# Why do folded dipoles have greater bandwidth than ordinary resonant dipoles?

This is something I've read in passing, but never encountered an explanation of why. For example, Wikipedia says:

Another common place one can see dipoles is as antennas for the FM band; these are folded dipoles. The tips of the antenna are folded back until they almost meet at the feedpoint, such that the antenna comprises one entire wavelength. This arrangement has a greater bandwidth than a standard half-wave dipole.

Wikipedia isn't the only one with this notion: see comments to this answer:

it was my understanding that a folded dipole is still full length, just doubled over to increase bandwidth. Am I incorrect?

antenna-theory.com, which I'd consider at least three times more reliable than Wikipedia, doesn't say anything about bandwidth, but does say this:

Because the characteristic impedance of twin-lead transmission lines are roughly 300 Ohms, the folded dipole is often used when connecting to this type of line, for optimal power transfer. Hence, the half-wavelength folded dipole antenna is often used when larger antenna impedances (>100 Ohms) are needed.

I could then see how, if you had to use 300Ω transmission line, you might get better bandwidth with a folded dipole as you wouldn't need a matching network, which might limit bandwidth or incur additional loss, but that's a long guess.

So, really why do folded dipoles have greater bandwidth? Or is that just an unsubstantiated rumor?

The increased bandwidth of a folded dipole is almost entirely due to the extra thickness. Two parallel elements behave as a thicker single element. There is a small contribution too from the combination of the reactances of the transmission-line mode and radiator-mode acting in opposite directions.

• G8HQP summarizes the fact folded dipoles and thick dipoles have essentially the same bandwidth, well within calculable tolerances, governed almost entirely by conductor width.
– JSH
Jul 26 '16 at 14:21

# Loaded $$Q$$-factor

Like any resonant circuit, the bandwidth of an antenna is determined by its loaded quality factor, defined by $$Q_{\ell}\overset{.}{=}\frac{X}{R}$$.

The lower the loaded $$Q$$-factor, the broader the antenna's bandwidth will be: $$BW_{-3dB}=\frac{f_{res}}{Q_{\ell}}$$, with $$f_{res}$$ the resonant frequency.

# Analysis of the loaded $$Q_{\ell}$$ of a folded dipole in free space

### Resistance $$R$$ (identical to that of an ordinary dipole)

In above formula for $$Q_{\ell}$$, the resistance $$R$$ in the loaded resonant circuit will be half the radiation resistance $$R_{rad}$$ when the antenna is perfectly matched. Hence, $$R=\frac{R_{rad}}{2}$$. Note that this value is not any different from that of an ordinary dipole, even though the input impedance $$R_{in}$$ of a folded dipole is four times that of an ordinary dipole. The higher input resistance $$R_{in}$$ is merely due to the impedance transforming property of closely spaced parallel wires. In conclusion, $$R$$ provides no explanation for the higher bandwidth.

### Reactance $$X$$ (lower than that of an ordinary dipole)

Now, let us evaluate reactance $$X$$ in above formula for $$Q_{\ell}$$. For a dipole in free space, $$X$$ is determined by the self-reactance of the antenna conductor. This will be a combination of self-inductance of the element conductor and self-capacitance between the two element halves.

Moreover, the closely spaced parallel conductors of a folded dipole should really be seen as one very thick conductor. Hence, its self-inductance will be lower and its self-capacitance higher than that of an ordinary dipole. Both effects result in a lower reactance $$X$$, hence lowering the loaded $$Q_{\ell}$$ and broadening the bandwidth of the folded dipole.

A little capacitive reactance is what gives you the greater bandwidth.

In a regular ½λ dipole, the current that flows along the conductors are in phase. When we add the second conductor in a folded dipole, what we are really doing is extending the dipole. As a result the current in the new section flows in the same direction as those in the original dipole. The currents along both the half-waves are therefore in phase and the antenna will radiate with the same characteristics as a regular simple ½λ dipole.

Perhaps an illustration will help (ARRL Extra Class License Manual): Remember, this is AC current. At points B and C (the ends of a simple dipole) it drops to zero. Since a folded dipole is like extending a dipole by a 1/4λ at each end, we will observe the current reversing at B & C as we start a new cycle.

Since the current is now evenly divided into the two sections, the impedance must increase according to Ohm’s Law (W = I^2R) by a factor of 4. This makes the folded dipoles an attractive option to hams that like to use twin lead or ladder line for feed lines.

Another way to look at the situation is that the impedance of the dipole appears in parallel with the impedance of the folded sections. At a frequency away from resonance, the reactance of the dipole is of the opposite form from that of the folded section and as a result there is some reactance cancellation at the feed point of the antenna.

So in essence you get some “free” matching right at the feedpoint of the antenna.

• Maybe you can elaborate on "As a result the current in the new section flows in the same direction as those in the original dipole." It's somewhat counter-intuitive to imagine that an antenna that makes a loop results in in-phase currents. Dec 2 '13 at 21:39
• I've added a picture and some more text explaining the current flow better (I hope). Dec 2 '13 at 23:58
• Better, though it's still a big mental jump to see how the current "turns around" at points B and C. The natural thing to do with this image is start at one half of the feedline and follow the arrows. They all point in the same direction, except the one at the top. This is something that's bothered me a long time, which I always thought unnecessarily difficult to learn -- I'm certain the solution is a picture, just not sure what. Animated, maybe. Dec 3 '13 at 0:03
• Perhaps the mental trap is this: if the current drops to zero at points B and C, then how does it get past those points? Or, why does the current drop to zero at points B and C? In an ordinary dipole it's easy to see: there's no place for it to go. In a folded dipole, that's not the case. Dec 3 '13 at 0:06
• Maybe what we really need is a question like How does a folded dipole work? Dec 3 '13 at 0:11

It may help to consider the two arms of the folded dipole as end-shorted (ROUGHLY quarter-wave) sections of 300 ohm transmission line. Any current which attempts to flow in opposite directions on the two conductors in each arm will not radiate (which is why the transmission line works - the two currents generate opposing fields). Only a "common-mode" current can radiate. Looking to either side of the antenna from the centre point, the impedance for differential mode currents across the two conductors is very high (because each arm acts to transform the short at its far end into a high impedance at the centre).

Now consider the feed point changing voltage differentially. If one side of the feed rises in voltage and (say) a current flows out, then the voltage and current on the opposite (folded section) conductor must follow it (differential currents are cancelled out!). The antenna is balanced and very nearly resistive near its centre, so the other feed point drops in voltage and pulls in a current of the same magnitude. This current is flowing in the same direction along the antenna as the current from the other feed-point (just as in a regular, non-folded dipole). Again, the transmission line effect forces the voltage and current in the conductor opposite to the second feed point to follow,, and this current is in the same direction as for the other arm.

There are effectively TWO current modes operating. A transmission-line mode (which does not radiate) and an antenna-like, radiative mode.

It should be noted that this concept should have the two arms slightly too short for an efficient half-wave radiator, since the two quarter-wave sections of line would have to be cut accounting for the velocity factor of the line (typically 0.85 or so). Some designs (like ARRL) make the antenna length similar to tat of a simple dipole but add shorting bars slightly in-board of the ends at the points which are electrically a quarter-wave along each arm from the feed point.

The original question asked about why the bandwidth is "wider" for a folded dipole than for a single-wire dipole. There are probably two effects to consider. The reactance shows a very sharp dip at resonance, but the resistive impedance is high (near 300 ohms), and varying with frequency. The combined complex impedance magnitude normalised to 300 ohms does not change so fast. The "non-radiative" mode tends to cancel the reactance changes. The reason may also be that the effective diameter of the folded dipole is increased compared to the single wire system. A "fatter" effective wire has a lower Q (its resonance is less sharp), and its effective bandwidth is wider. This effect is used on, for example, "caged" dipoles whose (non-folded) arms are made from multiple wires spaced with hoops.

• Great answer! Re: "the two quarter-wave sections of line would have to be cut accounting for the velocity factor of the line", I don't think the velocity factor is a factor here, the VF of open wire is 0.95-0.99. Nov 1 '16 at 18:57

As has been pointed out, one factor of the folded dipole that explains greater bandwidth is that it is effectively thicker. The main reason however is that it's impedance is effectively the parallel combination of a half wsve dipole and a 1/4 wave stub shorted at the far end. The reactance/susceptance of this stub varies in the opposite direction to that of the dipole, thus compensating it. The best way to think of it is in terms of even and odd modes. If you split the folfed dipole into two U:s, you can feed both legs of the U in parallel in phase as a non'folded dipole where the current in each leg of the U is in the same direction. This is the even mode and the impedance will be that of a fat dipole. 75 ohms say, or 150 for each of the two U legs in parallel. Now feed the two legs of the U in antiphase, with current in opposite directions. This is the odd mode and it doesn't radiate and has the impedance of a short circuited stub. Now feed both modes at once. You will have even+odd=2 on one U leg and even-odd=0 on the other. The 0 volt leg can be shorted across to the 0 volt leg of the other U completing the folded dipole. You now have a 300 ohm nominal feedpoint but the current is the sum of the even and odd mode currents, the odd mode current being that of the stub, the impedance of which is therefore in parallel with the dipole. You can get a similat effect by connecting a 1/4 S.C. stub across a non-folded dipole, except the impedance will not be multiplied by 4