IDEAL TRANSMISSION-LINE FUNDAMENTALS
3.4 TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE Earlier, we discussed how an electromagnetic wave propagates on a transmis-
3.4.6 Field Mapping
infinitely thin sheets of “virtual” conducting foil. Using the constant potential surfaces, the capacitance per unit length of a two-dimensional system can be found with reasonable accuracy from a carefulfield sketch. For example, con- sider the cross section of a coaxial line as shown in Figure 3-24, where electric fieldflux lines originate from a chargeqdistributed over a lengthl on the inner conductor and terminate on −q in the same length on the outer conductor. If equal potential lines are drawn between the inner and outer conductors, the total
C0
C0 C0
−q
+q
(a)
(b) Φ= V
∆C
Φ= 0
Φ= 0 Φ= V
Figure 3-24 Field map of a coaxial line showing (a) series capacitance between equal potential surfaces and (b) capacitance per unit cell.
capacitance is simply the series combination of the capacitance between equal potential lines as shown in Figure 3-24a. If the equal potential lines are chosen so that each capacitance is the same and ns denotes the number of elements in series, the total capacitance is
C= C0 ns
(3-92) Furthermore, each of the series capacitors in Figure 3-24a can be subdivided into parallel capacitance values with a value of C for each cell, as shown in Figure 3-24b. Assuming np parallel elements, C0=npC, yielding a total capacitance
C= np ns
C (3-93)
In order to use (3-93), we need to calculate the cell capacitanceC. Assuming that the chargeqand−q are present on the top and bottom of each cell wall, we can write the cell capacitance in terms of the potential difference between the boundaries using (2-76) and (3-1):
C= q
E·dl (3-94)
If the problem is simplified by drawing the field lines such that each cell is approximately the same size, the capacitance can be written in terms of the average electricfields,
C= q
Eavehave (3-95)
wherehave is the average height of a cell andEaveis the average electricfield for a cell.
Equation (3-95) can be simplified to eliminate the electric field using the integral form of Gauss’s law, (2-59), SεE·ds= SD ·ds=q:
C= q
Eavehave = εq
Davehave = εq
(q/ lwave)have = εwave
have l (3-96) where wave is the average cell width and ds=lwave is the area of the cell surface for the transmission line length l. If the cells can be drawn so that wave ≈have, the total capacitance is calculated by substituting (3-96) into (3-93):
C l =εnp
ns
F/m (3-97)
Example 3-2 Usefield mapping techniques to calculate the impedance of a coax- ial transmission line shown in Figure 3-25, where b/a=2 and the permittivity of the dielectric isεr =2.3.
a
b = 2a
1
1 1
1
1 1
1 1
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 ns =2 andnp=(4)(4)=16. Therefore, equation (3-97) yields
C=εrε0np
ns =(2.3)(8.85×10−12) 16
2
=162.8×10−12 F/m To calculate the impedance of the coaxial line, we must first recalculate the capacitance forεr =1 and calculate the inductance from (3-46),
Cεr=1=ε0np
ns =8.85×10−12 16
2
=70.78×10−12F /m L= 1
c2Cεr=1 = 1
(3×108)2(70.78×10−12) =156.9×10−9H /m yielding the characteristic impedance calculated from (3-33):
Z0,fieldmap=
156.9×10−9
162.8×10−12 =31.0
This can be compared directly to the results derived in Section 3.4.2.
C= 2π ε
ln(b/a) = 2π(2.3)(8.85×10−12)
ln(2) =184.5×10−12 F/m
Again, to calculate the impedance, we must first recalculate the capacitance for εr =1.
C= 2π ε
ln(b/a) = 2π(1)(8.85×10−12)
ln(2) =80.2×10−12 F/m L= 1
c2Cεr=1 = 1
(3×108)2(80.2×10−12) =138.5×10−9 H/m Z0=
138.5×10−9
184.5×10−12 =27.3
Comparing this result toZ0,fieldmapshows reasonable accuracy but demonstrates the inherent errors that are inevitable when using hand sketches.
Example 3-3 Use field mapping techniques to calculate the impedance and the effective dielectric permittivity of the microstrip transmission line shown in Figure 3-26, where w/ h=1 and the permittivity of the dielectric material is εr =4.0.
SOLUTION It is inherently more difficult to apply mapping techniques to a microstrip because the fields are difficult to draw accurately and the dielec- tric is not homogeneous. Figure 3-26 shows a drawing of the field lines. Note
1 1
0.5 1
1
1 1
w 1 1
0.5
1 0.5 1 1
1 1 1
1 1
1 1
1 1 1 1
1 1 1
1
1 1 1
1 1 1
1
0.5 1
h er= 1
er= 4
Figure 3-26 Field map used to calculate the effective dielectric permittivity and the impedance of a microstrip.
that the authors were not able to draw the cells to be all the same size; how- ever, we will still assume an average electricfield 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 thefields 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:
εeff≈ 9
9+27(1)+ 27
9+27(4)=3.25
The results from the two-dimensionalfield solver in Table 3-1 showsεeff=2.96 forw/ h=1, so the result above is not a bad approximation.
Thefield plot was divided intofive series sections, and seven parallel sections for one-half of the structure, yieldingns =5 and np=(2)(7)=14 (due to the symmetry). Using,εr =εeff=3.25, equation (3-97) yields
C=εrε0np
ns =(3.25)(8.85×10−12) 14
5
=80.53×10−12 F/m and the impedance is calculated as in Example 3-2:
Cεr=1=ε0np
ns =(8.85×10−12) 14
5
=24.77×10−12 F/m L= 1
c2Cεr=1
= 1
(3×108)2(24.77×10−12) =448.5×10−9 H/m Z0=
448.5×10−9
80.53×10−12 =74.6
Comparison to the numerically calculated results from the two-dimensionalfield solver in Table 3-1 show Z0=74.6 . The fact that the exact answer was obtained as the two-dimensional solver is serendipity.