Exercise 7.7.4: Choosing the substrate thickness for a 77 GHz radar

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 7.7.4

Choosing the substrate thickness for a 77 GHz radar

An automotive radar front end at $77 GHz$ is printed on RO3003 ( $ε_{r} = 3.00$ , $tan δ = 0.001$ ) with $50 Ω$ microstrip feed lines. The laminate comes in $h = 0.127$ , $0.254$ and $0.508 mm$ . (a) Find $w / h$ for $50 Ω$ (synthesis: which branch?), $ε_{eff}$ and $λ_{g}$ . (b) For each $h$ find $w$ , the TE $_{1}$ surface-wave cutoff and the transverse-resonance limit. Which thicknesses are usable at all? (c) Find $α_{c}$ (smooth copper) and $α_{d}$ for the usable ones, and the loss of a $10 mm$ feed. (d) Which thickness would you choose, and why?

Try it yourself first.

Strategy

$w / h$ for a given $Z_{0}$ is independent of $h$ , so $ε_{eff}$ and $λ_{g}$ are the same on all three laminates. The limits scale as $1/ h$ , the conductor loss as $1/ w \propto 1/ h$ , the dielectric loss not at all, and radiation from bends and steps as $(h / λ_{0})^{2}$ . The choice is a trade-off between these.

$A = \frac{50}{60} 2 + \frac{2}{4} (0.23 + \frac{0.11}{3}) = 1.1785 + 0.1333 = 1.312$ , and the first branch gives $8 e^{A} / (e^{2 A} - 2) = 2.52 \geq 2$ : the second branch applies. $B = \frac{60 π ^{2}}{50 3} = 6.838$ , $C = \frac{2}{6} (ln 5.838 + 0.39 - \frac{0.61}{3}) = 0.650$ and

\frac{w}{h} = \frac{2}{π} [B - 1 - ln (2 B - 1) + C] = \frac{2}{π} (5.838 - 2.540 + 0.650) = 2.514.

Analysis: $ε_{eff} = 2.416$ , $Z_{0} = 50.3 Ω$ . At $77 GHz$ , $λ_{0} = 3.896 mm$ and $λ_{g} = 3.896/1.554 = 2.506 mm$ .

$h$ [mm]	$w$ [mm]	$(h / λ_{0})^{2}$
$0.127$	$0.319$	$0.0011$
$0.254$	$0.638$	$0.0043$
$0.508$	$1.277$	$0.0170$

$h$ [mm]	TE $_{1}$ : $c / (4 h ε_{r} - 1)$	$c / [ε_{r} (2 w + 0.8 h)]$
$0.127$	$418$ GHz	$234$ GHz
$0.254$	$209$ GHz	$117$ GHz
$0.508$	$104$ GHz	$58.5$ GHz

The $0.508 mm$ laminate has its transverse resonance below $77 GHz$ : unusable. The other two are usable.

$R_{s} = 72.4 mΩ$ and $k_{0} = 1613 rad/m$ .

α_{d} = \frac{k _{0} ε _{r} ( ε _{eff} - 1 ) tan δ}{2 ε _{eff} ( ε _{r} - 1 )} = 1.10 Np/m = 9.6 dB/m

on both. $α_{c} = R_{s} / (Z_{0} w)$ : $4.51 Np/m = 39.2 dB/m$ for $h = 0.127 mm$ and $19.6 dB/m$ for $0.254 mm$ . A $10 mm$ feed loses $0.49 dB$ and $0.29 dB$ .

$0.127 mm$ . Its first higher-order effect is three times above the operating frequency, and it radiates four times less at every discontinuity than $0.254 mm$ , whose transverse resonance is only $1.5$ times above $77 GHz$ . The price, $0.2 dB$ more on a $10 mm$ feed, is small, and feeds at $77 GHz$ are kept short anyway.

Answer

\frac{w}{h} = 2.514 (second branch) ε_{eff} = 2.416, λ_{g} = 2.51 mm h = 0.508 mm unusable choose h = 0.127 mm (w = 0.32 mm)

Check

The two limits scale exactly as $1/ h$ : $418/2 = 209$ , $234/2 = 117$ . $α_{c}$ doubles when $h$ (and $w$ ) halves.

Remark

Real copper foils are rough on the scale of the $0.24 μ m$ skin depth at $77 GHz$ , which roughly doubles $α_{c}$ . That is why millimetre-wave laminates are clad with smooth (rolled or reverse-treated) copper, and why antennas and transitions to the radar chip are placed as close to each other as possible.

cf. Home exercise 7.7.1