If the input impedance of a resonant half wave dipole at the feed point is 75 + j0 ohms, then if an AC sine wave voltage V is applied to the feed point, how can the voltage at the center be zero ? Ohms law always applies correct ? So for the rms value then voltage is Vrms = Irms x Z right ? so Z = Vrms / Irms = Vrms / 75 +j 0 = 0.

The picture from Wikipedia below clearly shows that the voltage is always zero at the center. My understanding is that this picture shows the standing waves on a dipole. The standing waves are the result of the applied AC voltage across the feed point and vary in amplitude over time but have a fixed voltage profile in space along the length of the dipole. I thought that anywhere along the length of the dipole ohms law always applies.


Dipole standing waves animation

Is the 'voltage' in the picture the voltage potential with respect to the center of the dipole ? and the one which results from that applied to the feed point ? The applied voltage potential difference results in a current which is the movement of electrons in the dipole in one direction during the positive half cycles of the ac waveform and in the opposite direction during the negative half right ?

What am i missing.

Is the voltage not really zero but rather I / Z = I / 75 + j 0 which is not zero ?

And then, if the voltage is not really zero, how come it's ok to connect the center of the driven element in a yagi to the boom because the voltage at the center is zero ?

Someone please help before my brain implodes !!!! :)

  • $\begingroup$ Welcome to AR! Maybe it's worth noticing that it's impossible to have both conductors at the same point (a feedline with 0 distance between conductors will have 0 impedance. So this small distance already accounts for some voltage to be present at the feed "point". $\endgroup$ – Juancho Oct 26 '18 at 15:16
  • $\begingroup$ Is that animation correct? $\endgroup$ – Mike Waters Oct 26 '18 at 19:45
  • $\begingroup$ I don't see how it could be for a 1/4 wave dipole. If these red and blue areas are trying to represent instantaneous voltage and current, they are 90 degrees out of phase, meaning the feedpoint impedance is purely reactive, meaning it's impossible to transfer any real power to it. This antenna could never radiate! $\endgroup$ – Phil Frost - W8II Nov 3 '18 at 23:15
  • $\begingroup$ correction: I mean 1/2 wave dipole $\endgroup$ – Phil Frost - W8II Nov 3 '18 at 23:31
  • $\begingroup$ The animation is wrong, on several points. As pointed out in several companion posts, "voltage = 0 at the center of a half wave dipole" is a convenient analytical construct, wherein the voltage at any point arbitrarily close to the center is +/- V/2. $\endgroup$ – Brian K1LI Nov 4 '18 at 15:17

Look closer at the diagram. At the two wires coming from the source, the voltage is NOT always zero. The only way for the voltage at the center point to be zero is for the two source wires to occupy the same point which is impossible. Think about the voltage as an electric field which is indeed zero at the point halfway between the two wires.


The simple answer is that the graphic is not accurate. There is an RMS voltage present at the center feed point of the dipole that follows Ohm's law relative to the feed point impedance.

Most likely, the graphic is attempting to make the point that a center fed half wave dipole is fed at the current maxima. As such, the feed point voltage is therefore at its minima - but not zero.

  • $\begingroup$ Hi Glen, thanks. Does this mean then that the AC signal starts it's cycle at the feedpoint at a voltage of E = + (I / Z), where Z = 75 +j0, propagates to the end of the element, then is reflected back towards the feedpoint 180 deg out of phase and ends up back at the feedpoint at the end of that cycle and with a voltage of E = - (I / Z) ? $\endgroup$ – Andrew Oct 30 '18 at 3:54
  • $\begingroup$ That article and animation on Wikipedia give the impression that there is one waveform across the entire length of the dipole which is moving through zero volts at the center, when there are actually two separate waveforms each on it's own 1/4 wave element of the dipole, 180 deg out of phase with each other and both only ever reaching a minumum voltage magnitude of E = I / Z. $\endgroup$ – Andrew Oct 30 '18 at 3:56
  • 1
    $\begingroup$ @Andrew Keep in mind that the ~75 ohms impedance is a result of the standing wave that develops on each half of the dipole. If you could measure the "surge impedance" (the impdedance at the very instant a signal is applied) of the antenna, you would find a totally different value. This is essentially the same effect of the transformed input impedance of a transmission line due to reflections on the line. The surge impedance is the characteristic impedance (e.g. 50 ohms) of the line but the steady state impedance is quite different due to the reflection(s) from a mismatched load (and source). $\endgroup$ – Glenn W9IQ Oct 30 '18 at 11:13
  • $\begingroup$ It's more accurate to say there's an RMS electric potential difference at the feedpoint. But there is also a point of zero electric potential in the middle of the dipole. Otherwise, how could we go from a positive electric potential at one end to a negative potential at the other end with a finite electric field intensity? $\endgroup$ – Phil Frost - W8II Nov 5 '18 at 2:33
  • 1
    $\begingroup$ @Andrew I explain electric potential in my answer. $\endgroup$ – Phil Frost - W8II Nov 5 '18 at 13:04

"Voltage" is not a very specific term: it only means "the value of something measured in volts". You might as well ask, "How can a sheet of paper have a meterage of 0.1 mm when it's on Mt. Everest with a meterage of 8848 meters?" Of course comparing thickness to elevation doesn't make much sense, even though both are measured in meters. But if you'd never been introduced to the concepts of thickness and elevation because they were both commonly called "meterage", it would be quite confusing.

So maybe it will help to understand exactly what thing is being measured by "voltage".

electric potential difference (volts)

When someone says "the voltage of this battery is 1.5 volts", they are measuring electric potential difference. This kind of voltage is always a difference between two points. An electric potential difference of 1.5 volts means for each coulomb of electric charge moved from the first point to the second point requires 1.5 joules of work. It is similar to saying "this hill is 10 meters tall", which says something about how much work must be done to lift a rock from the base of the hill to the top.

The feedpoint in an antenna isn't a point in the mathematical sense. So when discussing the voltage at the feedpoint, someone could mean the electric potential difference between the usually two parts of the feedpoint, like the center conductor and shield of a coaxial connector.

electric potential (volts)

When someone says "the voltage at the center of a dipole is zero" they are measuring electric potential. Protons and electrons (and other fundamental particles) have an electric charge. Opposite charges attract, and like charges repel through the electromagnetic force. To pull two like charges apart increases their electric potential, just like pulling a rock away from Earth increases its gravitational potential.

Unlike electric potential difference, electric potential can be defined on just one point. You've no doubt you've seen visualizations of gravity wells where massive objects pull down a rubber sheet. We can define a similar visualization for the electromagnetic force, and since we have both positive and negative charge there can be wells and hills:

enter image description here

The height of this sheet is the electric potential. At infinity the sheet is flat, this is by definition an electric potential of zero volts. The colored lines are contours with constant potential: moving around them neither increases nor decreases electric potential. Moving up or down one volt requires 1 joule of work per coulomb of charge moved.

Notice there's one contour that's hard to see: it's a straight line directly on the y axis, equidistant between the two charges. Because it's equidistant from two equal but opposite charges, the potential here is zero volts. This is what "voltage is zero at the center of a dipole" means.

electric field intensity (volts per meter)

What is not zero is the electric field intensity. It's the gradient of electric potential, and measured in volts per meter. This value is not zero at the center of a dipole: as you can see there's actually a quite steep slope at the origin of the graph.

And here I believe lies the crux of your issue. If you've done any lumped circuit analysis, you've probably equated "voltage" to "electric potential difference". That difference gives rise to an electromotive force, and if there's a conductive path across the distance, current. But when making the jump from lumped element analysis to electric field analysis, electric field intensity is the thing analogous to what you've been calling "voltage" in that it relates to electromotive force, and thus current.

some facts about dipoles without the word "voltage"

The center of a dipole is a single point. At this point, the electric field potential is zero, and the electric field intensity is non-zero. The non-zero electric field intensity means the mobile electrons in the antenna are moved by the electromotive force they experience, thus current is also nonzero at the center. The feedpoint is most likely two points, as in a coax center conductor and shield. Since these two points are separated by space, there is a nonzero electric potential difference between them.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.