Ham Radio Forum

Ok - a bit of a basic one this - but relevant, nonetheless . . .

TX requires a 50-ohm load. I have a half-wave dipole, whose feed impedance is 75-ohms.

Do I use 50-ohm coax - and live with the mis-match at the antenna end ?
Do I use 75-ohm co-ax - and accept the mis-match at the TX end ?
Does the mis-match really matter anyway ?
And if it does, should I use a matching device? If yes, what ?

Your thoughts on this, please.

Al / G8DLH

G8DLH wrote:Ok - a bit of a basic one this - but relevant, nonetheless . . .

TX requires a 50-ohm load. I have a half-wave dipole, whose feed impedance is 75-ohms.

Do I use 50-ohm coax - and live with the mis-match at the antenna end ?
Do I use 75-ohm co-ax - and accept the mis-match at the TX end ?
Does the mis-match really matter anyway ?
And if it does, should I use a matching device? If yes, what ?

Your thoughts on this, please.

Al / G8DLH

The feedpoint impedance of a half wave dipole close to ground is somewhat lower than the 72 ohms you will see for a dipole in free space. You can expect it to be in the region of 40 to 60 ohms for a 80m dipole at around 10m in height, dependent on height and ground conditions.

I use a dipole for 40m that is cut short so as to cause a VSWR of 1.5 on a RG6 (75 ohm) feedline. Because of the VSWR, the impedance changes all the way along the feedline. There are points on that feedline where the impedance is 50 ohms resistive, so I cut the feedline at one of those points that is convenient to the transmitter and connect the 50 Ohm load directly to the transmitter for a perfect load.

More info at http://www.vk1od.net/7MDipole/7MDipole.htm .

Owen

If the half-wave dipole is installed less than 1/2 wavelength above earth, it probably has a radiation resistance less than 75 ohms and may match 50-ohm feed-line reasonably well without any type of impedance matching device. How high is it above ground and what is its approximate resonant frequency?

Assuming the dipole radiation resistance actually is 75 ohms:

1) If the feed-line has reasonably low loss per wavelength and will be less than a wavelength or so in length, there probably would be no practical difference in reported signal strengths with and without impedance matching, because the difference in radiated field strengths would be a small fraction of an S-unit and would be masked by the normal signal fading that occurs with ionospheric propagation.

2) There are issues other than radiated field strength to consider. One is transmitter loading. Most transmitters with 50-ohm outputs can be tuned, or will self-tune, to match a 75-ohm resistive load. So, if the dipole antenna is perfectly resonant at the operating frequency, so its feed impedance has no reactive component, your transmitter probably will load fine at that frequency if you use 75-ohm transmission line. However, the range of frequencies over which it will load correctly will be less than if the transmitter was working into a 50-ohm center-frequency termination. If you use 50-ohm transmission line without impedance matching, transmitter loading will depend critically on transmission line length.

3) As to the question of whether it is more important to match impedances at the antenna or in the shack, it depends on whether you are transmitting or receiving. A given transmission line has its least possible RF loss per unit length when it is terminated in its characteristic impedance. Many people falsely assume that "terminated" means both ends, but the "terminating" end is the far end. Therefore, when transmitting, a transmission line's characteristic impedance should match the impedance of antenna that terminates it. However, when receiving, the receiver antenna input impedance should match the transmission line impedance. Given that most receivers have plenty of excess RF gain, transmission line loss when receiving tends to be far less important than transmission line loss when transmitting and the distant received signal must compete with noise and/or other signals. So, if only one end of a transmission line is to be matched, the antenna-end tends to be most important.

3) Dipole antenna feed balance generally tends to be more important than transmission line impedance matching throughout the HF spectrum (except where transmission lines are unusually long). If the antenna feed impedance is truly 75-ohms, an ideal solution would be to use 50-ohm coax transmission line and a 50 to 75 ohm balun between the transmission line and the antenna. The impedance matching would make the antenna useful over a wider range of frequencies and the balanced antenna feed would result in a much more symmetrical antenna radiation pattern.

Most baluns have 1 to 4 impedance transformation ratios, such as from 75 to 300 ohms. However, a simple 50 to 75-ohm impedance transforming balun can be made by winding separate primary and secondary windings on a toroidial core. The impedance ratio of a transformer is equal to the square of its turns ratio, so a primary to secondary turns ratio of 1 to 1.225 will transform from 50 to 75 ohms. That exact ratio can't be achieved without using more turns than an HF transformer with a toroidial core should have, but a 1 to 1.2 turns ratio is practical and will transform 50 ohms to 72 ohms, which is plenty close enough.

pallen wrote:... A given transmission line has its least possible RF loss per unit length when it is terminated in its characteristic impedance.

Not quite true...

Although there are lots of graphs that show the "increased loss due to VSWR", they depend on assumptions that are not usually stated.

A short length of line with a high load load impedance can easily have loss under mismatched conditions that is less than the matched line loss.

Try the line loss calculator at http://www.vk1od.net/tl/tllc.php with 1m of RG58 at 1MHz with a 500 ohm load (VSWR=10). The mismatched loss is less than matched loss, and much less than predicted by the graphs I mentioned. Contrast it with the case of load=5 ohms (also VSWR=10).

Why is it so? (to borrow a phrase from Prof Julius Sumner Miller)

Owen

Power losses in a transmission line increase as SWR increases, because current and voltage down the line both oscillate above and below the values they would have with zero SWR and P=I^R and P=E^2/R. The nonlinear power relationships that power losses have to both current and voltage cause the increased losses in the high current and voltage sections of line to exceed the reduced losses in the low current and voltage sections, causing the net power loss to increase with increases in SWR.

John

John Cuthber wrote:Power losses in a transmission line increase as SWR increases, because current and voltage down the line both oscillate above and below the values they would have with zero SWR and P=I^R and P=E^2/R. The nonlinear power relationships that power losses have to both current and voltage cause the increased losses in the high current and voltage sections of line to exceed the reduced losses in the low current and voltage sections, causing the net power loss to increase with increases in SWR.

John

Yes, the voltage and current at points along the line are higher and lower than in a line with VSWR=1, and losses in an incremental length of line are the sum of I^2*R and E^2*G. For most practical lines at HF, R is small (is ohmic loss is small), and G is extremely small (ie dielectric loss is extremely small), so the loss is mainly caused by ohmic losses (the I^2*R component), and so loss is highest in the region of current maxima. In the region of a current minimum, the current, and therefore the loss, may be lower than for the case when VSWR=1 (for the same power transmission).

Regarding your last sentence which implies that under high VSWR there are sections of the line with simultaneously high voltage and high current and other sections with simultaneously low voltage and low current. If that was the inference, it is wrong, a current maximum occurs at approximately the point of a voltage minimum (and vice versa) for practical lines.

Owen

Power can be lost from RF transmission lines in these ways:

1) Electromagnetic radiation

2) Capacitive or magnetic induction from a single conductor

3) RF current flowing through series RF line resistance

4) RF current leaking through RF leakage resistance between the two conductors.

RF power losses due to the first two of those causes are negligible with twin conductors that are spaced very closely compared to a wavelength or with coaxial conductors in most practical applications throughout the radio spectrum. That leaves nearly all the power being lost due to the last two causes.

It is true that RF line series resistance tends to relatively low and that conductor-to-conductor line RF leakage resistance tends to be relatively high in most practical RF transmission line applications throughout the HF portion of the spectrum. Even so, nearly all the line loss that does occur is due to those two causes and power lost due to those two causes increase with increases in line standing wave ratio.

The last sentence in my previous post does not say and was not meant to imply that the same sections of line simultaneously have higher current and voltage with a high SWR. That of course does not occur. However, voltage nodes and current nodes both reach increasingly higher amplitudes as line SWR increases and that is the fundamental underlying reason why power losses increase with increasing SWR.

John

John Cuthber wrote:Power can be lost from RF transmission lines in these ways:

...

It is true that RF line series resistance tends to relatively low and that conductor-to-conductor line RF leakage resistance tends to be relatively high in most practical RF transmission line applications throughout the HF portion of the spectrum. Even so, nearly all the line loss that does occur is due to those two causes and power lost due to those two causes increase with increases in line standing wave ratio.
...
John

John,

A very general statement that avoids the fact that in the presence of very high VSWR, that the I^2*R loss in the series resistance in the region of a current minimum is lower than under matched conditions (because the current is less that for the same power with VSWR=1), and since that I^2*R loss dominates the total loss, total loss in the region of a current minimum may be less than matched line loss. This is the case for the example I gave above and it disproves the general statement that power lost always increases with increase in VSWR.

Many books showing a VSWR based formula for "additional loss due to VSWR" don't spell out the assumptions underlying the formula. Phillip Smith does in his book "Electronic Applications of the Smith Chart", he says "If a waveguide is one or more wavelengths long, the average loss due to standing waves in a region extending plus or minus a half wavelength from the point of observation may be expressed as a coefficient or factor of the one way transmission loss per unit length." and he gives the ratio as (1+S^2)/(2*S). Though the ARRL shows graphs and formulas they don't always (if ever) spell out the assmptions.

So, yes I assert that the Line Loss under mismatch conditions may be less than the Matched Line Loss.

Owen

pallen wrote:Most baluns have 1 to 4 impedance transformation ratios, such as from 75 to 300 ohms. However, a simple 50 to 75-ohm impedance transforming balun can be made by winding separate primary and secondary windings on a toroidial core. The impedance ratio of a transformer is equal to the square of its turns ratio, so a primary to secondary turns ratio of 1 to 1.225 will transform from 50 to 75 ohms. That exact ratio can't be achieved without using more turns than an HF transformer with a toroidial core should have, but a 1 to 1.2 turns ratio is practical and will transform 50 ohms to 72 ohms, which is plenty close enough.

Forgive me asking at this later time, but. . . . .
what sort of toroidal core and how does one work out the number of turns required with what type of wire?
[ I have a requirement for an HF dipole (mostly about 60m) for my local cadets; 100w output ]