NETWORK ANALYSIS FOR DIGITAL ENGINEERS - Advanced Signal Integrity for High-Speed Digital Design - page 401

Z odd = 50

50 50

∼

v

Z odd = 50 Z 0 = 50

50 50

∼

− v

I

∆

Figure 9-38 Unbalanced differential pairs cause energy to be converted from the odd
mode to the even mode, which looks like additional differential insertion loss.

energy is in the odd mode, which means that for all practical purposes, equation
(9-67b) is sufficient for design purposes.
Invariably, a question is asked: If there is significant energy in the even mode,
doesn't the common mode matrix need to be accounted for ? The answer is that
it is already included in the differential matrix of (9-67b). If an asymmetrical
four-port system is driven differentially, the energy that is converted to common
mode will simply look like differential insertion loss. For example, consider the
loss-free unbalanced differential transmission line pair being driven in the odd
mode in Figure 9-38. Since the system is being driven differentially (in the odd
mode), and the odd-mode impedance is equal to the termination, there will be no
reflections. However, since it is unbalanced, it is expected that a portion of the
energy will be transferred to the even mode. From a differential receiver's point
of view, this will look like extra channel loss.
To demonstrate this concept, consider the case where the difference in length
between the legs of the differential pair ( l) is 200 mils with a delay of 27.77 ps
(refer to Figure 9-38). If the initial phase difference between v + and v − at the
driver is of 180 ◦ , when the signals arrive at the receiver, the phase difference
will no longer be equal to 180 ◦ , due to the extra length on the second leg. The
actual phase difference at the receiver can be calculated using equation (9-22)
as a function of frequency. For example, at 1 GHz the phase difference at the
receiver will be

180 ◦ + θ = 180 ◦ + ( 360 ◦ )( 27 . 77 190 ◦ 10 − 12 )( 1 10 9 ) = × ×

meaning that since the signals are no longer exactly 180 ◦ out of phase, and some
of the energy is being dumped into the even mode.
As the frequency increases, the phase difference due to the extra delay will
approach 180 ◦ . When θ = 180 ◦ , the differential signal at the driver has been

fully transformed into a common-mode signal at the receiver. When the delay
difference between the legs of the differential pair is 27.77 ps, the frequency
where the signal is 100% common mode at the receiver is 18 GHz:

= ( 360 ◦ )( 27 . 777 180 ◦

10 − 12 )(f 180 ) =

θ ×

10 9 ⇒ f 180 = 18 ×