IDEAL TRANSMISSION-LINE FUNDAMENTALS - Advanced Signal Integrity for High-Speed Digital Design - page 109

1

1

1

1

1

1

1

1

a

b = 2 a

V

Figure 3-25 Field map used to calculate the capacitance in Example 3-2, showing one
quadrant divided up into two series sections and four parallel sections.

SOLUTION Since the cross section of the coaxial line is symmetrical, we can
take advantage of symmetry and draw the field lines for only one quadrant, as
shown in Figure 3-25. In this sketch, the quadrant was divided into two series
sections using a single equal potential line, and four parallel sections yielding
n s = 2 and n p = ( 4 )( 4 ) = 16. Therefore, equation (3-97) yields

10 − 12 ) 16
2

C = ε r ε 0 n p 10 − 12 F / m

n s = ( 2 . 3 )( 8 . 85

× = 162 . 8 ×

To calculate the impedance of the coaxial line, we must first recalculate the
capacitance for ε r = 1 and calculate the inductance from (3-46),

10 − 12 16
2

= ε 0 n p 10 − 12 F/m C ε r = 1

n s =

8 . 85 × = 70 . 78 ×

1
c 2 C ε r = 1

1

10 − 9 H/m L = =

10 − 12 ) =

156 . 9 ×

( 3 × 10 8 ) 2 ( 70 . 78 ×

yielding the characteristic impedance calculated from (3-33):

156 . 9

10 − 9

×

Z 0 , fieldmap = = 31 . 0

162 . 8 × 10 − 12

This can be compared directly to the results derived in Section 3.4.2.

10 − 12 )

×

2 πε
ln (b/a) =

2 π( 2 . 3 )( 8 . 85

10 − 12 F / m C = = 184 . 5 ×

ln ( 2 )