QST antenna article June 2015, page 30

[Opcom]

"Don't blow up your balun" -starting page 30. A somewhat lengthy article about where the losses are in an antenna system, including the coax from the TX to tuner, the tuner itself, the feed to to a 40M dipole, etc.

Proceeding from an assumption of a 40M half wave dipole (~66Ft on each side), and a power level of 1500W: What it says is that #12 open wire line can have 90+ watts of loss, an that the 450 Ohm ladder/window line can have 400+ watts of loss. I can't explain this, you have to see the article, but I thought one purpose of a so called balanced wire line was to cut down on power losses. It says that at 80M, the loss can be 1000W+

Is the loss in the wire line due to I2R at current nodes?

[KD6VXI]

Would a half wave 40 meter dipole be about half those dimensions?


And,  wouldn't those same losses be incurred in the actual elements themselves?   I2r is i2r, regardless of I'd it's in the feeling or the antenna element.

Taking skin effect into i2r?  The rf resistance of a piece of wire will change due to its diameter,  i2r and frequency,  I would think...

Again,  just conjecture,  haven't read the article.   I have burned up a #12 feeder before,  so it is entirely plausible.

--Shane
KD6VXI

[w1vtp]

<snip>

Proceeding from an assumption of a 40M half wave dipole (~66Ft on each side),

<snip>

That's a 1/2 wave 80 meter dipole

[AB2EZ]

Most people find antennas and associated transmission lines to be very confusing.

Focusing on just the transmission line (as per the question asked in Opcom's initial post)

If the SWR along the transmission line is not 1 (i.e. the transmission line is not terminated at the antenna end in its characteristic impedance) then:

At a point along the line where there is a current maximum*, the resistive loss per unit length (in decibels per 100ft) will be: the matched resistive loss per unit length (in decibels per 100ft) x the SWR

At a point along the line where there is a voltage maximum*, the dielectric loss per unit length (in decibels per 100ft) will be: the matched dielectric loss per unit length (in decibels per 100ft) x the SWR.

These relationships follow directly from the definition of SWR, and the definitions of forward power and reflected power.

Depending upon the physical length of the transmission line (and also where the current peaks and voltage peaks are located relative to the ends of the transmission line)... the total loss of the cable will be approximately: the nominal loss of the transmission line when matched (in decibels per 100 feet) x [0.25 + .375 x (SWR + 1/SWR)].  

For example, a transmission line with a loss of 0.9dB/100feet, when looking into a matched load, and fed by a transmitter with a tuner at the input end (SWR looking into the tuner=1), will have a loss of approximately (0.9 dB/100ft) x [0.25 + 0.375 x 12 + 0.375/12)] when the load causes the SWR to be 12. I.e., the loss will be approximately: 0.9dB/100ft x [0.25 + 4.5 + .03125] = 4.78dB/100ft. If the transmission line is 100feet long and if the input power is 1500 watts, then the output power will be 1500 x 10**-0.478 = 498 watts. The remaining 1002 watts will produce heating due to resistive losses and dielectric losses in the transmission line.

*Maxima in current and (separately) maxima in voltage occur every 1/2 wavelength along the transmission line... where the wavelength is the free space wavelength x the velocity factor of the transmission line (typically 1 for ideal open wire line, and between 0.8 and 0.67 for coaxial cable)


Stu
 

[KD6VXI]

Thanks for the breakdown,  Stu.   

Really makes you wonder if it was worth it to make all that owl...   Until you think the loss Kotex Cable would have under the same circumstances....

--Shane
KD6VXI

[Steve - K4HX]

A dipole with 66’ of wire on either side of center and the feed point, 50 feet over ground with average conductivity, and at a frequency of 7.2 MHz would have a feedpoint impedance of 5686.100 + j1366.400. That’s a pretty nasty impedance. The loss in 100 feet of 600 Ohm open-wire line would be only 0.23 dB. For the same conditions and 450 Ohm ladder line, the loss would be 1.12 dB. I haven’t read the QST article, but something doesn’t sound right. If you feed this dipole with RG-8 coax, the loss would be almost 9 dB. Here, you would burn up over a kW in the feedline.

[AB2EZ]

Steve

1.12dB of loss (i.e. the loss using 100ft of 450 ohm ladder line) implies that:

10**(-1.12/10) of the input power will be coupled into the antenna.

10**(-1.12/10) = 0.773

1500 watts x 0.773 = 1159 watts

1500 watts - 1159 watts = 341 watts lost (as heat) in the 450 ohm ladder line

Stu

[Opcom]

Thanks for the explanations. Mine is a 80M dipole, the articles is a 40 IIRC, so the confusion. The explanations make perfect sense though.

It is easy to see now why my antenna system has nasty behavior. EZNEC model I made of the antenna and feed agrees fairly reasonably with the ugly impedance Stu mentioned but I did not want to believe it.

The manual for the antenna says it has a nominal 450 ohm impedance and should be used with a tuner like a Johnson matchbox.  Am I right to assume I could also use a "balanced L network" (low pass or pi-network-like) tuner for this? Its a question of QRO-sized parts on hand. picture shows what I mean.

A balun is not recommended between a tuner and this antenna in the manual and indeed I have burned one up before with a piddly 300W carrier and a T network (high pass) MFJ tuner. The manual says not to use a balun because of no harmonic attenuation, but here there may be a practical or safety reason not to use one.

(EZNEC with >20 segments can't be properly saved in the ARRL version but an ARRL file can be modified and analyzed without saving, and screen captures made for one's reference)

[K6JEK]

66' total, not 66' each side. That's the antenna in the article. It's an eye opener about just how much loss there can be using a 40 meter dipole on 20 meters with various configurations of feed lines, matching units and baluns.

The old rule of thumb, make a 1/2 wave dipole for the lowest band, feed it with open wire line and you'll be fine turns out to be more or less true but stray from that configuration and your losses mount rapidly. In the best case, open wire to a Johnson Matchbox, 1325 of 1500 watts make it to the antenna. In the worst case, coax to ATU to coax to balun at feed point, only 122 watts make it out. The other 1378 are lost. Yikes. The window line case throws away 300 more watts than open wire. In all cases, 100' of feedline is assumed.

Author is Dean Straw, N6BV, former editor of the ARRL Antenna book.