Impedance of complanar traces on PCB surface?
in [Design]
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From: lemonjuice on 8 Feb 2005 16:29 On Tue, 08 Feb 2005 08:48:33 +0100, Andy <andy(a)nodomain.nod> wrote: >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 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. |