If there is a phase difference on a lossless transmission line between current and voltage, is it always 90deg? Assume the Load matches the transmission line characteristic impedance.

If not what are the conditions for the phase not being 90deg (load or no load)?

Anyway to show this orthogonality or phase difference in the equations?

enter image description here
Source: https://www.ibiblio.org/kuphaldt/electricCircuits/AC/AC_14.html

  • \$\begingroup\$ Do you know how hard it is to ask questions that haven't been covered already? When was the last time you wrote a question? \$\endgroup\$
    – Voltage Spike
    Sep 25 '19 at 19:24
  • \$\begingroup\$ I cant recall asking a question but rather than ask about something you already know the answer to, read a good book then ask about something with measurable issues that needs clarification in your mind so you learn something not easily found by a web search of key words. Who cares about points \$\endgroup\$ Sep 25 '19 at 21:18

A transmission line can support two waves, one travelling left, one travelling right. For each wave, the voltage and current are always strictly in phase, with the ratio of the line impedance.

When you add the two waves together to get voltages and currents at a particular point on the transmission line, which gets a bit non-intuitive as voltages add but currents subtract, then depending on the position on the line and the characteristics of the terminations, you can get essentially any phase, any ratio.

Some interesting line lengths are

  • Very short - an open or shorted line behaves like a capacitor or inductor respectively
  • Multiples of half a wavelength - the ratio is the same at each end
  • Odd multiples of a quarter wavelength - the most interesting - inverts the ratio at one end through the square of the line impedance - so open becomes short - 25 ohms becomes 100 ohms if the line is 50 ohms

Your diagram illustrates a standing wave, rather than simple wave-propagation on a long transmission line. The waves sent out by the 500KHz signal generator are being reflected by the infinite-Z located at the end of the unterminated line. The waveforms are dynamically changing over time (and hence the dotted lines in the diagram.) To get rid of this effect, either use an infinitely-long line, or add a 75-ohm terminating resistor.

Instead, if assuming an ideal infinite transmission line, the current and voltage are at zero phase, not 90deg, and the line behaves as a resistor (an energy-storage medium having characteristic impedance: volts times amps gives single-direction watts.) Fifty-ohm and 75-ohm lines have 0deg between their V and I waveforms, same as with 300-ohm twinlead, or that 'magical' 377 ohms of waves in empty space.

If the VI phase wasn't zero, this would produce some negative watts during part of the cycle, which is the same as having some wave-propagation in the backwards direction.

Note that something here is orthogonal: the e-field and b-field vectors of the flux surrounding the conductors is everywhere at 90deg. Simplified version: each conductor is surrounded by circular b-field, and each conductor is surrounded by radial e-field, where the flux-lines cross each other at 90deg. In non-tech explanations, this is often described as "EM waves have transverse E and M at 90deg." But it's not the waves that are 90deg, instead it's the flux-lines. Therefore, it applies to circuitry at DC, not just applying to waveguides and light waves. (Even in a flashlight, the propagating EM circuit-energy includes orthogonal flux lines, E-cross-M giving power density of unidirectional energy flow.)


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