IDEAL TRANSMISSION-LINE FUNDAMENTALS - Advanced Signal Integrity for High-Speed Digital Design - page 111

that the authors were not able to draw the cells to be all the same size; how-
ever, we will still assume an average electric field strength for each cell, which
will induce some error. Since some cells are small and some cells are large,
the errors induced by the cell size discrepancy will partially average out and
should give reasonable results. For this example, symmetry was used and only
half of the field lines were drawn. The distortion of the electric field lines at
the dielectric boundary was ignored, which will induce an additional, although
small error.
Initially, we must calculate the effective dielectric permittivity. In Section 3.3.3
we explained that the effective permittivity was due to the weighted average of
the fields fringing through the air and the dielectric. By counting the number of
cells in the air and in the board, we can use a weighted average to calculate
the effective permittivity. Counting the cells in Figure 3-26 yields nine cells in
the air and 27 cells in the dielectric. Note that some of the cells were divided
up between the air and the dielectric material. Therefore, the effective dielectric
permittivity is calculated:

9 27

ε eff ≈

27 ( 1 ) + 27 ( 4 ) =

3 . 25

9 + 9 +

=

The results from the two-dimensional field solver in Table 3-1 shows ε eff 2 . 96

for w/h = 1, so the result above is not a bad approximation.

The field plot was divided into five series sections, and seven parallel sections
for one-half of the structure, yielding n s =

5 and n p = ( 2 )( 7 ) = 14 (due to the

symmetry). Using, ε r = ε eff = 3 . 25, equation (3-97) yields

10 − 12 ) 14
5

C = ε r ε 0 n p

10 − 12 F / m

n s = ( 3 . 25 )( 8 . 85

× = ×

80 . 53

and the impedance is calculated as in Example 3-2:

10 − 12 ) 14
5

C ε r = 1 = ε 0 n p 10 − 12 F / m

n s = ( 8 . 85

× = 24 . 77 ×

1
c 2 C ε r = 1 =

1

10 − 9 H / m L =

10 − 12 ) =

448 . 5 ×

( 3 × 10 8 ) 2 ( 24 . 77 ×

448 . 5

10 − 9 ×

Z 0 = = 74 . 6

80 . 53 × 10 − 12

Comparison to the numerically calculated results from the two-dimensional field
solver in Table 3-1 show Z 0 74 . 6 . The fact that the exact answer was

obtained as the two-dimensional solver is serendipity.

=