transmission lines ch 2

7/27/2019 Transmission Lines Ch 2

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Chapter 2

Parameters of common

transmission lines

2.1 Introduction

In chapter 1 we have obtained the transmission line equations on the basis of a phenomenologicalmodel that contains four primary parameters: L (inductance per unit length, p.u.l.), R (resistancep.u.l.), C (capacitance p.u.l.), G (conductance p.u.l.). The expressions that yield these parametersas a function of the geometry of the structure require the solution of Maxwell equations for thevarious cases. In this chapter we limit ourselves to a list of equations for a number of commonstructures: the reader can consult the books in the bibliography for further details . In particular,we show only the expressions of the inductance and capacitance p.u.l. The parameters related tothe losses will be shown in chapter 4.

2.2 Coaxial cable

The coaxial cable is a transmission line consisting of two coaxial cylindrical conductors, separatedby a dielectric (see Fig. 2.1). The two conductors, here shown as homogeneous, are often made ofbraided small diameter copper wires.

If r denotes the relative permittivity of the insulator, the line parameters are given by:

C = 20rlog(D/d)

, L = 02

log

D

d

, (2.1)

Z =

00r

1

2log

D

d

60

rlog(

D

d), (2.2)

vf =cr, (2.3)

where the logarithms are natural (basis e). Fig. 2.2 shows a plot of Z, L e C versus the ratio ofthe conductor diameters. Fig. 2.1 shows the field lines of the electric and magnetic fields of theTEM mode, the fundamental one of this structure viewed as a waveguide. We can observe that

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2 Parameters of common transmission lines

Dd

Figure 2.1. Coaxial cable. The field lines of the electric field are shown by solid lines, thoseof the magnetic field by dashed lines.

Figure 2.2. Parameters of the coaxial cable vs.the geometrical dimensions.

the electric field configuration is that of a cylindrical capacitor, consistently with the fact that theTEM mode has zero cutoff frequency. If the operation frequency increases, a point is reached inwhich higher order modes start to propagate. The maximum frequency for which the coaxial cableis single mode is approximately

fmax = vf(D + d)

, (2.4)

The corresponding minimum wavelength is

min = (D + d). (2.5)

The electric field in the cable is radial and its magnitude is given by

E(,,z) =V(z)

log(D/d)

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where V(z) is the voltage. Hence the maximum electric field, not to be exceeded in order to avoidsparks, is on the surface of the inner conductor and has the value

Emax = V(z)log(D/d)

1d

Example

Compute the parameters of a cable, with inner conductor diameter d =1.6 mm, outer conductor diameterD = 5.8 mm, r = 2.3.

Applying the previous formulas we get L = 0.2576 H/m, C = 99.35 pF/m, Z = 50.92 , vf/c = 1/r= 65.9%, fmax = 8.5 GHz. The normalized maximum electric field is Emax = 485.3V/m if the voltage Vis 1V.

It is to be remarked that the coaxial cable is an unbalanced line, which means that the return conductor

is connected to ground. Hence the voltage of the other conductor is referred to ground.

2.3 Two-wire line

The two-wire line consists of two parallel cylindrical conductors. This structure has a true TEM mode onlyif the dielectric that surrounds the conductors is homogeneous and the formulas reported hereinafter referto this case. In practice, of course, the conductors are embedded in a thin insulating support structure,which causes the fundamental mode to be only approximately TEM.

The parameters of the two-wire transmission line, whose geometry is shown in Fig. 2.3 are:

C = 0rcosh1(D/d)

, L = 0

cosh1(D/d), (2.6)

Z =1

0

0rcosh1

D

d

120

rcosh1

D

d

, (2.7)

vf =cr

. (2.8)

It may be useful to recall that

cosh1 x = log(1 +

x2 1) log(2x), se x 1. (2.9)

D

d

Figure 2.3. Two-wire transmission line. The field lines of the electric field are shown solid,those of the magnetic field dashed.

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Example

Compute the parameters of a two-wire line, in which the wires have a diameter of 1.5 mm and a separationof 5.0 mm and are located in air.

we find that C = 14.84 pF/m, L = 750 nH/m, Z = 224.71, vf = c.

It is to be remarked that the TEM fields are non negligible up to large distance from the line itself,so that the two-wire line is never isolated from the other nearby conductors, which entails problems ofelectromagnetic compatibility. On the contrary, in a coaxial cable with sufficiently good outer conductor,the operation of the line is completely shielded from external interference. For this reason, the two-wireline is always used in a balanced configuration, i.e. the two wires have opposite potentials with respect toground.

2.4 Wire on a metal plane

This line consists of a single wire running parallel to a grounded metal plate, see Fig. 2.4a. If the metalplate were infinite, this line would be perfectly equivalent to a two-wire line, because of the image theorem(Fig. 2.4b). When the ground plane is finite, the equivalence is only approximate, but if its size is muchlarger than the distance h between the wire and the plane, the errors are negligible.

d

h

D= 2h

d

(a) (b)

Figure 2.4. (a) Wire on a metal plane and (b) equivalent two-wire transmission line.

The parameters of the two-wire line are:

C = 0rcosh1(2h/d)

, L = 0

cosh1(2h/d), (2.10)

Z =1

0

0rcosh1

2h

d 120

rcosh1

2h

d , (2.11)vf =

cr

. (2.12)

Example

Consider a wire with diameter d = 3.2 mm in air, placed at an height h = 5.74 cm on a ground plane.

We find C = 6.51 pF/m, L = 1.71 H/m and Z = 512.4 .

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2.5 Shielded two-wire line

To avoid the electromagnetic compatibility problems of the two-wire line, the structure of Fig. 2.5 can beused. Note that this is a three conductor line (two plus a grounded one). In this case there are two TEM

2h

Dd

Electric field

Magnetic field

Figure 2.5. Shielded two-wire line and field configuration of the symmetric (balanced) TEM mode.

modes, a symmetric (balanced) one where the potentials of the two inner conductors are symmetric withrespect to that of the outer one, connected to ground, and an asymmetric (unbalanced) one, with differentparameters. The parameters for the symmetric mode can be computed from the following equations:

C = 0rlog

2h(D2 h2)d(D2 + h2)

, L = 0

log

2h(D2 h2)d(D2 + h2)

, (2.13)

Z =1

0

0rlog

2h(D2 h2)d(D2 + h2) , (2.14)

vf =c

r. (2.15)

Example

Consider a shielded two-wire line with diameter of the outer conductor D = 100 mm, inner conductorswith diameter d = 15 mm e spacing 2h = 50 mm.

Using the previous formulas we get: C= 25.77 pF, L = 0.43 H, Z = 129.39 .

2.6 StriplineThe stripline consists of a metallic strip placed between two grounded metal planes (Fig. 2.6). This is clearlyan unbalanced structure, which is used only inside components and devices. Since the two planes have thesame potential, this is a two conductor line and the fundamental mode is TEM. The relevant parameterscannot be expressed in terms of elementary functions. We report below an approximate expression for thecharacteristic impedance, which is valid in the case the strip thickness is negligible:

Z 30r

b

weff + 0.441b(2.16)

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w

b

Figure 2.6. Stripline geometry .

where the equivalent strip width weff is computed from

weffb

=w

b

0 if w/b > 0.35,0.35 w

b

2

if w/b < 0.35(2.17)

The phase velocity, as for all TEM structures is given by

vf =cr

. (2.18)

The previous equations are appropriate in an analysis problem, in which the dimensions of the structureare known. For the design activity, in which the dimensions are to be determined in order for the lineto have a desired characteristic impedance, we can use the following equations, obtained by inversion of(2.16) and (2.17):

w

b=

x if

rZ < 120 ,

0.85 0.6 x ifrZ > 120 (2.19)

where

x =30rZ

0.441 (2.20)

ExampleDesign a stripline with characteristic impedance Z = 50 , separation between the ground planes b =0.32 cm, r = 2.2. Find then the value of the propagation constant and the wavelength at the frequencyf = 10 GHz and the delay = l/vf introduced by line lentgth l = 5 cm.

Since Z

r = 74.2 (< 120 ) we compute x = 0.830 by means of (2.20) and this is already the valueof w/b. Hence w = 0.266 cm. then the propagation constant is computed from

k =

vf=

c/

r=

2f

rc

= 3.1065 cm1

and

=2

k= 2.0212 cm, =

l

vf=

l

rc

= 0.247 ns.

Fig. 2.7 shows plots of the characteristic impedance of a stripline where the strip thickness f is nonnegligible.

2.7 Microstrip

A microstrip consists of a conducting strip deposited on a dielectric layer, whose lower face is covered witha metal ground plane, as shown in Fig. 2.8. Since the transverse cross section is not homogeneous, the

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Figure 2.7. Characteristic impedance of a stripline vs. its dimensions.

Ground conductor

Figure 2.8. Microstrip geometry.

fundamental mode is not rigorously TEM. In practice, the longitudinal field components are very smallwith respect to the transverse ones and the so called quasi-TEM approximation is used. Even in thiscase, only approximate formulas are available for the characteristic impedance. In an analysis problem, in

which the dimensions of the line are known, we compute first an equivalent dielectric constant eff, whichis a weighted average of the permittivities of air and of the substrate:

eff =r + 1

2

1 +

11 + 12h/w

. (2.21)

The phase speed is computed as always, but exploiting this effective permittivity

vf =ceff

(2.22)

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and the characteristic impedance is given by

Z =

60

efflog8h

w

+w

4h if w

h

< 1,

120effw

h+ 1.393 + 0.667 log

wh

+ 1.44 if w

h> 1

(2.23)

where natural logarithms are used.

For design these formulas are not convenient and the following are used instead. First of all, threeauxiliary quantities are computed:

A =Z60

r + 1

2+

r 1r + 1

0.23 +

0.11

r

(2.24)

B =377

2Z

r(2.25)

C = log(B 1) + 0.39 0.61r

(2.26)

Next

w

h=

8eA

e2A 2 ifw

h< 2,

2

B 1 log(2B 1) + r 1

2rC

if

w

h> 2

(2.27)

Example

Compute the width w and length l of a microstrip with characteristic impedance Z = 50 , whichintroduces a phase shift of 90 at the frequency f = 2.5 GHz. The substrate thickness is 1/20 and r =

2.2.We compute A = 1.159, B = 7.985 and C = 2.056. Moreover, from the first of (2.27) we get w/h = 3.125.Since this result is greater than 2, it is not acceptable. From the second, instead, we get w/h = 3.081,which is in the domain of vality of the equation and hence it is acceptable. From this w = 0.391 cm results.Next, from (2.21) the effective dielectric constant is computed, eff = 1.88. Then the propagation constantis given by

k =2f

eff

c= 71.87 rad/m = 41.18/cm.

If the phase shift must be kl = /2, we obtain l = 2.19 cm.

Fig. 2.9 and Fig. 2.10 show the plots of eff versus w/h in the two ranges of wide and narrow strip, forvarious values of r of the substrate. Fig. 2.11 and Fig. 2.12 show the analogous plots of the characteristicimpedance Z.

Note that the effective permittivity eff given by (2.21) does not depend on frequency, as it is to beexpected in the case of a TEM mode. If we desire a more accurate model, which takes into account thefrequency dispersion ofeff due to the longitudinal field components, we can use the approximate formula(Getzinger, 1973)

eff = r r eff(0)1 + (f2/f2p ) G

(2.28)

where eff(0) is the zero frequency value given by (2.21) and the other parameters are

fp = Z0/(20h) (2.29)

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orfp(GHz) = 0.398Z0/h(mm) (2.30)

andG = 0.6 + 0.009Z0. (2.31)

where Z0 is the zero frequency characteristic impedance (in ). The characteristic impedance at theoperating frequency is then computed by (2.23) with this value of eff(f).

Figure 2.9. Effective permittivity eff versus microstrip dimensions (wide strip approximation).

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Figure 2.10. Effective permittivity eff versus microstrip dimensions (narrow strip approximation).

Figure 2.11. Characteristic impedance Z versus microstrip dimensions (wide strip approximation).

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Figure 2.12. Characteristic impedance Z versus microstrip dimensions (narrow strip approximation).

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transmission lines ch 2

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