# What is output power exactly?

I'm trying to comprehend what one would see if one could observe what's happening in the antenna wire. Is the difference between a 1 watt and a 4 watt transmitter just that the amplitude/voltage on the antenna will be higher?

• Welcome to Amateur Radio SE. Be sure to take the tour at ham.stackexchange.com – SDsolar Apr 24 '17 at 21:14

Yes.

Just like any other electronic device producing a signal, the transmitter is “really” applying a voltage to the output, which causes a current, and so on.

But we talk about power more often because power is a much more useful figure to work with for RF engineering above the level of designing circuits; voltage and current can be traded off (while power stays constant) so there's an arbitrary scaling factor that can be avoided.

(And once you have a electromagnetic wave propagating through the air, the voltage depends on the distance you measure it across, so if you want volts you can only get volts-per-meter. That's a good unit for measuring field strength — but field strength is not the same thing as power (they're related by the gain of your antenna).)

What voltage you find on the conductors of the antenna will depend on exactly what two points on the antenna you measure. But it is always the case that the output power is proportional to the square of the voltage (measured in a consistent way), just as in more familiar cases where the usual formulas apply:

$$P = VI = V\left(\frac{V}{R}\right) = \frac{V^2}{R}$$

Within the antenna, $R$ is fairly impractical to calculate, but constant for any two points (if it weren't, it wouldn't be a linear system and you would get highly undesirable intermodulation). But on the feed line — the (typically coaxial) cable between the transmitter and the antenna — there is in fact a value you can use for $R$, the characteristic impedance of the cable (usually 50 or 75 Ω for coaxial cables).

(In RF systems we talk about impedance $Z$ instead of resistance $R$, but the difference is irrelevant to this particular question, so I stuck to basic-electricity concepts.)

• As far as I understand it, the field impedance around the antenna can be calculated at one point. Once you get far away from the antenna it becomes the impedance of free space, about 377 ohms. I got the impression from the second paragraph ("traded off") with distance, but voltage and current will always be in that ratio of 377 volts per ampere in free space. But the impedance can be whatever you want, if there's an antenna or feedline around. Might merit some rewording to avoid ambiguity. – Phil Frost - W8II Apr 25 '17 at 12:23
• @PhilFrost-W8II I've revised my answer to clarify I'm not talking about the impedance of free space ("voltage … on the conductors of the antenna"). Does that work? – Kevin Reid AG6YO Apr 25 '17 at 14:35
• LGTM, with some hair splitting -- "once you have a electromagnetic wave propagating through the air , the voltage depends on the distance you measure it across" -- this is also true when measuring voltages on the antenna. There's also a definition of voltage, or more precicely electric field potential, at just one point. Or you can look at the electric potential difference as the distance between two points approaches zero to find the gradient at one point in V/m like you say. Not sure if they lead to the same results WRT field impedance. – Phil Frost - W8II Apr 25 '17 at 16:39

@Kevin's answer is correct. Power is the product of voltage and current.

In your question you ask if it is just higher voltage, and it may well be, but more commonly they both go up at the same time to get more power (amplitude).

Since you asked how you can see what is happening on the wire, I would direct you to the Wikipedia entry, which has a great animation. (I can't copy the animation here so you will need to follow this link to see it):

Wikipedia - Dipole Antenna (Right-click on this and open it in a new tab)

Back in the old days, we would often use small light bulbs as dummy loads on low-power transmitters to see the amount of power being put out as we made transmitter adjustments. We would just solder them to an antenna plug (PL-259) and plug them in instead of an antenna.

Radios today do not need transmitter adjustments so that isn't really needed, but you definitely could see the difference in power output between 1 and 4 Watts with a small bulb.

For larger transmitters we would use regular 100-Watt light bulbs.

Antennas have a characteristic impedence, Z, with three components, inductance, capacitance and resistance. When the inductance equals (and thus cancels) the capacitance, it is called being in resonance. This means that the only component left is resistance.

The equations are complex, and for that reason it is best to talk about properly-tuned antennas. At that point the equation P = E x I works without any further math. Power (Watts) equals electromotive force (Volts) times current (Amps).

The light bulbs would present a purely resistive load, simulating a resonant antenna.

Those old tube-type transmitters have several adjustments that needed to be made to tune them properly for maximum power output and we could see that by the brightness of the bulb.

Of course, you can do the same with a simple Wattmeter. But we're hams so we had to be different and innovative.

When building a dipole it is very important that the Capacitance and Inductance truly cancel each other out. You want a resonant antenna. The way you do that is by adjusting the length.

While adjusting your antenna length you use a SWR (standing-Wave Ratio) meter, which shows the amount of power being reflected back to the transmitter. When you are at the correct length for the frequency you are using the SWR will be at its lowest as the antenna accepts the power and puts it out on the wire.

At that point the antenna is seen by the transmitter as a purely resistive load. Thus the other equation that @Kevin showed. p = E^2 / R - Power in Watts equals the voltage squared divided by the resistance.

Standing waves are similar to the animation in that Wikipedia entry, except they are happening in your feedline and basically turning into heat, wasting power that would go out to the antenna.

But more importantly for your answer, the patterns of current and voltage are way different in non-resonant antennas, and visualizing them is difficult and unique to each configuration.

Here is a good reference on dipole antenna theory, including visualizations of the radiation patterns:

Full-Wave Dipole

But you will notice that the current is low in the middle in that link.

Instead, we use half-wave dipoles where the current is highest in the middle so the voltage is the lowest. (This is where I am diverging a bit from your assumption in the question - because at the antenna feedpoint we want highest current at lowest voltage, and that creates higher amplitude. In other words, Amplitude is measured in terms of power, not just in terms of voltage)