Basic Antenna Theory (HF Dipole)

[@robpinter5431]

25 minutes of the best antenna theory teaching, than you so much for your efforts. The life of an electron is not an easy one....

[@n4hnhradio]

Wow, thanks! Indeed, we put the electrons through a tough “life”, making them work for us like slaves. There is an expanded version of all this in the Antennas collection on my Patreon site.

[@thepracticalhistorian]

I am just starting out and I been on an endless search to understand antennas. This was the most helpful video by far. Thank you so much for sharing.

[@n4hnhradio]

You are welcome! I’m glad you liked. I teach operating techniques and radio communications principles. There are over 50 playlists here, plus N4HNHradio.com and Patreon.com/N4HNH.

[@JingsWorld]

I purchased an MLA-30 plus antenna to used it on my Sangean ATS-909x2, didn’t got any signal. I m on a residential area . I mounted it on a 25 ft. Pvc pipe ,scan the whole shortwave frequencies and had maybe 3 SW stations picked up. Give up and returned it, now I’m convinced I’ll do it just like the way I did watch this video. Thanks , this looks pretty much a winner antenna☺️

[@n4hnhradio]

The dipole. Is a fundamental design that most antennas are based upon. The dipole creates its own counterpoise. No “grounding” or ground radials required.

[@ramlosaclash]

There are few that present this so well. Thank you, Sir!

[@n4hnhradio]

I’m glad you liked it. Thank you for letting me know. 73, de N4HNH

[@bourasabdennour8369]

Awesome informative video about antenna fundamentals 👍

[@n4hnhradio]

Glad you liked it.

[@TheRINOShow]

Thanks for the tutorial.

[@n4hnhradio]

You're welcome!

[@MrErik038]

Very good video. Can whe not wrap the feedline whit aluminiumfoil to not radiate..?

[@n4hnhradio]

Common-mode current would ride on the aluminum foil. It will find the outer portion of the outermost portion of feed-line.

[@CaptainRadioAdventures]

Great video Doug

[@n4hnhradio]

Thank you Cap’n Sir!

[@danmarquez3971]

GREAT tutorial!! Just a minor correction, though. The left-hand rule, not the right-hand rule, applies if you show the direction of electron flow. Benjamin Franklin originally thought that electrons flowed from positive to negative but he was in error. Oh well.

[@n4hnhradio]

I know that. I teach electron flow - negative to positive. Fortunately, we learned in engineering that the formulas work either way.

[@marks2254]

That was really good. Thank you.

[@n4hnhradio]

You’re very welcome! 73, de N4HNH

[@alexalonso2093]

good reminders ty for breaking down theory and demystifying these builds 7 3

[@n4hnhradio]

I kept it very basic, but that’s all you need to know in order to understand and build a dipole antenna. 73, de N4HNH

[@aquariumsmaintainedbyandy8408]

Good stuff

[@n4hnhradio]

Glad you liked it. 73, de N4HNH

[@9M6WST]

Thanks for sharing

[@n4hnhradio]

You're very welcome! 73, de N4HNH

[@JayLikesLasers]

06:00 If I'm not mistaken, conventional current flows opposite to the direction of the electrons. You could say that your positive charge carriers (or holes) flow in the direction of the current.

[@n4hnhradio]

It depends upon your engineering professor, or whose book you learned from. Some teach hole flow. But hole flow is imaginary. The formulas still work though. I was taught, and prefer, electron flow, because holes aren’t actually flowing. In the physical world, the electron is flowing and leaving a hole.

[@hugovanderzee4944]

Thanks for the video, it clarified many things for me. Impedance is still a bit a mistery for me. For example, where does the 73 ohm of a dipole come from? Is this some constant? Is it related to the impedance of open space (374 Ohm)?

[@n4hnhradio]

A thorough explanation, excerpt from my book, is available on my Patreon site at Patreon.com/N4HNH.

[@artgumbus]

Boy this was very good. Learned quite a bit. Do you think you could do a deep dive on the ZS6BKW. On that antenna, I just cannot get a grip on the need for 75 feet plus of coax to ladder line and what exactly the ladder line's function is in "matching".

[@n4hnhradio]

There is a ton of information on the Internet about the ZS6BKW. I shot a recent video unboxing the NI4L.com version of it. The ZS6BKW is not a standard dipole. It is a type of doublet antenna system. The ladder line is part of the antenna, serving as a matching section. The coax helps complete the match, transforming the impedance to a usable range. Reflected energy bounces back and forth between the source (radio) and the antenna. The length of coax helps dissipate the losses. That is admittedly a very simplistic way to look at it, but the 70-75 feet of coax helps get the impedance down to a usable/tunable range. The original G5RV ended up with around 90-Ohms impedance where the feedline met the coax. That was fine for back then, when hams sometimes used 75-Ohm coax. But this was only the case for 20 meters. The antenna could function on other bands with a tuner but Louis Varney, G5RV, only intended to use the antenna on 20 meters. Brian Austin, ZS6BKW, computer modeled the G5RV to come up with improvements that might allow the antenna to provide tuner-free operation on other bands, with 50-Ohm coax. He was successful in his quest and that is why his variant of the G5RV antenna is known as the ZS6BKW. 73, de N4HNH

[@sketchin1009]

This was great, the best Ive heard so far. And can you also explain how an unbalanced feedline (coaxial cable) negates creating magnetic waves.

[@n4hnhradio]

That’s not quite the case. Balanced feedline cancels the currents that might otherwise radiate, because the currents are equal in strength but opposite in polarity. Coaxial cable isn’t balanced. The current flowing on the center conductor is not equal and opposite of the current flowing on the shield. Common-mode current can ride the outer portion of the shield back into your shack when the coax is connected to an antenna that isn’t resonant for the operating frequency. This can result in the pesky RF that affects our audio, computer, etc. If you must use an “antenna tuner” with a coax-fed antenna, there will be common-mode current.

[@rrh2918]

Most informative

[@n4hnhradio]

Glad it was helpful!

[@redf7209]

Yes it was informative. Is the insulation coating on the the dipole wire not normally sufficient to prevent electric shocks from touching? I was thinking you would just get burns from the radiative energy. And what risk is there from touching the external surface of the coax connectors at the rig end of the feeder? I feel these are very basic questions i never see answers for.

[@n4hnhradio]

It isn’t quite the same as a direct electric shock. It’s a burn. RF energy can heat tissue. You don’t have to touch the antenna to receive a burn. Just get close enough with your hand to draw an arc. Of course it depends upon the frequency and the amount of power (P=EI) being radiated. 0.5W has been likened to a bee sting. 100W feels like heat from a cigarette lighter. 1,000W could cause hospitalization. Here is an interesting thread from a discussion forum related to this subject: www.reddit.com/r/amateurradio/comments/2yrts7/rf_burns/ And here is an article that will take you on a deep dive into the subject: www.lbagroup.com/resources/rf-shock-and-burn-radio-frequency-radiation-technical-note-124 73, de N4HNH

[@batooldalloul9397]

Thank you! but that is an open circuit, how are those wires (antennas) pulling current?

[@n4hnhradio]

The wires don’t pull current. Try not to think in terms of direct current. The transmitter produces alternating current, by producing a positive-going voltage that causes electron motion (current). While the transmitter produces a positive-going voltage on the center conductor of the coax, it simultaneously produces an equal and opposite polarity (negative-going) voltage on the shield of the coax. Remember that voltage is a measure of the potential for electron motion. Voltage attracts electrons out of the valence ring (outer orbit) of the copper atoms that make up the wires. The electrons are pulled through the wire toward the transmitter. Look at a dipole from left to right. Let’s say that the center conductor of your coax is connected to the left hand wire and the shield is connected to the right-hand wire. When the voltage produced by the transmitter is positive on the center conductor, the electrons in the copper atoms of left-hand wire are pulled toward the radio. But a split second later, the high voltage switches to the shield, which is connected to the other wire. The electrons in the valence ring of the copper atoms of the right-hand wires get pulled toward the transmitter. This back and forth happens 14 million, 200 thousand times per second at 14.200 MHz. The movement of the electrons within the wires causes an electromagnetic field that expands away from the wires according to the intensity of the electron movement (current flow). The electromagnetic field expands and contracts at the frequency of the alternating voltage and resulting current flow. This all results in your transmitted “radio wave” signal. During receive, the electromagnetic field off the antenna of the other station’s transmitter induces current flow into the copper wires of your dipole. Now, one of the measurements of a receiver is in how sensitive it is at detecting the alternating current that is induced into the antenna during receive. More current will be induced into the wires if they are each 1/4 wavelength long, based upon the frequency you wish to receive. The two wires add together to become a half-wavelength dipole. I hope this helps. 73, de N4HNH

[@wild-radio7373]

Thank you :) ♡♡♡

[@wild-radio7373]

I am going to watch this one twice♡

[@n4hnhradio]

You’re very welcome! I’m glad you liked it. 73, de N4HNH

[@wild-radio7373]

@@n4hnhradio I did very much! I am one of those people who enjoys seeing the maths ;)

[@joseduarte6729]

What determines the distance the dipole signal can travel? Or I guess how is that determined cause I understand that the dipole size is built depending on the frequency we are using but how far will that go?

[@n4hnhradio]

Elevation above the Earth beneath the antenna. 1/2 wavelength minimum above the Earth for a half-wave dipole will yield good DX results. Beyond that, it’s all about propagation. Where we are in the 11-year sunspot cycle and time of day are major determinants for signal distance.

[@paulhastings3109]

Where is is all the extras and thinking about POTR get out into some fresh air or chasing. Have g90 for that the antenna don't know what to get. 73 KQ4CD Paul ⚓

[@n4hnhradio]

POTA and SOTA are very enjoyable. For POTA you can use your mobile or set up at a picnic table. I use a EFHW from NY4GEFHW.com. I have featured his antennas on the channel. If you look at the SOTA playlist you will see my portable setup. 73, de N4HNH

[@DK5ONV]

👍👍👍

[@n4hnhradio]

I’m glad you found it informative. 73, de N4HNH

[@rsb762]

Good.

[@n4hnhradio]

I’m glad you found it helpful. 73, de N4HNH

[@keithviglucci5934]

What would the power factor of this antenna be?

[@n4hnhradio]

The gain of a dipole antenna, at the frequency it is cut for, is 2.15dBi or 0dBd.

[@keithviglucci5934]

@@n4hnhradio thank you for your reply. I am trying to understand the relationship between the applied voltage and current in the antenna. If they are 90 degrees out of phase then wouldn't the power be all reactive power?

[@n4hnhradio]

You might hear the term impedance when someone refers to an antenna feedpoint. It is approximately 73 to 74 Ohms at the point of peak current, the feedpoint, of a half-wave dipole antenna. This is the point where the capacitive and inductive reactances have canceled one another out. The E (Electric) and H (Magnetic) fields are in phase (zero reactance) at this point of resonance, resulting in radiation. Another interesting term you might want to explore is radiation resistance. It isn’t a “bad” resistance, such as Ohmic resistance. Ohmic resistance exists at the feedpoint as well. Ohmic resistance causes a loss of power in the form of heat. Radation resistance, on the other hand, allows for the development of an electromagnetic wave. I believe that an understanding of radiation resistance is perhaps as important, if not more important, than understanding impedance, when discussing antennas. 73, de N4HNH

[@keithviglucci5934]

@@n4hnhradio thank you! I need to think of the relationship between the E and H fields rather than the applied voltage and current. I was thinking in terms of AC circuit theory. 73 de KD2EVC

[@n4hnhradio]

Yes, Keith, you nailed it. AC circuit theory is different because there is a circuit. With a dipole, there are open ends. Remember too that if an antenna is too long, the reactance is inductive, opposing a change in current. If too short, the reactance is capactive, opposing a change in voltage. An antenna matching unit (aka antenna tuner) adds capactitance when the radio sees the antenna as too long for our desired operating frequency. Conversely, the antenna matching unit adds inductance when the radio sees the antenna as too short for our desired operating frequency. The point of the antenna matching unit is to balance the inductive and capacitive reactances that influence the impedance the radio sees. It is acting as a type of impedance transformer. Note that when an antenna matching unit is needed to “fool” the radio into thinking the antenna is the perfect length, the antenna system (feedline plus antenna) is still not perfectly resonant. But the antenna matching unit, acting as a type of impedance transformer, effectively makes lemonade out of lemons. 73, de N4HNH

[@robertbutwell5211]

It is possible. You just showed it.

[@n4hnhradio]

What is possible?

[@truthonly7699]

now look up Tartaria the old buildings that have steeples that were really antennas to collect free energy from the ether

[@n4hnhradio]

If I cannot connect a coax or ladder line to those steeples, and TX/RX RF energy, I have no interest in them.

[@kellybryson4319]

It’s actually V.S.W.R

[@n4hnhradio]

Yes, technically it is VSWR. But the focus of my channel is to help my viewers gain a practical understanding of communications concepts. Voltage Standing Wave Ratio relates maximum Electromotive Force (EMF) to minimum Electromotive Force (EMF) along a length of transmission line. EMF influences electron movement and is measured in Volts. But what we hams are actually looking for is ERP (Effective Radiated Power). So we conceptualize VSWR in terms of a reflection coefficient. We don’t need the V. Most of our meters read as SWR, not VSWR. So we want to know how many watts were reflected back by the antenna toward our transmitter and lost as heat. ERP is what we have after we factor in the power lost to heat and the power gained by the gain factor of our antenna. Maybe we lost 10 of our 100 watts due to the reflection coefficient. Maybe we also lost 10 watts due to the length of our transmission line. So only 80 of 100 watts makes it to the antenna. But perhaps our antenna provides 3dB of gain. That 80 watts becomes 160 watts of Effective Radiated Power. And that is our end goal.

[@linuxman0]

This is an informative video. However, I was confused by the use of imperial units; I think it would be better to get a metric tape measure.

[@n4hnhradio]

I am in the USA. We use feet and inches. Sorry. I’m not going to reshoot this video.

[@paulMW7WCS]

Stick with the feet and inches - I wish we had I understand those

[@CaptainRadioAdventures]

Actually open wire feedlines are more complex than that and the simple answer of cancellation may be overly simplistic. The characteristic losses of open wire feedlines are low when compared to coaxial feedlines of the same length especially under high SWR (up to 9:1). The matching is done at the rig using a balanced line tuner to match the radio's source impedance of 50 ohms and the losses at the tuner itself are relatively low compared to "lossy" coaxial cable. At very high SWR, the cancellation of return currents becomes less effective. So your statement is true up to a certain point. The answer then becomes "it depends...."

[@n4hnhradio]

Yes, I didn’t want to go into those weeds with this video. The open wire line, at 420 to 600 Ohms isn’t going to match up to a dipole that is cut according to the formula of 935/Frequency in MHz, with its 73-Ohm impedance. The math I use in this video assumes 50-Ohm coax connected to a dipole cut for the 20 meter band. This video targets those who are new to amateur radio and want to learn how to build a simple dipole. There are circumstances where the current isn’t completely equal and opposite, for either coax or open wire line, but I didn’t want to get into that in a basics video. I think I did put a comment mentioning that under certain circumstances the feedline can radiate. I prefer a BalUn even for a single band dipole, but I have a multiband ZS6BKW by NI4L that has no BalUn. It works perfectly on 12, 17, 20, 40, and the FM portion of 10 meters with no BalUn or tuner. The interaction of the 39 feet of window line with the 47 feet of wire on each side of the center insulator yields resonant points at harmonic intervals. The ZS6BKW is a great variation of the G5RV. There are many variables with antennas and feedline. With antennas there is the theory and then there is the reality brought about by the installation location. 73, Doug

[@CaptainRadioAdventures]

@@n4hnhradio I understand - open wire feedlines for a G5RV and ZS6BKW need to be matched to the radiating elements and open wire feedline versus can be a deep rabbit hole that one may not be able to come out of. I am actually trying to do a video as to the why's to the concepts you presented. You are correct in saying that the signals "effectively" cancel - trying to say though it is only true up to a point. Actually the return currents on any feedline is low - effectively zero - if the load impedance is a complex conjugate of the source impedance - ideally 50 ohms resistive with no reactive components. Remember that EFHW coaxial feedlines do radiate when the there is a mismatch and return currents present themselves on the shield of the coax. Being that we are QRP and we take great pains to match impedances - this is not a problem. I did get an RF bite touching the ground lug of my 120W rated EFHW transformer during field day while transmitting at 150 watts. That was a wake up call not to do that again.

[@n4hnhradio]

How dare you put 150W into a 120W UnUn. Hi! Hi!

[@CaptainRadioAdventures]

@@n4hnhradio It was field day and I wanted to do a torture test. Oh BTW it is not a balun but a transformer. A balun goes from unbalanced to balanced. It is more an Un-Un because it remains unbalanced and connects to a coaxial feedline which is unbalanced. It is not a pure un-un because there is a voltage tansformation from roughly 3200 ohms to 50 ohms - hence it is a transformer.

[@n4hnhradio]

I don’t want to get into EFHW yet. I was only asked to cover basic antenna theory. If I get asked about a EFHW I’m going to defer to you.

[@Golfer-rn7lh]

Why do Americans always seem to take ten minutes to say what only needs 10 seconds to say?

[@n4hnhradio]

This is not the channel you’re looking for. Goodbye.

[@terrymarshall5480]

Just can't make it simple

[@n4hnhradio]

Huh?

[@kn4mqr426]

Great video. Thank you for sharing. de KN4MQR

[@n4hnhradio]

You’re very welcome! 73, de N4HNH

[@СамійлоМикита]

How to produce the antenna?

[@n4hnhradio]

142.5/frequency in MHz = length of horizontal wire in meters. Cut in half. Connect wires to 1:1 choke Balun at the center, one on each side. Install “dog bone” insulator on opposite end of wires. Connect 50-Ohm coaxial cable to 1:1 choke Balun. Erect 1/2 wavelength minimum above the Earth.