[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
Mmmmmmm,
Well..............
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:
>Doug
>
>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
>
>
>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
>http://www.ultracad.com
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