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

Keep the outer radius $b$ fixed (it is set by size, weight and the frequency limit) and vary $x = b / a$ . With $a = b / x$ , Eq. (7.5.13) and Eq. (7.5.11) become

α_{c} = \frac{R _{s}}{2 η b} \frac{1 + x}{ln x}, P_{m a x} = \frac{π b ^{2} E _{br}^{2}}{η} \frac{ln x}{x ^{2}} . (7.5.14)

Minimum loss: $\frac{d}{d x} \frac{1 + x}{ln x} = \frac{ln x - ( 1 + x ) / x}{ln ^{2} x} = 0$ gives $ln x = 1 + 1/ x$ , solved by $x = 3.59$ , $Z_{0} = 76.7/ ε_{r} Ω$ .
Maximum power: $\frac{d}{d x} \frac{ln x}{x ^{2}} = \frac{1 - 2 ln x}{x ^{3}} = 0$ gives $ln x = 1/2$ , $x = e = 1.65$ , $Z_{0} = 30/ ε_{r} Ω$ .
Highest voltage ( $E_{a}$ smallest for a given $V$ , or $V_{m a x} = E_{br} b ln x / x$ largest): $ln x = 1$ , $x = e$ , $Z_{0} = 60/ ε_{r} Ω$ (Exercise 3.3.10 of Section 3.3).

The optimum coaxial lines (fixed b)

optimum	condition	$b / a$	$Z_{0}$ in air
minimum loss	$ln x = 1 + 1/ x$	$3.59$	$76.7 Ω$
maximum power	$ln x = 1/2$	$1.65$	$30 Ω$
highest voltage	$ln x = 1$	$2.72$	$60 Ω$

$50 Ω$ (air: $b / a = 2.30$ ) is the compromise.

Loss relative to its minimum (blue, dashed) and peak power relative to its maximum (orange) of a coaxial line with fixed outer radius, against b/a. The loss is least at 3.59, the power largest at 1.65. Both optima are broad, and the 50\ \Omega air line (2.30) loses only 10\,\% more than the best and carries 86\,\% of the largest power. — **Figure 7.5.4.** Loss relative to its minimum (blue, dashed) and peak power relative to its maximum (orange) of a coaxial line with fixed outer radius, against $b / a$ . The loss is least at $3.59$ , the power largest at $1.65$ . Both optima are broad, and the $50 Ω$ air line ( $2.30$ ) loses only $10 %$ more than the best and carries $86 %$ of the largest power.

Why 50 and 75 ohms

An air line at $50 Ω$ loses only $10 %$ more than the minimum-loss line and carries $86 %$ of the maximum power. With polyethylene the minimum-loss line has $76.7/1.5 = 51 Ω$ : $50 Ω$ is also very nearly the lowest-loss solid cable. $75 Ω$ , the television standard, is close to the air-line optimum $76.7 Ω$ , and receiving systems care about loss, not power.

Quick check

Why does a fat inner conductor give high power but not low loss?

A fat inner conductor spreads the surface field over a large radius, so $E_{a}$ is small for a given voltage and much power can pass before breakdown. But a fat inner conductor means a small $ln (b / a)$ , a low $Z_{0}$ and therefore a large current for a given power, $∣ I ∣^{2} = 2 P / Z_{0}$ , and the loss grows with $∣ I ∣^{2}$ . A thinner one raises $Z_{0}$ , so the same power needs less current, until the $1/ a$ growth of the inner-conductor loss takes over at $b / a = 3.59$ .

7.5.5 The best ratio b/a: minimum loss, maximum power