# 2.5 The Lossless Microstrip Line

Because its geometry is well suited for fabrication on printed circuit boards, the microstrip line is the most common interconnect configuration used in RF and microwave circuits. It consists of a narrow, very thin strip of copper (or another good conductor) printed on a dielectric substrate overlaying a ground plane (Fig. 2-10(a)). The presence of charges of opposite polarity on its two conducting surfaces gives rise to electric field lines between them (Fig. 2-10(b)). Also, the flow of current through the conductors (when part of a closed circuit) generates magnetic field loops around them, as illustrated in Fig. 2-10(b) for the narrow strip. Even though the patterns of E and B are not everywhere perfectly orthogonal, they are approximately so in the region between the conductors, which is where the E and B fields are concentrated the most. Accordingly, the microstrip line is considered a quasi-TEM transmission line, which allows us to describe its voltages and currents in terms of the one-dimensional TEM model of Section 2-4, namely Eqs. (2.26) through (2.33).

# Figure 2-10 Microstrip line: (a) longitudinal view, (b) cross-sectional view, and (c) circuit example.

(a) Longitudinal view

Figure 2 10-a Full Alternative Text

(b) Cross-sectional view with E and B field lines

Figure 2 10-b Full Alternative Text

(c) Microwave circuit

Courtesy of Prof. Gabriel Rebeiz, U. California at San Diego

The microstrip line has two geometric parameters, the width of the elevated strip, w, and the thickness (height) of the dielectric layer, h. We will ignore the thickness of the conducting strip because it has a negligible influence on the propagation properties of the microstrip line so long as the strip thickness is much smaller than the width w, which is almost always the case in practice. Also, we assume the substrate material to be a perfect dielectric with σ=0 and the metal strip and ground plane to be perfect conductors with σ c≈∞. These two assumptions simplify the analysis considerably without incurring significant error. Finally, we set μ=μ 0, which is always true for the dielectric materials used in the fabrication of microstrip lines. These simplifications reduce the number of geometric and material parameters to three, namely w, h, and ε.

Electric field lines always start on the conductor carrying positive charges and end on the conductor carrying negative charges. For the coaxial, two-wire, and parallel-plate lines shown in the upper part of Fig. 2-4, the field lines are confined to the region between the conductors. A characteristic attribute of such transmission lines is that the phase velocity of a wave traveling along any one of them is given by

u p=c ε r,(2.36a)

where c is the velocity of light in free space and ε r is the relative permittivity of the dielectric medium between the conductors.

# 2.5.1 Effective Permittivity

In the microstrip line, even though most of the electric field lines connecting the strip to the ground plane do pass directly through the dielectric substrate, a few go through both the air region above the strip and the dielectric layer (Fig. 2-10(b)). This nonuniform mixture can be accounted for by defining an effective relative permittivity ε eff such that the phase velocity is given by an expression that resembles Eq. (2.36a), namely

u p=c ε eff.(2.36b)

Methods for calculating the propagation properties of the microstrip line are quite complicated and beyond the scope of this text. However, it is possible to use curve-fit approximations to rigorous solutions to arrive at the following set of expressions:*

ε eff=ε r+1 2+ε r−1 2 1+10 s−x y,(2.37a)

where s is the width-to-thickness ratio,

s=w h,(2.37b)

and x and y are intermediate variables given by

x=0.56 ε r−0.9 ε r+3 0.05,(2.38a)
y=1+0.02 ln(s 4+3.7×10−4 s 2 s 4+0.43)+0.05 ln(1+1.7×10−4 s 3).(2.38b)

# 2.5.2 Characteristic Impedance

The characteristic impedance of the microstrip line is given by

Z 0=60 ε eff ln 6+(2 π−6)e−t s+1+4 s 2,(2.39)

with

t=30.67 s 0.75.(2.40)

Figure 2-11 displays plots of Z 0 as a function of s for various types of dielectric materials.

# Figure 2-11 Plots of Z 0 as a function of s for various types of dielectric materials.

Figure 2 11 Full Alternative Text

The corresponding line and propagation parameters are given by

R′=0 (because σ c=∞),(2.41a)
G′=0 (because σ=0),(2.41b)

C′=ε eff Z 0 c,(2.41c)
L′=Z 0 2 C′,(2.41d)
α=0 (because R′=G′=0),(2.41e)
β=ω c ε eff.(2.41f)

# 2.5.3 Design Process

The preceding expressions allow us to compute the values of Z 0 and the other propagation parameters when given values for ε r, h, and w. This is exactly what is needed in order to analyze a circuit containing a microstrip transmission line. To perform the reverse process, namely to design a microstrip line by selecting values for its w and h such that their ratio yields the required value of Z 0 (to satisfy design specifications), we need to express s in terms of Z 0. The expression for Z 0 given by Eq. (2.39) is rather complicated, so inverting it to obtain an expression for s in terms of Z 0 is rather difficult. An alternative option is to generate a family of curves similar to those displayed in Fig. 2-11 and to use them to estimate s for a specified value of Z 0. A logical extension of the graphical approach is to generate curve-fit expressions that provide high-accuracy estimates of s. The error associated with the following formulas is less than 2%:

(a) For Z 0≤(44−2 ε r) Ω,

s=w h=2 π{ (q−1)−ln(2 q−1)+ε r−1 2 ε r[ ln(q−1)+0.29−0.52 ε r ] }(2.42)

with

q=60 π 2 Z 0 ε r,

and

(b) for Z 0≥(44−2 ε r) Ω,

s=w h=8 e p e 2 p−2,(2.43a)

with

p=ε r+1 2 Z 0 60+ε r−1 ε r+1 0.23+0.12 ε r.(2.43b)

The above expressions presume that ε r, the relative permittivity of the dielectric substrate, has already been specified. For typical substrate materials including Duroid, Teflon, silicon, and sapphire, ε r ranges between 2 and 15.

# Example 2-2: Microstrip Line

A 50 Ω microstrip line uses a 0.5-mm thick sapphire substrate with ε r=9. What is the width of its copper strip?

Solution:

Since Z 0=50>44−18=32, we should use Eq. (2.43):

p=ε r+1 2×Z 0 60+(ε r−1 ε r+1)(0.23+0.12 ε r)=9+1 2×50 60+(9−1 9+1)(0.23+0.12 9)=2.06,s=w h=8 e p e 2 p−2=8 e 2.06 e 4.12−2=1.056.

Hence,

w=s h=1.056×0.5 mm=0.53 mm.

To check our calculations, we use s=1.056 to calculate Z 0 to verify that the value we obtained is indeed equal or close to 50 Ω. With ε r=9, Eqs. (2.37a) to (2.40) yield

x=0.55,y=0.99,t=12.51,ε eff=6.11,Z 0=49.93 Ω.

The calculated value of Z 0 is, for all practical purposes, equal to the value specified in the problem statement.