Home › Forums › temp › Loading coil chart for 10-ft broadcast AM vertical antenna › RE: Re: .45% radiation?

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[quote=kk7cw]Could you possibly mean .45 X 100 = 45%? [/quote]

The equation to calculate antenna system radiation efficiency is simply Ohm’s law for a series circuit containing several resistance elements. The tx supplies some value of r-f current to it, which has to flow through all of those resistances. Part of the total power of the tx is dissipated in each of those resistances, in accordance with the ratio of each resistance to the total resistance. The power consumed by the antenna in my example really is 0.004525 times the total power applied, which I expressed as a percentage (~0.45%).

[quote=kk7cw]Otherwise, the typical Part 15 antenna system that can reach a mile or more is radiating less than 1/2 percent of the 50-60 percent (efficiency of the PA) of the 100 milliwatt input to the final amplifier. … That would indicate (Ohms Law) a very high impedance at the feedpoint, which you correctly explained is not true.[/quote]

The impedance at the base of a 3 meter Part 15 AM antenna IS extremely high. It consists of the radiation resistance and reactance of that radiator on that frequency. The radiation resistance will be very small (~0.1 ohm or less) and the reactance will be very high (several thousand ohms). Only the radiation resistance can dissipate power (which radiates), but no practical tx can deliver much power into such a highly reactive load. That is the reason for the loading coil — to cancel the high capacitive reactance of the short radiator so that the tx has a usable load for its output power.

With the coil, the feedpoint impedance at the coil input contains the original radiation resistance plus the added r-f loss in the loading coil, but with ~ zero reactance. This is a load impedance that the tx can deliver power into. Now we revert to my comments just above to see how much of that total power of the tx actually is radiated — ie, to determine the radiation efficiency of that antenna system.

[quote=kk7cw]Broadcast AM stations determine the efficiency of antenna systems by 2 methods. … No computers or software needed to determine the efficiency. Pencil and paper work fine. I have built and consulted several AM stations (etc) [/quote]

Pencil and paper also work fine using the method I posted above. And our methods are the same, anyway.

The impedance measured at the base of a tower of a licensed AM broadcast station includes all the parameters I have described in this thread: radiation resistance and reactance, matching network resistance and reactance, and r-f ground loss.

The methods you described for determining AM broadcast station performance would work equally well for Part 15. It’s just that test equipment to do that is expensive, and probably not justifiable for most hobbiests. And likewise, the method I described also would work equally well for licensed AM broadcast stations if the impedance values of the various elements of the antenna system are separately known.

[quote=kk7cw]And by the way, increased surface area or diameter does decrease the impedance of the antenna, allowing more current to flow in the conductor.[/quote]A conductor with a large OD is no more efficient a radiator by itself than one with a much smaller OD, when both are resonant, and have the same current flowing in them. I think this point is not being fully recognized by your statement, although I expect that you understand this yourself.

Anything that increases the amount of current flowing in *any* radiator will increase the fields it produces. One way of doing that is to improve the impedance match between the tx and the antenna feedpoint. But in the case of a larger OD radiator, the increased radiation field is not related to the larger surface area in a direct way — it is a secondary effect.

For calibration, here are calculated values for two 3-meter conductors of two ODs, on 1,700 kHz:

Radiation resistance > Reactance > VSWR referenced to 50 ohms

1/2″ OD: 0.12 ohms > -2,498 ohms > over 1,000,000:1

4″ OD: 0.12 ohms > -1,335 ohms > about 300,000:1

So in neither case will a tx be able to supply much power into such mismatches. A loading coil will be necessary, but the one for the 4″ OD radiator probably will have less loss than the one for the 1/2″ OD radiator, and that can give some relative improvement in radiated field.

[quote=kk7cw]Additionally, the bandwidth of the antenna system is improved making the program audio in the sidebands louder and, ultimately, the signal louder and more listenable.[/quote]The bandwidth of the tx is set by its modulating characteristics. How much of that modulation gets radiated is set by the bandwidth of the antenna system. It won’t matter if the tx can be modulated at 15 kHz if the match of the antenna system at those sideband frequencies rejects them.

[quote=kk7cw]The largest challenge Part 15 stations face is the lack of radiation resistance in the super short antenna. Even with a greatly enhanced Part 15 ground radial system, the increase in signal strength is minimal in decibels.[/quote]

Yes, the low radiation resistance is a big contributor to the coverage performance of Part 15 AM. But the effect of the r-f ground is not really minimal, either.

The r-f ground resistance of a Part 15 AM antenna system using a “broadcast-type” radial ground system might be on the order of 2 ohms. If the buried ground system consists only of a few ground rods at the base of the 3-meter radiator, that r-f ground resistance might be ~ 50 ohms (or more).

Using the method from my earlier post about a 3-meter Part 15 radiator, here’s how those numbers crunch when determining radiation efficiency with those two ground systems.

Broadcast radial ground system ( 2 ohm ground resistance):

0.1/4.1 = ~0.0244 = ~2.44%

A few ground rods (50 ohm ground resistance):

0.1/52.1 = ~0.00192 = ~0.192 %

The power ratio of 0.0244 and 0.00192 is about 11 decibels, which is very significant — not minimal.

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