I have been on a historical kick recently reading original technical works penned at the dawn of the electrical age. If you plan any quiet time this winter sitting around the old fireplace, I heartily recommend devoting a few evenings to Benjamin Franklin, "Experiments and Observations on Electricity", recently republished in 1996 by the "Classics of Science Library" (original date 1769). The scope of knowledge gleaned by that man from a few well-planned experiments is breathtaking.
A review of those old topics has also brushed up my knowledge of Ampere’s law and also Gauss’, making it possible for me to pen this analysis of crosstalk between differential vias.
Crosstalk - Differential Vias
Silence DoGood writes:
My board has lots of hot differential digital signals, plus some sensitive analog circuitry (also differential).
My CAD tools predict the level of crosstalk from differential digital traces to differential analog traces. That's fine, but how about the crosstalk from differential digital vias to differential analog vias? How does that work and how big is it?
Thanks for your interest in High-Speed Digital Design.
The expression for the inductive component of inductive crosstalk (the lion's share of crosstalk in a differential-via coupling situation) is fairly straightforward. If I understand your question correctly, you have a multi-layer board, with (at least) two solid planes. A digital differential signal passes from the top layer to the bottom, traversing the inter-plane cavity, on a differential via pair. Somewhere nearby, another via pair conveys an analog differential signal. You seek an expression for the crosstalk caused by fields trapped between the planes (via crosstalk). Here's how that situation works.
Imagine your board lying on a table, with the Z-axis sticking up vertically out of the board. The thickness of the board is H. What I mean by H, precisely, is the inside-surface separation between the two solid planes.
Place the digital vias at locations x1 and x2. Each of these position indicators comprises a two-dimensional location on the board. Place the analog vias at locations y1 and y2. I will assume that the distance dx between the vias of each pair appears small compared to the separation between the analog and digital worlds. You will see if you read the mathematical appendix at the end of this message how those assumptions come into play.
The mutual inductance between the aggressor via pair and the victim via pair is approximated (worst case) by:
I like this simple approximation because it shows that differential crosstalk, in the absence of any ground vias, falls off quadratically with separation R. If you pepper your board with interplane ground vias, crosstalk falls off even faster (the subject of my EDN article, "Differential Vias with Grounds").
Now let's play around with via orientation.
To grasp these concepts, first bisect the line separating x1 and x2 and pin the center of this line to the board, so the vias can rotate freely in pinwheel fashion to any orientation, but the centroid of the pinwheel remains fixed, as does the separation between those two vias.
Do the same for vias y1 and y2.
Holding vias x1 and x2 in any fixed position, let vias y1 and y2 pinwheel slowly about their central axis. Common sense tells you that reversing the two via locations y1 and y2 must invert the received crosstalk. Therefore, as the two vias y1 and y2 pinwheel slowly around each other, the polarity of crosstalk must fluctuate from positive to negative and back again once each revolution. From this simple principle, you may deduce that there must be a pinwheel position that nulls the crosstalk to zero. This position aligns y1 and y2 tangent to the local pattern of magnetic field lines. Ninety degrees away from this position maximizes the crosstalk (the maximizing assumption is built into the formula given above).
Given mutual inductance Lm, calculate the coupled noise voltage Vn in the victim circuit by multiplying the rate of change of the source current times the mutual inductance.
- Vs is the differential signal swing of the source (differential peak to peak),
- Z0 is the differential characteristic impedance of the source,
- Tr is the 10-90% rise time of the source, and
- Lm is the estimate of mutual inductance given in the previous formula, or using the more accurate expression in the mathematical appendix.
Half of the coupled noise voltage Vn propagates in either direction away from the victim via pair. In a both-ends terminated architecture, the resulting noise voltage at the receiver equals half of Vn. In a receiver with poor source termination, the signal initially propagating backwards eventually reflects off the driver and returns to the load, possibly doubling the received signal to a level equal to Vn.
For digital purposes, crosstalk at the level of a few millivolts is usually not a problem unless you aggregate together lots of aggressors, but for analog applications, even 1 mV can be a killer.
Hint: Dividing your reference planes into distinct analog and digital regions prevents the flow of waves of current across the dividing boundaries. These waves, with their attendant magnetic fields, are the root cause of via crosstalk. Separating analog and digital planes, therefore, reduces via crosstalk.
Another hint: analog applications that incorporate low-pass filters in the analog circuitry may substantially benefit from the reduced signal amplitude resulting from low-pass filtering of the crosstalk.
I hope these comments are helpful to you.
Dr. Howard Johnson
I have hidden the details down here for those of you interested in such things.
The magnetic field intensity between the planes in a pcb via problem remains constant in the Z-axis direction, simplifying the problem enormously. Even more simplifications follow from the fact that the field pattern from each via is circularly symmetric about the via. NOTE: this discussion assumes infinite planes with no boundaries. If you encounter plane boundaries then the fields bounce off these discontinuities, making the situation more complicated.
With digital vias at locations x1 and x2, I will assume the via at position x1 carries a current of one amp. An equal but opposite current flows at x2. Ampere's law gives the steady-state magnetic field intensity at a distance r removed from via x1 as μ0/2πr, where μ0 equals the magnetic permeability of free space and r is the distance. If r has units of inches, the constant μ0/2π works out to 5.08 nH/inch.
Given the stated field pattern emanating from the via at position x1, and an equal but opposite pattern, slightly offset, emanating from the via at x2, you can calculate the value of mutual inductance Lm between the digital and analog via pairs, and from that value estimate the crosstalk.
To find Lm, you must calculate the total integrated field intensity passing between the vias y1 and y2 at a time when vias x1 and x2 carry a current of one ampere (down one via and back on the other). The procedure comprises the fundamental definition of inductance, a direct consequence of Gauss’ law for magnetism.
The integration is not as difficult as it sounds; there is a closed-form expression for the result. To begin, Figure 1 illustrates the pattern of magnetic field lines that result from current flowing in via x1. The circular symmetry of the situation plays a big role in establishing the solution. Because the magnetic lines of force from via x1 flow in a circularly symmetric pattern about the via, the total flux penetrating through the imaginary purple line linking vias b1 and b2 precisely equals the total flux associated with the two vias in alternate positions c1 and c2. As a result, the azimuthal positions of any two vias b1 and b2 do not enter the equation for total integrated flux. All you need to determine the total integrated flux penetrating through the purple line between b1 and b2 are the radial distances r1 and r2.
Figure 1—Because magnetic flux from via x1 flows in a circularly symmetric pattern about the via, the total flux passing between c1 and c2 (four arrows shown) precisely equals the flux passing between b1 and b2.
In performing these calculations I assume the vias are small compared to the other dimensions of the problem so that it makes no sensible difference whether we refer to the radial distance from x1 to the near side of a via, its center, or its far side. To simplify the description I shall define all distances relative to the via centers.
If the field intensity as a function of radial distance from x1 equals μ0/2πr, and you seek the total integrated flux falling into the gap between two objects located at distances r1 and r2 away from x1, then the flux is given by:
This flux expression must be further integrated in the Z direction from floor to ceiling of the inter-plane cavity. Since the flux is uniform in the Z direction you need only multiply  by the cavity height H (units of inches), leading to the final form of the expression of the partial flux coupling Lm,1 accounting for the interaction of the single via x1 and the via pair y1, y2:
Equation  substitutes for the distances r1 and r2 the expressions |x1-y1| and |x1-y2|, respectively. It also replaces the value μ0/2π with 5.08 nH/inch. Please note that you cannot use expression  to calculate crosstalk from single-ended vias without first establishing where the return current associated with the single-ended via might be found, and accounting for its effect. This is a partial inductance and only makes sense when the other component of current flow is included in the equations, as we shall do next.
The second via x2 produces a second term with form similar to , only substituting the position x2 position for x1, and subtracting the result (because current in the second via appears negative with respect to the first). Combining the subtracted terms inside the logarithm produces the following expression for the total mutual inductance Lm.
In this last section, I will show an upper bound for crosstalk, calculated using all the points in a straight-line orientation (like an astronomical syzygy). This orientation is a worst-case example. In the straight-line case, without loss of generality, assign x1 and y1 to the inside two locations. Define their separation to be R. Let dx represent the separation |x2–x1| and dy the separation |y2–y1|. From those rules you may define four substitutions:
|x2–y1| = R+dx
|x1–y1| = R
|x2–y2| = R+dy+dx
These substitutions, suitably inserted, lead to:
Now remembering that each distance dx, dy measures far less than R (one of our earliest assumptions), expression  must sensibly differ very little from:
Finally, approximate the function ln(1+α) by its argument α, a rule that holds for small arguments α, to obtain:
Expression  represents the worst case under all possible orientations of x1, x2, y1 and y2, and serves as a good general guide to differential via coupling. Expression  overestimates crosstalk in cases of close proximity or unusual orientation. For better accuracy, use .
Appendix B: Via Orientation
I shall quote here without proof other interesting facts about via orientation. These apply to vias located well away from the edge of the planes.
- Align x1 and x2 so the line a connecting them passes through the centroid of y1 and y2, and orient y1 and y2 perpendicular to line a. This configuration nulls the crosstalk to zero.
- Alternately, Align y1 and y2 so the line b connecting them passes through the centroid of x1 and x2, and orient x1 and x2 perpendicular to line b. This configuration also nulls the crosstalk to zero.
- Turn both pinwheels so the vias appear in a straight line (a syzygy). This orientation maximizes the crosstalk.
- Turn both pinwheels so the via pairs face each other broadside. This orientation also maximizes the crosstalk.
Questions & Comments: all students who attend our High-Speed Digital Design seminars have the opportunity to talk directly with Dr. Johnson about signal integrity issues.