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Coupling's Effect on Impedance: Why is Zdiff Less Than 2*Zo?

Assume we have a trace (T1) with a controlled impedance equal to Zo. Now bring a second trace (T2) near it (and parallel to it) so that the signal on T2 couples into T1.

Statement: The impedance of T1 (Zo) changes as a result of that coupling.

Consider this statement. It is not qualified. Nothing is said about the type of signal on T2, i.e., whether it might be a differential or common mode signal or anything else. It is simply a flat statement. The impedance of a trace changes if a signal on a parallel trace couples into it. Period.

This raises two questions:

1. Why is this statement true?
2. If it is true, why do we only (seem to) worry about it in the case of differential or common mode signals, but not any other kind of signal?

Coupling Affects Impedance

Consider Figure 1. It illustrates a typical single-ended transmission line. It is driven by a voltage, V1. It has a characteristic impedance Zo. By Ohm’s Law the current on the trace is:

i1 = V1 / Zo  (Equation 1) 

Figure 1. Single-ended trace.

Now consider Figure 2. It illustrates a pair of traces, T1 and T2, that are parallel to each other with signals coupling into each other. Trace T1 is driven by a voltage V1. It has a characteristic impedance equal to Zo. The current from V1 is therefore, by Ohm’s Law, i1 = V1 / Zo. Similarly trace T2 is driven by a voltage V2 and it has a characteristic impedance we will call Zo. (It is not necessary that these two traces have the same characteristic impedance, Zo. It is just convenient.) The current on T2, i2, is also given by Ohm’s Law. 

Figure 2. Parallel coupled traces.

Since the traces are coupled together, each voltage couples into the other trace with a coupling coefficient, k. We may not know the magnitude of k, but we do know the situation is symmetrical, so each trace couples into the other with the same coupling coefficient, whatever it is.

Now, here is an important point: Impedance is a point concept. The impedance can be different at different points along the trace (Note 1.) At the beginning of the trace the driver sees an impedance of Zo, as described above. But the coupling along the trace changes the relationship between the voltage, current, and impedance. Specifically, the voltage at any point on trace T1 is the voltage contributed by the driver (V1) plus the coupled voltage (k*V2) or the voltage at any point (x) is:

Vx =  V1 + k*V2  (Equation 2)

where k is the coupling between the traces.

So the impedance at any point x, (Zx), is the voltage divided by i1, or:

Zx = Vx/i1 = V1/i1  + k*V2/i1 (Equation 3)

Multiply the right-hand term by V1/V1 to get

Zx = V1/i1 + k*V2*V1/(i1*V1)  orZx  = Zo  + k*Zo*V2/V1 ,  orZx = Zo * (1 + k*V2/V1)  (Equation 4)

We can think of k*V2/V1 as being the effect of the coupling of trace T2 on trace T1.

Equation 4 is the impedance of the trace at any point, x, which includes the end point of the trace. To prevent reflections, we would want to terminate trace T1 in a resistor whose value would equal Zx.

Thus, the statement, “The impedance of a trace changes if a signal on a parallel trace couples into it” is true. The correct impedance, and therefore the correct terminating resistance, of the trace is given by Equation 4.

When does it matter?

So if the statement is true, why don’t we talk about it very much? The reason is that it doesn’t matter under most circumstances. In most cases, the adjustment factor, k*V2/V1, is either small or not knowable. Here are some examples:

1. If the traces are far apart, k is near zero.
2. If V2 is small or approximately equal to zero, the adjustment factor is small.
3. If V2 < V1, then the adjustment factor is very small.
4. If V2 is a constant (DC), then k is zero (i.e., only changing voltages couple.)
5. If V2 is unrelated to V1, in the sense of a random variable, then the adjustment factor may be significant, but its average value would be near zero, so no adjustment is possible (see Note 2.)

But there are two situations where (a) the coupling is significant and (b) there is direct correlation between V1 and V2. That is when

1. V2 = - V1 (the case of differential signals, or more correctly odd-mode signals.)
2. V2 = V1  (the case of common-mode signals, or more correctly even-mode signals.)

Differential and Common Mode

Let’s let V2 = -V1. This is the case of differential (odd mode) signals when the return current is on trace T2. This has the effect of reducing the adjustment factor to, simply, -k. Remember, k is simply the coupling coefficient between the traces. In this case, Equation 4 becomes:

Zx = Zo (1 - k) (Equation 5)

Equation 5 describes the appropriate terminating resistor for the signal trace (trace T1), one trace of the differential pair. The other trace (trace T2) also needs to be terminated with Zx. This situation is illustrated in Figure 3a. More commonly we use a single terminating resistor between trace T1 and trace T2 whose value is 2*Zx, as shown in Figure 3b. This single resistor is usually referred to as having the value Zdiff, the differential impedance of the trace-pair. Zdiff then, becomes:

Zdiff = 2*Zo ( 1-k) (Equation 6) 

Figure 3. Terminating a differential pair.

We recognize Equation 6 as being the typical expression for the differential impedance of a differential pair. It is 2 times the single-ended impedance reduced by a factor related to the coupling between the traces. This is how two 50-Ohm traces in a differential pair end up with a differential impedance something less that 100 Ohms (maybe 80 ohms, or so.)

The parallel situation occurs if there is perfect (positive) correlation between the signals on both traces, this time with V2 = V1. This is the common mode (even-mode) case. In this situation, Equation 4 reduces to:

Zx = Zo (1 +k)  (Equation 7)

but now the coupling is significant and matters.  We typically terminate common mode signals (if we terminate them at all) as shown in Figure 4. Each resistor has the value given in Equation 5. But this time, the resistors are effectively in parallel, so the effective termination is one-half Zx. This is where the expression:

Zcommon mode = (1/2)*Zo (1+k)  (Equation 8)

originates from.

Figure 4. Terminating common mode signals.

How to Determine the Value of k

The value for k cannot exceed 1.0 (perfect coupling) nor can it be less than zero (zero coupling). Howard Johnson says that the coupling coefficient (k) is approximately proportional to 1 / [1 + (D/H)2], where H is the height of the trace above the plane and D is the center-to-center distance between the traces (see Note 3). Based on studies I have done, I believe this is a probably a worst-case relationship. Since the relationship also depends on other nearby factors, particularly the influence of any other nearby planes, the value of k is very difficult to determine.

I know of no formulas, other than Howie’s, for estimating k. Otherwise, we must rely on field solver techniques for dealing with coupling and its effects on trace impedance.

Summary

Any time we have coupling between traces, the coupling affects the impedance of the trace. In most cases it either doesn’t matter (because the effect is so small) or we can’t do anything about it (because the effect is pretty much random.) There are two specific cases, however, where such coupling does matter. Those are the differential (odd) mode case and the common (even) mode case. In each of these two cases there is perfect correlation between the signals on the two traces. In these two cases we can adjust our terminating resistors to match the changed impedance and eliminate reflections from the end of the trace.

Notes

1. Our traces on circuit boards have different impedances at different points if we do not control the geometry well enough. That’s the whole point of a “controlled impedance” trace.
2. The signal V2 is unrelated to V1 in a statistical sense if the two signals are uncorrelated. Then, over time, the value of k*V2/V1 averages out to zero, even if the coupling itself is significant. This case should be recognized as the crosstalk case. There is in fact significant coupling between the traces, but since that coupling is random not much can be done about it.
3. Howard Johnson, High Speed Digital Design: A Handbook of Black Magic, Prentice Hall, 1993, page 192. See various articles I have written on crosstalk effects at www.ultracad.com.

Douglas Brooks has an MS/EE from Stanford University and a Ph.D. from the University of Washington. He has spent most of his career in the electronics industry in positions of engineering, marketing, general management, and as CEO of several companies. He has owned UltraCAD Design Inc. since 1992. He is the author of numerous articles in several disciplines, and has written articles and given seminars all over the world on Signal Integrity issues since founding UltraCAD. His book, Printed Circuit Board Design and Signal Integrity Issues was published by Prentice Hall in 2003. Visit his Web site at www.ultracad.com.