W5DXP

CECIL A MOORE

TYLER, TEXAS

Cecil A. Moore was first licensed in the early 1950's as WN5DXP. A year later, he upgraded to Conditional and was eventually grandfathered into the General class.

He received an Electrical Engineering degree from Texas A&M in 1959 and spent 40 years in industry, primarily as a digital systems engineer and microcontroller applications engineer.

When he moved to CA from TX, his W5DXP call was replaced with W6RCA by the FCC. He upgraded to Advanced class in the 70s at the FCC office in San Francisco.

Cecil is retired and enjoys riding his 100th anniversary 2003 Harley Road King. He upgraded to Extra in 2000 and got his original W5DXP call back in 2001.

His primary interests in amateur radio are antennas and transmission lines and regularly contributes to various web/internet technical newsgroups.

Why the G5RV is a good antenna on 75m and 40m

G5RV 102 ft. dipole with 28 ft. 300 ohm matching section

The following chart shows the optimum length for the matching section per HF band according to EZNEC. It's obvious why the ~30 foot length works well on 80m, 40m, 20m, 12m, and not bad on 15m. If some method is devised to vary the length of the 300 ohm series matching section on the G5RV, it can be made to work well on all HF bands.

W5DXP's No-Tuner All-HF-Band Antenna

W5DXP's No-Tuner, All-HF-Band, Horizontal, Center-Fed Antenna

The No-Tuner, All-HF-Band, Horizontal, Center-Fed Antenna is our old friend, the 80 meter halfwave dipole dressed up a bit. By varying the length of the 450 ohm ladder-line feeding the antenna, we can achieve an SWR of less than 2:1 on all frequencies on all HF bands with the exception of the lowest part of 80m. On 75m, we are feeding the antenna with a half-wavelength of ladder-line. On 40m, we are feeding it with 3/4 wavelength of ladder-line.

The Ladder-Line Length Selector actually does tune the antenna system so no conventional "antenna tuner" is needed - no coils and no capacitors. Switches or relays (remote control) can be used for the switching function and should be sized according to the RF power levels involved. W5DXP presently uses ten DPDT Knife switches attached to a piece of plexiglas mounted in the hamshack window. For portable or backpacking use, the length selector function can be performed simply by 1/2/4/8/16 foot pieces of ladder-line with mating connectors on the end. The proper length of ladder-line is selected to cause resonance in the antenna system.

Here's a table that explains it all. The transmission line always consists of a matching section and from zero to six halfwavelengths of ladder-line. The impedance at the antenna is shown along with the 450 ohm SWR and the impedance at the transmitter is shown along with the 50 ohm SWR, i.e. the SWR seen by the transmitter.

..Freq-MHz..	..T-line length = Matching Section + 1/2WL's..	..Impedance at XMTR..	..50 ohm SWR..	..Impedance at Antenna..	..450 ohm SWR..
3.8	109.5' = 109.5' + 0	69 ohms	1.4:1	71+j84	6.6:1
7.2	92.0' = 30.5' + 1x61.5'	40 ohms	1.2:1	4939-j716	11.2:1
10.125	99.4' = 12.0' + 2x43.7'	50 ohms	1.0:1	116-j510	9.1:1
14.2	110.2" = 16.6' + 3x31.2'	53 ohms	1.1:1	2120+j1886	8.5:1
18.14	101.9' = 4.3' + 4x24.4'	81 ohms	1.6:1	111-j267	5.5:1
21.3	94.8' = 11.6' + 4x20.8'	70 ohms	1.4:1	1210+j1378	6.4:1
24.95	94.1' = 5.35' + 5x17.75'	65 ohms	1.3:1	186-j593	6.9:1
28.4	102.8' = 9.2' + 6x15.6'	87 ohms	1.7:1	721+j1009	5.2:1

Here are the ten DPDT switches mounted on a piece of plexiglas that mounts in W5DXP's hamshack window. It shows the ten DPDT switches with the one foot, two feet, and four feet loops installed. The eight feet and 16 feet loops are not installed yet in this picture. RF flow is right to left from banana socket set to banana socket set. When installed in the hamshack window, the switches are on the inside and the loops of ladder-line are on the outside.

Here's a close up view of the one foot section. The RF flow is right to left into the banana sockets. The switches are shown in the shorted position, i.e. the one foot loop is floating completely out of the circuit to avoid capacitive effects. The bare copper wires in the center are the short. When the switches are thrown into the other position, the one foot loop is inserted into the circuit and the short is completely out of the circuit. This is the cleanest mechanical configuration W5DXP could think of but there might be a better way.

This is a plot of all the current maximum points between the antenna and W5DXP's shack. The transmission line is 90 feet long and the Ladder-Line Length Selector can add in an additional zero to 31 feet for a total of 90 feet to 121 feet. 90 feet matches the antenna on about 7.3 MHz and 121 feet matches the antenna on about 3.6 MHz. The matching points for all the other HF bands lie between these two extremes. Note that if a fixed length of ladder-line needs to be chosen for best results with this antenna, that length should be around 100 ft. which should work with internal autotuners. Caution: Do not expect a similar antenna erected in a different location to exactly match W5DXP's results. The antenna environment has a large effect on the antenna characteristics so W5DXP's results are only approximations when applied to other antenna locations and environments. Mounting this antenna in an inverted-V configuration, for instance, is likely to change the characteristics by an unexpected amount. "450" ohm ladder-line characteristic impedance varies all the way down to 375 ohms for the #14 stranded configuration and velocity factor varies among the different manufacturers and batches of ladder-line.

Who says a full-wave dipole is hard to match? Here's what EZNEC predicts will be the 50 ohm SWR across the 40 meter band for W5DXP's No-Tuner All-HF-Band Antenna given the chart lengths of ladder-line. Similar SWRs occur in similar patterns on the other HF bands.

For those who don't have the space for a 130 foot antenna, here's a "Shorty" version designed to work on all HF ham frequencies above 7 MHz. Like the bigger version, the 50 ohm SWRs predicted by EZNEC are below 2:1 for the bands of interest. This antenna will work on 75 meters at reduced efficiency with a matching network or tuner.

Here is the physics that makes it all possible. Any 450 ohm SWR between 4.5:1 and 18:1 will result in a 50 ohm SWR of less than 2:1 IF the antenna system is fed at a current maximum point. Moral: Make your center-fed HF antenna system at least a half- wavelength long at your lowest operating frequency and feed it at a current maximum point on the ladder-line.

Optimum Length For A Matching Section

This graph shows the optimum length for a matching section when feeding a center-fed horizontal dipole. The bottom of the chart is normalized to wavelengths so it works for most HF frequencies and most popular lengths of center-fed wire dipoles. The left side of the chart indicates the optimum wavelength for a 450 ohm ladder-line matching section for connection to coax or connection to a multiple of half-wavelengths of 450 ohm ladder-line.

Example: Assume a 102 ft dipole on 7.2 MHz. 102/(936/7.2) equals 0.785 wavelengths on 7.2 MHz. Reading the matching section length from the graph yields 0.3 wavelength. A wavelength of 450 ohm ladder-line on 7.2 MHz is 886/7.2= 123 ft. 0.3 times 123 equals 36.9 ft for the 7.2 MHz matching section. Add 123/2 = 61.5 ft if 36.9 ft is too short for a total of 98.4 ft.

The following BASIC program approximates the optimum feedline lengths given the length of a horizontal dipole and the frequency. It works for both 300 ohm and 450 ohm ladder-line by assuming a velocity factor of 0.8 for the 300 ohm and 0.9 for the 450 ohm. The results are only approximations based on EZNEC and must be fine-tuned to perfection in reality. This program can be cut and pasted to Notepad and stored in the BASIC directory as Imax.bas. Or an unzipped, ready to run, 30kB DOS "imax.exe" file can be downloaded: Download imax.exe

Note: This BASIC program only works for horizontal dipoles, not for inverted-V's or any other folded antenna.

5 CLS
10 PRINT "This program calculates optimum ladder-line"
20 PRINT "lengths given dipole length and frequency"
31 PRINT "for dipoles that are at least 2/5 wavelengths"
32 PRINT "long at the lowest frequency of operation": PRINT
33 PRINT "Enter Break to exit this program at any time.": PRINT
40 INPUT "Enter Frequency in MHz ", freq
50 INPUT "Enter Dipole Length in Feet ", diplenft
51 length = 375 / freq
55 IF length > diplenft THEN
56 PRINT : PRINT "********** Warning! **********": PRINT
57 PRINT "Dipole Length Too Short. For This Frequency"
58 PRINT "It Needs To Be Longer Than"; length; "Feet"
59 PRINT : PRINT
60 GOTO 40
61 END IF
65 INPUT "Enter either 450 or 300 for Z0 ", Z0
70 IF Z0 = 450 THEN LLWL = 886
80 IF Z0 = 300 THEN LLWL = 787
90 dipwl = diplenft / (936 / freq)
100 IF dipwl < .5 THEN dipwl = dipwl + 1
110 IF dipwl < 1.5 THEN GOTO 140
120 IF dipwl > 1.5 THEN dipwl = dipwl - 1
130 GOTO 110
140 fedlinwl = .25 - (TAN(2.5 * (dipwl - 1))) / 12.02
150 fedlinft(0) = (LLWL / freq) * fedlinwl
160 FOR i = 1 TO 7
170 fedlinft(i) = fedlinft(0) + i * ((LLWL / 2) / freq)
180 NEXT i
190 PRINT "Imax points (Current Loops) at"
200 FOR i = 0 TO 7: PRINT fedlinft(i), : NEXT i
210 PRINT : PRINT : GOTO 40
220 END

Half-Extended Double Zepp Dualband Antenna

First published in WorldRadio, March 2007, and reproduced with permission

The Half-Extended Double Zepp (HEDZ) Antenna - (A dual-bander for 80M and 40m)

Background Information: A Zepp is an end-fed ½ wavelength antenna. A Double Zepp (DZ) is a one-wavelength center-fed dipole, i.e. Double the Zepp. An Extended Zepp (EZ) is an end-fed 5/8 wavelength antenna. An Extended Double Zepp (EDZ) is a 5/4 wavelength center-fed dipole, i.e. Double the Extended Zepp. This article will introduce the Half-Extended Double Zepp (HEDZ) which has characteristics that a lot of amateur radio operators should find quite interesting. Like the G5RV, it has a certain dipole length and a certain matching section length that works for 80m, 40m, and possibly 17m. It also minimizes the existence of any coax in the system. Please note that this article is about an antenna system which includes the tuned feeder transmission line as an integral part of the system which transforms impedances to useful values.

The author has for many years been telling the amateur radio community that a 130 foot dipole will work on all HF bands without an antenna tuner by simply varying the length of the 450 ohm ladder-line feedline. W5DXP's No-Tuner All-HF-Band Antenna System.

One of the nagging problems with varying the length of the ladder-line is that the knife switches (or relays) must be changed when changing bands. However, the author has stumbled upon a partial solution to that problem for 80m, 40m, and 17m, the author’s favorite bands. The HEDZ is designed for operation at 7.15 MHz and the other bands simply fall out from that choice.

The results of EZNEC modeling will be used here although the actual dimensions at the author’s actual QTH are very close to the results predicted by EZNEC. A free demo version of EZNEC is available at: CLICK ON >>> EZNEC <<< CLICK ON 

A one-wavelength Double Zepp dipole has about 1.5 dB gain over a ½ wavelength dipole in free space. The ladder-line feedline length needs to be a ¼ wavelength section plus a number of ½ wavelength sections. Such a dipole would be 130 feet on 7.15 MHz and would require approximately 92 feet of 450 ohm ladder-line. That makes the antenna system resonant at 4.065 MHz on 80m, i.e. too high and out of band.

A 5/4 wavelength Extended Double Zepp dipole has about 3 dB gain over a ½ wavelength dipole in free space. The ladder-line feedline length needs to be a ~0.2 wavelength section plus a number of ½ wavelength sections. Such a dipole would be 166 feet on 7.15 MHz and would require approximately 85.5 feet of 450 ohm ladder-line. That makes the antenna system resonant at 3.65 MHz on 80m, i.e. lower than most hams would want.

EPIPHANY TIME!!! If a 40m ladder-line fed DZ is also resonant on 4.07 MHz and if a 40m ladder-line fed EDZ is also resonant on 3.65 MHz, wouldn’t somewhere in between be resonant on 3.8 MHz? - thus the birth of the Half-Extended Double Zepp. Halfway between 130 feet and 166 feet is 148 feet. Let’s see what a 148 foot ladder-line fed dipole will do on 40m and 80m.

A 148 foot dipole fed with ~90 feet of 450 ohm ladder-line has a gain of about 2 dB over a ½ wavelength dipole on 7.15 MHz. But its real claim to fame is that it is also resonant on ~3.85 MHz with a 3:1 SWR bandwidth of approximately 200 kHz. That means that this single antenna system is resonant on both ~7.15 MHz and ~3.85 MHz at the same time. This is something for which a lot of amateur radio operators have been wishing. Using my IC-756PRO’s internal tuner allows me to tune the entire 40m band plus 200 kHz of 80m without changing anything.

An additional bonus is that the 17m band is also within the tuning range of my IC-756PRO’s built-in autotuner. So I have seamless switching among three bands, 80m, 40m, and 17m which are my favorite bands. I couldn’t be happier with this 148 foot dipole fed with approximately 90 feet of 450 ohm ladder-line. Of course, a good 1:1 choke-balun is required at the ladder-line/coax junction.

Here’s the procedure for tuning one’s own 148 foot dipole for 80m, 40m, and hopefully also 17m.

Install the 148 foot dipole fed with 100 feet of 450 ohm ladder-line. Check the SWR on 7.15 MHz which will be too high. Start trimming the ladder-line one foot at a time until a minimum SWR is achieved - (It will not be 1:1 unless you are extremely lucky.) Now you have a Half-Extended Double Zepp resonant on 7.15 MHz.

Check the resonant frequency on 80m. It should be close to 3.85 MHz. If it is not the desired frequency, lengthening the dipole and shortening the ladder-line will lower the resonant frequency on 80m. Shortening the dipole and lengthening the ladder-line will raise the resonant frequency on 80m – all while keeping the antenna resonant on 7.15 MHz. Somewhere between a 135 foot dipole and a 175 foot dipole will allow one to pick one’s favorite 80m resonant frequency. EZNEC says a 135 foot dipole fed with 92 feet of ladder-line will be resonant on both 7.15 MHz and 3.99 MHz. A 145 foot dipole fed with 90 feet of ladder-line will be resonant on both 7.15 MHz and 3.85 MHz. A 155 foot dipole fed with 88 feet of ladder- line will be resonant on both 7.15 MHz and 3.74 MHz. A 165 foot dipole fed with 86 feet of ladder-line will be resonant on both 7.15 MHz and 3.66 MHz. The trend is obvious. Ten feet dipole increments shifts the 75m resonant frequency by about 100 kHz while maintaining the 7.15 MHz resonant reference frequency.

Hopefully, this will solve the dual-band 80m/40m antenna system problem that so many amateur radio operators seem to experience. As can be seen from the radiation patterns below, the 80m pattern is essentially a ½ wavelength dipole pattern and the 40m pattern is akin to an EDZ pattern. On 17m and 15m, the antenna will have a multi-lobbed "cloverleaf" pattern with gain over average ground in excess of 10 dBi and a take-off-angle of ~16 degrees.

Since the article was written, W5DXP has taken a look at the ladder-line lengths required for all HF operation of the 148 foot dipole at 40 feet as indicated by EZNEC. The 90 foot length of ladder-line used in this article should be close to system resonance on 3.8, 7.15, 17m, and 21.4 MHz. ~82.5 feet should work for 30m, 12m, and 28.4 MHz. Finally, ~75.5 feet is required for resonance on 14.2 MHz according to EZNEC. Of course, local conditions can and will affect those typical lengths.

40M Triangular Full-Wave Vertical Loop Antenna

Add the Missing Leg to that Inverted-L Antenna 

Note the 2x80 degree vertical beam width.

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Note the 2x102 degree horizontal beam width. 

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 My Dipole Vs Vertical on 40M

130 ft. Dipole at 40 ft. Vs 33 ft. vertical with 8 radials at 20 ft.

Used on 40 Meters

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Rotatable Dipole for 40m-6m

 

First published in WorldRadio, June 2000, and reproduced with permission

Who Needs a Beam?

This article describes a simple, inexpensive, dipole antenna that will rival the performance of a ten-meter beam. It's our old friend, the Extended Double Zepp but this one has been dressed up to achieve some benefits that are not readily apparent at first glance. Here are its characteristics:

 

1. ~10 dBi gain with a take-off angle of 15 degrees on 10 meters
2. Horizontal Beamwidth of 31 degrees on 10 meters
3. Rotatable using an inexpensive TV rotor for coverage in any direction
4. Low wind resistance - less than three square feet of loading
5. More gain than a 1/2WL dipole on 20m, 17m, 15m, 12m, and 6m
6. Performs close to a resonant dipole on 30m (and is rotatable)
7. Performs well enough on 40m (and is rotatable)
8. Fed with low-loss 300-ohm Ladder-Line or TV Twinlead
9. Requires 1:1 current balun (choke) between the coax and ladder-line
10. Mounted on a 33 foot tall pole or tower (but higher is better)

The Extended Double Zepp (EDZ) has some well-known features - the most well known being the approximately 3-dB gain achieved over a dipole. It accomplishes that gain because it has a narrower horizontal beamwidth than a dipole. For rotatable antennas, the EDZ will outperform the half-wavelength dipole by approximately half an S-unit.

 

EDZ's have a bad reputation for being hard to match. They are hard to match with coax but not with ladder-line. If the EDZ is fed at a current maximum point on the ladder-line, one doesn't even need an antenna tuner.

 

For details on this no-tuner matching technique please reference: "W5DXP's No-Tuner All-HF-Band Antenna" at http://www.qsl.net/w5dxp/notuner.htm.

Let's take a look at some "magic" lengths of dipoles.

 

3/8 wavelength - the minimum effective dipole length recommended by Walter Maxwell in his book, Reflections. Shorter than this length, the resistance decreases rapidly and the capacitive reactance increases rapidly resulting in a very high SWR. A 3/8 wavelength dipole is 48 feet long on 40 meters.

1/2 wavelength - the most popular dipole length - associated with resonance. 1/2 wavelength on 30 meters is 46 feet.

 

Double Zepp - The Zepp is an end-fed 1/2 wavelength antenna that once was trailed from a Zeppelin airship. A Double Zepp is simply a one-wavelength dipole. A Double Zepp on 15 meters is 44 feet long.

 

Extended Double Zepp - A 1.25 wavelength dipole giving the maximum broadside gain possible from a dipole, about 3dB gain over a 1/2 wavelength dipole. An EDZ on 10 meters is 41 feet.

 

If one wants best performance possible on ten meters, then 41 feet is the length of choice. If one wants best performance on 40 meters then 48 feet is the length of choice. This article describes a compromise between 10-meter performance and 40-meter performance. The compromise dipole length is 44 feet.

 

The antenna is assembled from telescoping aluminum tubing and is a total of 44 feet long center-fed with 300-ohm ladder-line. The aluminum tubing is available from Texas Towers or MFJ. Please reference the The ARRL Handbook and/or The ARRL Antenna Book for the rotatable dipole mechanical and electrical design details.

 

The length of the feedline is important if one wants to avoid/satisfy an antenna tuner. The 300-ohm ladder-line used for these simulations (Wireman #562) has ~0.8 velocity factor. Other 300-ohm feedline may vary from this value. 450-ohm ladder-line could just as easily have been used and usually has a velocity factor around 0.9.

 

Let's say we have our 44 foot, 10-meter rotatable EDZ fed with 88 feet of 300-ohm ladder-line mounted and operational on a 33-foot tower. It has essentially the same gain as a two-element Yagi. What else can we do with it? Believe it or not, there’s not much else we can’t do with it all the way from 7 MHz to 54 MHz.

 

The 44-foot dipole is resonant on approximately 10.6 MHz so when it is used on any HF frequency above 10.6 MHz, it exhibits more gain than a dipole. That includes all HF amateur frequencies from 20m through 10m. And 44 feet is long enough to do a good job on 30m. The gain is that of a dipole on 30m and is somewhat lower than a dipole on 40m. Besides, it is rotatable and thus will beat a fixed 40m dipole most of the time.

 

Figure 1

Figure 1 is what EZNEC tells us about the horizontal radiation pattern of the 44-foot dipole on 10m. Mounted at a height of 33 feet, it has a whopping 10.4 dBi gain over average ground, approximately 3 dB better than a half-wavelength dipole.

 

Band - Gain - Take-Off-Angle - Feedpoint Impedance
40m: 5.8 dBi @ 65 degrees   27.5 - j422 ohms
30m: 6.0 dBi @ 42 degrees   72 - j70 ohms
20m: 7.5 dBi @ 29 degrees   194 + j379 ohms
17m: 8.8 dBi @ 23 degrees   931 + j1207 ohms
15m: 8.9 dBi @ 20 degrees   2334 - j1141 ohms
12m: 9.4 dBi @ 17 degrees   419 - j959 ohms
10m: 10.4 dBi @ 15 degrees   126 - j471 ohms
6m: 9.7 dBi @ 8 degrees (Cloverleaf)   110 - j241 ohms

 

The length of the ladder-line is of utmost importance if one wishes to avoid the use of an antenna tuner or make the built-in tuner in a transceiver happy. EZNEC predicts the following necessary lengths of ladder-line for system resonance on the various bands:

 

7.200 MHz - ~.001uf capacitor in parallel at 86 feet = 50 ohms
10.125 MHz - 81 feet or 120 feet = 68 ohms resistive
14.200 MHz - 75 feet or 102 feet = 68 ohms resistive
18.140 MHz - 77 feet or 99 feet = 35 ohms resistive
21.300 MHz - 83 feet or 101 feet = 31 ohms resistive
24.950 MHz - 85 feet or 101 feet = 32 ohms resistive
28.400 MHz - 88 feet or 102 feet = 35 ohms resistive
52.500 MHz - 84 feet or 99 feet = 65 ohms resistive

 

These lengths are only approximate and the actual length in practice must be varied to achieve a near-perfect match without an antenna tuner or with a built-in antenna tuner.

 

However, if one wishes to use a fixed feedline length, it appears that 85 feet or 101 feet would be good fixed feedline lengths to use with this antenna and an antenna tuner. A good 1:1 choke/balun is an absolute necessity. Such a choke is positioned between the coax and the 300-ohm ladder-line.

 

The 300-ohm SWR is reasonable on all bands except 40m where it is about 30:1. 30:1 is not much to worry about as far as losses are concerned but it does mean that not as good a match is possible at the low impedance point (approximately ten ohms) on the transmission line as occurs on other bands. The remedy for the 40m matching problem is a capacitive stub or actual discrete capacitor. It is estimated that a ~.001uf capacitor in parallel with the 300-ohm ladder-line about 86 feet from the 44-foot dipole will result in a near perfect match. Of course, an open stub can also be used. An antenna tuner will also work without the capacitor on 40m but at reduced efficiency.

 

This 40m-6m rotatable dipole was modeled using EZNEC, an antenna-modeling program available from Roy Lewallen, W7EL. Please feel free to visit his web site and download the free demo version of ELNEC from: CLICK ON >>> EZNEC <<< CLICK ON

 

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G5RV 40m Beam Antenna

Adding a 28 ft. piece of vertical wire to one end of a 102 ft. center-fed dipole turns it into a 40m beam with a very wide beamwidth. From my QTH in Texas, I can cover virtually all of the USA.

The feedpoint impedance changes when the vertical wire is added. To achieve a better match, W5DXP reduces the length of the 300 ohm matching section to 17 feet when configured as above. Also, since the antenna is unbalanced by the single vertical wire, a good choke is necessary at the twinlead/coax junction. As a bonus, 17 ft. is also a good length for the matching section when the standard G5RV is used on 30m.

 

 

 

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W5DXP's IC-706 Screwdriver Antenna Tuning Circuit

The circuitry could not be more simple. There is one 10k ohm pullup resistor and one ~12vdc relay. Any small ~12vdc relay will work but I used the Radio Shack 275-248a relay priced at about $3. The operation of the circuit is as follows: If the present SWR needs to be known, the tuner button on the IC-706 is depressed. For a moment, the transmitter is put into 10 watt CW transmit mode and the SWR can be read from the display. Otherwise, the IC-706 need not be touched. When the screwdriver antenna tuning switch is depressed in either direction, the 12vdc relay is enabled thus supplying GND to the TKEY terminal. Anytime the screwdriver motor has 12vdc applied, the IC-706 turns on automatically in 10 watt CW transmit mode. The SWR can be observed anytime the screwdriver motor is powered and the tuning switch can be released when minimum SWR is achieved. W5DXP searched the web for similar designs and decided to roll his own. For about $5 worth of parts, this is the most convenient design available. Save about $25 and roll your own. W5DXP mounted the relay on a small PC board that plugs into the IC-706. The four-pin Molex plug is also available at Radio Shack.

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75m Mobile Shootout Results

 

Following are the summarized results of three 75m mobile antenna shootouts held in California during the 1980's.

0 dB - (Reference) Bugcatcher or Screwdriver with large top hat

-2 dB - Bugcatcher or Screwdriver with no top hat

-5 dB - 8.5' whip with bugcatcher base loading coil

-6 dB - Bugcatcher with Stainless Steel Loading Coil

-8 dB - Hustler High Power system

-9 dB - Outbacker

-12 dB - Hamstick

-12 dB - 11.5' whip with SGC-230 autotuner

-14 dB - CB whip with SGC-230 autotuner (estimated, not measured)

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Current Distribution in Antenna Systems

How can the current "flowing" out of the top of a mobile loading coil be greater than the
current "flowing" into the bottom of the coil? Scroll down to the bottom of this page.

Current Distribution on a Trapped Dipole

The above graphic was copied from Antennas For All Applications, Third Edition, by John D. Kraus and Ronald J. Marhefka. It is from page 824, Figure 23-21(a). The accompanying quote is: "At the frequency for which the dipole is 1/2WL long, the traps introduce some inductance so that the resonant length of the dipole is reduced." So on 75M, the 40M traps introduce inductance and function as loading coils. It is clear from the diagram that there exists a current drop at the location of the inductance on 75m.

Four In-Phase 1/2WL Elements with Phase-Reversing Coils

This graphic was copied from the same reference above. It is also from page 824, Figure 23-21(b). The accompanying quote is: "A coil can also act as a 180 degree phase shifter as in the (above) collinear array ... Here the elements present a high impedance to the coil which may be resonated without an external capacitance due to its distributed capacitance. The coil may also be thought of as a coiled-up 1/2WL element." In a phase-reversing coil, the current is flowing into both ends of the coil at the same time (which doesn't violate Kirchhoff's laws). It just means that a lumped circuit analysis is not valid for a distributed network problem.

What EZNEC Says About Current Distribution Using Inductive Loading Stubs

There's not a lot of difference between inductive loading stubs and loading coils. The EZNEC current distribution is virtually identical to the current distribution illustrated by Kraus in the example above, i.e. there is a current drop across the inductance.

Why the Net Current is not Constant Through a Loading Coil

Some say that the current through a loading coil must be constant according to Kirchhoff's laws. What they are missing is that there are two currents flowing through a loading coil in a standing-wave antenna, a forward current and a reflected current. Speaking on standing-wave antennas, Kraus says (page 187 in the above reference): "A sinusoidal current distribution may be regarded as the standing wave produced by two uniform (unattenuated) traveling waves of equal amplitude moving in opposite directions along the antenna." Balanis, in Antenna Theory, second edition, page 489, agrees: "Standing wave antennas, such as the dipole, can be analyzed as traveling wave antennas with waves propagating in opposite directions (forward and backward) and represented by traveling wave currents, If and Ib, in Figure 10.1(a)."

A transmission line has a distributed inductance and distributed capacitance that causes a delay through it. A real-world loading coil has a distributed inductance and distributed capacitance that causes a delay through it. At a single frequency, this delay can be specified in degrees. Let's use the same assumptions as Kraus. Consider the following loading coil with all four currents having equal magnitudes, i.e. |If1|=|If2|=|Ir1|=|Ir2|.

Let's assume the coil is located at the base of a mobile antenna and that If1 and Ir1 are in phase at zero degrees. The net current on the left side of the coil will equal 2*|If1| or 2*|Ir1| at a phase angle of zero degrees. Also assume that the coil causes a 45 degree delay at the resonant frequency. It follows that If2 will lag If1 by 45 degrees and that Ir2 will lead Ir1 by 45 degrees. The following phasor diagram shows what happens.

It is obvious that the net current on the right side of the coil will equal 1.414*|If2| or 1.414*|Ir2|. The phase angle of the net current hasn't changed through the coil but the magnitude certainly has. In fact, the delay in degrees through the coil can be had from the angle whose cosine is I2net/I1net = 1.414/2 = 0.707.

In reality, the magnitudes of If1, If2, Ir1, and Ir2 are not equal so this exercise is completely accurate only for thin-wire antennas. However, the ballpark conclusions from that assumption are apparently good enough for John D. Kraus.

The moral to this exercise is to avoid using a lumped circuit analysis on a distributed network problem. That includes all problems where forward and reflected waves exist as they do on standing-wave antennas and transmission lines with an SWR greater than 1:1.

The following graphic explains why the current is different at the top and bottom of a loading coil. The magnitudes and phases of the currents at each end of the coil depend simply upon its physical location within the standing wave environment.

What Does EZNEC® Have To Say About Currents Through a Coil?

The following graphic (on the left) is of a base-loaded mobile antenna for ~60m operation. It is 8 feet tall from the feedpoint to the tip and contains a loading coil occupying the space between one foot and two feet from the base. The loading coil was generated using the "Wires - Create - Helix" option available within EZNEC. The antenna is resonant on 5.89 MHz and shows a typical current distribution at that frequency with a greater magnitude of standing wave current at the bottom than at the top of the loading coil. Some might say it is logical for more current to be "flowing" into the bottom of the coil than out the top. Let's question that logic by adding 1/4 wavelength of wire to the top of the antenna on the left to obtain the antenna on the right.

Observe what happens to the currents between the two antennas. In the antenna on the right, is it possible to have 1.29 amps of current "flowing" into the bottom of the coil and 2.068 amps "flowing" out of the top? Of course not! The technical facts are that standing wave current doesn't flow in the normal sense of current flow. The equation for a normal traveling wave current is K*Func(kz ± ωt). The equation for standing wave current is K*Func(kz)*Func(ωt). These two currents are quite different in form and function. EZNEC displays the net standing wave current, not the underlying forward current and reflected current. Both antennas illustrated above are Standing Wave Antennas!

Moral: There is no useful phase information contained in the standing wave current phase measurement. Therefore, the standing wave current phase measurement alone cannot be used to determine the percentage of a wavelength that is occupied by the loading coil. Loading coils occupy tens of degrees of a wavelength but measuring that length is quite a technical challenge. The estimated number of degrees occupied by the coil in the above examples is estimated to be ~60 degrees since the self-resonant frequency of the coil is approximately 9 MHz. A very rough estimate of the electrical length of the coil can be obtained using an arc-cosine function on the standing wave current amplitudes.

Hint: The only phase information in a standing wave is embedded in its amplitude, not in its phase.

Download the EZNEC files for the above antennas:       Download test316.EZ.        Download test316c.EZ.

There are two very interesting web pages that shed light on this issue. Please pay close attention to the limitations of the lumped circuit model when applied to large loading coils:

http:www.ttr.com/corum/index.htm

http://www.ttr.com/TELSIKS2001-MASTER-1.pdf

What Does EZNEC® Have To Say About Traveling-Wave Antennas?

We have seen above that standing-wave current phase cannot be used to measure delays through a coil because that phase is virtually unchanging over the entire length of an antenna, whether a coil is present or not. Therefore, the phase measurements reported using standing wave current phase are meaningless. One way to actually measure the delay through a coil would be to eliminate reflections and measure the forward traveling wave delay through the coil. Such an antenna is called a "traveling wave antenna". Following is what EZNEC tells us about such an antenna.

The coil, which models out to be self-resonant around 13.7 MHz, is being used at 5.89 MHz, just as in the above earlier example. EZNEC says the delay through the coil is 15.68 degrees at 5.89 MHz. This configuration could be easily duplicated and measured in the real-world by anyone so equipped.

Download the EZNEC file for the above antenna:       Download test316y.EZ.

A free demo version of EZNEC is available at CLICK ON >>> EZNEC <<< CLICK ON

Energy Analysis - Transmission Lines

First published in WorldRadio, Oct 2005 - Jan 2006, and reproduced here with permission. (Revision History [10])

An Energy Analysis at an Impedance Discontinuity in an RF Transmission Line

Where Does the Power Go? [1] There continue to be many differing responses to the question within the amateur radio community and, so far, no one else has presented the facts of the physics of electromagnetic energy as understood from the field of optics. Those technical facts from optics have been known and understood for decades and are consistent with the laws of physics and the equations governing the behavior of RF transmission lines. Light and RF waves are both composed of electromagnetic energy.

Most of the following information comes from Optics [2]. In the field of optics, irradiance is the same thing as the power flow vector in an RF transmission line. Irradiance, like a power flow vector, has the dimensions of energy per unit time per unit area. The 1/4 wavelength thin-film deposited on glass to obtain a non-reflective surface performs in a virtually identical way to a 1/4 wavelength series matching section in a transmission line. Single-source RF energy in a transmission line and laser light are both coherent electromagnetic energy waves that obey the laws of superposition, interference, conservation of energy, and conservation of momentum. The power density terms in the irradiance equation have been multiplied by the unit area of the transmission line to obtain the resulting power equation in watts.

My Historical Perspective

My first memories of the answer to "Where does the power go?" are articles published in QST written by Walter Maxwell, W2DU, some quarter of a century ago. Mr. Maxwell later compiled the information into a book titled, Reflections, which quickly became the bible for Amateur Radio applications involving stub matching, transmission lines, and forward and reflected energy flow. Mr. Maxwell coined the terms, "virtual short" and "virtual open", as a shorthand description of what rearward-traveling reflected energy encounters at a match point in a transmission line resulting in 100% re-reflection [10]. He also explained the function of destructive wave interference and constructive wave interference in achieving a match point on a transmission line [8] which is what a large part of this article is about.

Sometime after the publication of Reflections, some people questioned the validity of Mr. Maxwell's concepts. In particular, Dr. Steven Best, VE9SRB, took Mr. Maxwell to task in a series of articles published in QEX [3]. Simply put, Dr. Best disagreed with Mr. Maxwell that reflected power is 100% "re-reflected" in a matched system. Before publication of his Part 3 QEX article, Dr. Best sent up trial balloons for his ideas on the usenet newsgroup, rec.radio.amateur.antenna. My opinion was that Dr. Best's future article contained numerous errors which were pointed out to him. However, the article as published still contained the alleged errors. My determination to resolve the conflicts between the concepts presented by Walter Maxwell and the ones presented by Dr. Best culminated in this present article.

In a nutshell, Walter Maxwell's "virtual short" is a two step process. The reflected wave from the load encounters the impedance discontinuity at the match point. A re-reflection occurs that equals the incident reflected power multiplied by the power reflection coefficient at the match point (the square of the voltage reflection coefficient). This re-reflected energy joins the forward wave traveling toward the load. That first energy re-reflection is not the only energy that joins the forward wave. That fact is what Dr. Best missed in his article. Interference of any kind was never mentioned in Dr. Best's QEX article.

The part of the reflected wave that is not re-reflected is transmitted back through the impedance discontinuity at the match point and attempts to flow toward the source. We know the reflected energy doesn't make it to the source in a matched system, so where does it go?

The answer is mentioned in Reflections II [8]. What Mr. Maxwell is describing is wave cancellation due to total destructive interference between two reflected waves. The first wave is the part of the source forward wave that is initially reflected back toward the source from the match point. The second wave is the part of the reflected wave from the load that is transmitted through the match point toward the source. These waves are equal in magnitude and opposite in phase so, as Mr. Maxwell asserts in Reflections II, the two wavefronts cancel to zero at the match point thus eliminating reflections between the match point and the source. The canceling of these two waves to zero is the second step in Mr. Maxwell's virtual short process of 100% "re-reflection" (actually reflection plus redistribution [10]).

Voltages can cancel and currents can cancel but energy cannot cancel. What happens to the energy that existed in the waves before they were cancelled? Since we know that all the energy in a matched system winds up flowing toward the load, the answer is a no-brainer. There are only two directions in a transmission line. If energy that was previously flowing toward the source isn't flowing toward the source anymore, it must necessarily be flowing toward the load. The conclusion is inescapable. Not only is 100% of the reflected energy redistributed back toward the load at the match point, but wave cancellation is the cause of part of that redistribution. This is a well understood phenomenon in the field of optics [9] but not well understood in the field of RF engineering.[10]

An RF engineer will tell us that there are three things that can cause 100% reflection. Those are a short-circuit, an open-circuit, or a purely reactive impedance. But there is another phenomenon that can cause the reflected energy to reverse direction and flow toward the load - wave cancellation.

In general:

The destructive interference energy resulting from wave cancellation at an impedance discontinuity becomes an equal magnitude of constructive interference in the opposite direction. Since there are only two directions in a transmission line, wave cancellation redistributes the energy in the opposite direction. [9] This redistributed energy joins the forward wave just as the re-reflected energy does.

The General Case Qualitative Analysis

The following discussion of a generalized impedance discontinuity in an RF transmission line, operating under steady-state conditions, is not intended to replace a conventional quantitative voltage analysis. It is intended instead as a conceptual qualitative energy analysis that extends the voltage analysis and answers that original question: Where Does the Power Go?

The following diagram is of a generalized impedance discontinuity in a typical RF transmission line. P_for1 is the forward power and P_ref1 is the reflected power on the source side section of transmission line having a characteristic impedance of ZØ₁. Likewise, P_for2 is the forward power and P_ref2 is the reflected power on the load side section of transmission line having a characteristic impedance of ZØ₂.

Every power component has an associated voltage component (and an associated current component). For instance, P_for1 is associated with the RMS values of V_for1 (and I_for1) in the following way:

We will continue that wave reflection model convention for this paper. Thus for any wave(x) or component wave(x) in a transmission line, there exists Vx and Ix such that:

In the following figure, every power or component power has an associated voltage (and current) obeying the above rules. The Greek letter 'θ' will be used in the following equations to represent the relative phase angle between the two voltages, V₁ and V₂. V₁ is associated with P₁ and V₂ is associated with P₂. For the purpose of this paper, we will consider powers to be scalar values without associated phase angles. When power expressions become a function of a phase angle, that phase angle will be taken as the relative phase angle between the corresponding voltage waves.

Figure 2 shows an imaginary transverse plane (viewed on edge) dividing the transmission line into two segments, each with its own characteristic impedance. On each side of that plane, there exists a relationship between the energy flows and the transmission line that are governed by the characteristic impedance on that particular side of the impedance discontinuity. However, the imaginary plane has no width, so in Figure 2, all energy flows coexist at that same transverse plane, i.e. all superposition happens at the plane, not some distance away from the plane.

The following equations govern the distribution of energy at the impedance discontinuity, which, of course, is constrained by the conservation of energy principle. The Greek letter, 'ρ', will be used to designate the voltage reflection coefficient.

Note that 100 times the power reflection coefficient is the percentage of power that is reflected. Likewise, 100 times the power transmission coefficient is the percentage of power that is transmitted.

While the voltage reflection coefficients are of opposite signs on the two sides of the junction, the power reflection coefficient is always positive on either side of the junction. The same concept applies to the power transmission coefficient. Since the impedance discontinuity is an imaginary plane and contributes no losses, all the power that is not reflected at the impedance discontinuity plane is transmitted through the impedance discontinuity plane.

So far, nothing has been presented that differs from the wealth of information available on reflections in a transmission line. Anyone who is confused at this point should review a good reference on the reflection model for RF transmission lines. The remainder of this paper will answer the question:

"Where Does the Power Go?"

The following equations are paraphrased from similar relationships published in Optics. The symbol 'I' representing irradiance in a light wave has been replaced by the symbol 'P' representing the power flow vector in an RF wave (or RF wave component) in a transmission line. Note that these equations apply to coherent waves. V₁ and V₂ are superposed at the load side of the impedance discontinuity, using phasor math, to obtain V_for2 where 'θ' is the relative phase angle between V₁ and V₂. Note that cos(θ) = cos(-θ), so for an energy analysis, it doesn’t matter if V₂ is leading or lagging V₁.

This equation is identical to Dr. Best's (Eq 12) [3], which was presented without any reference to interference in his QEX articles. From Optics, we can identify the third term in the equation as the interference term. [4] The interference term accounts for the fact that (V₁+V₂)² is not equal to (V₁² + V₂²), which, again, follows directly from the theory of superposition. These subjects are discussed in detail in Optics, chapters 7 and 9.

The following is the key equation that Dr. Best neglected to include in his QEX article. It is the other half of the energy equation and brings all of the energy into balance as required by the conservation of energy principle. It can be shown that the relative phase angle between V₃ and V₄ is (180 degrees minus θ) and since cos(θ) = -cos(180-θ), the following equation applies when V₃ and V₄ are superposed at the source side at the impedance discontinuity:

Note the negative sign of the last term! Steady-state energy is supplied into the impedance discontinuity by two waves, one associated with P_for1 (from the source) and the other associated with P_ref2 (reflected from the load). Steady-state energy leaves the impedance discontinuity as two waves, one associated with P_for2 (toward the load) and the other associated with P_ref1 (toward the source). Anyone familiar with an S-parameter analysis will recognize those four terms. In fact, Eq 1 and Eq 2 can be independently developed from the S-parameter equations.[5] Since the power supplied to the impedance discontinuity is (P_for1 + P_ref2) and none of it is stored, (other than the energy being handed back and forth continuously between the electromagnetic and electrostatic fields), the power entering the impedance discontinuity must equal the power exiting the impedance discontinuity such that:

In a matched system, P_ref1 equals zero but in a mismatched system, P_ref1 can be any value between a small value and P_for1. The conservation of energy principle requires that the total destructive interference energy equal the total constructive interference energy such that:

This is the equation that explains everything about power at an impedance discontinuity in a transmission line. Note that there is a dotted line for interference energy in Figure 2. The destructive interference energy resulting from wave cancellation at the impedance discontinuity becomes an equal magnitude of constructive interference in the opposite direction. [6] This can happen in either direction in a mismatched system but will happen in only one direction in a matched system. The following cases are all the possibilities referenced to the relative phase angle, 'θ', between V₁ and V₂:

If ( 0 ≤ θ < 90 ) then there exists constructive interference between V₁ and V₂, i.e. cos(θ) is a positive value. Therefore there exists an equal magnitude of destructive interference between V₃ and V₄ where cos(180-θ) is a negative value. A positive sign on the interference term indicates constructive interference. A negative sign on the interference term indicates destructive interference.

If θ = 90 deg, then cos(θ) = 0, and there is no destructive/constructive interference between V₁ and V₂. There is also no destructive/constructive interference between V₃ and V₄. Any potential destructive/constructive interference between any two voltages is eliminated because θ = 90 deg, i.e. the voltages are superposed orthogonal to each other (almost as if they were not coherent).

If (90 < θ ≤ 180) then there exists destructive interference between V₁ and V₂, i.e. cos(θ) is a negative value. Therefore there exists an equal magnitude of constructive interference between V₃ and V₄ where cos(180-θ) is a positive value.

One note of importance is that, in the case of a mismatched impedance discontinuity, reflected power is not 100% re-reflected and redistributed. Dr. Best was right about that.

Now we are in a position to discover something that falls out from the conservation of energy principle. A simple mathematical manipulation of equation 3 above will show that:

Therefore the power resulting from constructive interference is:

This equation tells us where the extra energy comes from that allows (V₁+V₂) to superpose to P_for2 > |P₁|+|P₂| in a typical matched RF system. The destructive interference event (involving the superposition of V₃ and V₄) feeds energy into the constructive interference event (involving superposition of V₁ and V₂).[6]

Why not turn this qualitative analysis into a quantitative analysis? If we know the forward power and reflected power on each side of the impedance discontinuity and the reflection coefficient at the impedance discontinuity, we can certainly do a quantitative analysis. The only problem is that there are two solutions. Without additional information, one cannot tell whether V₂ is leading or lagging V₁ and therefore there exists two possible solutions. In order to confine the results to one unique solution, one would need to know the number of wavelengths between the discontinuity and the load and the reflection coefficient at the load. But as we shall discover in Part III: For a ZØ-matched system, the two-solution problem disappears because the phase angle between V₁ and V₂ is always zero degrees (ZØ₂ > ZØ₁) or 180 degrees (ZØ₂ < ZØ₁).

Note: It cannot be over-emphasized that "wave cancellation" does not imply energy cancellation. "Wave cancellation" refers to the cancellation of two coherent EM voltage/current waves traveling the same path in the same direction. The energy components in the cancelled waves cannot be destroyed so the energy must seek another path, i.e. it is redistributed in the opposite direction in a transmission line.

The Special Case ZØ-Matched Analysis

Someone might ask, why bother with a special case? As it happens, this special case applies to all matched systems for which P_ref1 = 0, and is the most common case within amateur radio. One might say the matched case is not all that 'special'. Here's a diagram for the ZØ-matched system.

Note that interference energy flows in only one direction for a ZØ-matched system and that direction is always toward the load. When the system is matched, P_ref1 equals zero, i.e. there are no reflections flowing toward the source and a quantitative analysis becomes possible. For a matched system, the phase angle between V₁ and V₂ is zero degrees and the phase angle between V₃ and V₄ is 180 degrees. In addition, the magnitudes of V₃ and V₄ are equal and therefore the magnitudes of P₃ and P₄ are equal. V₁ is arbitrarily assigned a reference phase angle of zero degrees. V₂ will therefore, possess a phase angle of zero degrees. For a positive ρ, V₄ will be at zero degrees, and V₃ will be at 180 degrees. If ρ is negative, V₄ will be at 180 degrees and V₃ will be at zero degrees. This is in agreement with the rules governing wave reflection.

When two coherent waves of equal magnitude and opposite phase encounter each other while moving in the same direction in a transmission line, complete cancellation of the two waves occurs. The total voltage goes to zero and the total current goes to zero in the direction of the canceled waves. V_ref1 (and I_ref1) go to zero at the match point. That's entirely logical since the reflected power flow vector toward the source equals zero in a matched system.

In a ZØ-matched system with reflections, total destructive interference occurs at the source side of the ZØ-match point and eliminates reflections toward the source. Following the principle of conservation of energy, the destructive interference energy previously associated with the two cancelled reflected waves becomes total constructive interference energy flowing toward the load as part of P_ref2.

The energy equations governing the behavior of a ZØ-matched system are simplified because the phase angle between V₁ and V₂ is zero degrees for total constructive interference. The phase angle between V₃ and V₄ is 180 degrees for total destructive interference.

The constructive interference event of Eq 5 is called total constructive interference.[7]

The destructive interference event of Eq 6 is called total destructive interference.[7]

For a ZØ-matched system, in which all reflections are cancelled toward the source, it is necessary for P₃ to equal P₄. From that fact and knowing that (P₁)(P₂) = (P₃)(P₄), it can be shown that, in a matched system:

Which is what a lot of people have been saying for a lot of years.[8] Note: The absolute value marks are included to indicate that these powers are scalar values, not power flow vectors.

Conclusion: There are two steps leading to the total redistribution of reflected energy back toward the load in a ZØ-matched system.

1. P₂ = P_ref2 ( ρ² ) is the first re-reflection event and occurs when the reflected wave associated with P_ref2 encounters the impedance discontinuity.

2. P₃ and P₄ are associated with two waves involved in total destructive interference. Since the related voltages, V₃ and V₄, are equal in magnitude and 180 degrees out of phase, the energy components in P₃ and P₄ cease to exist on the source side of the impedance discontinuity, and instead are redistributed toward the load as total constructive interference. P₃ and P₄ are power flow vectors associated with two reflected waves that cancel, so according to the principle of conservation of energy, their combined energy (and momentum) must change direction. Since P₃ and P₄ are reflected energy components that end up flowing toward the load, the interference event can be considered to be a redistribution [9] of reflected energy. Combining steps 1 and 2 above, it is apparent that 100% of the reflected energy is re-reflected and redistributed in a matched system. P₂, P₃, and P₄ are all reflected wave components.

Note that the author previously used the word "reflection" for both actions involving a single wave and the interaction between two waves. Now the word "reflected" is being used only for single waves and the word "redistributed" is being used for the two wave interference scenario.

Step 2 above, is a somewhat new concept in the field of RF engineering although it has existed for decades in the field of optics.[9] We amateur radio operators can add an item to the list of things that can cause a redistribution of the energy in incident waves: 1. Short-circuit, 2. Open-circuit, 3. Pure reactance, 4. Wave cancellation.[10] Wave cancellation cannot occur at a single load fed by a single source through a single transmission line but it can occur in a transmission line when waves are incident upon an impedance discontinuity from both directions. Although harder to understand and prove, wave cancellation (and therefore 100% redistribution of canceled wave energy) can also occur at (or inside) a source when RF energy is coming from both directions. For instance, reflected wave cancellation will occur in a tube-type final amp when the pi-network is tuned for system resonance. In that case, the reflected wave cancellation point would be the Zg-match point inside the transmitter.

Note: The steady-state existence of the P₃ wave can be inferred from P₃ = P_ref2(1- ρ²) where P_ref2 and (1-ρ²) do exist. Given that the P₃ wave exists, then the existence of the P₄ wave is necessary for wave cancellation. Unfortunately, superposition happens faster than humans can observe, even with their fastest instruments.

And that is where the power goes; (actually, it is the energy that does the "going").

A Simple Example

Consider the following lossless system with a 1:1 choke at point 'x':

100W XMTR---50 ohm coax---x---300 ohm twinlead---load
		Pfor1-->	Pfor2-->
		<--Pref1	<--Pref2

The source is supplying 100 watts. The SWR meter reads 1:1 on the coax. With the information given, can we calculate the forward power, reflected power, and SWR on the twinlead? How about voltages and currents on the twinlead?

The power reflection coefficient is [(300 - 50)/(300 + 50)]² = 0.51 and the power transmission coefficient is (1 - 0.51) = 0.49. For the system to be matched, these coefficients must also exist at the load. So the forward power on the twinlead must be 100W/0.49 = 204.1 watts. That makes the reflected power on the twinlead equal to (204.1 - 100) = 104.1 watts. From these two power values, we can calculate V_for2 = 247.4V, I_for2 = 0.825A, V_ref2 = 176.7V, I_ref2 = 0.589A , SWR(300) = 6:1

The SWR can be calculated in any number of ways. VSWR(300) = (V_for2 + V_ref2)/(V_for2 - V_ref2)

If we know the physical length and velocity factor of the 300 ohm twinlead, we can actually calculate the feedpoint impedance of the load (antenna).

The author has endeavored to satisfy the purists in this series of articles. The term "power flow" has been avoided in favor of "energy flow". Power is a measure of that energy flow per unit time through a plane. Likewise, the EM fields in the waves do the interfering. Powers, treated as scalars, are incapable of interference. Any sign associated with a power in this paper is the sign of the cosine of the phase angle between two voltage phasors. A plus sign indicates constructive interference (or energy flow toward the load) and a minus sign indicates destructive interference (or energy flow toward the source).

I would like to thank Mr. Robert E. Lay, W9DMK, for his substantial contributions to this article.

References

[1] Bloom, Jon, "Where Does the Power Go?", QEX, Dec. 1994

[2] Hecht, Eugene, Optics, Fourth Edition, (c)Aug. 2001, Addison-Wesley, ISBN 0805385665

[3] Best, Steven R., "Wave Mechanics of Transmission Lines, Part 3", QEX, Nov/Dec 2001

[4] "Interference term", Optics, Eugene Hecht, Fourth Edition

Section 7.1 The Addition of Waves of the Same Frequency: It follows ... that the resultant flux density is not simply the sum of the component flux densities; there is an additional contribution 2E₀₁E₀₂cos(α₁-α₂), known as the interference term.

Section 9.1 General Considerations: The 'interference term' becomes I₁₂ = 2*SQRT[(I₁)(I₂)]cos(σ)
(where 'SQRT' replaces the square root sign.)

[5] S-Parameter Techniques, Hewlett Packard Application Note 95-1, available on the web. The S-Parameter normalized voltage equations are:

b1 = (s11)(a1) + (s12)(a2) and b2 = (s21)(a1) + (s22)(a2)

The squares of all those terms are related to power as explained in the application note. It is left as an exercise for the reader to square both sides of both equations above and observe that the resulting equations contain the interference term that agrees with Eq 1 and Eq 2 in the body of this paper.

[6] Optics, Eugene Hecht, Fourth Edition

Section 3.3 Energy and Momentum, "One of the most significant properties of the electromagnetic wave is that it transports energy and momentum." [Note from W5DXP: Energy and momentum must be conserved. The direction of the energy and momentum associated with reflected waves must be reversed for a match to occur.]

Section 4.11 Photons, Waves and Probability, "The principle of conservation of energy makes it clear that if there is constructive interference at one point, the 'extra' energy at that location must have come from somewhere else. There must therefore be destructive interference somewhere else. "If two or more electromagnetic waves arrive at point P out-of-phase and cancel, 'What does that mean as far as their energy is concerned?' Energy can be distributed, but it doesn't cancel out."

Section 7.1 The Addition of Waves of the Same Frequency, "The superposition of coherent waves generally has the effect of altering the spatial distribution of the energy but not the total amount (of energy) present."

[7] Optics, Eugene Hecht, Fourth Edition

Section 9.1 General Considerations, "A maximum irradiance (power) is obtained when cos(σ) = 1. ... In this case of total constructive interference, the phase difference between the two waves is an integer multiple of 2π, and the disturbances are in-phase. ... A minimum irradiance (power) results when the waves are 180 degrees out-of-phase, ... cos(σ) = -1, ... and is referred to as total destructive interference."

[8] Maxwell, Walter, Reflections II, (c) 2001 Worldradio Books, ISBN 0-9705206-0-3 page 4-3, "The destructive wave interference between these two complementary waves ... causes a complete cancellation of energy flow in the direction toward the generator. Conversely, the constructive wave interference produces an energy maximum in the direction toward the load, ..." page 23-9, "Consequently, all corresponding voltage and current phasors are 180 degrees out of phase at the matching point. ... With equal magnitudes and opposite phase at the same point (point A, the matching point), the sum of the two (reflected) waves is zero."

[9] Quotes from two web pages from the field of optical engineering:

www.mellesgriot.com/products/optics/oc_2_1.htm

"Clearly, if the wavelength of the incident light and the thickness of the film are such that a phase difference exists between reflections of p, then reflected wavefronts interfere destructively, and overall reflected intensity is a minimum. If the two reflections are of equal amplitude, then this amplitude (and hence intensity) minimum will be zero." (Referring to 1/4 wavelength thin films.)

"In the absence of absorption or scatter, the principle of conservation of energy indicates all 'lost' reflected intensity will appear as enhanced intensity in the transmitted beam. The sum of the reflected and transmitted beam intensities is always equal to the incident intensity. This important fact has been confirmed experimentally."

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/waveinteractions/index.html

"... when two waves of equal amplitude and wavelength that are 180-degrees ... out of phase with each other meet, they are not actually annihilated, ... All of the photon energy present in these waves must somehow be recovered or redistributed in a new direction, according to the law of energy conservation ... Instead, upon meeting, the photons are redistributed to regions that permit constructive interference, so the effect should be considered as a redistribution of light waves and photon energy rather than the spontaneous construction or destruction of light."

Note from W5DXP: In an RF transmission line, since there are only two possible directions, the only "regions that permit constructive interference" at an impedance discontinuity is the opposite direction from the direction of destructive interference.

[10] Revision 1.1, Feb. 20, 2008 - In the original version, the redistribution of energy due to wave cancellation was called a "reflection", a common practice in amateur radio circles. W5DXP has changed that description in favor of a "redistribution" as described by the FSU web page. The word "reflection" is reserved for describing the event when a single wave encounters an impedance discontinuity. This is accordance with The IEEE Dictionary definition of "reflected wave". The word "redistribution" of energy is adopted for describing what happens to the energy when two or more waves interact. In like manner, since interference can occur with or without permanent wave interaction, interference alone is necessary but not sufficient to correspond to the permanent redistribution of energy.

Energy Analysis - Simple Voltage Source

An Energy Analysis of a Simple Ideal Source, Part 1: Zero Average Interference

Introduction

The debate concerning energy events inside a source when reflected energy is incident upon that source is decades old and seemingly without resolution. [1][2] Part 1 of this series of articles will take a close look at what occurs inside a simple ideal source under special case conditions of zero average interference. Part 2 will address destructive interference at the source resistor and Part 3 will cover constructive interference. Perhaps the simple concepts presented in this elementary three-part series will help to "shed some light" on the subject. [3]

Please note that any power referred to in this paper is an AVERAGE POWER. Instantaneous power is beyond the scope of this article, irrelevant to the following discussion, and "of limited utility" according to Eugene Hecht. [4]

Zero Average Interference Example

Please note that it takes only one example to prove an explicit assertion to be false, e.g. "Reflected energy incident upon a source is always re-reflected back toward the load." The following example will prove that explicit assertion to be false.

Everything presented in this first article will be based on the following example chosen as a special case from the infinite number of possibilities. This example is so simple that almost everyone should be able to understand "where the power goes". More complex examples will be presented in Parts 2 and 3.

Some characteristics of this example are very interesting:

1. The magnitude of the transient forward power is equal to the magnitude of the steady-state forward power, i.e. the forward power, Pfor, is always equal to 50 watts no matter what the value of the load resistor.

2. After the forward wave is incident upon the load resistor, the reflected power, Pref, from the load is constant as long as the load resistance is unchanged. The voltage reflection coefficient at the load is: rho = (R_L-50)/(R_L+50) so Pref = 50w(rho²) watts, where rho² is the power reflection coefficient.

3. Steady-state is achieved at the time that the reflected energy wave reaches the source because R_S = Z0 = 50 ohms. Thus the reflected wave is not re-reflected at the source.

4. Since the transmission line is 1/8 wavelength (45 degrees) long and the load is purely resistive, the reflected wave incident upon the source resistor will be 2(45) = 90 degrees out of phase with the forward wave at the source resistor. This is the necessary and sufficient condition to produce zero average interference at the source resistor. Although instantaneous interference exists within a cycle, it averages out to zero over each complete cycle.

This last characteristic is interesting because it ensures that interference [4] will not exist anywhere in the example. Therefore, the subject of interference between two RF waves will be postponed until Parts 2 and 3. The complete absence of interference is what makes this simple example so interesting.

Where does the reflected energy go?

Remembering that this special case was chosen specifically to allow us to track the reflected energy through the system, let’s see where the reflected energy goes.

When R_L = 50 ohms, the system is matched and reflected power equals zero. In the matched case, an equal amount of power is dissipated in R_S and R_L, i.e. 50 watts in each resistor and maximum power transfer takes place. Recall that the forward power is 50 watts no matter what the value of the load resistor. This allows us to assert that 50 watts of the dissipation in the source resistor is due to the forward wave and is constant for the above example.

In this special case, as the reflected power is increased above zero by varying the load resistance away from the matched-case 50 ohms, the power dissipated in the source resistor, R_S, is exactly equal to the matched-case 50 watts plus the magnitude of the reflected power. Thus (for this special case) the following equation applies - the power dissipation in the source resistor, R_S, is:

P_Rs = 50w + Pref       Eq. 1

In the following table, random values of R_L are chosen to illustrate the validity of Eq. 1.

Please note that although we have not calculated a single voltage or current, any valid voltage and/or current analysis of the above example will yield identical results in agreement with this energy analysis. Eq. 1, above, will be valid for any voltage or current analysis performed on this particular special-case example.

For this special case, it is obvious that the reflected energy from the load is flowing through the source resistor, R_S, and is being dissipated there. But remember, we chose a special case (resistive R_L and 1/8 wavelength feedline) in order to make that statement true and it is usually not true in the general case. What we have proved false is the assertion that: "Reflected energy is never dissipated in the source." What we have NOT implied or proved is the equally false statement that: "Reflected energy is always dissipated in the source." In the next two parts of this article, we will cover the general cases for ideal class-A voltage (and current) sources. What we will discover is that: "The part of the reflected energy that is not dissipated in the source resistor is redistributed back toward the load as part of the forward wave." The part of the reflected energy that is redistributed back toward the load varies from 0% to 100%, depending upon system configuration.

Notes and References

[1] Bruene, Warren B., W5OLY, "On Measuring Rs", QEX, May 2002

[2] Maxwell, Walter, W2DU, http://www.w2du.com/BrueneRebuttal.pdf

[3] Since the "light" being "shed" comes from the field of optical physics, this seems to be an appropriate play on words.

[4] Hecht, Eugene, Optics, Fourth Edition, (c)Aug. 2001, Addison-Wesley, ISBN 0805385665, Chapter 9: "Interference". This reference lays the groundwork for Parts 2 and 3 of this article.

FUTURE ARTICLES

Dual-Z0 Shortened Stubs

Introduction

Shorted and open-circuit stubs, located along a transmission line, have long been a method of achieving a match or notch through the impedance transformations, the inductive reactance, or the capacitive reactance that stubs provide. Such stub techniques are discussed in The ARRL Handbook and The ARRL Antenna Book. (A detailed discussion of stub applications and stub matching techniques is beyond the scope of this article.) What is presented here is a method of shortening those matching stubs by as much as 66% by taking advantage of the phase shift between two pieces of transmission line with differing Z0s (characteristic impedances).

The topic that seems to have been ignored in amateur radio publications is the physical length advantage to be gained by using dual-Z0 shortened stubs with two differring Z0 characteristic impedances. Since the pure reactance circle on a Smith Chart is associated with a phase angle and the normalized value of reactance is a tangent function of that phase angle, we can use the ARCTAN function to determine the phase shift in degrees at an impedance discontinuity in a stub.

(This lossless phase shift at an impedance discontinuity point in a stub seems to be a free lunch granted to us mere mortals by the transmission line gods.)

Assuming a reactance of 0-j300 ohms and normalizing to Z01 = 600 ohms and Z02 = 50 ohms:

ARCTAN(300/600) = 26.6 deg = 0.0739 wavelength

ARCTAN(300/50) = 80.5 deg = 0.2236 wavelength

If the impedance is 0+j300 ohms at an impedance discontinuity between Z01 = 600 ohm line and Z02 = 50 ohm line, a steady-state phase shift of 80.5 - 26.6 = 53.9 deg will occur at that point when the impedance is purely reactive, i.e. an infinite SWR exists. The maximum possible shortening of a stub occurs at the highest Z0H/Z0L ratio and at the shorter physical lengths. (Z0H is the high-Z0 section and Z0L is the low-Z0 section.) For instance, when Z0H/Z0L = 12/1, one can obtain a 30 degree electrical length from an ideal stub that is only 5 degrees long physically, i.e. 1/6 of the physical length of a single-Z0 stub.

For the purpose of this theoretical paper, all transmission lines and stubs will be considered to be ideal and lossless. The velocity factor, VF, of the feedlines is assumed to be 1.0 but will obviously require consideration in an actual real-world implementation when the VF is not 1.0. This paper applies only to stubs used at the single frequency of interest. Unlike single-Z0 stubs, a 1/4 wavelength dual-Z0 shortened stub does not act like a 1/2 wavelength stub at double the frequency of interest. It can be seen from the graph in Fig. 4 that the electrical length to physical length function is not a straight line when two different Z0s are used.

It is unlikely that the information contained in this paper is original but the author has been unable to locate any references in amateur radio publications that explain and present information about dual-Z0 shortened stubs. However, what transpires at an impedance discontinuity in a transmission line has been covered in a previous article. [1] This paper is an extension of the principles covered in that earlier article.

The electrical length of a stub varies with frequency. Rather than using a particular frequency for the examples, all examples have been normalized to degrees. They could just have easily been normalized to radians or wavelength. Here are the relationships between frequency, degrees, radians, velocity factor, and wavelength.

Free space wavelength in feet = 984/frequency in MHz

Transmission line wavelength in feet = VF*984/frequency in MHz

Where VF is the velocity factor of the transmission line, e.g. 0.66 for RG-213.

One wavelength = 360 degrees = 2 pi radians

Most amateur radio operators have heard of stubs and stub matching techniques. Unfortunately, useful lengths of stubs may not be feasible on the longer wavelength HF bands, e.g. where 1/4 wavelength is ~65 feet on 3.8 MHz. This paper will present information about a method of shortening stubs by using two lengths of transmission lines of differing characteristic impedances, Z0H and Z0L. The above ~65 feet on 3.8 MHz can be shrunk to ~25 feet using Z0H = 600 ohms and Z0L = 72 ohms transmission line stock (assuming VF = 0.9).

Dual-Z0 Stubs

The relative locations of the two sections of feedline are important. The following figure indicates the correct locations for the high-Z0 line and the low-Z0 line for open stubs and shorted stubs. (Stubs with other than open or shorted terminations are beyond the scope of this paper.)

For open stubs, the low-Z0 section must be placed at the open (bottom) unconnected end of the stub. The high-Z0 section becomes the top (connected) end.

For shorted stubs, the high-Z0 section must be placed at the shorted (bottom) unconnected end of the stub. The low-Z0 section becomes the top (connected) end.

Let’s assume that we have two types of transmission lines available with differing characteristic impedances, Z0H ohms and Z0L ohms (where Z0H > Z0L) and we want to make a physically shortened stub that is electrically longer than the two physical lengths combined.

It appears that the most efficient result occurs when the two section lengths are equal in degrees. If the VFs of both sections are equal, the two sections will also be equal in physical length.

It turns out that the actual values of the characteristic impedances are not important. It is the ratio of Z0H/Z0L that is important. The frequency is also not important as everything is expressed in degrees which can be converted to wavelength by dividing by 360. The equation that governs such physically shortened stubs is:

Where: TELD is the Total Electrical Length in Degrees of the Dual-Z0 stub. TPLD is the Total Physical Length in Degrees of the stub and assumes that the two different Z0 sections are equal in number of physical degrees. Z0H/Z0L is the ratio of Z0H to Z0L.

As an example, let's assume that we create an open stub made from Z0H = 600 ohm feedline and Z0L = 72 ohm feedline such that Z0H/Z0L = 8.3333. Assume that each section is physically 19.11 degrees long making the total stub 38.22 degrees long physically. What will be the electrical length in degrees?

Need a 1/4 wavelength open stub that is 43% of the length required by a normal 1/4 wavelength open stub? We have created that electrically long 1/4 wavelength stub using only 19.11 + 19.11 = 38.22 physical degrees of transmission line in a dual-Z0 configuration. The secret is that the impedance discontinuity provides a 51.87 deg phase shift in the middle of the stub as illustrated in the following graphic.

For those who can read a Smith Chart, here is a graphical representation of this stub.

If we take the identical stub above and swap ends such that the Z0H becomes the shorted bottom end of a shorted stub and the Z0L end becomes the top connected end, the stub becomes an electrical 1/4 wavelength shorted stub. (If the ends are not swapped when going from open-stub to shorted-stub, the desired shortening effect of the dual-Z0 stub will not be realized.) See Fig. 1 above.

The stub can also have other electrical lengths besides 90 degrees. What electrical length would we get from a shorted-stub with the following characteristics?

TPLD = 25 degrees (Total physical degrees assuming equal length sections)

Z0H/Z0L = 6 (Z0H = 300 and Z0L = 50)

TELD = 25/2 + ARCTAN[(6.0)TAN(25/2)] = 65.56 degrees = 0.182 wavelength

The physical length is 25 degrees. The electrical length is 65.56 degrees. The dual-Z0 stub is 38% of the length of a single-Z0 stub because of the 40.56 degree phase shift at the impedance discontinuity.

1/4 Wavelength Stubs

For the special case of 1/4 wavelength stubs, the following equation applies:

If Z0H = 600 ohms and Z0L = 72 ohms, each of the two sections is 19.11 degrees which is the example given above in Fig. 2. Use the above equation for creating shortened 1/4 wavelength stubs.

One interesting combination of characteristic impedances would be Z0 = 600 ohm open-wire line combined with a side-by-side parallel run of 50 ohm coax which has a Z0 of 100 ohms. With the Z0H/Z0L ratio of 600/100, a 1/4 wavelength dual-Z0 stub will be almost exactly one half the length of the single Z0 stub.

Graphical Solution

The following graph allows one to choose the total electrical length required for a stub using a particular Z0H/Z0L ratio to determine the total physical length in degrees for the shortened stub using equal lengths (degrees) of sections of Z0H and Z0L feedlines.

It should be obvious that the above curves converge where the total electrical length and the total physical length are both equal to 180 degrees, i.e. 1/2 wavelength. At that convergence point, The Z0s of the two sections doesn’t matter at all as long as they are equal.

1/2 Wavelength Transmission Lines

The above technique can also be used to shorten transmission lines. For instance, assume one has a 75m dipole with a feedpoint impedance of 50+j0 ohms at resonance. If we want to match that 50 ohms by feeding the dipole with 1/2 wavelength of Z0 = 600 ohm feedline, we will need 116.5 feet on 3.8 MHz assuming a velocity factor of 0.9. Can we shorten the feedline? Of course we can by using a piece of Z0 = 100 ohm feedline in the middle of a Z0 = 600 ohm feedline section. (The Z0 = 100 ohm feedline can be two side-by-side runs of 50 ohm coax.)

Using the above graph, we find that the physical length of a shortened 1/4 wavelength stub when the Z0H/Z0L ratio is 6:1 is approximately 45 degrees. So if we make each end of the feedline equal to ~22.5 degrees of Z0 = 600 ohm feedline and make the middle section equal to ~45 degrees of Z0 = 100 ohm feedline, then we will have created a 1/2 wavelength (180 deg) shortened feedline that is physically approximately 1/4 wavelength (90 deg) long.

If the Z0 = 600 ohm feedline has a VF of 0.9, the two lengths will be close to 14.6 feet each. If the Z0 = 100 ohm feedline has a VF of 0.66, the center length will be close to 21.4 feet. Thus we have created an electrically 180 degree long feedline using only about 36 feet of feedline. Our shortened electrical 1/2 wavelength feedline is only about 31% of the length of the original 1/2 wavelength of Z0 = 600 ohm feedline.

Note that the Z0H = 600 ohm line is connected to the low-Z, 50 ohm antenna. This is in accordance with the requirements in Fig. 1, above.

References

[1] "An Energy Analysis at an Impedance Discontinuity in an RF Transmission Line", by Cecil Moore, W5DXP, Worldradio, Oct 2005 - Jan 2006. Also available at: Energy Analysis

Rev. 1.1, April 28, 2009

Current Through a 75m Bugcatcher Loading Coil

Many experiments and measurements have been made on loading coils using net standing wave current. A lack of understanding of the nature of standing wave current has resulted in some strange and magical assertions about current through a loading coil.[1] The equation for standing wave current is of the form:

For any point location 'x', it can be seen that the standing wave current is not "flowing" in the ordinary sense of the word but rather, is just oscillating in place at that fixed point. EZNEC confirms that the phase of standing wave current is essentially constant all up and down a typical HF mobile antenna and therefore cannot be used to make a valid measurement of the phase shift (delay) through a loading coil (or even through a wire.) The validity of that statement is obvious if one understands the implications of the standing wave current equation above. In fact, we can just as easily write the standing wave current equation as:

We can reverse the direction of rotation of the standing wave current phasor and still have the same value of current. Standing wave current really doesn't have a direction of flow. In order to make valid measurements of the phase shift (delay) through a loading coil, we need to use a traveling wave current with an equation of the form:

If we can cause a traveling wave current to flow through a loading coil and eliminate (or minimize) reflections, we can make a valid measurement of the phase shift through that loading coil. The key is the elimination of the reflected current.

 

The method for eliminating reflected current is borrowed from transmission lines. Somewhat like a transmission line, a large helical coil has a characteristic impedance in the range of a few thousand ohms. If we install a resistor from the top of the coil to ground and vary the resistance, we can find the characteristic impedance of the coil. When the impedance seen by the source at the bottom of the coil is equal to the value of the load resistance, we have found the characteristic impedance. Something akin to this exercise will yield the unknown characteristic impedance of a short piece of transmission line.

 

A piece of transmission line also has a velocity factor such that the actual velocity of an EM traveling wave through the transmission line is VF(c) where 'c' is the speed of light in a vacuum. In like manner, a loading coil has a velocity factor. If we can suceed in eliminating reflected current and thus measure the phase shift through a loading coil, we can calculate the velocity factor which is typically in the range of 0.01-0.03 for mobile loading coils. The speed of an EM wave through a typical loading coil is slowed by a ballpark factor of roughly 50 to one compared to a vacuum.

 

Can EZNEC be used to shed some light on the "current through a loading coil" controversy? EZNEC faithfully reports the standing wave current in a standing wave antenna and faithfully reports the traveling wave current in a traveling wave antenna. The goal of this exercise for EZNEC used with a mobile loading coil, is to eliminate the reflected current through the coil so EZNEC will display the forward traveling wave current rather than the standing-wave current. Using that technique, the phase shift (delay) through a loading coil can be readily observed using the EZNEC "loads" feature.

 

A large loading coil, like a 75m Texas Bugcatcher loading coil, has a characteristic impedance (like a transmission line) and a velocity factor (like a transmission line). An IEEE white paper [2] indicates that a 75m Texas Bugcatcher coil should have a characteristic impedance of a few thousand ohms and a velocity factor between ~0.01 and ~0.03. If we treat the 75m Bugcatcher loading coil as a transmission line and load it with a resistance equal to its characteristic impedance, we can minimize (if not eliminate) the reflected current leaving only the forward traveling current. Let's see if we can use EZNEC to accomplish that task.

 

The EZNEC files generated for this application can be downloaded from:
coil505s.EZ  which is the coil at its 1/4 wavelength self-resonant frequency of 7.96 MHz and
coil505u.EZ  which is the coil at its user frequency of 3.8 MHz. Here's what it looks like:

If we load the coil with its characteristic impedance, reflected current is eliminated (or at least minimized). The Z0 of the Texas Bugcatcher coil is apparently ~2745 ohms at 7.96 MHz and ~1975 ohms at 3.8 MHz. The VF (velocity factor) is apparently ~0.016 on 7.96 MHz and ~0.018 on 3.8 MHz. The coil is six inches in diameter, 7.5 inches long, and is 1/4WL self-resonant on 7.96 MHz. VF = 0.5'/(246'/7.96 MHz) = 0.01618

 

Here is the graph of the currents through the coil as reported by EZNEC:

Now let's make the following assumptions:
1. Assume there is no current amplitude "drop" through the coil for the forward current or for the reflected current. The amplitude of the forward current is the same at both ends of the coil. The amplitude of the reflected current is the same at both ends of the coil. The coil has been modeled as lossless so this assumption seems fair.
2. Assume that the amplitude of the reflected current is equal to the amplitude of the forward current. This seems like a fair assumption in a lossless system and works for lossless stubs.
3. Since the forward current and reflected current are in phase at the base of the coil and the phasors are rotating in opposite directions, assume the reflected current phase is the mirror image of the forward current phase. This seems to be an accurate assumption since these two phasors are reported by EZNEC and Kraus to sum to very close to zero phase.
4. Assume 0.5 amps of forward current flowing into the bottom of the coil and out of the top of the coil. Assume 0.5 amps of reflected current flowing into the top of the coil and out of the bottom of the coil.
The following graphic assumes all of the above.

Remembering that there is zero loss in our coil, let's phasor-add the forward current and the reflected current. The result is the following graphic:

This last graph is pretty close to measurements and EZNEC simulations. In fact, if we take the coil, remove the load resistor, and install a nine foot whip, we will bring the system to resonance. EZNEC reports that the graph of the current in that case is identical to Figure 4. That EZNEC file can be downloaded at: coil505t.EZ  Apparently, the primary factor in the net current at the bottom and top of a loading coil is the superposition of the forward and reflected currents, not the losses in the coil, not the radiation from the coil, and not even the local environment of the coil. The effects of these last factors appear to be secondary compared to the superposition of the forward and reflected currents.

 

Note that the measured net current amplitude "drop" through the coil is an illusion caused by the phasing of the forward and reflected currents. There is no drop in either of these component currents, i.e. they are both of the same amplitude at each end of the coil. Please compare the above graph to N7WS's Figure 7 [3]. The current "drop" through the coil in Figure 7 is also mostly an illusion since the current is mostly net standing wave current.

Standing wave current cannot be used to directly measure either a valid amplitude change or a
valid phase shift through a loading coil. All of the reported conclusions [1] based on loading coil
measurements using standing-wave current on standing-wave antennas are conceptually flawed.

[1]   3 nS delay through a 100 turn loading coil???
[2]   RF Coils, Helical Resonators and Voltage Magnification by Coherent Spatial Modes
[3]   N7WS Coil Article

Information provided here as a courtesy for all of our ham radio friends