Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 7.7.1

A 50 Ω microstrip line on FR-4

A $50 Ω$ microstrip line is made on FR-4 with $ε_{r} = 4.4$ , $h = 1.6 mm$ , $tan δ = 0.02$ , copper conductors ( $σ_{c} = 5.8 \times 1 0^{7} S/m$ ), thin strip.

Find $w$ with the synthesis formula and check $Z_{0}$ and $ε_{eff}$ with the analysis formulas.
At $2.4 GHz$ find $β$ , $λ_{g}$ , $v_{p}$ , the length of a quarter-wave section, and the delay per cm.
Find $L^{'}$ and $C^{'}$ .
Find $α_{c}$ and $α_{d}$ at $2.4 GHz$ in dB/m.
How wide is a $100 Ω$ line on the same board?

Try it yourself first.

Strategy

Synthesis first (Wheeler), then analysis (Hammerstad) to confirm. In the quasi-TEM range the line is an ordinary transmission line in a uniform medium of permittivity $ε_{eff}$ , so $β$ , $λ_{g}$ , $v_{p}$ , $L^{'}$ and $C^{'}$ follow from $ε_{eff}$ and $Z_{0}$ alone.

1. Bookkeeping

At $2.4 GHz$ : $λ_{0} = 125 mm$ , $k_{0} = 50.27 rad/m$ and

R_{s} = \frac{π f μ _{0}}{σ _{c}} = \frac{π \times 2.4 \times 1 0 ^{9} \times 4 π \times 1 0 ^{- 7}}{5.8 \times 1 0 ^{7}} = 12.78 mΩ .

Wheeler's $A$ for $Z_{0} = 50 Ω$ :

A \frac{w}{h} = \frac{50}{60} 2.7 + \frac{3.4}{5.4} (0.23 + \frac{0.11}{4.4}) = 1.369 + 0.161 = 1.530, = \frac{8 e ^{1.530}}{e ^{3.060} - 2} = \frac{36.94}{19.32} = 1.912 < 2,

so the first branch is the right one and $w = 1.912 \times 1.6 = 3.06 mm$ . Analysis with $w / h = 1.912$ ( $p = 0$ ):

ε_{eff} F Z_{0} = 2.7 + 1.7 (1 + 12/1.912)^{- 1/2} = 2.7 + 1.7 \times 0.3707 = 3.330, = 1.912 + 1.393 + 0.667 ln 3.356 = 4.112, = \frac{377.0}{3.330 \times 4.112} = 50.2 Ω.

$ε_{eff} = 1.825$ :

β λ_{g} v_{p} delay = 50.27 \times 1.825 = 91.7 rad/m, = \frac{125}{1.825} = 68.5 mm, \frac{λ _{g}}{4} = 17.1 mm, = \frac{c}{1.825} = 0.548 c = 1.64 \times 1 0^{8} m/s, = \frac{ε _{eff}}{c} = 60.8 ps/cm .

C^{'} L^{'} = \frac{ε _{eff}}{c Z _{0}} = \frac{1.825}{3 \times 1 0 ^{8} \times 50.2} = 121 pF/m, = \frac{Z _{0} ε _{eff}}{c} = \frac{50.2 \times 1.825}{3 \times 1 0 ^{8}} = 306 nH/m .

α_{c} α_{d} = \frac{R _{s}}{Z _{0} w} = \frac{0.01278}{50.2 \times 3.06 \times 1 0 ^{- 3}} = 0.083 Np/m = 0.72 dB/m, = \frac{50.27 \times 4.4 \times 2.330 \times 0.02}{2 \times 1.825 \times 3.4} = 0.831 Np/m = 7.21 dB/m,

a total of $7.9 dB/m$ , dominated by the dielectric.

For $Z_{0} = 100 Ω$ : $A = 2.7386 + 0.1606 = 2.899$ , $w / h = 8 e^{A} / (e^{2 A} - 2) = 0.443 < 2$ , and $w = 0.71 mm$ . Check with the $w / h \leq 1$ formulas: $p = 0.04 (1 - 0.443)^{2} = 0.0124$ , $ε_{eff} = 2.7 + 1.7 (0.1887 + 0.0124) = 3.042$ and

Z_{0} = \frac{60}{3.042} ln (\frac{8}{0.443} + \frac{0.443}{4}) = 34.40 \times 2.899 = 99.7 Ω.

Answer

w = 3.06 mm, ε_{eff} = 3.33, Z_{0} = 50.2 Ω λ_{g} /4 = 17.1 mm, 60.8 ps/cm C^{'} = 121 pF/m, L^{'} = 306 nH/m 0.72 + 7.21 = 7.9 dB/m, w_{100 Ω} = 0.71 mm

Check

$L^{'} C^{'} = 306 \times 1 0^{- 9} \times 121 \times 1 0^{- 12} = 3.70 \times 1 0^{- 17} s^{2} / m^{2} = ε_{eff} / c^{2}$ , and $L^{'} / C^{'} = 50.3 Ω$ .

(a) Z_0(w/h) with the two designs. — (a) $Z_{0} (w / h)$ with the two designs.

(b) The two strips to scale. — (a) $Z_{0} (w / h)$ with the two designs.

Remark

The narrow line has the lower $ε_{eff}$ ( $3.04$ against $3.33$ ) because more of its field is in the air. The parallel-plate formula $η_{0} h / (ε_{r} w)$ would give $94 Ω$ for the $50 Ω$ strip: fringing cannot be ignored.

Problem 7.7.1: A 50 Ω microstrip line on FR-4