Impedance of complanar traces on PCB surface?

Hello:

I want to lay out a couple of traces on the FR4 PCB, to provide a transmission line of specific impedance for a differential-mode signal.

The impedance calculation formulas that I have found so far, for the common-mode and differential impedance, concern the traces with the ground plane underneath. I think I don't want the ground plane underneath, to minimise the common-mode capacitance, and I am looking for the formula for the coplanar traces on one side of the PCB, with no metal (copper) in the other PCB layers in the vicinity of the trace.

Can somebody help?

Thank you.

-- Andy

This is quite normal as there is usually a ground plane somewhere about - if just the screening box. Make the calculation with the ground plane at an appropriate distance for this, and all should be well.

I don't know how tight your coupling is, and the calculations really aren't terribly good much tighter than about 8dB, but you will probably need a couple of iterations or a bit of scalpel scraping to get it right.

d

Pearce Consulting

You need to think about the common mode impeadance and the differential impedance. Unless the ground plane is very far away, both will impact your result. Consider a thin twisted pair inside a sewer pipe. In most practical PWB stackups, the common mode impedances dominate because you cannot get the two differential conductors close enough to each other relative to their distance from the ground planes.

Mark

Do you have a specific reason to minimize the even-mode capacitance? In the case of a typical geometry, deleting the bottom ground plane won't make a lot of difference, unless the board is thin.

To reduce the even-mode capacitance and keep your target differential impedance, you could just make the traces skinnier and closer together.

John

Cranking up Txline, it recommends 0.32 mm trace widths to get 124 diff z (62 odd mode impedance), with an opposite ground plane and your other dims. Even mode impedance is 203 ohms. I doubt that eliminating the ground plane would change these numbers a lot.

John

signal.

no

Differential impedance Assume for a moment that you have terminated both traces in a resister to ground. Since i1 = -i2, there would be no current at all through ground. Therefore, there is no real reason to connect the resisters to ground. In fact, some people would argue that you must not connect them to ground in order to isolate the differential signal pair from ground noise. So the normal connection would be a single resister from Trace

1 to Trace 2. The value of this resister would be the sum of the odd mode impedance for Trace 1 and Trace 2, or Zdiff = 2 * Zo * (1-k) or 2 * (Z11 - Z12) Calculations: To say that Zdiff is 2*(Z11 - Z12) isn't very helpful when the value of Z12 is unintuitive. But when we see that Z12 is related to k, the coupling coefficient, things can become more clear. . National Semiconductor has published formulas for Zdiff that have become accepted by many. Zdiff = 2*Zo[1-.48*exp(-.96*S/H)] (Microstrip) Zdiff = 2*Zo[1-.347*exp(-2.9*S/H)] (Stripline) where S is the distance between adjacent traces and H is the height of the board. Zo is as traditionally defined

common mode impedance differs only slightly from the above. The first difference is that i1 = i2 (without the minus sign.) Thus Eqs.

V1 = Zo * i1 * (1+k) k is the coupling coefficient V2 = Zo * i1 * (1+k) and V1 = V2, as expected. The individual trace impedance, therefore, is Zo*(1+k). In a common mode case, both trace terminating resisters are connected to ground, so the current through ground is i1+i2 and the two resisters appear (to the device) in parallel. Therefore, the common mode impedance is the parallel combination of these resisters, or Zcommon = (1/2)*Zo*(1+k), or Zcommon = (1/2)*(Z11 + Z12) Note, therefore, that the common mode impedance is approximately ¼ the differential mode impedance for trace pairs.

As I remember it, a distant ground plane was not one of the limit conditions that gave rise to large errors. Areas of importance are close traces, and finite thicknesses of traces.

He could always do a finite element analysis based on Laplace's equation. Most CAD packages offer some sort of field evaluator that works that way. Mostly they are a bit clunky, and you need to calculate odd and even modes separately, but they do give the right answers.

d

Pearce Consulting

Provided the FR4 is really thin, you can almost ignore it. Performance is better that way too, as you work in true TEM mode, rather than the quasi TEM that happens if odd and even mode velocities are not the same. Most CAD packages will treat mixed mode dielectrics in some reasonable fashion, though.

d

Pearce Consulting

Thank you. I will try the formulas that are given for PCBs with the ground plane. But how about the presence of two different materials - the FR4 and the air?

-- Andy

The signal source is a differential current source, with each output loaded with a resistance (R) connected to a common DC reference point. I wanted this combined source see mostly the differential load, with the impedance Z=2R. A microstrip with a ground plane would present the differential mode impedance significantly different from two times the common mode one, unles one distances the strips by several millimeters - which is not practical in my case. A possibility there is to try to match the source to the microstrip impedance, by attaching a resistance between the differential outputs, but I don't like that much as I don't know what effect this would have on the accuracy of the signal.

I will study this. But the first try is not promising: with 0.2 mm-wide, 0.2mm-spaced traces with the ground 1.6mm away across the FR4, the common Z is about 136Ohm, and the differential - 154 ; I need 124 Ohm differential one.

Thanks.

-- Andy

I have had a problem with my computer (motherboard failure), back on-line only recently.

In fact, I cannot approach the required impedance with the ground plane on the opposite PCB side removed.

A nice program, that Txline.

Thank you!

-- Andy