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. [email protected]
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.
>