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

Problem 7.5.1

TEM fields of a precision air line

A precision air line (laboratory standard) has an inner conductor of diameter $2 a = 3.04$ mm and an outer conductor of inner diameter $2 b = 7.00$ mm. A TEM wave of $f = 1$ GHz with voltage amplitude $V_{0} = 100$ V travels along it.

Find $Z_{0}$ , $L^{'}$ and $C^{'}$ , and check $v$ .
Write the phasors $E_{ρ}$ and $H_{ϕ}$ , and find the current $I_{0}$ and $∣ E ∣$ at $ρ = a$ and $ρ = b$ .
Find the power by integrating the Poynting vector, and compare with $V_{0}^{2} /2 Z_{0}$ .
Air breaks down at $3$ MV/m. Find the largest voltage amplitude and power the line can carry (matched).

Try it yourself first.

Strategy

Every TEM quantity follows from $V_{0}$ and the single number $ln (b / a)$ . The field is that of a cylindrical capacitor, $E_{ρ} \propto 1/ ρ$ , so it is strongest at the inner conductor, where breakdown starts.

Bookkeeping

Air: $η_{0} = 120 π Ω$ , $v = c$ . $a = 1.52$ mm, $b = 3.50$ mm, $ln (b / a) = 0.8341$ . $f = 1$ GHz: $k = 2 π f / c = 20.94$ rad/m. $V_{0} = 100$ V.

a) Line constants

Z_{0} L^{'} C^{'} \frac{1}{L ^{'} C ^{'}} = 60 \times 0.8341 = 50.0 Ω, = \frac{μ _{0}}{2 π} ln \frac{b}{a} = 2 \times 1 0^{- 7} \times 0.8341 = 166.8 nH/m, = \frac{2 π ε _{0}}{0.8341} = 66.6 pF/m, = 3.00 \times 1 0^{8} m/s = c .

b) Fields and current

$V_{0} / ln (b / a) = 119.9$ V and $119.9/ η_{0} = 0.318$ A:

E_{ρ} = \frac{119.9}{ρ} e^{- j 20.94 z} V/m, H_{ϕ} = \frac{0.318}{ρ} e^{- j 20.94 z} A/m (ρ in m) .

Ampère's law around the inner conductor gives $I_{0} = 2 π ρ H_{ϕ} = 2 π \times 0.318 = 2.00$ A $= V_{0} / Z_{0}$ . At the conductors $∣ E (a) ∣ = 119.9/ (1.52 \times 1 0^{- 3}) = 78.9$ kV/m and $∣ E (b) ∣ = 119.9/ (3.50 \times 1 0^{- 3}) = 34.3$ kV/m.

c) Power

P \frac{V _{0}^{2}}{2 Z _{0}} = \int_{a}^{b} \frac{∣ E _{ρ} ∣ ^{2}}{2 η _{0}} 2 π ρ d ρ = \frac{π \times 119. 9 ^{2}}{120 π} ln \frac{b}{a} = 99.9 W, = \frac{1 0 ^{4}}{100.1} = 99.9 W .

d) Breakdown

The field at $ρ = a$ reaches $E_{br}$ when

V_{m a x} P_{m a x} = E_{br} a ln \frac{b}{a} = 3 \times 1 0^{6} \times 1.52 \times 1 0^{- 3} \times 0.8341 = 3.80 kV, = \frac{380 3 ^{2}}{2 \times 50.0} = 145 kW .

Answer

Z_{0} = 50.0 Ω, L^{'} = 166.8 nH/m C^{'} = 66.6 pF/m, I_{0} = 2.00 A ∣ E (a) ∣ = 78.9 kV/m, ∣ E (b) ∣ = 34.3 kV/m P = 99.9 W V_{m a x} = 3.80 kV, P_{m a x} = 145 kW

Check

$∣ E (a) ∣/∣ E (b) ∣ = 78.9/34.3 = 2.30 = b / a$ . At $100$ V the inner-conductor field is $1/38$ of the breakdown field, so the line can carry $3 8^{2} \approx 1450$ times more power: $1450 \times 99.9$ W $\approx 145$ kW.

(a) |E_\rho| across the gap. — (a) $∣ E_{ρ} ∣$ across the gap.

(b) The TEM field lines. — (a) $∣ E_{ρ} ∣$ across the gap.

Remark

$145$ kW is a peak-field limit for a matched line. A standing wave raises the peak voltage, and long before breakdown the inner conductor overheats: the average power rating of such a small line is a few hundred watts. Both the breakdown and most of the copper loss start at the inner conductor, where $∣ E ∣$ and $∣ H ∣$ are $b / a = 2.30$ times larger than at the outer one.

Problem 7.5.1: TEM fields of a precision air line