We just had snow and ice overnight and my wire dipole was coated in ice. Why does the ice covering negatively affect the SWR of the antenna?
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asked Apr 16, 2014 at 13:05
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$\begingroup$ A basic question, obviously one should shake off the ice coating from the wire to solve the high SWR. But its a good seed question for the beta site, and I look forward to getting a good technical answer. $\endgroup$– Ron J. KD2EQSApr 16, 2014 at 13:06
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1$\begingroup$ Precipitation effects the effective Ground level, changing the SWR. Temperature changes antenna length, especially noticeable at VHF & UHF, also changing SWR. $\endgroup$– OptionpartyApr 16, 2014 at 23:33
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$\begingroup$ @Optionparty Copper has a coefficient of linear thermal expansion of $16.6 \mu \mathrm m/\mathrm m$. So, an antenna that is 20m long at -10°C and is then heated 40°C will become $40 \cdot 16.6 \cdot 20 = 13280 \mu \mathrm m \approx 13 m \mathrm m$ longer. VHF and UHF antennas are shorter, and their lengths change correspondingly less because there is less metal to expand. I doubt change in length due to temperature has a significant effect on SWR. $\endgroup$– Phil Frost - W8IIApr 19, 2014 at 11:59
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$\begingroup$ @Optionparty is that due solely to changes in antenna length, or also temperature dependence in component values throughout the transmitter? Given that TV broadcast operates at much higher power, and also at 100% duty cycle, I'd think the transmitter components get pretty toasty too, and have to deal with a larger operating range than just the antenna. $\endgroup$– Phil Frost - W8IIApr 20, 2014 at 12:32
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2$\begingroup$ Not just ice, but volcanic Ash can cause SWR problems too. Back during the 1980 Mt. St. Helens eruption many ham antennas were covered with damp volcanic ash, especially Yagi beams with larger cross-section for such ash. Although I was not operating at the time, I heard a number of stories of Ash problems and one guy I know even hosed off his big beam antenna. At my QTH we had a number of days with volcanic dust settling on everything outside (cars, patio table, etc.). $\endgroup$– K7PEHNov 18, 2014 at 18:21
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4 Answers
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A heavy coating of ice or snow is likely to short the feedpoint of the antenna. Compare:
simulate this circuit – Schematic created using CircuitLab
Even if the wires and feedpoint are protected by insulation and the ice isn't directly touching, a thick coating of ice will make a tube around the wires, and the resulting capacitive coupling is a low impedance at RF. (ANT3)
Snow or ice isn't a great conductor, but it's a much better conductor than the air or PTFE insulator that was between the halves of the antenna. Clearly the antenna impedance, and thus the SWR, will be affected.
The change in temperature, dielectric constant, and other things mentioned in other answers do affect the antenna impedance, but not to a very significant degree. I'd wager that 80% of the times you sit down in the morning after a storm to find the SWR is all messed up, it's because there's water in the feedpoint. The other 20% of the time, it's because the antenna is sagging into a tree or the ground or broken from the additional weight.
answered Apr 19, 2014 at 12:14
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The increasing SWR means that some of the transmitted power is reflected back to your transmitter. There are several possible explanations for the effect:
The ice is not completely dry and it partially short circuits (through some resistance) the two arms of your dipole. This way the impedance seen at the feed port of the antenna is lower than it is meant to be making the matching worse.
The feed port and the dipole arms are properly insulated from the melting ice but there is still some conductance in the ice layer: this prevents the radiated field from being transmitted to the free space increasing the power reflected towards the transmitter. This would more or less correspond to a situation where you have a (poorly) conducting box around your antenna.
The third possible explanation is related to the reactive near fields surrounding the antenna (see the image below). The impedance of a dipole is determined by the radiation resistance, capacitance between dipole arms, and the inductance of the dipole wire. (See my answer to a related question). If you coat your dipole with ice ($\epsilon \approx 3$) the capacitance between the dipole arms is increased dragging the resonance frequency lower.
It is not easy to say what is the correct explanation for this specific case. If you can plot the SWR as function of frequency with and without the ice and see a clear shift in the resonance frequency, number three would be my suspect. If the ice is near melting and the SWR is poor at all frequencies, the explanation 1 would be my guess.
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Ice and water has very big dielectric permittivity. If the antenna is in a dielectric, so the resonance frequency changes with the square root of it. So for ice the dielectric permittivity may change due to its density. Water has dielectic permittivity of 80. So the antenna resonance frequency changes with the ratio 1/squareroot (80). It doesn't guarantee that the antenna works well at the new frequency, because 80 is very high permittivity wich can create high reflection, but it works better at this frequency. That's why the antenna doesn't work when it has ice on it. It is not about the change of antenna length or anything else because of temperature. If you want to solve this problem you must have heat resistance wires on it in order not to have ice on it.
answered Dec 3, 2014 at 22:01
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$\begingroup$ So if it's raining, my antenna which was resonant at 7 MHz is now resonant at $7\:\mathrm{MHz}/\sqrt{80} = 783\:\mathrm{kHz}$? That does not seem right at all. $\endgroup$ Dec 18, 2014 at 16:24
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2$\begingroup$ The antenna must be covered by a dielectric for this principle, the electric field vector between two poles of the antenna must go inside the dielectric. This is known as dielectric loaded antennas in the literature. Rain is a different situation, most part of electric field vectors between two poles doesn't go inside the dielectric in rain situation. $\endgroup$ Dec 18, 2014 at 22:48
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$\begingroup$ I think you should edit the answer to clarify that. The way it's written, it sounds like if there's any water at all, the resonant frequency of the antenna will go down absurdly. Some formatted math with mathjax and some links to references would also help a lot. $\endgroup$ Dec 19, 2014 at 3:46
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$\begingroup$ Small typo fix the original poster needs to fix wich should be which. $\endgroup$ Feb 16, 2017 at 19:20
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Assuming we aren't worried about water intrusion into the center insulator or any of the feedline connections, any of which could raise the SWR, the increased SWR is probably due to water ice acting as an insulator on the wire antenna.
This, of course, detunes the antenna. The dielectric constant of normal wire insulation (polyethylene) is about 2.1 whereas water has a dielectric constant of about 5. This means the antenna is suddenly too long for the intended frequency because the velocity factor of the conductor is now higher than it was when the antenna was cut.
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$\begingroup$ Conductors don't have velocity factors: transmission lines do. Put another way, if the issue is an altered dielectric constant, and a dielectric is an electrical insulator that can be polarized by an applied electric field, then where's the electric field and what's polarizing the dielectric? If all you have is a conductor, then the field is between that...and what? $\endgroup$ Apr 17, 2014 at 0:32
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$\begingroup$ Although not very common, I think you can model, or think, a wire antenna structure as a group of transmission lines carrying currents and having velocity factors. A traveling wave antenna is an extreme example of this, but the same ideas apply for shorter wires as well. $\endgroup$– OH2FXNApr 17, 2014 at 4:27
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$\begingroup$ @OH2FXN You can, and if you are going to consider the effect of water as a dielectric on an antenna that's the way to go about it. The problem is that "velocity factor of the conductor" doesn't invoke mental images of transmission lines and electric fields: it suggests that if you buy a spool of wire, then somewhere on the datasheet should be "velocity factor", which there isn't. Not because it's missing, but because it's undefined until you make a transmission line with that wire, by making an electric field between it and another wire, or the ground. $\endgroup$ Apr 17, 2014 at 11:20
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$\begingroup$ Of course you are correct, the term "velocity factor of the conductor" makes no sense. $\endgroup$– OH2FXNApr 17, 2014 at 14:50
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$\begingroup$ A conductor, fed with high frequencies, is frequently considered a transmission line. A thin wire has a velocity factor of about 0.98. Tubing on higher frequencies have velocity factors between 0.90 and 0.98. There was an excellent article in Ham Radio (2 parts, April '77 and May '77) which explores this 'analog' between antennas and transmission lines. As the dielectric constant changes with the ice, one could argue that velocity factor changes. $\endgroup$– jcoppensNov 18, 2014 at 16:27