The two traces comprising a differential pair, when routed close together, share a certain amount of cross-coupling. This coupling lowers the differential impedance between the traces.
For example, when two 50Ω traces are well-separated, the differential impedance between them should be precisely 2Z0=100Ω. When you jam these traces close together, as you might do to improve routing density, the differential impedance will be a lower value, perhaps something like 82Ω.
If the coupling effect lowers the differential impedance too much for your taste, you can fix it. Just reduce the trace widths, thus raising their impedance. The traces remain coupled, but you can push the differential impedance back up to 100Ω.
What happens to a tightly coupled pair when it traverses an obstacle, such as a via (Figure 1)? If you have room to route both traces on the same side of the obstacle, maintaining their constant separation, no special problems arise. If, on the other hand, you separate the traces to pass by, then the differential impedance in the separated region reverts to its original, uncoupled value of 2Z0. If you have thinned the trace widths to produce exactly 100Ω in the coupled state, then the reverted, uncoupled impedance with skinny traces will exceed 100Ω .
To calculate (approximately) the effects of such a mismatch, let's assume you have a long, uniform transmission line of impedance, Z0. Add to it a short section of mismatched impedance, Z1, having length (in time) T. Further assume that T is much less than the signal rise time (or fall time) TR, so the whole mismatch section acts as a simple lumped-element circuit.
Analysis begins by computing the L and C values corresponding to the mismatched section: L= Z1T, and C =T/Z1. Next, mentally break the full value of inductance L into two pieces, L =L'+LF, with L'=Z02C.
The beauty of breaking up L is that the natural impedance of the combination of L' and C is now exactly Z0, as you can see by forming the ratio of one to the other: L'/C =Z02. Inductor LF represents the excess inductance in the mismatch region above and beyond L'. In other words, you have modeled the mismatch region as a short transmission line of impedance Z0 (comprising L' and C) plus a series inductance, LF. You will next model the reflection generated by component LF.
When a fast step input hits an inductive discontinuity of size LF, you get a reflected pulse. The pulse duration equals the rise time of the incoming step. You can approximate the reflection coefficient RL (ratio of reflected pulse height to the incoming step size) as follows:
RL (1/2)(LF/Z0) (1/TR)
Substituting your known expression for LF yields:
Further substituting your basic expressions for L and C gives:
RL (1/2)((TZ1–Z02T/Z1)/ Z0)(1/TR )
And consolidating the terms provides:
That's about the best simple approximation you will find for the case of a short separation between the elements of a differential pair. If you want more accuracy, use a simulator.
In the example of Figure 1, taking the ratio of Z1/Z0 to be (100Ω ) /(82Ω)=1.22, the equation for RL reduces to:
If this amount of signal degradation is troublesome, try thickening the traces in the separated region to match the impedance of the thinner, more highly coupled traces elsewhere.
If you've kept T/TR less than 1/6, you can expect an accuracy of a couple of percentage points from my approximation. If you try to stretch the approximation to a ratio of T/TR as big as 1/3, expect the approximation to be good only to about 20, and beyond that, at T/TR=1/2, it will probably fall completely apart, delivering totally erroneous answers.
The same approximation works for BGA layouts, where signals escaping from the inner rows neck down to pass through the BGA ball field. The neck-down region raises the local trace impedance in a small region.