l impedance vs common mode impedance

Dear community, can someone explain to me some misunderstandings. There is a differencial pair in a PCB. Simulation returns: Impedance 65.5 ohm, differencial impedance 98 ohm, common mode impedance 82 ohm..I wonder what is the difference between 'impedance' and 'common mode impedance'? Just 'impedance is (as I see it) the impedance of a trace, if that trace was alone.Well.But what common mode impedance is? Any links or tips would be appreciated. Slav.

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This should be explained in the manual of your simulator. In case it is not, ask google: Searching for "common mode impedance" gives as hit no. 12

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which should get you started.


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Georg Baum

"Impedance" should be the characteristic impedance of one trace of the differential pair if it were used in isolation somewhere as, e.g., a microstrip line. "Differential impedance" is the impedance seen by a fully differential signal traveling down the pair (e.g., 1V on one trace, -1V on the other). "Common mode impedance" is that seen by a common mode signal traveling down the pair (e.g., 1V on both traces... or -1V on both traces).

Differential impedance is also known as "odd mode" impedance, common mode impedance is also known as "even model" impedance.

The link Georg gave is a good one. In general, differential impedance will be something LESS than TWICE the "regular" impedance because the coupling between the traces "help" one another (since the currents in each one travels in opposite directions -- from a lumped circuit perspective, the mutual inductance of the two traces is negative) -- they'd really like to get closer together if they could! The twice comes about since each traces has its own terminating resistor that ends up in series between +/-1V. Likewise, common mode impedance will be something MORE than HALF the regular impedance, where the "half" comes from the terminations appearing in parallel and the "more" coming from the two traces "fighting" each other since they're now carrying now the same current (mutual inductance is positive) and they'd prefer to get away from one another if possible.

In your case, we can compute the coupling coefficient between the traces as

0.252 (see the link for the math), so your odd mode impedance should be about 41 ohms. For some reason, you tool is giving you twice that -- 82 ohms; this is probably just a convention your particular tool uses.

Any old signal across two traces can be viewed as a sum of a common mode and differential mode signal, where Vcm = (V1+V2)/2 and Vdm = (V1-V2)/2.

If you know that your system is using differential signalling all the time, you'd termiante your differential pair in the differential impedance your tool is giving you. However, if that "differntial pair" sometimes carries common mode signals are well (USB, for instance, does this), there's no particularly simple way to properly termiante the 'pair for both modes, so sometimes people will terminate in the geometric mean of the even and odd mode impedances as a starting point (63.4 ohms in your case) and then take a look at the results. (Although, for USB, 99% of the time the signals really are differential, so they terminate much close to the differential impedance...)

A coupling coefficient of 0.252 is actually pretty high for most applications, so you probably will have to experiment a little to find the best termination possible.

---Joel Kolstad

Reply to
Joel Kolstad

I realized I made an error here...

This should be sqrt(2)*63.4 = 89.6 ohms, since the common mode impedance needs to be doubled to get the total series value of the terminators.

Reply to
Joel Kolstad

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