Hello Craig,
This response is a bit long but it should answer your questions. I
assume that your dual stripline configuration looks like this:
1/2 oz. Cu gnd plane
--------------------------------------
dielectric layer thickness = th
------------- -------------
|1/2 oz. Cu | |1/2 oz. Cu |
------------- -------------
dielectric layer thickness = th
--------------------------------------
1/2 oz. Cu gnd plane
Conductor width is still 6 mils, and all conductors are copper 58.14e6
[S/m]. Since the 1/2 oz Cu is equivalent to 0.7 mils and you give the
total gnd-gnd distance as 16.7 mils I can calculate dielectric
thickness as:
th = (16.7 - 0.7)/2 = 8 mils
conductor separation was not given.
I assumed that FR-4 has 4.7 dielectric constant. The loss tangent of
0.018 was specified in your letter.
Frequency is now 1 GHz.
Using these assumptions I get for a single stripline:
====== C r o s s - S e c t i o n R e p o r t ======
Program: OptEM ID Version 4.3
File name: strip50.rpt
Analysis frequency: 1 GHz
Results included in this cross-section report are:
[L] [R] [C] [G] Td,v a,b [Z] [Y] Zo
Results normalized to length: 1000.0 mil
[L] inductance matrix in H
9.2560e-9
[R] resistance matrix in Ohm
1.0108
[C] capacitance matrix in F
3.7092e-12
[G] unit conductance matrix in S
419.50e-6
a,b attenuation in dB, phase in deg
178.88e-3 66.704
Zo characteristic impedance vectors in Ohms, polar coordinates
magnitude
49.954
phase
310.13e-6
=================================
The dielectric may cause losses by "leaking" charge (like porous
materials permit leakage of liquids) and by polarization losses. These
effects can be easily separated by sweeping frequency. Leakage is not
frequency dependent whereas dielectric polarization loss is
proportional to frequency and can be best represented by the loss
tangent. For a single frequency, the power loss of a "leaky
dielectric" and a "lossy dielectric", which, though of different
origin, are indistinguishable. In practice, at the GHz range leakage
of most dielectric used in interconnect manufacturing is negligible in
comparison with polarization loss and can be neglected (Gdc=0).
You asked for a formula for calculation of attenuation. Our solver
calculates this value automatically, but you can also calculate it
manually (yes, it is possible for a single conductor gnd combination)
by calculating the real part of the propagation factor (w-is the
omega):
attenuation [Np] = Re ( sqrt((R+jwL)*(G+jwC))
or
attenuation [dB] = 20 log(e) * attenuation [Np] = 8.68589 *
attenuation [Np]
For manual calculation of differential impedance you can start with
the circuit representation LRCG, then connect it for differential
signal and calculate equivalent LRCG for differential pair. Then use
the formula:
Z= sqrt ((R+jwL)/(G+jwC))
or if you have the characteristic impedance matrix, calculate the
inverse (admittance matrix) then assemble an equivalent circuit and
calculate the value from the circuit - which in simple terms should
give you:
Zdiff=2*(Z11-Z12)
You said that you would like to use a differential driver. Using a 50
Ohm stripline configuration you can build a dual differential circuit
with characteristic impedance 0<Z<100 - function of separation. I
selected 80 Ohms differential impedance and our solver suggested a
separation of 4.544 mils and then calculated the following results per
inch:
====== C r o s s - S e c t i o n R e p o r t ======
Program: OptEM ID Version 4.3
Analysis frequency: 1 GHz
Wire Type Wire Name Conductors
SIGNAL T1 T1
SIGNAL T2 T2
GROUND ref plane1_plane2
Circuit Number Circuit Name Source/[Return] Wires
C1 circuit01 T1
C2 circuit02 T2
Results normalized to length: 1000.0 mil
[L] inductance matrix in H
C1 C2
C1 9.1374e-9 1.7076e-9
C2 1.7076e-9 9.1374e-9
[R] resistance matrix in Ohm
C1 C2
C1 1.0491 51.465e-3
C2 51.465e-3 1.0491
[C] capacitance matrix in F
C1 C2
C1 3.9018e-12 -741.85e-15
C2-741.85e-15 3.9018e-12
[G] unit conductance matrix in S
C1 C2
C1 441.28e-6 -83.901e-6
C2 -83.901e-6 441.28e-6
Zo characteristic impedance vectors in Ohms, polar coordinates
magnitude
Zg(grounded) Zu(user def.) Zc(calculated) Ze(effective)
Zf(floated) load = 0 load = Zu
load = Ze load = open
C1 47.541 50.000 48.422 48.408
49.292 C2 47.541 50.000 48.422
48.408 49.292
=================================
defining circuits as even and odd you can also get:
====== C r o s s - S e c t i o n R e p o r t ======
Program: OptEM ID Version 4.3
Analysis frequency: 1 GHz
Circuit Number Circuit Name Source/[Return] Wires
C1 odd T1 [T2]
C2 even T1 T2
a,b attenuation in dB, phase in deg
C1 399.10e-3 133.74
C2 345.03e-3 133.29
Zo characteristic impedance vectors in Ohms, polar coordinates
magnitude
C1 40.001
C2 58.582
=================================
Sincerely yours
Juliusz
"Juliusz Poltz" <jpoltz@optem.com>
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OptEM Engineering Inc. Ph: (403) 289-0499
100 Discovery Place Fax: (403) 282-1238
3553 - 31st Street N.W. Email: info@optem.com
Calgary, AB Canada T2L 2K7
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