What is differential impedance? Differential impedance is the ratio of voltage to current on a pair of transmission lines when driven in the differential mode (one signal positive and the other negative).
For example, the circuit in Figure 6.10 drives a signal x ( t ) differentially into a pair of uncoupled transmission lines [48] . Because the lines are symmetrical, the voltage splits evenly and you see voltage ½ x ( t ) on the top line and  ½ x ( t ) on the bottom. The current through each load must therefore equal ½ x ( t )/ Z . This current flows through both lines, and through both loads, in the direction shown.
[48] Traces separated by more than four times the trace height h interact so slightly that for purposes of impedance analysis, one usually ignores the effect of one on the other. Such traces are said to be uncoupled . The exact degree of separation required to produce an inconsequential interaction of course depends upon what you consider to be inconsequential.
Figure 6.10. The differential input impedance of a pair of noninteracting transmission lines is 2 Z .
The differential impedance (ratio of differential voltage to current) is
Equation 6.13
where 
Z DIFF is the differential impedance of a pair of uncoupled transmission lines ( W ), 
Z is the characteristic impedance of either line alone, also called the uncoupled impedance ( W ), 

x ( t ) is the differential voltage applied across both lines (V), and 

½ x ( t )/ Z is the current driven into each line by the source (A). 
The differential impedance of two matched, noninteracting transmission lines is double the impedance of either line alone .
If the lines are coupled , the situation changes. As an example of coupling , think of two parallel pcb traces. These traces will always exhibit some (perhaps very small) level of crosstalk. In other words, the voltages and currents on one line affect the voltages and currents on the other. In that way the two transmission lines are coupled together.
With a pair of coupled traces, the current on either trace depends in part on crosstalk from the other. For example, when both traces carry the same signal, the polarity of crosstalk is positive, which reduces the current required on either trace. When the traces carry complementary signals, the crosstalk is negative, which increases the current required on either trace. Figure 6.11 illustrates the situation. The characteristic impedance of each line to ground (including consideration of the adjacent trace) is represented as a single lumpedelement component Z 1 . The impedance coupling the two transmission lines is depicted as a single lumpedelement device Z 2 .
Figure 6.11. Driving point impedance of two different twowire scenarios.
In the first scenario SigB is driven identically with SigA . The voltages on the two wires are therefore the same at all points in time, so (conceptually) no current flows through the coupling impedance Z 2 . The impedance (ratio of voltage to current) measured for each wire under this condition equals Z 1 . This is called the evenmode impedance of the transmission structure.
In the second scenario SigA and SigB are driven with complementary signals. This is called differential signaling, or sometimes antipodal signaling. Under these conditions the midpoint of impedance Z 2 remains at a virtual ground. An AC analysis of this circuit reveals that the impedance measured for each wire under this condition equals Z 1 in parallel with half of Z 2 . This value of impedance is called the oddmode impedance. The oddmode impedance is always less than the evenmode impedance .
The terms evenmode impedance and oddmode impedance are closely related to the terms commonmode impedance and differentialmode impedance . Commonmode impedance is measured with the two wires driven in parallel from a common source. Commonmode impedance is by definition half the evenmode impedance.
Differentialmode impedance is twice the oddmode impedance. It is measured using a wellbalanced source, computing the ratio of the differential voltage (twice the oddmode voltage) to the current on either line.
Combining your knowledge about common and differential mode impedances, you should be able to prove that the commonmode impedance always exceeds onefourth of the differentialmode impedance.
Just as an example, suppose we have two 50ohm, uncoupled transmission lines. As long as the lines are far enough apart to remain uncoupled, the evenmode and oddmode impedances for these two lines will be the same, and equal to 50 ohms. Should you connect these two lines in parallel, the commonmode impedance would be 25 ohms. Should you connect these two lines to a differential source, the differential input impedance of the pair of lines would be 100 ohms.
POINTS TO REMEMBER
6.9.1 Relation Between OddMode and Uncoupled Impedance
I should now like to discuss the relation between the oddmode impedance and the uncoupled impedance . The uncoupled impedance Z C is what you would measure if the same transmission lines were widely separated, so they couldn't interact. What you need to know is simple: The oddmode impedance of a coupled transmission line is always less than the uncoupled impedance. The evenmode impedance is always greater . The closer you place the line, the more coupling you will induce, and the greater a discrepancy you will see between the oddmode, uncoupledmode, and evenmode impedances.
Let's codify this into a differential impedance principle:
Coupling between parallel pcb traces decreases their differential (or oddmode) impedance .
When implementing tightly coupled differential traces on a pcb, one normally reduces the width of the lines within the coupled region in order to compensate for the expected drop in differential impedance.
6.9.2 Why the OddMode Impedance Is Always Less Than the Uncoupled Impedance
The proof relies on the construction of a thing called an equipotential plane midway between two differential traces. Due to symmetry, all the electric fields in the oddmode situation will lie perpendicular to this plane. Therefore, the potential everywhere along the equipotential plane will be zero. If the potential everywhere along the plane is zero, I could replace the imaginary equipotential plane with a real, solid copper wall and it wouldn't make any difference. The oddmode characteristic impedance is not affected by the wall. What's really neat about this construction is that once the wall is in place , the problem is partitioned into two noninteracting zones. This gives us a way to evaluate the oddmode impedance using a single trace and a solid copper wall instead of two traces. If you think about the impact of the wall on the characteristic impedance of a single trace, it's pretty obvious that the wall will add capacitance and decrease the impedance. The net result of this argument is that the oddmode impedance of a coupled structure is always less than the uncoupled impedance Z .
6.9.3 Differential Reflections
High Speed Digital Design Online Newsletter , Vol 2, Issue 21 John Lehew writesIn the HighSpeed Digital Design book and in a few other places it states the fractional reflection G coefficient caused by a mismatch in impedance is Equation 6.14
This formula is typically used to calculate reflections that happen in a single transmission line that is referenced to a ground plane. Does this formula also apply to differential or balanced lines? 
ReplyThanks for your interest in HighSpeed Digital Design . Aside from the complications introduced by unbalanced modes, differential transmission lines behave pretty much like singleended ones. Equation [6.14] applies to both. Assume I have a section of differential transmission line with a differential impedance of Z 1 . Assume I couple that into a load with differential impedance Z 2 (it doesn't matter whether Z 2 is a lumpedelement load or another section of differential transmission line with characteristic impedance Z 2 ). The size of the signal that bounces off the joint, in comparison to the size of the incoming signal, is given by your equation for the reflection coefficient G . Let's do an example using unshielded twistedpair cabling (UTP). Suppose I couple a section of category 5, 100 W (nominal) UTP cabling to another section of category 4 120 W (nominal) UTP cabling (such cable is available only in France). The reflection off the joint will be of (nominal) size: Equation 6.15 Now, what could go wrong with this simple example? If the cable is inherently un balanced (i.e., more capacitance from one side to ground than on the other), then you have a more complicated situation. In general, there are four modes of propagation involved in the problem, one differential mode and one common mode for each of the two cables. The complete problem is described by a 4x4 coupling matrix whose entries vary with frequency. Imperfections in the balancing of the cable result in crosscoupling between the differential modes and the common modes at the joint, which is one of the things that creates EMI headaches . 
POINT TO REMEMBER
Fundamentals
Transmission Line Parameters
Performance Regions
FrequencyDomain Modeling
Pcb (printedcircuit board) Traces
Differential Signaling
Generic BuildingCabling Standards
100Ohm Balanced TwistedPair Cabling
150Ohm STPA Cabling
Coaxial Cabling
FiberOptic Cabling
Clock Distribution
TimeDomain Simulation Tools and Methods
Points to Remember
Appendix A. Building a Signal Integrity Department
Appendix B. Calculation of Loss Slope
Appendix C. TwoPort Analysis
Appendix D. Accuracy of Pi Model
Appendix E. erf( )
Notes