[SI-LIST] Re: Question regarding return current in a differential pair

  • From: Doug Brooks <doug@xxxxxxxxxx>
  • To: Scott McMorrow <scott@xxxxxxxxxxxxx>
  • Date: Fri, 21 Apr 2006 09:50:28 -0700

Draw a picture of an *ideal* trapezoidal repetitive waveform and then think 
about what you said.
During transition time, there is a high di/dt component.  (hence a high 
coupling component to the plane).

During the part of the waveform where the signal has *stabilized*  the 
instantaneous di/dt is very low or zero (hence a lower coupling component 
to the plane).

Nevertheless that part of the signal changes polarity twice a cycle. If you 
want to think there is no AC component there, that is your prerogative. I 
happen to think there is.

Doug Brooks, PhD

At 11:16 AM 4/20/2006, Scott McMorrow wrote:
>I beg to differ.  If the signal has "stabilized" there is therefore no AC 
>component.  If there is no AC component, there is nothing to radiate.
>Scott McMorrow
>Doug Brooks wrote:
>>In my humble opinion, and not counting common mode currents:
>>During the signal rise and fall times, the return current tends to flow on
>>the reference plane, just as signals on single-ended traces do.
>>During the time that the signal is "stabilized," there is no coupled signal
>>on the plane and the loop is around from one trace of the differential pair
>>to the other.
>>It is during this latter phase of the signal that loop area (as in EMI)
>>might be an issue. During my signal integrity seminars I show some
>>animations that illustrate this pretty clearly.
>>Doug Brooks
>Check out UltraCAD's differential impedance and skin effect calculators at 

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