21
Dispelling the Myth
To begin with, the typical HF SWR meter does not have the ability to separately sample the forward and reverse power, voltage, or current. Any description of the device or its circuitry that suggests this capability is flawed. We can show this empirically with two different experiments.
Experiment 1
Connect a 100 ohm resistor directly ...
12
Remember that carriers are selling bandwidth, that’s it. Actually, they are overselling it, exactly like cable modems (but that's another story). If everyone in the coverage area of a cell tower tried to make a call at once, most of them would be unhappy.
The carriers paid a boat load of money for their chunk of spectrum and their overriding goal is to ...
11
What is dBm?
dBm stands for decibels relative to one milliwatt. Decibels represent multiplicative factors, or ratios; by establishing a specific reference level they can instead be used as absolute values: 0 dBm is 1 milliwatt, 3 dBm is approximately 2 milliwatts, etc.
How do I convert it to watts?
Convert the decibel value to a scale factor and ...
11
Electromagnetic hypersensitivity does not exist. There is no confusion. It has been rigorously refuted by science.
One watt is orders of magnitude below safe exposure limits, and indeed probably less than exposure you are likely to get from cell phones, commercial broadcast stations, WiFi, and unintentional radiation from all kinds of electronics.
10
Yes, crystal radios convert radio signals to audio without any additional power.
But it is true that there is not much energy in a radio signal at any significant distance from the station, so a few tricks are involved to produce audio from that weak signal without any additional electricity:
The earphones are special models that are high impedance and ...
10
A cross-meter is capable of showing you three measurements simultaneously:
Output Power
Reflected Power
SWR
From this image by Axel Schwenke on Wikipedia, you can see that the needle on the left indicates forward power, and the needle on the right indicates reflected power. The observed intersection of the two needles can be used to indicate the SWR of the ...
10
Here's 47 CFR § 97.313:
§ 97.313 Transmitter power standards.
(a) An amateur station must
use the minimum transmitter power necessary to carry out the desired
communications.
(b) No station may transmit with a transmitter power exceeding 1.5 kW
PEP.
So sorry, you're only allowed a maximum of 1.5 kW per station, not per antenna.
I can anticipate a further ...
9
Coverage maps are a thing that is notoriously difficult to do right. There are some very complex programs (Radio Mobile is the one I know the most about) which can generate coverage maps if you're prepared to wade through the technical details and the quirks of the software. Unfortunately it's not as easy as drawing a circle of a certain size around each ...
9
It depends on the frequency range the termination is rated for, and the power handling required. High power and high frequency is very difficult.
At HF, any old non-inductive 50 Ω resistor will do. For higher power, we'd put many resistors in series and parallel, perhaps on a heat sink or in a can of oil.
At VHF and UHF I've made very good terminations ...
9
The cheapest antenna you can set up is a speaker wire dipole. A spool of speaker wire long enough to split the conductors and make a center fed half wave dipole ought to cost around ten dollars US (or less). You can get useful radiation with supports made from cheap stud lumber from the home improvement store (8-10 dollars per support to get 5 meters off ...
8
You are making an unfair comparison. HTs and amateur repeaters use the same technology now as they did forever ago. If you want to compare HT performance with cell phones, compare it with one of these:
In that light, I think HTs fare favorably. Modern HTs have grown more modern batteries (NiCd -> Ni-MH -> Li-ion), but the RF bits have seen about as much ...
8
Trying to work it out I did:
20W (in) = 10log10(20W/1W) = 13dB(W)
15W (out) = 10log10(15W/1W) = 11.76dB(W)
so, component power loss = 13dB(W) – 11.76dB(W) = 1.24dB(W)
Your error is merely in the units, not in the calculation. Taking the difference of two dBW (or any two absolute dB values using the same reference level) gets you dB.
If you want to think ...
8
I convert the 27dBm and 3dBm -> 10^2.7-10^0.3. But how can we simply subtract the two dBm
If you convert to exponential form then you must simultaneously replace subtraction with division (or addition with multiplication), so you have $10^{2.7} / 10^{0.3}$ instead of $10^{2.7} - 10^{0.3}$. Then you will see that you get the same result:
$$2.7 - 0.3 = 2.4$$
...
7
The short answer:
\begin{equation}
\frac{V_{p-p}}{V_{rms}} = 2\sqrt{2}
\end{equation}
The long answer, or how to derive the above:
As noted on the Wikipedia page for root mean square, the RMS of a sine wave is equal to its amplitude divided by the square root of two. (You can also derive this by doing the integral over a sine wave yourself.)
\begin{...
7
It's about transmission duty cycles.
For example, FM has a very high duty cycle during transmission: the transmitter's output power is constantly at 100% regardless of the amount of modulation. AM varies between 50% (carrier only) and 100% (full modulation). SSB varies with modulation between 0% and 100%. CW is like FM, 100% power output during transmission,...
7
There are a lot of ways to approach this problem, but here's one: we can calculate the power density of that field, and determine the area from which the antenna captures power, and multiply them together.
Calculating the power density
If we know the electric field to be 1 mV/m, and the transmitting antenna is distant, then the magnetic field must be ...
7
Decibels are all "ratios" at their core. A unitless dB is a simply a ratio of one number to another, perhaps input power relative to output power. We can also use decibels for absolute values, by fixing the denominator to a standard reference — e.g. one milliwatt in dBm. But the most convenient thing about decibels is that, although they are ratios, because ...
6
A properly calibrated cross-needle power meter such as e.g. the MFJ-842 actually tells you something more than just the forward and reflected power, which as you point out can just as easily be indicated by two separate instruments.
The intersection of the needles gives you a pretty good indication of the actual standing wave ratio or SWR because the SWR is ...
6
Sure, there are plenty. Unfortunately, they all seem to be selling something.
The scientific consensus is quite clear: no known risk, beyond the obvious risk of being cooked which MPE limits are set to avoid.
6
I think you pretty much summarized limiting factors in the general case. I'd add that there can be dielectric losses or breakdown at the ends, as well as the feedpoint. Common antenna designs (quarter-wave verticals, center-fed half-wave dipoles) are fed at a low-impedance node, and the ends are high-impedance nodes, so I'd expect dielectric breakdown (...
6
If you search for "automatic power off car ham" you'll find a number of solutions which don't require additional wiring. When the engine is running the voltage supply is higher than when the engine is off, so these work by sensing when the voltage is high enough, and turning the radio on.
6
While many transceivers and scanners can be powered by a cigarette lighter adapter, you typically want to run at most one transceiver in this way, and should make absolutely certain that you know where that lighter is getting powered from. Most car fuseboxes are relatively easy to access, and most owners manuals will tell you what fuse is associated with ...
6
This is not commonplace. Most operators in the UK get along fine with the 400W limit. Given the path losses inherent in Earth-Moon-Earth - hundreds of decibels - the antenna is a much more important factor than is the power. You need every extra bit of push you can get, but if you're not making it at 400W, 1500W isn't likely to be much better.
Remember that ...
6
Think of the "power" setting on the radio as more of a guideline than a rule. It should be more or less accurate, given reasonable inputs.
Few radios actually measure the output power and adjust gain to achieve the desired power. More likely, the power setting simply adjusts the gain of the radio's power amplifier with no feedback whatsoever.
Ideally, ALC ...
6
The link is actually balanced, because the base station receiver has diversity gain.
2G and 3G cellular systems use two antennas at the base station (for each coverage direction). They are either spaced a few metres apart, or more commonly, +45 and -45 degree polarisation, in the same housing. With two antennas and two receivers, the base station has a much ...
6
Here's what you are thinking: if the radar station and the target are 1km away, then the distance there-and-back is 2km. If the distance doubles to 2km, then the distance there-and-back is 4km, just twice what it was. So the there-and-back distance also doubles, so you should need only a 4x increase in power, right?
If the radar target were a flat reflector ...
6
I'd speculate a large part of the justification is that amateur stations, unlike commercial broadcast stations, are not periodically certified to meet spurious emission regulations. Or any regulations at all. While the station operator is responsible for ensuring the proper operation of his or her station, realistically amateurs will not always have the ...
6
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 ...
6
Think about it this way.
27 dBm means 27dB above a milliwatt.
Take 6dB away.
Now you have something that's 21 dB above a milliwatt.
Or, 21 dBm.
6
If the frequency counter is measuring -5.3 dBm, then before 20 dB of attenuation the power was 20 dB more than that, so 14.7 dBm.
"dBm" means decibels relative to 1 milliwatt. 14.7 decibels can be converted to a ratio like so:
$$ 10^{14.7/10} = 29.5 $$
So 14.7 dBm is 29.5 milliwatts.
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