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

About the further exercises of Section 7.7

Five further exercises on microstrip lines, in the spirit of Problem 7.7.1 and the home exercises, graded from easy to harder: reading a time-domain reflectometer record, the manufacturing tolerance of a controlled-impedance line, the quality factor of a microstrip resonator, the choice of substrate for a $77 GHz$ radar, and a measurement of the substrate permittivity. They use only the relations listed at the head of this subsection.

Exercise 7.7.2

Will the board house meet $50 Ω \pm 10 %$ ?

A $50 Ω$ microstrip on a $0.2 mm$ FR-4 prepreg (Table 7.7.1: $w = 0.38 mm$ , $ε_{r} = 4.4$ ) is ordered with $\pm 10 %$ impedance tolerance. The process gives: etched width $\pm 25 μ m$ , dielectric thickness $\pm 20 μ m$ , $ε_{r}$ between $4.2$ and $4.6$ (glass weave). (a) Find $Z_{0}$ and $ε_{eff}$ at the nominal values. (b) Find $Z_{0}$ at the two extremes of each tolerance alone. Which one dominates? (c) Find the two worst-case combinations. Is $45$ to $55 Ω$ met? (d) The board house can hold the thickness to $\pm 10 μ m$ . Does that suffice, and what should the nominal width be?

Try it yourself first.

Strategy

$Z_{0}$ depends on $w / h$ and on $ε_{r}$ only. A narrower strip, a thicker dielectric and a lower $ε_{r}$ all raise $Z_{0}$ , so the worst cases combine them. Hammerstad's analysis formula ( $w / h \geq 1$ branch) is evaluated at each corner.

$w / h = 0.38/0.20 = 1.90$ : $ε_{eff} = 3.329$ , $Z_{0} = 50.4 Ω$ .

change	$w / h$	$Z_{0}$ [Ω]	deviation
$w = 0.355$ / $0.405$ mm	$1.775$ / $2.025$	$52.5$ / $48.5$	$+ 4.1$ / $- 3.8 %$
$h = 0.18$ / $0.22$ mm	$2.111$ / $1.727$	$47.3$ / $53.3$	$- 6.2$ / $+ 5.8 %$
$ε_{r} = 4.2$ / $4.6$	$1.90$	$51.5$ / $49.4$	$+ 2.1$ / $- 2.0 %$

The dielectric thickness dominates: $\pm 10 %$ in $h$ is a larger change of $w / h$ than $\pm 25 μ m$ in $w$ ( $\pm 6.6 %$ ), while $ε_{r}$ enters only through $1/ ε_{eff}$ .

High corner ( $w = 0.355$ , $h = 0.22 mm$ , $ε_{r} = 4.2$ , $w / h = 1.614$ ): $Z_{0} = 56.6 Ω$ . Low corner ( $w = 0.405$ , $h = 0.18 mm$ , $ε_{r} = 4.6$ , $w / h = 2.25$ ): $Z_{0} = 44.5 Ω$ . Both corners lie outside $45$ to $55 Ω$ .

With $h = 0.20 \pm 0.01 mm$ the corners become $55.1$ and $46.1 Ω$ : still $0.1 Ω$ too high, because the nominal $50.4 Ω$ is not centred. A nominal width of $0.385 mm$ ( $Z_{0} = 50.0 Ω$ ) gives corners of $54.7$ and $45.7 Ω$ , inside the tolerance.

Answer

Z_{0} = 50.4 Ω nominal; h dominates 44.5 to 56.6 Ω (fails) h \pm 10 μ m, w = 0.385 mm : 45.7 to 54.7 Ω (passes)

Check

The worst-case spread ( $+ 12.3 %$ , $- 11.7 %$ about nominal) is close to the sum of the single spreads ( $4.1 + 5.8 + 2.1 = 12.0 %$ upwards, $3.8 + 6.2 + 2.0 = 12.0 %$ downwards): for small changes the effects add, as a first-order sensitivity analysis predicts.

Remark

This is why controlled-impedance boards are specified with a tolerance on $Z_{0}$ , not on $w$ : the board house adjusts the etched width to the measured thickness and permittivity of the actual laminate, and checks the result on a TDR coupon (Exercise 7.7.1).

cf. Application 7.7.1

Exercise 7.7.2: Will the board house meet 50\,\Omega\pm10\,\%?