr. Bruce† Archambeault.