[SI-LIST] Re: differential signaling (common-mode)
- From: steve weir <weirsi@xxxxxxxxxx>
- To: mike_bihan@xxxxxxxxxxxx, si-list@xxxxxxxxxxxxx
- Date: Sat, 30 Jul 2005 02:17:58 -0700
Bi,
Whether you model it or not in a real chip and board there will be a CM=20
return path. You would be well advised to control it within the package so=
=20
that it can match what is achievable on the board. Then you can terminate=
it.
In unshielded twisted pair applications the transformer buys a huge CMR.
Steve.
At 11:32 AM 7/30/2005 +0800, Bi Han wrote:
>Hi, SI-listers:
>
>The design task is to design on-chip differential transmission line to=20
>send signal. According to a lot of papers, edge coupled stripline is used.
>
>The cross-section is as below graph(might be lost). One key thing is that=
=20
>there is no underneath common-mode return path.
>
>___________________________________________________
>
> ______ ______
>dielectric |__+__| |__-___|
>___________________________________________________
>
>
>Hi-Z substrate
>
>___________________________________________________
>
>
>Then what is going to happen to common-mode signal? Common mode should not=
=20
>be a propagating mode, however, its S21 parameter after several=20
>reflections should be as below.
>
>S21=3DA*Hx*T/(1-R2*R1*Hx*Hx)
>
>R1 is source reflection coefficeint;
>
>R2 is load reflection coefficeint;
>
>Hx is propagation function, Exp(-gamma*L);
>
>Common mode impedance will be quite high if there is no good return path.=
=20
>Therefore, both source and load are not well terminated. Source could be=20
>very like short, and load could be very like open.
>
>Then R1=3D-1, R2=3D1; the S parameter could go infinity if Hx(w)=3D-1; It=
could=20
>resonate;
>
>It looks that twisted pair could survive this condition and goes well,=20
>why? Also several IEEE papers also used above configuration on-chip and=20
>goes fine, why?
>
>
>
>
>---------------------------------
>DO YOU YAHOO!?
> =D1=C5=BB=A2=C3=E2=B7=D1G=D3=CA=CF=E4=A3=AD=D6=D0=B9=FA=B5=DA=D2=BB=BE=
=F8=CE=DE=C0=AC=BB=F8=D3=CA=BC=FE=C9=A7=C8=C5=B3=AC=B4=F3=D3=CA=CF=E4
>
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