Radio TX Power: 1A @ 1V, or?

[Ablice]

Heylo,

Had a quick question, and don't see a General Discussion forum for this type of question.

When I key in on a two-way, let's use my Tekk NT-10s for instance, I am transmitting 2 watts of power into space.

Is that ERP 2 amps at 1 volt? 50 volts at .04 amps? How is the power proportioned with transmiting power? What about those TX only broadcasting towers with 1500W power? Are those 3000v at half-amp?

Because power (watts) is voltage x amperage, I was just wondering how it was divided.

 

[davidgcet]

you know the impedance(resistance) and the power, do the math. V= sq rt of PxR. so (if my math is correct at this late hour)10 volts for 2 watts of power into a 50 ohm antenna. 2W/10V=0.2A. so in your example 10v and 0.2 A. you can also google some dbm conversion calculators, 2w is 33dbm.

 

[Ablice]

That's a good response, except I have no idea of the resistance of the antenna. Measuring it (standard duck) gives me 0 resistance. That's not quite logical, but who am I to argue with a volt-ohm meter?

 

[JSixpack]

In many instances the voltage is limited. A mobile unit is probably running on 12V, a handheld on 6V etc. I suspect that max voltage is used to reduce current load on individual components

The bottom line is probably that power = energy/time. 2A at 1V and 0.04A at 50V probably produce the same results.

Would imagine that for high powered units there are optimal voltage / current considerations. (What they are is beyond my expertise).

Not to get too far from the subject, another thing to consider is that transmit power is generally "effective power". Different antenna configurations vary in regard to power efficiency and this is taken into consideration when stating power ratings. There's a good discussion on this somewhere in (I believe), the Antenna and Coax Forum.

Your voltmeter isn't wrong. DC resistance is probably zero. Impedence needs to be measured at your transmit frequency.

PS: I thing that davidgcet meant to say voltage = sqrt(power / impedence). (Neglecting the phase angle of course)!

 

[Ablice]

Huh. Maybe someone with a ham rig can check it out? You'd only need to measure two variables to find out the other.

No matter how I figure it, it doesn't quite compute. 50v/.04a means the similar thing to high voltage power lines over long distance, in regards to high voltage/low amps compared to low voltage/high amps. (The high voltage style nets less loss for the distance).

 

[davidgcet]

to find voltage it is the square root of (power TIMES resistance). as you stated, unless you want to do a lot of math you neglect phase angle and have to just substitue impedance for resistance. 2 way radios use 50 ohm impendance and the values i gave above are for an isotropic antenna at 50 ohnm impendance. gain antennas will be different, as will mismatched antennas.

ever seen a low band radio light a flourescent tube from 3 feet away? that will give you an idea of the voltage from a 100w low band radio! if the antenna is right you can light it up several feet away, if it has high SWR you may almost have to touch the bulb to the antenna rod to get full brightness. and no, this is not an approved method of checking SWR!

 

[zz0468]

Is that ERP 2 amps at 1 volt? 50 volts at .04 amps?

ERP means "effective radiated power", which is a comparison to the intensity of a signal between a theoretical reference antenna, and the actual antenna being used. Probably not the term you need to use here.

What's missing in your calculations above is the impedance of the antenna. In most cases, the target impedance is 50 ohms. You could use an rf ammeter and measure current directly, and voltage, and calculate the power. You could measure the current and voltage, and calculate the antenna impedance. And so on...

But it really gets more complicated than that. There are variations of ohms law that deals with ac circuits, so you should be using those formulas, not the dc ohms law formulas. The dc ohms law formulas can only be used on resistive circuits, and would represent the effective values of the ac waveform, not the peak values.

So, you could use that to determine the required value of, say, a dummy load, but you could be misleading yourself by grossly underestimating the required breakdown voltage in a capacitor to be used in an antenna matching network.

How is the power proportioned with transmiting power? What about those TX only broadcasting towers with 1500W power? Are those 3000v at half-amp?

That's not unreasonable. In a high power transmitter, you will see very large ac voltages and high currents in the antenna system.

Because power (watts) is voltage x amperage, I was just wondering how it was divided.

Essentially correct, but there are several ways of expressing an ac voltage... Peak, peak-to-peak, RMS, average... with DC, it's just one expression - volts. A complex ac waveform such as a modulated radio signal generated by a transmitter could be correctly expressed in a number of ways. Which way you calculate could have vastly different answers and all be correct, provided you understand the difference.

 

[kb2vxa]

The question as it stands is unanswerable as it depends on phase angle, or in other words where along the transmission line or antenna measurements are taken. Power measurements are usually taken at the 50 ohm point using an RF ammeter as the reading easily plugs into the formula but that's a simplistic answer, you have to be a serious RF engineer to understand the finer points that are WAY beyond the scope of an internet discussion forum. Besides, the board software can't handle the Greek symbols of complex math. Ohm's Law only works for DC and VAR only for AC power frequencies, RF is a whole 'nuther ball game.

 

[rfradioconsult]

To answer your question you need the requisite prior 2 years of EE courses to be able to feel comfortable understanding transmission line theory and have basic knowledge of a Smith chart. DC circuit theory is different than AC circuit theory, while RF theory is different from the first two.

 

[Ablice]

Man...I thought it would be that simple, like calculating the speed an engine given the ratios of the 'boxes, the size of the tires and the speed. (And the gear selected)

Doesn't anyone have that answer somewhere buried in the internet? Or does it vary from Tx'er to Tx'er, such that I should fill in that question in my brain as "It is really complicated, and it depends"?

 

[davidgcet]

I should fill in that question in my brain as "It is really complicated, and it depends"?

that is your answer right there. in reality unless you are an engineer you don't have a need to even contemplate it.

 

[zz0468]

Doesn't anyone have that answer somewhere buried in the internet?

Depends on what you're really looking for. The generalities are there. Yeah, there are formulas for calculating RF currents, voltages, powers, and impedance's.

Or does it vary from Tx'er to Tx'er, such that I should fill in that question in my brain as "It is really complicated, and it depends"?

Yeah, the specific details will vary from transmitter to transmitter. There are too many variables. The known formulas help you figure out the details. Where it gets complicated is the quantity of details that need to be known in order to account for everything. It's information that one gets by reading lots of thick, dry, technical books... not some hidden web page full of shortcuts. And some of the details are gathered by measurements of the actual device in question.

So, in the end, if your question is "is it possible to calculate the voltage and current in an antenna", yes, absolutely. If the question is "is it fast and easy?", then the answer is 'probably not'.

 

[rfradioconsult]

Thick and dusty, especially when you trying to work through it with the help of Pickett. The first handheld calculator came on the market about 4 years after I was out of school and you could do all 4 functions; a bargain at $99.00.

 

[gmclam]

P = e2 / z

Man...I thought it would be that simple, like calculating the speed an engine given the ratios of the 'boxes, the size of the tires and the speed. (And the gear selected)

EVERYTHING in life except death and taxes is complex, and I'm beginning to think they're complex too. And, especially with the Internet, lots of people try to make complex things look simple. If it is a GENERAL answer you want, OK. But I am not sure just what you will be able to do with the information.

We're talking RF (Radio Frequencies) here, so it is AC, NOT DC. That means we're talking IMPEDANCE, not resistance. You can't directly measure impedance with an ohm meter, so don't expect to see "50" when check your antenna with one.

A GENERAL antenna impedance for the transmitting equipment and scanners I own is 50 ohms. My TV antenna equipment is 75 ohms, but that's another discussion.

Power can be calculated by: P = EI (or power equals voltage times current)
or by P = E2 / Z (or power equals voltage squared divided by impedance).

What about those TX only broadcasting towers with 1500W power?

1500 = E2 / 50, so 1500 * 50 = E2, so E (voltage) = 273.8 volts.

Are those 3000v at half-amp?

No. 1500 = 273.8 * I, so I = 5.47 amps.

 

[zz0468]

Power can be calculated by: P = EI (or power equals voltage times current)
or by P = E2 / Z (or power equals voltage squared divided by impedance).

1500 = E2 / 50, so 1500 * 50 = E2, so E (voltage) = 273.8 volts.

No. 1500 = 273.8 * I, so I = 5.47 amps.

Yes.

But don't forget the part that many broadcasting towers are something other than 50 ohm's resistive. THEN it gets complicated.

 

[gmclam]

But don't forget the part that many broadcasting towers are something other than 50 ohm's resistive. THEN it gets complicated.

And whether or not they're using several towers to get a directional pattern and the phasing between them and so on.

I thought perhaps the OP was talking about amateur or public safety transmissions, but then I wonder about the 1500W example he's asking about.

 

[Ablice]

Cool lads, I wasn't too interested in the broadcasting towers, I knew that those would be a different matter simply by the power figure.

I was mainly interested in handhelds and mobile units. So given that gmclam posted some very basic figures...

2watts = (volts)2
.................50z
so volts = 100w/z= v2 ; thus v = 10 (how nice!)

power (watts) = volts x amps
2w = 10v • xamps
x = .2 amps

Bam. My Tekk NT-10 is running approximately 10 volts at .2amps at the 50z impendance point in the antenna. A 5W GMRS unit will likely be running close to....

250w/z = v2 ; v= 5√10 which is about 15.8 volts.
5w=15.8 • xamps ; so about .316 amps.

Cool. May not be precisely accurate, but gives us a general idea about what the ratios for the power are. I won't ask the same question about microwave oven power (but I figure it's very high voltage), so rest easy.

 

[kb2vxa]

What a bunch of show-offs! (;->)