I have two conductors forming a transmission line like in the figure below (left). Only B is grounded. What is the voltage between B and D at any given moment? If I use a transmission line model (right) then I can compute the voltage between A and B (v(t_0,0)) and the voltage bewteen C and D (v(t_0,L)) using transmission line equations but I can't find the voltage between B and D.


Edit 1: System connection (if it makes a difference): Let's say that A and B are connected to a voltage source and that C and D are a) left in open circuit b) connected to a resistance of value R

  • \$\begingroup\$ Thanks I just added the external connections \$\endgroup\$
    – Ken Grimes
    Commented Mar 6, 2020 at 16:35
  • \$\begingroup\$ Thanks Ken, then the only way to affect the B to D conductor is just electrostatic since there is no current flow in the A to C conductor. Dr. Hortons paper should help. Sorry if I’m misunderstanding - I’m thinking 60Hz. \$\endgroup\$ Commented Mar 6, 2020 at 16:45
  • \$\begingroup\$ You'll have a simple resistor ladder, consisting of resistances AC, R, and BD. Voltage across each resistor should be in proportion to its resistance. When R is infinity, then obviously all the voltage drop happens in R, and the AC and BD voltage drops will be zero. \$\endgroup\$ Commented Mar 6, 2020 at 19:51
  • \$\begingroup\$ @Harper, since we're calling it a transmission line we should consider that it is reasonably long compared to the wavelength of whatever signals are exciting it. Say, at least 0.1 wavelengths, and possibly many wavelengths long. \$\endgroup\$
    – The Photon
    Commented Mar 7, 2020 at 6:35
  • \$\begingroup\$ Thanks @Harper. What I perhaps didn't mention was that I'm dealing with signals with fast rise-times (~5ns) on a relatively long PCB. I get your suggestion of considering it a resistor ladder, that gives me an idea of the I*R drop, but how to account for the distributed L and C on the transmission line? \$\endgroup\$
    – Ken Grimes
    Commented Mar 7, 2020 at 10:55

1 Answer 1


There's rarely any real reason to need an answer to this question. Presumably B and D are far apart so there's no way to measure the voltage between them, and really no reason why the voltage between them will matter to the operation of the circuits attached at the B end or the D end of the line.

Obviously if the circuit at the D end is itself grounded somewhere, the design will need to somehow isolate the connection to the transmission line to avoid ground loops (i.e. undesired currents flowing due to ground potential differences between the two ends of the transmission line).

If you really do want to know, Then you can treat the B-D wire an antenna, with the A-C wire as a parasitic antenna element. And the voltage difference between D and B depends what signals are impinging on it from all outside sources.

  • \$\begingroup\$ Thanks for your comment. I'm trying to understand an EMC problem on a PCB working with rise-times of ~5 ns, so in my case I can really measure that voltage. \$\endgroup\$
    – Ken Grimes
    Commented Mar 7, 2020 at 10:53
  • \$\begingroup\$ @KenGrimes, if this line is all on a single PCB, why not just ground both ends? \$\endgroup\$
    – The Photon
    Commented Mar 7, 2020 at 16:23
  • \$\begingroup\$ If you're interested in playing tricks with this kind of structure, google "Marchand balun". But if you're working with wideband signals (as implied by you're mentioning the rise time) then you won't get a clean result, but something very messy to analyze. \$\endgroup\$
    – The Photon
    Commented Mar 7, 2020 at 16:28
  • \$\begingroup\$ thanks for the comments. I didn't ground the other side because I was afraid of ground loops. I'll check out that "Marchand balun" :) \$\endgroup\$
    – Ken Grimes
    Commented Mar 8, 2020 at 20:48

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