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

Problem 7.5.2

Design of a PTFE $50 Ω$ cable

A flexible $50 Ω$ cable is to be built with a PTFE dielectric ( $ε_{r} = 2.1$ , $tan δ = 2 \times 1 0^{- 4}$ ) and copper conductors ( $σ_{c} = 5.8 \times 1 0^{7}$ S/m). The inner diameter of the outer conductor is $2 b = 4.0$ mm.

Find the inner-conductor diameter $2 a$ , and $L^{'}$ , $C^{'}$ , $v$ .
Find $α_{c}$ , $α_{d}$ and the total loss in dB/m at $3$ GHz.
Up to which frequency is the cable single-mode?
At which frequency would $α_{d} = α_{c}$ ? Comment.

Try it yourself first.

Strategy

Invert $Z_{0} = \frac{60}{ε _{r}} ln \frac{b}{a}$ for $b / a$ . Then the loss formulas need only $a$ , $b$ , $η$ and $R_{s}$ . For part (d) write both losses as functions of frequency: $α_{c} = A f$ (skin effect), $α_{d} = B f$ .

Bookkeeping

$ε_{r} = 2.1$ : $η = 377/ 2.1 = 260.1 Ω$ , $v = c / 2.1 = 2.07 \times 1 0^{8}$ m/s, $b = 2.0$ mm. Copper at $3$ GHz: $R_{s} = 14.29$ m $Ω$ .

$b / a = exp (50 2.1 /60) = e^{1.2076} = 3.346$ , so $2 a = 4.0/3.346 = 1.20$ mm ( $a = 0.598$ mm), and

L^{'} C^{'} v = 2 \times 1 0^{- 7} \times 1.2076 = 241.5 nH/m, = \frac{2 π ( 2.1 ) ε _{0}}{1.2076} = 96.6 pF/m, = \frac{1}{L ^{'} C ^{'}} = 2.07 \times 1 0^{8} m/s .

$1/ a + 1/ b = 1673 + 500 = 2173$ m $^{- 1}$ :

α_{c} k α_{d} = \frac{0.01429}{2 \times 260.1} \times \frac{2173}{1.2076} = 0.0494 Np/m = 0.429 dB/m, = \frac{2 π f}{v} = 91.1 rad/m, = \frac{91.1 \times 2 \times 1 0 ^{- 4}}{2} = 0.0091 Np/m = 0.079 dB/m .

Total $0.51$ dB/m at $3$ GHz.

The first higher mode is TE $_{11}$ :

f_{c} \approx \frac{v}{π ( a + b )} = \frac{2.07 \times 1 0 ^{8}}{π \times 2.598 \times 1 0 ^{- 3}} = 25.4 GHz

(the exact root of Eq. (7.5.16) gives $26.1$ GHz).

With $f$ in GHz, $α_{c} = 0.429 f /3 = 0.248 f$ dB/m and $α_{d} = 0.0791 f /3 = 0.0264 f$ dB/m. They are equal when $f = 0.248/0.0264$ , that is at $f = 88$ GHz, far beyond the single-mode limit of part (c). Copper dominates the loss in the whole usable band.

Answer

2 a = 1.20 mm, L^{'} = 241.5 nH/m C^{'} = 96.6 pF/m, v = 2.07 \times 1 0^{8} m/s α = 0.429 + 0.079 = 0.51 dB/m at 3 GHz f_{c} \approx 25 GHz α_{d} = α_{c} at 88 GHz

Check

$L^{'} / C^{'} = 241.5 \times 1 0^{- 9} /96.6 \times 1 0^{- 12} = 50.0 Ω$ . In circuit form $R^{'} = \frac{R _{s}}{2 π} (\frac{1}{a} + \frac{1}{b}) = 4.94 Ω/ m$ and $α_{c} = R^{'} /2 Z_{0} = 0.0494$ Np/m, the same.

(b) The mode that ends the band. — (a) The losses against frequency.

Remark

A better dielectric would hardly help, a bigger cable would. Above $25$ GHz a second mode with its own speed would carry part of the power and distort the signal: to go high, the cable must stay thin, and therefore lossy.

Problem 7.5.2: Design of a PTFE 50\ \Omega cable