I don't understand Watts

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scanmanmi

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Watts is amps x volts. How can watts be measured when there is no completed circuit. I understand a circuit with a 50 ohm resistor but the circuit becomes open with a normal antenna. If I'm reading 50 watts where are my amps?
 

dlwtrunked

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Watts is amps x volts. How can watts be measured when there is no completed circuit. I understand a circuit with a 50 ohm resistor but the circuit becomes open with a normal antenna. If I'm reading 50 watts where are my amps?

The circuit is not "open" with a proper functioning antenna antenna. The watts are not just electrons in wire, they are the power which goes through the wire to radiate from the antenna into space as electromagnetic radiation. So it is not "open" the power just changes form from being electrons in wire to electromagnetic radiation in space.
 

wx5uif

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To DC that appears an open. However, when applying AC at the resonate frequency, the impedance will be near 50 ohms.
 

Project25_MASTR

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To DC that appears an open. However, when applying AC at the resonate frequency, the impedance will be near 50 ohms.

And with AC the impedance is equal to resistance.
To answer the OP though, P=I^2*R so therefore, 50W with 50 Ohm impedance is equal to 1A (and subsequently, 50 VAC).
 

N9PBD

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Watts is amps x volts. How can watts be measured when there is no completed circuit. I understand a circuit with a 50 ohm resistor but the circuit becomes open with a normal antenna. If I'm reading 50 watts where are my amps?

Grab the antenna while you are transmitting, and you'll find your amps. OK, just kidding, seriously, never do that. RF burns hurt. Really.
 

jonwienke

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Watts is amps x volts. How can watts be measured when there is no completed circuit. I understand a circuit with a 50 ohm resistor but the circuit becomes open with a normal antenna.

An antenna completes an AC circuit the same way a capacitor does, for the simple reason that it IS a type of capacitor. The ground plane (or bottom half of a dipole) is one half of the capacitor, and the vertical element forms the other half.
 

SpugEddy

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So, this thread made me think of a question (stupidity level = 4)

Is there a difference between Watts used by the radio from the
power supply and the amount of watts transmitted out of the antenna?
 

jwt873

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So, this thread made me think of a question (stupidity level = 4)

Is there a difference between Watts used by the radio from the
power supply and the amount of watts transmitted out of the antenna?

Yes.. An amplifier (radio or audio) isn't 100 percent efficient so 100 Watts from the power supply will produce less than 100 Watts output from the radio.

There are various classes of amplifiers with different degrees of efficiency. http://www.electronics-tutorials.ws/amplifier/amplifier-classes.html
 

Project25_MASTR

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And power must be conserved so what is drawn and isn't transmitted will generally become heat.
 

paulears

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The often quoted Watts = amps x volts is not an absolute, because this assumes a resistive circuit, and AC circuits, especially radio frequency ones, are reactive - with inductance and capacitance a critical feature - so power factor gets added into the equation. Google power factor if you want to get confused even more. You can't really think about RF circuits like DC ones. If you think about a quarter wave antenna - stick a meter on it and it is an open circuit. Stick it across a folded dipole and it's a dead short. Both work fine as an antenna.
 

prcguy

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Watts used by the radio is easy to measure using basic DC circuit measurements of Voltage X Current. Watts produced by the radio to the antenna can be looked at by the transmitter producing about 70.7 volts of AC at whatever frequency its transmitting on, and with an antenna or load impedance of 50 ohms, the load will draw about 1.4 Amps of current to equal 100 Watts of energy consumed by the antenna or load.

As others have mentioned, a transmitter or amplifier is not 100% efficient and you may consume well over 200 Watts of power from 12 Volts to produce 100 Watts of power to the antenna. A typical example is most 100 Watt HF transmitters consume about 20 Amps of current at 13.8 Volts, which is 276 Watts of DC power to produce 100 Watts of RF energy into a 50 ohm load.

As another example, if the same transmitter was capable of putting out 70.7 volts into any load then into an antenna or dummy load with 2 ohm impedance you would have V2/R = 2,500 Watts of power (70.7 Volts X 35 Amps!) consumed by the load and so on. That is obviously not going to happen with a 100 Watt rated transmitter, which has a fairly narrow design range and current supplying capacity around a 50 ohm load impedance.
prcguy

So, this thread made me think of a question (stupidity level = 4)

Is there a difference between Watts used by the radio from the
power supply and the amount of watts transmitted out of the antenna?
 

wb6uqa

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watts

Many more watts come out of the power supply than the transmitter. Make sure your power supply has enough amps to handle it.Transmitters burn up a lot of watts in heat.
 

ko6jw_2

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One horsepower = 746 watts. However, you should exercise care when connecting an antenna to your horse. The horse must be properly grounded or, at least, tied up securely.
 

jim202

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The bottom line to remember is that your radio transmitter is not 100 percent efficient.

There are some losses in the effort to take the DC power that is going into the transmitter and making the RF output that we all measure in watts. Most of this loss is turned into heat. Some is lost in the transfer from each stage of the transmitter. Some is lost in the ability to take the transmitters ability to pass along the generated power in the matching section of the output to the antenna connector on the radio.

Then you will have losses in the coax cable going to the antenna. Some people use a ladder line type transmission system on the HF frequencies to get the transmitter energy up to the antenna.

Trying to wrap your thoughts around all this, basically comes out as not all the applied DC power is passed along in a direct relation to the amount of RF watts you can measure at the transmitter antenna connector.

One way of getting a first hand experience of this is to talk on your mobile radio or portable and then put your hand on the heat sink. Show yourself that the heat sink has warmed up.

Don't know of any better way to let you learn just what this is all about.
 

prcguy

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It doesn't really dissipate energy, radiated RF behaves much like light and goes on pretty much forever if it doesn't run into something. After you get into the far field RF levels will follow a predicted pattern of dropping off 6dB (or 75%) every time the distance is doubled.

If you can imagine a light bulb illuminating the inside of a beach ball, then measure how bright a small area of the inside of the ball is, like 1" square. If you double the size of the beach ball, all the light that was hitting the 1" square of the inside surface is now spread out over 4" square in the larger ball and will appear dimmer.

If you add up all the light in the 4" square of the larger ball it will equal the light that was hitting a 1" square in the smaller ball. The same thing happens to radiated RF, it spreads out as the distance is increased and appears smaller or weaker when measured with the same size antenna at further distances.
prcguy

It's high frequency AC. They go half a wavelength forwards and then come back the same half wavelength but they dissipate their energy on the way. :)
 
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