antenna-theory.com

[tensor20]

Hello Forum,
there is a great tutorial on baluns at http://www.antenna-theory.com/definitions/balun.php

Check it out.
I am not clear why, once the current IB reaches the end of the arm of the dipole antenna (the arm that is attached to the outer shield of the coax),
the current ends up having two choices: flow completely back along the inner surface of the coax and/or flow on the outer surface of the coax shield, as IC....
Why? Why does the current IB not flow only on the top surface of the grey horizontal arm of the dipole?
Why is the current IC created?

"...In the folded balun tutorial it is said that the current IC will depend on what impedance is looking down the outside of the coax cable relative to the impedance of the dipole arm..."

Can anyone clarify that? I know that current likes to flow where there is less resistance and does not flow at all where the resistance is huge (open circuit). ...

thanks
tensor20

[bigSteve]

Baluns are tricky, but should be understood. If you hook up a transmission line to a short antenna, and there is no balun, the transmission line itself will become part of the antenna. I'll try to clarify some of the discussion.

The first thing to understand is this:
1) The central conductor of the coaxial cable (pink arm) of Figure 1 on the balun page is one surface. Current can only flow on the outside of it.
2) The dipole arms where IA and (IB-IC) flow are each one surface. There is no inside or outside, topside or bottomside to the dipole arm. The current flows across the whole arm.
3) The shield of the coaxial cable is two surfaces. Current can independently flow on the inside and the outside. This is because the skin effect. RF current flows on a surface, and the two surfaces are isolated from each other at RF.

Now, when the current IB gets to the end of the coaxial cable, it has no way of knowing what is the antenna and what is the outer shield of the coax. As far as IB is concerned, these are two independent paths in which it can flow. IC is very real, not a theoretical construct.

In terms of "IC will depend on what impedance is looking down the outside of the coax cable relative to the impedance of the dipole arm..." This means that when IB gets to the end of the coaxial cable, it sees two paths, one on the dipole and one down the rear of the coaxial cable. This IC path is often more attractive to the electric current, because there will be a long transmission line and a receiver, and so it can see a loss resistance, and this is where the current will like to flow (remember, it is looking for 50 Ohms in general).

The purpose of the balan is to make this impedance (looking down the outside of the cable) infinite (or huge). In this manner, IC is choked (blocked) and all the current IB must flow on the dipole arm.

[tensor20]

Thanks BigSteve,

I am starting to understand. What confuses me

In a coaxial cable there is one current, I1, flowing on the surface of the inner conductor and there is a a current I2 (same magnitude but opposite direction) flowing on the inner surface of the coax shield. The outer surface of the coax shield has no current on it (if it did it would radiate which is undesired). The thickness of the coax shield is enough (enough skin depths) so that I2 does not leak on the outer surface....

Now, in the case of the dipole arm attached to the coax shield, once the current I2 enters the dipole arm it distributes itself on the whole surface of the arm and as the cycle reverse and the current changes direction, some the current I2 ends up flowing also on the outer surface of the coax shield...

Is that what you meant?
thanks,
tensor20

[tensor20]

Thanks BigSteve,

I am starting to understand. What confuses me

In a coaxial cable there is one current, I1, flowing on the surface of the inner conductor and there is a a current I2 (same magnitude but opposite direction) flowing on the inner surface of the coax shield. The outer surface of the coax shield has no current on it (if it did it would radiate which is undesired). The thickness of the coax shield is enough (enough skin depths) so that I2 does not leak on the outer surface....

Now, in the case of the dipole arm attached to the coax shield, once the current I2 enters the dipole arm it distributes itself on the whole surface of the arm and as the cycle reverse and the current changes direction, some the current I2 ends up flowing also on the outer surface of the coax shield...

Is that what you meant?

You say that "This IC path is often more attractive to the electric current, because there will be a long transmission line and a receiver, and so it can see a loss resistance, and this is where the current will like to flow (remember, it is looking for 50 Ohms in general). "

why would the rear of the coax shield be more attractive for the current to flow instead of returning completely to the interior surface of the shield? I don't get why it likes to see a "loss resistance". Again, I think that current likes to flow where the impedance is the smallest....

thanks,
tensor20

[bigSteve]

"Now, in the case of the dipole arm attached to the coax shield, once the current I2 enters the dipole arm it distributes itself on the whole surface of the arm and as the cycle reverse and the current changes direction, some the current I2 ends up flowing also on the outer surface of the coax shield..."

Hmm....no. The key point is that the current does not know what is the dipole arm and what is the outside of the coaxial cable. It could be that virtually no current at all flows on the dipole arm, rather it just travels down the rear of the coax.

"I think that current likes to flow where the impedance is the smallest.... "

The problem here is you are thinking in terms of low frequency circuit theory. For transmission lines with characteristic impedance Z0 (usually 50 Ohms), the current wants to flow where it sees 50 Ohms (then there will be no reflection).

To dive a little deep into the theory....
Gauss's law forces the current on the inside and outside of the coaxial transmission line to be equal. So it is not possible for the current on the outer conductor to be completely reflected and that of the inner conductor to travel to the dipole arm. Hence, any current that flows to one dipole arm must be balanced by current flowing into the transmission line on the other side. This is a consequence of the rules of electromagnetics.

Now, the amount of current that flows will be a function of the impedance seen by the dipole (which happens to have one arm in parallel with the outer shield impedance). This impedance won't be zero or infinite, and if the lengths are about a half-wavlenght the impedance will be reasonably close to 50 Ohms. As a result, current will flow from the transmission line to the dipole. And without a balun, well it won't be balanced.