To perfectly terminate these coupled transmission lines, regardless of what
the signal is, you need to connect Rx from one line to the other, and connect Ro
from each line to ground, where
Rx=2*Zeven*Zodd/(Zeven-Zodd)
Ro=Zeven
However, if only differential mode is excited, ground is mid-way across the two
lines, the effective termination becomes Rx in parallel with two Ro's in series.
Working out the math, the effective termination becomes 2*Zodd across the two
lines which is the differential impedance of the coupled transmission lines.
This translates into Zodd from each line to ground. In other words, differential
impedance of a symmetric coupled transmission lines is the characteristic
impedance (in the sense of perfect termination) between the two lines run in
differential mode.
Note that if the two lines are loosely coupled, M<<L and Cm<<C and the
differential impedance would become 2*Zo=2*sqrt(L/C)
Hope this helps.
Regards,
Wai-Yeung Yip
CPU Module Engineering
Sun Microsystems
> X-Sender: [email protected]
> Date: Thu, 05 Feb 1998 15:10:50 -0800
> To: [email protected]
> From: Mark Nass <[email protected]>
> Subject: [SI-LIST] : More on differential impedance calculations
> Mime-Version: 1.0
>
>
> From the advice I heard so far I used a microstrip topology with
> 2 - 5 mill wide traces, 5 mills apart and 10 mills over a plane.
> Using XFX I get 3 impedances:
>
> 85.77 is the trace to plane impedance
> 109.75 is the even mode impedance
> 60.13 is the odd mode impedance
>
> This appears to be what I want. Any comments?
>
> Also it seems to me that I want to have some amount of coupling between
> the two traces so that any noise that gets injected into these traces will
> have a canceling effect.
>
> Mark
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