# What is the maximum and minimum of antenna gap of dipole antenna?

Consider the picture below is design of half wave length (λ) dipole antenna. I read some explanations saying that it is the best design to get the optimum transmit power. Half wave length (λ) is from tip to the tip of the two elements (you know it).

But my concern here is, what is the minimum or maximum distance of the gap as in the picture? Gap between the two elements which the feeder line is connected? Some say that it should be as small as possible. But unfortunately, I don't have mathematical justification for that reason even it make sense. I expect that the explanation is in λ. But if you really need the used frequency, then just put 2,100 MHz. If needed, the feeder line is RG6 75 ohm.

• Is RG6 suitable for 2.1GHz? How long is the transmission line? I have no experience at these high frequencies, but dimensions tend to get small and tolerances tend to get tight. – Chris K8NVH Aug 6 '19 at 10:55
• It is the cable used for C-Band parabola, which the frequency is a bit higher. Probably, you are interested with this. This measurement and calculation I did with 2.1GHz as I intended to make external antenna of 3G mobile phone. – Sitorus Aug 6 '19 at 11:15
• For my purpose, the cable length will be around 10. But my C-Band tv antenna is using around 20m length RG6-75 ohm. – Sitorus Aug 6 '19 at 11:20
• I'm curious as to how you're even getting a gap using a coax cable. Surely at the feed point you've got an SO-239 connector, with the two legs of your dipole soldered to it? Or are you transitioning to a ladder line at some point? – TMN Oct 14 '19 at 18:55
• @tmn Some of us build our own dipoles without any connectors. – Mike Waters Oct 14 '19 at 23:37

The gap is neglectable in terms of λ. Let's say you are making a 20m (14 MHz) dipole and you decided to use a large gap, let's say 10cm. This is only 0.0005λ. For 2m band (144Mhz) it's 0.005λ. If you choose an even larger gap it means there will be some wires that will connect the arms of the dipole to the feed line. These wires will just work as parts of the arms. Once again - there is no significant gap except the distance between the center of the coax and the shield of the coax.

In other words just connect the arms to the coax the way it's comfortable and then trim the length of the arms to get minimum SWR in the center of the band.

[Richard Fry is the author of the following paragraph and graphic.] Below is a NEC4.2 study of a 40m, 1/2WL, center-fed, free space dipole with a 0.2m gap for an insulator at the feedpoint. NEC sources themselves are applied at a single point on a conductor, but this approach or a variation of it might lead to a reasonable, practical solution for most amateur radio operators and applications.

• As I have mentioned the frequency going to be used that is 2,100MH (3G Network) and the cable is coaxial RG6 75 ohm. then better if you match your calculation to that data as it is very var. The half λ will be 7.14cm. – Sitorus Aug 6 '19 at 10:07
• Apart from it, so where is actually impedance of an antenna system come from? Appreciate if you may help. – Sitorus Aug 6 '19 at 10:08
• The impedance depend on the antenna and it's form. A dipole has an impedance about 73 Ohm, thus it should be matched well with your RG6 coaxial cable. If you bant it in the inverted-V shape the impedance will be about 50 Ohms for 90-120° angle. It worth noticing that RG6 has losses ~10 dB per 30 meters on 2.1 GHz frequency, thus you should probably keep the cable short. The cap is quite noticable on 2.1GHz, thus you should keep it as small as possible. – Aleksander Alekseev - R2AUK Aug 7 '19 at 11:58

Looking at this sentence,

But unfortunately, I don't have mathematical justification for that reason

The minimum gap is zero and the maximum gap is $$\lambda/2$$. The mathematically ideal gap is 0, as explained in several books on antenna theory; there is an integral over the dipole length as charge is collected.

Any gap between 0 and $$\lambda/2$$ will work, but it will work a little "better" if the gap is closer to --or equal to-- 0.

However, with the way you are laying out your dipole, adding a zero gap will cause a short circuit and the dipole will not function. Adding an infinitesimally small gap will allow sparks to jump across the gap, which is also bad. So add a gap to prevent a short circuit and try to make it big enough to prevent sparks. Mathematically, you want this gap to be as small as possible to maximize the aforementioned integral.

If you accidentally make it too short, that is a problem, but an easy solution is to add some insulator to fill in the gap, or just make the gap bigger.

It really is that simple.

• Very interesting and reasonable. Please have a look at this post if you don't mind. – Sitorus Aug 6 '19 at 11:16

The point where the feedline separates marks the center of the dipole. That length of conductor matters. It radiates, carries current, and effects the behavior of the antenna.

If the conductors feeding the "dipole" wires don't diverge at 90 degrees and are "significantly long", it gets more complicated to calculate, and the NEC software many hams use can help to calculate the impact, and let you model variations.

• There are many recommend to use the NEC software. Let me see it. – Sitorus Aug 4 '19 at 1:25
• As far as I know NEC doesn't allows to specify the gap in the feed point. – Aleksander Alekseev - R2AUK Aug 6 '19 at 9:37
• Been a while since I used it. I thought the element geometry was coded and then ideal feed points were set. If so, then there would be the dipole elements, the coax shield after the split, and the coax center conductor after the split. The end points of the coax would be separated by the center conductor insulation thickness. The virtual feedpoints are the open ends of the coax conductors. The point is that the dipole elements are always separated by the coax insulator thickness. The RF doesn't know which conductor is carrying it, and everything that is unbalanced radiates. – cmm Aug 6 '19 at 13:27

Actually, that gap is fixed within cable. That is distance between inner conductor and shield.

Part of the cable from the point where inner conductor exits out of the shield is actually part of the antenna.

So, it does not matter what is distance between antenna wires.

However, minimum distance is important, as if wires are too close, it might cause sparking. It depends of transmitter power.

• Do you have reference for it, please? – Sitorus Aug 3 '19 at 7:37
• @Sitorus hm, "Ohm's law and Maxwell's equations" would be the ultimate reference for that. Together with your other question: I think you need to take a step back and learn complex current theory, linear electrical networks, a bit of electrodynamics and a few pages of antenna design – you seem to be jumping into the middle of a complex topic! – Marcus Müller Aug 3 '19 at 10:17
• A short, but possibly useful and practical answer requiring no math is that the gap can be any value that doesn't produce arcing across that gap when transmitter power is applied. Gaps wider than a very small fraction of a wavelength will change the dipole feedpoint impedance and radiation pattern/gain, however. MoM software such as 4nec2 can calculate those parameters. – Richard Fry Aug 3 '19 at 11:01
• @RichardFry, if the operating frequency is 2,100MH (wave length will be around 14.2cm, and half wave length will be 7.1cm) to transmit power around 1 mWatt, would be save to just put 1mm plastic insulator in between of the two elements? Think there will be no spark as only very small power. – Sitorus Aug 3 '19 at 17:56
• I really see no need for reference. Shielded transmission line (cable) has function until signal leaves shielded part of the cable. If you remove shielding, for example, to make short wiring contacts to attach dipole antenna, those parts of cable that are not shielded are part of antenna not part of transmission line. – Pedja YT9TP Aug 3 '19 at 23:31