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Short Wire Antenna Impedance




 
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N4LTA
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« on: January 15, 2010, 10:21:26 PM »

Anyone have an idea what the impedance of a 30 foot long wire antenna would be at  about 1 mhz?  I know that a typical long wire should be 1/2 wave - but what if it is very short/

My antenna analyzer says  " greater than 680 ohms" but I would have thought it would be very low, 1 ohm or so like a short vertical..

Pat
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Ed/KB1HYS
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« Reply #1 on: January 15, 2010, 10:26:46 PM »

End Fed, free space       21.0 - j91770 ohms.

Center fed, Free space    90.68 - j11690 ohms.

From EZNEC.


Sorry, I had left a load in the line from a previous model and that effected the calculations. Then I posted the results with out really checking.  Removing the load shows a resistance in agreement with the very low values posted below.

A red face...  Embarrassed
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73 de Ed/KB1HYS
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« Reply #2 on: January 15, 2010, 10:31:54 PM »

Thanks,

Not what I expected.

I am trying to get a decent match (or a lousy match) to a 15K ohm tube output stage on a wireless broadcaster.

Pat
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« Reply #3 on: January 16, 2010, 10:40:04 AM »

End Fed, free space       21.0 - j91770 ohms.

Center fed, Free space    90.68 - j11690 ohms.

From EZNEC.

I'm not so sure about those numbers.

The figures for the resistive component seem awfully high for a 30 ft. wire @ 1 mHz.  At 1 mHZ, λ = about 240 ft; 30 ft. comes out to 1/32λ. This would be equivalent to 7.5 ft @ 4 mHz, the length of a typical mobile whip for 75, whose impedance would be at most a couple of ohms.  90.68Ω would exceed the resistive component at the feedpoint of a full size half wave dipole.
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« Reply #4 on: January 16, 2010, 12:20:33 PM »

I agree with Don

When I ran EZNEC, I obtained the following:

A) Dipole, 15 feet on each side, height set to 30 feet, real ground, frequency = 1MHz)

Input impedance = 0.005734 ohms - j10710 ohms

B) Dipole, 15 feet on each side, height set to 60 feet, real ground, frequency = 1MHz)

Input impedance = 0.02237 ohms - j10710 ohms

C)  Dipole, 15 feet on each side, height set to 245 feet (1/4 wavelength high), real ground, frequency = 1MHz)

Input impedance = 0.2254 ohms - j10710 ohms

D)  Dipole, 15 feet on each side, free space, frequency = 1MHz)

Input impedance = 0.1963 ohms - j10710 ohms

E) Single ended (fed at 1.36 feet from one end and 28.64 feet from the other end, 30 feet long, height set to 30 feet, real ground, frequency = 1MHz)

Input impedance = 0.008073 ohms - j66550 ohms.

Note that, for the case of a dipole (doublet), since the ratio of the antenna length to 0.5 wavelengths is 30 feet / 492 feet = .061, I would expect the radiation resistance to be ~ 70 ohms x .061 x .061 = 0.26 ohms. This corresponds reasonably well (for a rule of thumb) to what EZNEC predicts for a dipole of this length if it were at a height of 245 feet. Naturally, I don't expect that it would be up that high... so the radiation resistance will be even lower (as predicted, above, using EZNEC)

Stu



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« Reply #5 on: January 16, 2010, 01:14:38 PM »

I've built some small 1 mhz antennas and found 2 ways to increase efficiency.
Multiple wires in parallel spread in the middle with all coming to a point at the far end. Matching a tapped inductor in series with the antenna to ground feed tap close to the ground connection. We used pot cores with medium perm. We were buildind 2 foot tall verticals and set up a test range to check them.
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« Reply #6 on: January 16, 2010, 04:04:06 PM »

Yes - I would expect it to have been very low.

I am not trying to get real good efficiency but the circuit for this transmitter  as published has a pi net with a 680uH inductor and a 400pf plate tune and a 15 pf load tune which is really weird.

I would think a transformer coupled output network would be better. Any ideas on how to get a realistic match of the plate of the output tube? The tube is running at about 150 volts at a few ma.

Pat
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« Reply #7 on: January 16, 2010, 04:19:12 PM »

There are two types of people:

People who believe that you can increase the radiation efficiency of a short antenna, over real ground, by tweaking the configuration, adding inductors, adding capacitors etc.

People who believe that the radiation efficiency of a short antenna over real ground is limited by (i.e. the best you can do, if you use a matching network): the ratio of the radiation resistance of the antenna to the effective resistance of the ground. This limitation is a consequence of physics, i.e., it can be proven by using the physical laws that currently are believed to govern electromagnetic phenomena.

Why do some people claim that they have obtained improved (or greatly improved) radiation efficiency, when using short antennas in conjunction with various tweaks of the antenna configuration:

a. They place the short antenna next to a large physical object that is a good conductor (low ohmic losses)... and that large physical object becomes part of the antenna (just like a dish that is illuminated by a feed horn).

b. They attempt to measure the "radiated power" in the near field... instead of measuring the power that actually radiates (in the far field)

A short antenna, regardless of the details of its configuration, couples very weakly to radiated (propagating) modes. In the presence of real ground (i.e. loss associated with high near field currents flowing through ground and other objects in the near field which have ohmic losses) a short antenna will couple most of its power into the objects in the near field. The shorter the antenna, the larger the ratio of: near field power that is lost (converted to heat) to radiated power.

See the attachment, for a graphical representation of the above.

Stu
 


* Short Antenna Model.jpg (60.59 KB, 960x720 - viewed 384 times.)
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« Reply #8 on: January 16, 2010, 04:45:08 PM »

Pat

I sounds to me like this is a transmitter that is designed to meet the FCC rules for unlicensed operation in the AM broadcast band (e.g. "talking house" transmitters):

"The total length of the antenna, feedline, and ground lead (if used) will not exceed 3 meters" [Note that a 30 foot long antenna does not comply with these rules]

This rule ensures that the antenna will radiate practically none of its power into the far field (i.e. into propagating modes), and essentially all of its power into the near field (which can be picked up by nearby receivers).

Correcting my earlier version of this post:

If you convert a series combination of a resistor of value R and a capacitor of value C into its equivalent parallel circuit, you get the following:

A capacitive admittance of j x 2pi x f x C / K1, in parallel with conductance of value (1/R) / K2

Where: f is the frequency (1 MHz in this case);, and

K1 = 1 + (2pi x f x R x C)(2pi x f x R x C); and K2 =  1 + 1/[(2pi x f x R x C)(2pi x f x R x C)]

In this case K1 is a number very, very close to 1, because 2pi x f x C ~ 1/(10,000 ohms), and R ~ .005 ohms
In this case K2 is a very, very large number ~ 1 + 1/ [(.0000005)(.0000005)]

The 15 pF "loading" capacitor has an admittance that is ~ j/(10,000 ohms) at 1 MHz... which is approximately equal to the parallel admittance of the antenna. Those two admittances add.

Now, converting back to a series equivalent circuit, one obtains a load on the combination of the tuning capacitor and the inductor that equals approximately the following

Z(load) = -j x (1/2) x 10,000 ohms + (1/4) x .005 ohms

In order to resonate this load, the inductor must have an impedance of greater than j5,000 ohms. The 680 uH inductor has an impedance (at 1 MHz) of 680 x 2 x pi ohms = 4272 ohms.

So, one cannot bring the pi output network into resonance.

The "tuning" capacitor has a value of 400pF, which corresponds to an impedance (at 1 MHz) of -j397 ohms.

The plate current will split between the "tuning" capacitor and the series combination of the inductor and the load.

The series combination of the inductor and the load has an impedance of approximately -j5,000 ohms + j4272 ohms = -j728 ohms.

Therefore, about 397/(397 + 728) ~ 35% of the plate current (the component of the plate current at 1 MHz) will flow into the load, and about 65% of the plate current (the component of the plate current at 1 MHz) will flow into the tuning capacitor

To increase the fraction of the plate current that flows into the load, one could increase the inductance of the pi network inductor and/or reduce the capacitance of the "tuning" capacitor and/or increase the value of the "loading capacitor" (up to a point... if you increase the value of the "loading" capacitor too much, you will begin to get less current flowing in the load).  


The purpose of the short antenna restriction is to make the far field power practically nil, and to keep the near field power very low (so that you don't interfere with radios that are outside your house, and within a few wavelengths of the antenna).

If you use more inductance and/or less "tuning" capacitance and/or a better optimized "loading" capacitance... you will produce an improved match between the output stage and the load... and therefore couple more power into the antenna. This will increase both the near field power and the radiated power.

Stu
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« Reply #9 on: January 16, 2010, 06:58:07 PM »

Thanks Stu,

I had no intent to get a good match to free space. I just wanted to get a few tens of milliwatts into the wire to get a range  of a couple hundred feet.

I am not a believer in good radiation resistance in a small physical antenna. Maybe with superconducting inductors and perfect capacitors?

I am tending to think that the best output circuit would  be a transformer coupled output with the primary resonant and using a decent resonably  high Q  transformer primary.

I would think that the 15 pf cap would be nearly useless in improving the match in the network and testing has  proved it to be pretty much useless.

Sort of like hanging a 90 K resistor across a 1 ohm load.

Pat
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« Reply #10 on: January 16, 2010, 07:27:23 PM »

Pat

The comment about coupling to radiating (free space) modes was a general comment... not directed to you. I apologize if I, inadvertantly implied that it was.

Please see my revised post, above. The purpose of the "loading" capacitor in this output network is to reduce the total capacitive reactance that the inductor needs to resonate against. However, the combination of the loading capacitance and the antenna capacitance, in parallel, must be adjusted to achieve resonance. A fixed value of 15pF is unlikely to be the correct value, because the resonance is fairly sharp. Separately, you can also improve the output power by reducing the value of the "tuning" capacitor.

In any event, you need to be careful to stay within the 100mW FCC input power restriction.

Make sure that you are using a balanced 30ft antenna (a 30 foot doublet). From the EZNEC calculation, a 30 foot (total) doublet has much less capacitive reactance than a 30 foot single ended antenna. [Note that the FCC Part 15 rules for unlicensed AM transmitters limit the total length of the antenna, feedline and ground conductor (if used) to a maximum of 3 meters. Thus, a 30ft antenna does not comply]

As a separate "heads up": the load impedance on the transmitter is so high (because of the capacitive reactance of the load) that the transmitter will try to push r.f. power into the AC line that is powering the transmitter. This is why one often includes extra 2.5 mH inductors on each of the wires leading to the AC line.

A consequence of pumping r.f. power into the AC line is that the house wiring can become the principal source of r.f. radiating into the house. Another consequence is that, due to various non-linear components in the path between the transmitter and the AC house wiring, the 1MHz r.f. power radiated by the house wiring will be modulated at 60 Hz and/or 120 Hz. Thus, if the house wiring becomes the dominant source of r.f., the signal picked up by the receivers in and around the house will have lots of hum on it.

All of the above is based on my own experience with a 100mW AM broadcast band transmitter that I purchased to use in demonstrations (for students) of software-defined radio technology. I use a very simple SDR receiver to pick up off-the-air AM radio signals.... showing the spectrum of the received signal, and to produce demodulated audio output.  But, sometimes, the metal structure of the building (at the university) blocks even nearby AM signals. If that happens, I use the 100 mW transmitter and an iPOD to create my own AM signal for the demonstration.

Best regards
Stu
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« Reply #11 on: January 16, 2010, 07:49:28 PM »

Thanks for the input. It gived me some ideas for experimenting.

My background is electrical power distribution. I wish I had paid more attention in electromagnetic fields back in 1972.

Below is a photo of the little transmitter. It was designed by a fellow on the Antique Radio Forum. I built a PC Board and built it for fun.



Pat
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« Reply #12 on: January 16, 2010, 09:12:07 PM »

Stu we knew of the low efficiency but wanted the best match to coax to avoid additional losses. I remember the pot core we used had about 30 turns of wire on it so plenty of L and the tap was at 1 or 2 turns up from the ground.
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« Reply #13 on: January 16, 2010, 09:42:30 PM »

As long as your input power is limited to 100 mW, and you limit your antenna length to 10 feet or "3 Meters", you do not have to comply to any ERP or field strength limits for the AM Broadcast band.

The 10 foot wire on your broadcaster will get you around the house and out into the yard. A properly designed copper pipe 10 footer with an efficient loading coil would get you out a mile or so ground mounted. And you can do better than that with a multiple wire deal like frank was suggesting or a capacity hat and a center load and using some questionable tricks like mounting the antenna on the garage roof. The FCC does not say anything about elevating the antenna.

Use a field strength meter or better yet get hold of an RF milliammeter and insert it in the antenna lead and change your loading until you get maximum current into the antenna.

Couple of good links

http://www7.brinkster.com/yvesroy/10ft-antenna.asp

http://www.sstran.com/pages/sstran_ant_dwg_pl.html
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« Reply #14 on: January 17, 2010, 09:36:35 AM »

Pat

Yesterday I revised my two prior posts in this thread to clarify how the output circuit works.

Today, I did a simulation of the output circuit using LTSpice

From the simulation, it is clear that the combination (sum) of the capacitance of the "loading" capacitor, and the series capacitance of the antenna (whose reactance at 1 MHz is about  -j10700) must resonate with the 680 uH inductor.


The value of 15pF for the loading capacitor is "in the ballpark"... but the resonance is sharp (high Q)... and by using a variable capacitor for the loading capacitor, you should be able to produce a significant increase (at resonance) in the output power of the transmitter.

The attachments show an LTspiceIV simulation of this circuit.

Note the significant increase in the antenna current when the "loading" capacitor is changed from 15pF (attachment 2) to 21.5pF (attachment 3). The antenna current increases from 16mA to 48mA. This implies a 9.5 dB increase in output power. This illustrates the importance of adjusting the "loading" capacitor. Also note that (making certain assumptions) the input power is much lower with the 21.5 pF capacitor (8 volts peak rf plate to ground voltage) than with the 15pF capacitor (26 volts peak rf plate to ground voltage)... thus, with a 100 mW input power constraint, an apples to apples comparison would indicate an even larger improvement in output power using the 21.5 (optimal) "loading" capacitor.

Assuming class C operation and plate modulation, to stay within the 100mW input power limit, the product of the peak rf plate voltage at the fundamental frequency (which equals the B+) and the average plate current (half of the peak amplitude of the fundamental frequency component of the plate current) must be less than 100 milliwatts

In the case where the loading capacitor is optimized (21.5 pf), the example here corresponds to 8 volts of peak plate to ground voltage at the fundamental frequency (= B+ voltage) x 50mA of average plate current (half of the 100mA peak current at the fundamental frequency) = 400 mW. So, the average plate current and the corresponding B+ will both have to be scaled back by a factor of 2. I.e. use 50ma (instead of 100mA) of peak plate current at the fundamental frequency; and 4 volts of B+ (instead of 8 volts of B+). Doing that with a vacuum tube will be a challenge. 

Stu


* Slide1.JPG (57.9 KB, 960x720 - viewed 376 times.)

* Slide2.JPG (114.92 KB, 960x720 - viewed 378 times.)

* Slide3.JPG (126.5 KB, 960x720 - viewed 384 times.)
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« Reply #15 on: January 17, 2010, 09:54:48 AM »

Stu,

Thanks again. I plan to wind a reasonable Q antenna coil and will try to get a peak with both caps. I have a few HF30s and that should be a decent cap for the output. I'll start with a single gang 365 pf cap go with an inductor of at least j6000 ohms  - one that I can remove turns if I have to.

I'll let you know how thing go.

Thanks again

Pat
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« Reply #16 on: January 17, 2010, 11:41:09 AM »

Pat

I apologize for so many posts on my part:

Using a separate inductor in series with the antenna to tune out the capacitive reactance of the antenna is a fine, alternative approach.

Please remember that if you do that... then the existing output circuit needs to be modified:

For this application (i.e. antenna with very low impedance, after resonating out the capacitive reactance) you will need to

a) resonate the existing 680 uH inductor with the correct value tuning capacitor. I.e. the 680uH tuning capacitor is no longer serving the purpose of canceling (most of) the capacitive reactance it was looking into previously.  At 1 MHz, you will need approximately 37 pF of tuning capacitance (which includes the output capcitance of the tube).

In this situation, exact resonance is not necessarily the best setting... but the 680uH inductor is a lot larger than it needs to be... and the existing 400uF tuning capacitor is no longer the right value to use with this inductor.

You might want to reduce the 680uH inductor to something closer to 68 uH... which will resonate with approximately 370 pF of tuning capacitance. Again, exact resonance is not needed (or even best), but you need to get close to resonance in order to put a reasonable rf load on the tube.

b) remove the 15pF "loading" capacitor (since the antenna's impedance / resistance is so low, this "loading" capacitor will now serve no useful purpose)

Attached are the LTspiceIV simulations of this approach. Note the significant increase in antenna current if you use a 68uH inductor and a 370pF capacitor (attachment 3) instead of a 680uH inductor and a 37pF capacitor (attachment 2). The current in the antenna increases from 67mA peak to 297mA peak. This corresponds to a 13 dB increase in output power.

Naturally, all of this has to be scaled to stay within the 100 mW input power constraint.

With the 100 mA fundamental frequency plate current in this example, and with the 68uH inductor in the pi network (attachment 3), the input power (assuming class C operation and plate modulation) would be 160 volts of B+ x 50mA of average plate current = 8 watts. To stay within the 100mW limit, the average plate current would have to be reduced to from 50 ma to 5.6 mA. The corresponding B+ would be reduced from 160 volts to 17.9 volts. 

Good luck

Stu


* Slide2.JPG (101.66 KB, 960x720 - viewed 417 times.)

* Slide1.JPG (117.39 KB, 960x720 - viewed 401 times.)

* Slide3.JPG (127.37 KB, 960x720 - viewed 382 times.)
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« Reply #17 on: January 17, 2010, 09:20:15 PM »

There is no power limitation but rather a field strength limitation. Part 15 limits the field to 24000/F(kHz) uV/m measured at 30 meters in the AM Broadcast band. The allowable field strength at HF is 30 uv/m.
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« Reply #18 on: January 18, 2010, 07:48:25 AM »

Actually, there appear to be restrictions on both the input power and the field strength (as well as other restrictions). These restrictions are copied from the Part 15 low-power transmitter frequency table on page 12 of:

http://www.fcc.gov/Bureaus/Engineering_Technology/Documents/bulletins/oet63/oet63rev.pdf


Frequency band: 525-1705 kHz

Type of use: Any

Emission limit / reference section of FCC part 15 rules:

1. 100 mW input to final RF stage / 15.219
2. 24000/f(kHz) μV/m @ 30 m /  15.209
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« Reply #19 on: January 18, 2010, 11:21:03 AM »

Stu,

I am going to wind a 68 uh inductor for tuning with a 400pf variable cap. and something in the 1600 uh range for a loading coil and see what I get initially. I'll let you know how it works out but it looks to be much better than the original circuit.

That 1700 uH one will be a lot of #30 wire but I would a power choke for a transmitter last week and it was
450uH , so I won't be winding blind.

Thanks,

Pat
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« Reply #20 on: January 18, 2010, 03:56:34 PM »

Wound the two inductors this afternoon. The large one is 1598 uH on my AADE meter. The small one is 111 uH  which will probably will resonate fine with a single gang 365 pF variable at 1 MHz.

This thing is going to look like a tesla coil project.



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« Reply #21 on: January 18, 2010, 10:49:34 PM »

Actually, there appear to be restrictions on both the input power and the field strength (as well as other restrictions). These restrictions are copied from the Part 15 low-power transmitter frequency table on page 12 of:

http://www.fcc.gov/Bureaus/Engineering_Technology/Documents/bulletins/oet63/oet63rev.pdf


Frequency band: 525-1705 kHz

Type of use: Any

Emission limit / reference section of FCC part 15 rules:

1. 100 mW input to final RF stage / 15.219
2. 24000/f(kHz) μV/m @ 30 m /  15.209

True sort of:

Most AM broadcaster experimenters operate under the 15.219 specific power input and antenna length limits, not the 15.209 field strength limits. 15.209 radiation can not be easily measured by most people, but they can comply to 15.219; .1 watt and the antenna length limit is 3 meters. You do not have to guarantee compliance to both.

That said, if an interference complaint is received, the FCC can shut down any Part 15 station pronto.

You can even get a Part 15 device certified under 15.219 vs. 209, if you wanted to sell a broadcaster design.

So I still say that there is no range or height limit for a Part 15.219 transmitter. Like a lot in Part 15, you only need to comply to the least restrictive and most specific rules for a service that you claim you are operating under and there is lots of room for interpretation.


Mike WU2D
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« Reply #22 on: January 18, 2010, 10:58:45 PM »

There is a lot of room for politics and interpretation in the FCC rules. In one of my stints I did quite a bit of intentional radiator certification work. We kept some darned expensive FCC lawyers in retainer and we consulted them at least three times a year on FCC matters. The rules are tough to interpret:

But this is a good one: http://www.kencradio.com/press.php?page=2

Mike Wu2D
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« Reply #23 on: January 19, 2010, 08:50:01 AM »

My guess would be that you can use most anything that is within reason (small mW type transmitter) - but if you interfere - you will be ordered to shut it down. If you check your radiation and it stays in the bounds of your property - you are not going to have a problem.

Pat
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