[SI-LIST] Re: Mode Conversion

Neven,

Thank you for the input. I have embedded
my response in your mail.


--- npischl <neven@xxxxxxxx> wrote:
> 
> For the novices, I believe that some fundamental
> mechanisms of 
> differential to common-mode (CM) conversion are well
> graphically 
> explained in handouts of a presentation I had for
> the Santa Clara 
> Valley IEEE EMC Chapter some time ago. It can be
> downloaded from
> 
> http://www.scvemc.org/scvemc/cm_probe.pdf
> 
> The relevant pages of the pdf. document are 3,4, and
> 5.
> 
> I would like to point out something that I feel is
> important and 
> somehow it seems to be missed so far. When talking
> about CM in 
> differential signaling, as most replies to this
> thread point out, it 
> can be caused by any imbalance in the differential
> path, including 
> imbalance in the signal waveform (e.g. difference in
> Tr and Tf of the 
> complementary signals in the driver). In a little
> broader sense, 
> differential to CM conversion can also be caused by
> ground (or power) 
> bounce, due to signal currents passing through the
> IC's power and 
> ground lead-inductance (see the drawing on the top
> of page 4 of the 
> handouts). In such case, CM-voltage (-LdI/dT) will
> drive the virtual 
> reference of the driver

A doubt, I have here is, since we are discussing
differential lines and if we assume differential
drivers then is there still a possibility of
ground bounce. Because when the differential
driver switches no doubt the current changes
direction at the output of the device but as
far as Vcc and Gnd is concerned there shouldn't
be any change in current(atleast ideally wrt
to single ended device outputs.)

> 
> There is one large difference between these two
> cases. If CM (shown 
> by red arrows in my drawings) is caused by ground
> bounce (or also by 
> coupling), then CM currents will flow more less
> equally on both lines 
> of the differential pairs, and will cause even-mode
> propagation.
> 
> However, if the conversion is due to imbalance, it
> will NOT 
> necessarily cause even mode propagation. The CM
> signal in this case 
> will propagate only on one of the two conductors in
> the pair, as 
> shown. Therefore, CM-propagation in this sense does
> not necessarily 
> mean even-mode propagation. This is a typical
> situation on PCBs, with 
> differential pairs routed adjacent to the reference
> planes.

By this explanation common mode propagation can never
occur due to imbalance on the differential line.
Imbalance due to (1) impedance change (2) propagation
delay variations.

Because for common mode to propagate we need two
conductors of the differential line to be driven
with signals in phase or (+x,+x) / (-x,-x)
I am using x because we are refering to 
signal transitions in case of digital signals
and common and differential mode propagations
only occur during transitions. Where 
high level > x > low level

Therefore for diferential in case of imbalance we
have if initially the differential line is (-1,1) then
(+x,1);(+x',-y);...;(1,-y');(1,-1) where 
high level > x,x',y,y' > low level but x != y at any
instant of time. Ofcourse it can also be 
represented as follows if the transition starts
on the other conductor starts first.
(-1,-y);(+x,-y');...;(+x',-1);(1,-1)

Therefore can we assume that pure common
mode propagation cannot occur on differential
lines due to imbalance. What occurs due to
imbalance is a skewed differential mode
of propagation.

Can somebody correct me if I am wrong,

Regards,
Ben


> 
> If the pair gets launched from a PCB into a twisted
> pair however, 
> then "one-sided" CM may eventually become
> "both-sided" CM due to the 
> mutual inductance between the wires, and eventually
> the real even-
> mode CM signal will start propagating.
> 
> In both cases the impact to CM emission from the
> circuit is same, but 
> understanding the cause and mechanisms may help
> remedy possible 
> problems.
> 
> Neven
>  
> 
> 
> --- In si-list@xxxx, "Tabatchnick, Justin"
> <justin.tabatchnick@xxxx> 
> wrote:
> > 
> > I sort of came in on the tale end of this - What
> is meant by mode
> > conversion- is it TEM to TE or TM or differential
> to common mode or 
> even
> > mode impedance to odd mode impedance or there
> abouts ?
> > 
> > Thanks
> > 
> > Justin 
> > SI engineer, Intel, Sacramento
> >
>
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