I think the 50 ohm value is realistic. Even if the ground conductor is connected
to system ground only at the source, it is still the primary and lowest impedance
path for the return current when compared to a metal enclosure or other far away
structures.
Another way of looking at it is to consider the signal on the top and the ground
at the bottom as a differential line. In this case, there would be no need for a
reference ground. The differential impedance of this configuration would be ~50
ohm.
Thanks,
Vinu
alterra@adnc.com wrote:
> Hi Shawn,
>
> I did the same calculation you did and got the same answers, but am
> concerned that these results may be misunderstood.
>
> Case 1:
> This is the problem you did. The signal and ground leads are 12 mils wide
> and 1.4 mils thick. The pitch is 31.5 mils. The substrate is 5 mil kapton
> or polyimide with a dielectric constant of 3.5
>
> GGGGG SSSSS GGGGG
> _________________________________
>
> 5 mil kapton, k=3.5
> _________________________________
> GGGGG
>
> You noted earlier that the main interaction was between the signal lead on
> top and the ground on the bottom. This is true for the capacitive coupling
> but not for the inductive coupline. The capacitive coupling is more than an
> order of magnitude greater to the bottom ground than to the side grounds,
> but the inductive coupling is only about twice as great.
>
> It is more correct to treat the flex circuit as a multiconductor
> transmission line. On the assumption that the 'ground' leads are only
> connected to system ground at the ends of the flex circuit, then they are
> not really ideal ground conductors nor can they be considered as a single
> ideal ground conductor. They can be (and probably will be) at different
> potentials from one another along the length of the flex circuit.
>
> If I make the same assumption you did that the ground conductors form one
> single ideal ground then I get the same value for the characteristic
> impedance of the signal lead, namely 49.70 ohms (low frequency limit). But
> this assumption is not realized in the actual physical circuit so I think it
> is misleading to say that this forms a 50 ohm line.
>
> In reality the actual reference ground will be somewhere outside the cable,
> usually far away compard to the conductor spacing...something like the metal
> enclosure containing the circuit or a large ground on a PCB. In this more
> realistic case, you really have 4 signal conductors which are described by a
> 4 x 4 impedance matris and not a single value.
>
> Case 2:
> This is the same problem but with two additional signal conductors, like this:
>
> GGGGG SSSSS GGGGG
> _________________________________
>
> 5 mil kapton, k=3.5
> _________________________________
> SSSSS GGGGG SSSSS
>
> With a distant ground, this is described by a 6 x 6 impedance matrix which
> is too large to give here. In the ideal ground case where all the 'ground'
> leads are really grounded, the three remaining signal leads are described by
> a 3 x 3 impiedance matrix which is
>
> left center right
> left 51.048 2.561 1.692
> center 2.561 49.604 2.569
> right 1.692 2.569 51.028
>
> The characteristic impedance of the center signal lead is not affected
> significantly by the additional signal traces, but there is some coupling
> between signal leads (about 5%). However, just as in case 1, this is an
> unrealistic situation. In the realistic case of a distant ground, the
> impedances and couplings are much different...for example the self-impedance
> of the center signal lead is 163 ohms or more than 3 times greater than the
> value in the ideal ground case.
>
> In conclusion, I urge great caution in extapolating the results of two
> conductor transmission line calculations to multiconductor lines like the
> flex circuit considered here. A safer approach is to use the complete
> description of the multiconductor line in a SPICE model and evaluate it's
> performance with realistic ground connections.
>
> Best regards,
>
> Eric Wheatley
>
> PS The above results were calculated using Ansoft's 2D Parameter Extractor
> tool.
> ---------------------------------------------------------------
> Eric Wheatley Ph.D. (760) 942-9426 (phone)
> Alterra Technology Co. (760) 942-2366 (fax)
> 435 Dunsmore Ct. alterra@adnc.com
> Encinitas, CA 92024 US
> ---------------------------------------------------------------
>
> At 11:21 PM 3/27/98 -0500, you wrote:
> >Good question. I tried it out, and added a twist that I didn't
> >consider before--the physical thickness of the conductors.
> >
> >With physically thick metal models, you get a 50 ohm
> >system with G = S = 12 mils (I had previously reported 13 mils,
> >but that was with a physically thin metal model applied to the EM
> >simulator).
> >
> >With a large ground plane (microstrip-like configuration) you reach
> >a 50 ohm system with a signal line width of just under 10 mils
> >(considering
> >the thickness of the metal). If you used this value for your signal and
> >finite ground conductor widths, you'd end up with a 59 ohm system -- off
> >by
> >almost 20%.
> >
> >BTW, I didn't find that the co-planar ground traces affected the
> >impedance
> >significantly for these dimensions. But maybe they help with the
> >isolation between the signal traces a bit.
> >