Home |
Search |
Today's Posts |
#18
![]() |
|||
|
|||
![]()
It's actually easy to predict the (resistive) impedance seen looking
into a shorted quarter wavelength line, or any odd multiple. The impedance is simply the Z0 of the line divided by the loss in nepers. One neper is about 8.7 dB, so the impedance is about 8.7 * Z0 / dB loss. All else being equal, the impedance gets higher as frequency increases. That's because the length of a quarter wave stub decreases in inverse proportion to frequency, while loss (up to 1 - 10 GHz or so, where conductor loss dominates) increases only as the square root of frequency. So the impedance of a stub increases as the square root of frequency. For example, a quarter wave stub, made from solid polyethylene dielectric coax (VF = 0.66) at 3.5 MHz is about 46 ft. That length of RG-58 has a loss of about 0.3 dB, so the impedance looking into a quarter wave stub of RG-58 at 3.5 MHz is about 1450 ohms. Quite a far cry from the textbook's example of 400 k ohms or Richard's extrapolation to 33 k ohms! An RG-58 stub at 350 MHz, or 100 times the frequency, would have an input impedance of about 14,500 ohms. A more typical VHF example would be a quarter wave of RG-8 at two meters. It would be about 13.4 inches long and a loss of about 0.03 dB, for an input Z of about 14,500 ohms. Incidentally, the formula I'm using is actually on the same page of King et al's text as the 400 k ohm value Richard quotes. They say the 400 k value is for "a reasonably low-loss line" -- to get 400 k ohms with a 600 ohm line, the loss would have to be about 0.013 dB. The input impedance of an open circuited quarter wavelength line or shorted half wavelength line is Z0 times the loss in nepers, or about Z0 * dB loss / 8.7. I actually ran into a case where the finite resistance of an open stub became a problem, and it illustrates the hazard of blindly following a "rule of thumb" without checking to see under what conditions it's valid. The "Field Day Special" antenna, similar to a ZL special, can be fed at the center of either element. I connected a one wavelength transmission line to the center of each element, and fed one or the other to switch directions, leaving the other line open circuited. When RG-58 was used, the current diverted into the finite resistance of the open stub disturbed the element current enough to very significantly degrade the front/back ratio. The lines were one wavelength at 14 MHz, or about 46 feet. Loss was a seemingly trivial 0.8 dB, but that means that the input impedance was only about 540 ohms! 400,000 or even 33,000 would be an awfully poor estimate! Changing to 300 ohm twinlead solved the problem. (Although 300 ohm twinlead can easily be as lossy as RG-58 when wet, the higher Z0 resulted in an adequately high stub impedance even when it was wet.) Roy Lewallen, W7EL Richard Harrison wrote: . . . From King, Mimno, and Wing, "Transmission Lines, Antennas, and Wave Guides" page 29: "A short-circuited line, one-quarter wavelength long at the desired output frequency may be connected across the output terminals of a transmitter or across the antenna feeder at any point without placing much load on the transmitter at this fundamental or desired output frequency, since at this frequency such a section has an impedance ideally infinite, actually about 400,000 ohms." Since I = E/Z, how much current do you think will flow into 400,000 ohms? King, Mimno, and Wing`s impedance might scale down to only 33,333 ohms on a 50-ohm line, still high, as they may have been considering a 600-ohm line. . . . |