EXCELLENT video, Greg. THIS should be mandatory viewing for all hams. TNX and 73 de DJ0IP.
@ve6wo
Жыл бұрын
Thank you.
@Simon-mz7sf
Жыл бұрын
Thankyou. You really clear the forest from the trees in your approach to explaining things. Appreciated
@MicheIIePucca
Жыл бұрын
Finally... someone from Alberta on KZitem :)
@ve6wo
Жыл бұрын
Yessir! :-)
@MicheIIePucca
Жыл бұрын
@@ve6wo I did cheat though and did a call sign lookup.
@ve6wo
Жыл бұрын
No harm in that. I usually have QRZ up on a screen while I’m playing radio so I can see where signals are coming from. Gives me an idea of band openings and where I can hope to communicate to.
@IZ0MTW
9 ай бұрын
Your work on the subject is remarkable. As said on another comment should be mandatory to the ham operator ti understand these concepts. Keep up doing great videos, I’m watching the others. Congratulations and 73. Diego
@ve6wo
9 ай бұрын
Thank you for the kind words.
@neilhankey2514
Жыл бұрын
Until now I have never seen someone get this exactly right. Thank you for the video. Keep up the great work.
@ve6wo
Жыл бұрын
Thank you for the kind words.
@ve3dvy
Жыл бұрын
Wow. I have yet to see an explanation this well done. And thank for providing your refferences. This qualifies this video. If OK I will be using it as extra study when we start Basic classes. Thanks!!! VE3SD
@ve6wo
Жыл бұрын
Yes, please use the video :-)
@KilianGosewisch
Жыл бұрын
i am currently getting into ham and had a bunch on previous "engineering" and electronics knowledge from previous hobbies and many things did not make sense to me. for example how people throw around the word balun and how it is often interchangeably used with choke etc. this video cleared up a lot and prooved and i am not stupid ;) thank you very much!
@ve6wo
Жыл бұрын
You’re welcome :-)
@2013Davey
10 ай бұрын
Fantastic explanation .... thank you
@DarkoObretan
9 ай бұрын
Current on the outside can flow only because of miss-match or imbalance of left and right dipole. Current inside coaxial cable is by definition equal and opposite (I1=-I2) and even if impedance of coax outside would be 0, there would not be any current flow outside of the coaxial cable if dipole would be balanced. (Which it can't be if using coax, because both dipole sides are having some capacitance to nearby coax shield, but voltage polarity is opposite, so they see different impedance, and that's why there is unbalance in current flow). (Dipole in practice also have different distance to nearby trees, buildings and that also affects it's balance...)
@ve6wo
9 ай бұрын
At the end of the coax where the antenna connects will be two paths for current to flow. One will be on the leg of the dipole, the other will be on the outside of the coax shield. Even with a perfectly balanced antenna, this is still true. The result will be that some current will always be present on the outside of the coax, no matter how perfect the antenna is. The best way to minimize current on the coax shield, and maximize current on the antenna is to install a common mode choke at the antenna feedpoint to present a high impedance to currents that would otherwise flow along the coax shield.
@DarkoObretan
9 ай бұрын
@@ve6wo Yes, I agree with everything, I was just bothered with what you said at 20 minutes (that current on the outside flows, no matter how well matched antenna is. Maybe it was misunderstanding from my side, because I understood matching also as 'how balanced antenna is' on the feed point).
@phystimn
10 ай бұрын
That's all a pretty nice explanation of the currents on 'inner' and 'outer' surfaces of the shield. However, wouldn't it be just more persuadive to organize an experiment (using nanoVNA or similar device for example) to demonstrate that the currents indeed flow in the described manner and that the balun indeed changes how do they flow? After all, the purely engineering objection to the given explanation could be that the shield is made up of individual wires, it has multiple 'holes', so that many connections between its inner and outer surfaces exist.
@ve6wo
10 ай бұрын
I was just recently thinking about this.. a person could use a clamp on RF current meter to demonstrate physically.
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