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Above is a connection of half-wave dipole antenna. The book states that the current profile is same at any given time instant on both the radiators, each of quarter wavelength length.

Now my confusion is clearly the coaxial cable is connected to both radiators. Say I am sending the signal through centre wire and the braid of the coaxial cable is always at ground. Now at any given time instant, only left end of the cable is going to have non zero potential, while the right end is always at $0V$. Then how come both radiators would have same current profile at any given time when the excitation is different for left and right radiators!

To be very specific, say at any given time instant the voltage at the left end is 2.3V, so standing wave is generated in the left radiator, but on the right radiator the voltage is zero (braid is at ground), so how any excitation would happen on the right radiator.

Support will be greatly appreciated. I am a total newbie in antenna systems.

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  • $\begingroup$ The answer below is correct about using a balun. If you don't, then you need to take into account the length of the feedline (typically a half-wave) for the normal 468/f equation to give you minimum voltage at the feedpoint. $\endgroup$
    – SDsolar
    Jan 20, 2017 at 19:37

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You have aptly discovered why a balun is necessary when feeding a dipole with coax. You are right to think the book is wrong, because it is. With a coax feed and no balun, the current distributions on the dipole are not equal because some share of the current that should be on the right half (connected to the shield) of the dipole is instead flowing on the feedline common mode.

The book would be accurate had it used a balanced feedline instead, or included a balun.

The coax shield is in fact not at 0V relative to ground, because without a balun there will be common-mode currents on it. The shield being at 0V is not a law but rather an assumption. That assumption is based on an antenna design with a properly designed feed and an absence of common-mode currents, which this design has not.

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  • $\begingroup$ If current source on the left side (of dipole) comes from the transmitter, would you kindly identify / demonstrate source of the "common mode" current. I am under the understanding that "common mode" current is current on the OUTSIDE of the transmission line shield (coax) as opposed to current on the INSIDE of the shield - both due to combined physics of EMW and skin effect. $\endgroup$
    – JulyJim
    Jan 23, 2017 at 4:33
  • $\begingroup$ @JulyJim ham.stackexchange.com/a/6985/218 might address some of your concerns. If you'd like a more detailed response would you mind asking a new question? Due to length restriction it's difficult to explain much in comments. $\endgroup$
    – Phil Frost - W8II
    Jan 23, 2017 at 15:59
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In the interconnect between a coaxial line of Zc = 75 Ω and a half-wavelength dipole of Zin = 73 Ω, the reflections are much more severe than one would predict. This is because the field and the current distributions in the coaxial line and at the input of the wire dipole are very different (see figure below).

The unequal currents on the dipole’s arms unbalance the antenna and the coaxial feed and induce currents on the outside of the coax shield which are the reason for parasitic radiation. To balance the currents, various devices are used, called baluns (balanced-to-unbalanced transformer).

currents in dipole and coax

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