What makes current flow in a transmission line?

**Bob S** · Nov7-09, 02:08 PM

Originally Posted by stephen163

I don't think that explanation is particularly useful. It's a lot more complicated than that. The voltage is between the forward conductor and the return conductors. To a good approximation, no voltage along the conductor is necessary. The signal propagates as a wave and Maxwell's equations that govern the wave's movement. This is true, as I understand, for all conductors, even at low frequency and DC. Look up the lumped element approximation and study how reflections in a transmission line act to give the illusion that there is an electric field along the conductive path.

Good. In the limit, consider a transmission line using superconducting wire. So the only voltage drop along the conductor is, in the EE notation, V = L dI/dt, where L is the series inductance. It is useful here to consider what the series inductance really means. Comparing the EE and physics formulas for inductive energy storage:

E = (1/2) L I² = (1/2) ∫B H dV =(1/2 μ₀) ∫B² dV,

Where B and H represent the magnetic field energy between the conductors, and dV is the volume. So the signal propagation in a transmission line is actually the propagation of electric energy (the electric field) and the magnetic energy (the magnetic field) between the conductors. The flow of power is actually represented by the Poynting vector cross product of these two fields between the two conductors:

S = E x H

For this reason, the transmission line problem is best treated using Maxwell’s equations.

Bob S

**cabraham** · Nov10-09, 09:14 AM

Originally Posted by Bob S

Good.
If we delve deeply into the magnetic fields H of the current, and the electric fields E of the voltage between the wires, the power flow is actually a TEM (transverse electric magnetic) wave between the wires given by the Poynting vector

P = ∫E x H dA

where A is a surface between the wires. The currents in the wires support the fields, just like the currents in the walls of a waveguide support the E and H fields inside. The power is in the fields inside, not in the waveguide walls.

A good example of a transmission line is the 300-ohm TV lead-in wire from the antenna. sqrt(L/C) = Z = 300 ohms.

Bob S

Very well stated. Power flow is indeed the cross product of E & H integrated over the area, known as the Poynting vector. It "points" in the direction of power propagation. Without I, there would be no power. All V w/ no I requires that Z0 be infinite, requiring infinite spacing between conductors. Likewise, all I w/ no V would need Z0 = 0, an impossibility.

E, H, I, V, B, & D, as well as P, the power, can only exist mutually under time-varying conditions. That is Maxwell's equations in a nutshell. When incident power impinges on an antenna, dipole for example, which is a modified t-line, I & V, as well as E & H, are mutually present. I, V, E, & H proagate at transmission speed less than c (light speed). They arrive at any point in unison. The current arrives at the same time as the voltage. Likewise for E & H.

Claude