Mismatch doesn't affect shielding. If you transition from your balanced openwire line to the pair of coaxes, there's no reason why there should be any currents on the outside of the coax shield. If there were, you could always choke it with the usual appropriate ferrite or powdered metal cores.
Observed mismatch on a feed line (at the transmitter end) might be a symptom of currents on the outside (e.g. The TL is acting as part of the antenna), but the converse isn't true (consider an example case: a shorting plug at the far end of a piece of coax... It won't radiate any more or less than the same coax with a terminating load, and that's an infinite mismatch). The case of feeding a 40m antenna on 20m is an interesting one. If the line is balanced, there's no real reason why it would be carrying any common mode current and radiate. Of course, *real* antennas are hardly ever balanced in installation, so they will couple unevenly to the balanced pair, and having the voltage peak at the feedpoint just aggravates it. However, I don't think a section of paired coax will make this appreciably worse, at least not from the "bump" in the impedance of the TL down the line. Voltage breakdown IS a potential issue with large mismatch, just like happening to have the coax pair at a current peak and getting large loss (which can melt the dielectric or center conductor). On 2/2/09 6:33 AM, "Dennis" <radio...@charter.net> wrote: > Hi James, > > I understand your comments. My response was more focused on using coax to > replace tuned feeders and by doing so believe you are Shielding the tuned > feeders from radiating; not so.... The standing waves on both coax lines will > radiate along the line (assuming an unmatch situation like using a 40m dipole > on 20m for example). Another issue is breakdown voltages of the coax in such > a situation. The shielding effect of the coax will not be realized like it > would in a single ended installation where the coax is properly match. > > Dennis, k0eoo > > > ---- "Lux wrote: >> >> >> >> On 2/2/09 5:13 AM, "Dennis" <radio...@charter.net> wrote: >> >>> This is only true if the coax is properly matched, on both ends, to its >>> characteristic impedance, whether you're using it single ended or >>> differential.... So, using coax to replace tuned feeders is very >>> problematic >>> (looses go up) unless both ends of both coax cables are matched to their >>> characteristic Z. >> >> Not necessarily. It depends on how long the piece of coax is, and whether >> you happen to be at a voltage or current peak. It's good to remember that >> those published loss numbers (whether matched, or recalculated for a >> mismatch and including the circulating currents) are essentially the >> "average" over a wavelength. If you're looking at substantially less than a >> wavelength (and 5 ft is substantially less than a wavelength, even on 10 >> meters), the loss could be higher or even lower. >> >> For HF, the loss is mostly the IR loss in the center conductor, and, so, is >> proportional to the current squared in that section of coax. A line with a >> big mismatch will have a current that varies along the line, but which >> (geometrically) averages to something in between. If you look up the >> "mismatched line loss" or calculate it, you'll be getting that average. At >> some places the loss will be less: potentially LESS than it would be in the >> matched case, if the current is particularly low. >> >> >> Here's a sort of special case. Consider a piece of transmission line that's >> exactly a quarter wavelength long, and shorted at the far end. The SWR is >> infinite. At the source end, the voltage is a maximum, but the current is >> almost zero, so the losses per unit length are very small. Move out along >> the line, and the voltage drops (eventually to zero) and the current >> increases and the IR losses per unit length increase. >> >> (This is aort of special case, because there's not any power actually >> flowing in the line.. BUT, imagine, instead of a short, that a quarter >> wavelength of 50 ohm line is terminated in, say, 10 ohms, and it's driven by >> a 250 ohm source... E/I = 250 at the source end, E/I = 10 at the destination >> end) >> >> This kind of thing is a royal pain to calculate by hand, so resorting to >> some sort of program or spreadsheet that does the transmission line >> equations is definitely worthwhile. I like XLZIZL, which is an excel >> spreadsheet that lets you put in up to 5 "stages" in a ladder, where each >> stage is either a transmission line, some reactive components, or a >> transformer. There's other programs too. I don't know if ARRL's TLW does >> multiple transmission lines. >> >> >> Jim >> > > _______________________________________________ FlexRadio Systems Mailing List FlexRadio@flex-radio.biz http://mail.flex-radio.biz/mailman/listinfo/flexradio_flex-radio.biz Archives: http://www.mail-archive.com/flexradio%40flex-radio.biz/ Knowledge Base: http://kc.flex-radio.com/ Homepage: http://www.flex-radio.com/
