The AM Forum

In another topic here in the Technical Forum (60' flat-top for 160 and 75), we saw via computer simulations of the antenna and the transmission lines that a short dipole being driven on 160 meters may present an impedance of 0.78 - j41 Ohms at the shack/tuner end of the transmission line.

Just for kicks, I decided to see how many simple L-C matching networks I could come up with to match that impedance to 50 Ohms, and then simulate the networks with realistically estimated component losses to see what the total losses of each network are.  The circuits were simulated at 1.9 MHz.

All of the capacitors used in the matching networks were modeled with a Q of 2000 (D = 0.0005).  All of the inductors used in the matching networks were modeled with a Q of 350 and then 700.  The networks are driven at 50 Ohms with 500 Watts.  The power developed across the 0.78 Ohm resistor of the load model is the output power.  A schematic of the simulation is attached.

A table was made of the results:

500 Watts input    Network 1             Network 2                Network 3
Q = 350              Two inductors!      Low-pass L-C              Hi-pass L-C

Power loss                  68 Watts               80 Watts                  77 Watts
Total Loss, dB.            0.63 dB.                0.75 dB.                    0.72 dB.
Current in L                 3.0 Amps              23.2 Amps                26.2 Amps           
Current in C (or L2)      20.6 Amps            23.1 Amps                2.9 Amps


500 Watts input      Network 1             Network 2                Network 3
Q = 700                 Two inductors!      Low-pass L-C             Hi-pass L-C

Power loss                 36 Watts                43 Watts                  42 Watts
Total Loss, dB.          0.33 dB.                 0.40 dB.                     0.38 dB.
Current in L               3.1 Amps               24.2 Amps                 27.2 Amps
Current in C (or L2)    21.4 Amps             24.0 Amps                  3.0 Amps

Conclusion – I came up with 3 networks.  By far, the most component loss is in inductors.  One surprise was that in spite of this, Network 1 which is made of 2 inductors has the lowest loss!  Also, note the extremely large shunt capacitor value in Network 2.  It doesn’t seem worthwhile to change your tuner from one configuration to another based on the results, but to utilize the highest Q inductors possible. 

- - - - - - - -
Appendix – Basic Derivation of the 3 Networks (May be confusing!)

The impedance of 0.78 – j41 Ohms is the impedance of a node (at one frequency) expressed in series rectangular coordinates; a resistance and a series capacitive reactance.  This impedance can also be expressed in terms of two parallel components; a resistance and a parallel reactance.  The Q of the two equivalent circuits is the same, but the R and X values are different.  The parallel equivalent of 0.78 Ohms resistance in series with 41 Ohms of reactance is a 2156 Ohms resistor in parallel with 41.01 Ohms reactance.

Series-to-parallel conversion and parallel-to-series conversion are used as we step through the design of the matching networks.

Network 2 – Start with the load model of 0.78 Ohms resistance in series with 41 Ohms of capacitive reactance.  First let’s add a series inductance of 41 Ohms to cancel out the 41 Ohms of capacitive reactance.  This leaves the 0.78 Ohms resistor.  Let’s put in a low-pass L-C network to step from 0.78 to 50 Ohms.  Now an additional series inductive reactance is needed in series with the 0.78 Ohms which has a parallel equivalent of 50 Ohms resistance with some parallel reactance value.  The parallel (or shunt) reactance needs to be of the opposite sign of the series inductive reactance.  Plugging these numbers into the series-to-parallel conversion formulas, the series reactance needed is 6.19 Ohms and a shunt capacitor of 6.29 Ohms (across the 50 Ohm output).  The 6.19 Ohms inductor is in series with the first 41 Ohms inductor, so this adds to be a 47.19 Ohms series inductor.

Network 3 – Starting with the load model of 0.78 Ohms resistance in series with 41 Ohms of capacitive reactance, convert to the parallel equivalent.  This is a 2156 Ohms resistor in parallel with 41.01 Ohms capacitive reactance.  First add a 41 Ohms shunt inductor across the parallel equivalent circuit to cancel out the 41 Ohms capacitor.  The capacitor and inductor are now parallel resonant and essentially disappear.  We are left with the 2156 Ohms resistor.  We need a shunt reactance now with the 2156 Ohms resistor that will have a series equivalent of 50 Ohms in series with some reactance.  Plugging these numbers into the parallel-to-series conversion formulas, the values needed are 332.2 Ohms for the parallel reactance and 324.5 Ohms for the final series reactance to the 50 Ohm port.  Let’s use a high-pass L-C to convert to 50 Ohms because the shunt inductor needed will nicely combine with the first 41 Ohm inductor it will be in parallel with.  A 332.2 Ohm inductor goes in parallel with the first 41 Ohm inductor to become one 36.5 Ohm shunt inductor.  Then a series 324.5 Ohms capacitor goes in series to the 50 Ohm output port.

Network 1 – Similar to Network 3; follows the description of the previous paragraph until the 3rd to last sentence where I chose a high-pass L-C for the final step to 50 Ohms.  Let’s use a low-pass L-C network instead from this point on.  The 41 Ohm inductor is now in parallel with a 332.2 Ohm capacitor forming one net inductor of 46.78 Ohms and a series inductor of 324.5 Ohms to the 50 Ohm output port.

Schematic file below:

Tom,
I would like to do a balanced balanced tuner that will work on all bands using open wire line. I just fixed my tuner by changing the input from 1:2 BB transformer to 1:1 current balun wound with RG393.
I would like one configuration to work for all bands even if I have to make it a dual pi rather than dual I.
I simulated a balanced balanced tuner by floating the signal source and the ground reference at the load SWCAD ran fine.  I have not tried anything lower than 10 ohms though.
I have not tried series caps so need to study this further.
Again very good information
Happy New Year and thank You for all your good work

Great stuff. The current through the capacitors should be a real eye opener for some.
It would be interesting to see what Frank actually reads for antenna current at his feeders. This is the easiest way to trace changes also.
   I almost hate to say it Tom but this would benefit plenty of ops if the model were for longer dipoles like 100 to 150 feet. Just think of all the lot limited hams who could get on 160. That means more work at the drawing board though. Still it's really cool to hear Frank getting on with a 40 metre skyhook.
Too bad none of this stuff will work with the new "pro" radios. You have to build something.

interesting stuff Tom. Now we just have to figure out how to build a coil with a Q of 700!

Tom,
I just took the values from the worst loss network and put them into a balanced balanced
with a pair of 2 uh inductors. Since the configuration was closest to the B/B configuration. I didn't do anything with added resistors simulating loss but quickly found the L value was too low. I changed to a pair of 2.5 uh inductors with a 3700 pf cap to resonate at 1.9 mhz.  This showed the highest output level with an input of 50 ohms.
I did the input as a pair of 25 ohm resistors off the floating source and one side of the load grounded. (SwCAD)
This is interesting because I would love to build a tuner with my pair of HD 12 uh inductors and be able to cover 160 meters. I find my full size antenna needs all 22 uh on 160 and the Q is pretty high.
I agree with HUZ the Q numbers are pretty high. I wonder what a 1/2 inch copper tubing inductor will give you for Q? Not sure of broadcash mica cap q ratings.

Just for an example on coil Q’s, I was working with this 1.7 microHenry coil for Class E experiments at 7.37 Mhz.  It’s not the range of inductance you would run into for the 160 meter tuners, but I did fire up the Heathkit Q meter on this coil.  It is made with 5 ½ turns of #6 AWG solid copper wire and is 2” i.d.

Q = 200 @ 6.3 MHz., the lowest frequency the Q-meter would resonate at with only its internal capacitors.
Q = 210 @ 7.2 MHz.
Q = 295 @ 14.3 MHz.

Photo of coil below:

Tom,
Check out how the Q dives as the turns come closer to each other. gfz

Here are 2 more coils with slightly surprising results, looks like turns spacing is very important.

1.  2 1/2" dia. airdux-type of stock, ~ 12 AWG, 10.5 uH., Q = 500 @ 2.2 MHz.

2.   3" PVC form (3.5" o.d.), #10 stranded, 9.6 uH., Q = 240 @ 2.3 MHz.

1/2" copper pipe... the turn diameter would be fairly large.  At some point, measured Q would be reduced not because of IR loss, etc, but because of radiation!  That could be good or bad, depending on where the leakage radiates, in what phase, etc.   And there could be inductive coupling to lossy things nearby, which would be bad.

The loss due to overly close windings is essentially caused by skin effect.  The current produces a magnetic field that tries to push the electron flow out of the conductor, and with the conductors too close together, the magnetic field in each wire pushes the electron flow away from the surrounding conductors as well.  The result is more IR loss and lower Q.

I thought the close turn Q loss was because the C between turns goes too high.

I was expecting that the PVC coil with the larger diameter and larger wire would do better than the airdux, but it didn't.  I only JS-slapped the PVC coil together.  Turns wound up touching each other.  With winding care, it could do better.

The airdux uses about 2 1/2 wire diameter spacing between turns.

The air capacitors formed between turns should be high Q.  I think the distributed capacitance is a misnomer as Bacon says, being easy to correlate to because it is directly related to the turns spacing which is the factor.  Just like same magnet poles pushing apart, the current going in the same direction is forced away from the adjacent surface areas.

It may be of interest if you could wind the pvc coil 'nicer'. Then lengthwise cut out windows into the pvc. I'm wondering how the pvc changes da q ???  one of these days i'll try the hombrew air dux thing - two sheets sheets of plexi  connected to form a + with  holes drilled for wire to be wound through...   klc

Tom,
Are you saying the surface resistance increases due to less area where current is flowing? interesting

No not exactly.  The surface resistance stays the same, the net effective resistance per length is increased since a fair percentage of the surface is not used, forcing more current in the rest of the surface, higher I2R.  Similar to what you are saying, have to be picky here.

For higher Q in transmitter inductors, I've been playing with larger diameter single turn inductors made from sheet copper.  The idea is that with a single turn there is no inter-turn capacitance which lowers Q and sheet offers lots of surface area to reduce skin-effect.  Copper tubing has the issue that the outside diameter is greater than the inside diameter causing an imbalance in inductance resulting in some eddy currents.

I haven't got carried away with silver plating yet...

Tom, We are saying the same thing but you may be closer to English

In an Article entitled "Notes on HF transmitting Coils" in Sep 2005 "Amateur Radio" magazine, Drew, VK3XU mesaured the Q of a number of coils would with different materials. The coils were all 70 mm dia, 90mm long with 18 turns.and 2mm dia wire. (The stranded wire was, I presume, of equivelent cross section)

Wire                            L       Q at 3.6Mhz   
Plain bright copper     12.8u   350
Enammeled Cu.           12.8    340
Silver plated Cu        12.8    350
Tinned Cu               13.3    340
Oxidised Cu             12.8    330
Insulated stranded wire 13.9    240

The coils were all nice neat imitation airdux type .

It would appear that the insulated stranded wire has problems of its own regardless of how it is wound. I'm assuming the insulation in VK cable is simiar to W, but it may be thicker due to higher mains voltage (240V)

                                                Ian VK3KRI

Getting a Q of 700?

Hmmm, we have capacitance multipliers in power supplies for really large effective filter caps using power mosfets
and we have pw Q multipliers in receiving circuits, mainly the old 455kc variaties.

So how about a Power Q multiplier? beef up the circuit's current handling capabilities electronically, coupled to a really stiff bypass. I'll have to think about it.

Yes, I can see it.
"Power Q multiplier here OM, into the 10ft. vertical."

I'm running my 10 foot super conductor antenna

I was thinking of some sort of switching driver to apply power to such a bizarre reactive load, with no loading reactances.  This might get around the narrow-bandwidth issue, but there is lots of voltage and/or current to deal with.

Superconducting? Well, There you are...

A "HelioTron," tm.    - (pun intended)

I rewound the #10 stranded, 9.6 uH., on the 3” PVC form with better spacing.  This time I got about 8.8 uH and the Q = 274 @ 2.4 MHz.  This is up from a Q of 240 in my previous posting.

I had about 6 feet of solid #6 AWG left over from the 40 meter FET evaluation and I wound that on the PVC form.  I have a 2.4 uH coil with 6 turns.  The spacing is about 5 – 6 wire diameters between turns. The Q = 354 @ 4.75 MHz.

Interesting. I wonder how edge wound inductors figure in.

There's hope for those superconductor coils!

http://www.physorg.com/news95525363.html

steve,
How about that oil thing?

Which?

the link front page telling about oil reserves .....then wanting to sell them in fine print.

Still not sure what you are talking about. I was refering the the article, not anything else.

When I opened your link I first came to the ad about oil reserves out West that were very large...then in fine print the get in on the ground floor.

That was a sign from God - your chance to get rich!!

maybe I should cash in the scrap copper and invest....