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

Problem 7.5.3

Interactive: Transmission Line Designer

Problem 7.5.3

The optimum ratios $b / a$ with numbers

An air-filled coaxial line for a transmitter feed has a fixed outer-conductor radius $b = 5$ mm, copper conductors, and the breakdown field of air $E_{br} = 3$ MV/m. Four inner radii are compared: $b / a = 1.65$ , $2.30$ , $e$ , $3.59$ .

Find $a$ and $Z_{0}$ for each design.
Find the largest voltage amplitude and power (matched line).
Find $α_{c}$ at $1$ GHz in dB/100 m.
Find the TE $_{11}$ cutoff of each. Which design would you pick for a high-power feed, which for a low-loss link, and why is $50 Ω$ the usual choice?

Three of the designs to scale (same b). — Three of the designs to scale (same $b$ ).

Try it yourself first.

Strategy

With $x = b / a$ and $b$ fixed, every quantity is a simple function of $x$ : $a = b / x$ , $Z_{0} = 60 ln x$ , $V_{m a x} = E_{br} a ln x$ (field at the inner conductor), $P_{m a x} = V_{m a x}^{2} /2 Z_{0}$ , $α_{c} = \frac{R _{s}}{2 η _{0} b} \frac{1 + x}{l n x}$ and $f_{c} \approx c / [π (a + b)]$ . One column is worked out in full, the others follow the same way.

Bookkeeping

Air, $b = 5$ mm, $E_{br} = 3 \times 1 0^{6}$ V/m. At $1$ GHz $R_{s} = 8.25$ m $Ω$ and $R_{s} / (2 η_{0} b) = 2.188 \times 1 0^{- 3}$ Np/m.

Column x = 2.30

$a = 5/2.30 = 2.17$ mm, $Z_{0} = 60 ln 2.30 = 50.0 Ω$ ,

V_{m a x} P_{m a x} α_{c} f_{c} = 3 \times 1 0^{6} \times 2.171 \times 1 0^{- 3} \times 0.834 = 5.43 kV, = \frac{543 3 ^{2}}{2 \times 50.0} = 295 kW, = 2.188 \times 1 0^{- 3} \times \frac{3.30}{0.834} = 8.67 \times 1 0^{- 3} Np/m = 7.53 dB/100 m, \approx \frac{3 \times 1 0 ^{8}}{π \times 7.17 \times 1 0 ^{- 3}} = 13.3 GHz .

a)–d) All four designs

$x = b / a$	$a$ [mm]	$Z_{0}$ [ $Ω$ ]	$f_{c, TE_{11}}$ [GHz]
$1.65$	$3.03$	$30.0$	$11.9$
$2.30$	$2.17$	$50.0$	$13.3$
$e = 2.72$	$1.84$	$60.0$	$14.0$
$3.59$	$1.39$	$76.7$	$14.9$

$x = b / a$	$V_{m a x}$ [kV]	$P_{m a x}$ [kW]	$α_{c}$ [dB/100 m]
$1.65$	$4.55$	$345$	$10.06$
$2.30$	$5.43$	$295$	$7.53$
$e = 2.72$	$5.52$	$254$	$7.07$
$3.59$	$5.34$	$186$	$6.83$

High power: $x = 1.65$ ( $30 Ω$ ). Low loss: $x = 3.59$ ( $77 Ω$ ). Highest voltage: $x = e$ ( $60 Ω$ ). The $50 Ω$ line carries $86 %$ of the best power with only $10 %$ more loss than the best: it is the compromise.

Answer

P_{m a x} : 345 kW (30 Ω) α_{c} : 6.83 dB/100 m (77 Ω) 50 Ω : 295 kW, 7.53 dB/100 m

Step through the four designs in the Example menu below and read $P_{m a x}$ and $α_{c}$ off the Best $b / a$ plot.

InteractiveLoading simulation…

Check

The optima of Section 7.5.5: $ln x = 1 + 1/ x$ is satisfied by $x = 3.59$ ( $ln 3.59 = 1.278$ , $1 + 1/3.59 = 1.279$ ), and $ln x = 1/2$ by $x = 1.65$ . Ratios: $295/345 = 0.86$ and $7.53/6.83 = 1.10$ . The exact TE $_{11}$ roots are $1$ to $3 %$ above the approximate cutoffs of the table.

Problem 7.5.3. Loss relative to its minimum and peak power relative to its maximum against b/a at fixed b, with the four designs marked (black dots). Both optima are flat and far apart. A fat inner conductor spreads the surface field over a large radius (high power). A thinner one raises Z_0, so the same power needs less current (low loss), until the 1/a growth of the inner-conductor loss takes over at 3.59. — **Figure 7.5.9.** Problem 7.5.3. Loss relative to its minimum and peak power relative to its maximum against $b / a$ at fixed $b$ , with the four designs marked (black dots). Both optima are flat and far apart. A fat inner conductor spreads the surface field over a large radius (high power). A thinner one raises $Z_{0}$ , so the same power needs less current (low loss), until the $1/ a$ growth of the inner-conductor loss takes over at $3.59$ .

Problem 7.5.3: The optimum ratios b/a with numbers