## Common Mode Analysis of Skew

Holding your arms out straight in front of you, let your hands represent two
signals, *a* and *b*. The vertical position of each hand represents
the voltage present on each signal. Now flap your arms up and down in an
opposite, or *antipodal*, fashion. This motion depicts the behavior of a
differential-signal pair.

The distance between your hands represents the *differential* component
of the signal pair [*a*, *b*], and the average position of your hands
represents the *common-mode* component.

To make obvious the meaning of the common-mode component, first find a stick about three feet long. Mark the midpoint of the stick. Now, holding opposite ends of the stick in each hand, repeat the arm-flapping exercise. The midpoint of the stick, which is at all times the mechanical average of the positions of your two hands, represents the common-mode component. The inclination of the stick represents the differential component.

Moving your arms exactly opposite each other such that the midpoint of the stick remains fixed in space creates a perfectly differential-signal pair with no common-mode component. It may help to have someone else hold the midpoint fixed so you can get the feeling of a perfectly antipodal motion. You will find it difficult to create a purely differential signal. Any asymmetry in the amplitude or timing of your arms' motions creates a common-mode component.

In
an electrical circuit, you care about the common-mode component of signal pair
[*a*, *b*] because that component creates far more radiation and
crosstalk than the differential component.

The relation between signal skew and the common-mode signal amplitude is
fairly easy to derive. Suppose signal *b* is the same as negative *a*,
only delayed by a small skew, *Δt*. You can then express the common-mode
signal *c*≡(*a*+*b*)/2 as

Those of you familiar with calculus will recognize that the numerator of the
right-hand term, for sufficiently small *Δt*, approximately equals *Δt* multiplied by the time derivative *da/dt*. Caution-this
approximation works only when *Δt* remains smaller than the signal rise
time.

The peak value of *c* coincides in time with the peak value of the
derivative of *a*. Assuming a 10 to 90% rise time, *t*_{r},
and a signal swing, *Δv*, on each wire, you can approximate the peak
value of the derivative of *a*(*t*) as *Δv*/*t*_{r}. After suitable rearrangement of terms, this
calculation leads to the following expression for the peak common-mode
amplitude:

This derivation assumes that signal *a* swings from zero to *Δv*,
and signal *b* swings from *Δv* to zero. The differential signal
(*a*-*b*) therefore swings from +*Δv* to -*Δv*.
Therefore, the *peak-to-peak* *amplitude* of the differential signal
is 2*Δv*, the *peak amplitude* of the differential signal is *Δv*, and the peak amplitude of the odd-mode signal (defined as half the
differential-mode signal) is *Δv*/2.

Look at the peak amplitude of the common-mode signal as a percentage of the
peak amplitude of the differential-mode signal, *Δv*. Also, imagine the
skew, *Δt*, to be a percentage of the rise time, *t*_{r}.
With these two concepts in mind, you can reduce this entire discussion to the
following rule: A 20% skew creates a 10% common-mode component.