The SWR seen in a RF transmission system is caused by reflections due to impedance mismatch, for example if an antenna is not matched to a transmission line then the reflection is caused at the point of connection between the line and the antenna. Part of the forward wavefront is reflected back toward the transmitter and from the forward V or I and reflected V or I the SWR is determined.
In a system where the other than 50 ohm antenna is connected to a transmitter "designed for a 50 load" where does the reflection take place?
What I am getting at is what does a transmitter "designed for a 50 ohm load" mean in terms of how the transmitter is designed?
Neil
SWR is Standing Wave Ratio. That is, the ratio of forward to reflected power. If you want to understand Standing Waves, check this out:
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If you tie a ten foot rope to the wall, and whip it up and down, you can create standing waves...fascinating study...
Doug
The antenna can be further adjusted to achive the matching 50 ohm impedance resulting in a 1:1 SWR.
Most of the real value of a Part 15.219(b) antenna feedpoint impedance consists of r-f losses in the loading coil (wherever it is located) plus the path to r-f ground.
The other component is the radiation resistance of that monopole, which for a compliant system won't exceed about 0.12 ohms.
So there is little effect available from "adjusting the antenna" to achieve a 50 ohm match over that 30 to 100 ohm range.
Also - transmitters commonly drive a load impedance of 50 ohms, but this doesn't mean that the source impedance of that transmitter should be 50 ohms. If it was, half of the power output of the transmitter would be dissipated within the transmitter.
Rather, most transmitters have a very low source impedance -- on the order of a few ohms. Then most of the available output power can be dissipated in whatever load the transmitter drives.
This is a difficult question to answer w/ a short reply because of the many factors involved in both creating a scenario with impedance mismatch and measuring the resultant SWR.
Assuming a transmitter w/ 50 ohm output impedance:
1. What is the impedance of the feedline?
2. What is the physical length of the feedline?
3. What is the electrical length of the feedline (i.e. the feedline velocity factor)?
4. What are the feedline losses at the operating frequency?
5. What is the operating frequency?
6. What is the antenna feedpoint impedance at that frequency?
7. Where is the SWR measured? At the transmitter? At the antenna feedpoint? somewhere in between?
Typically the SWR is measured at the transmitter output. Reflected power (indicated as high SWR) could be caused by any of the factors listed above. Hunting down the actual cause of the reflection has to take into account all the factors.
First, a comment about post 17, SWR is by convention VSWR which is a voltage (an amplitude) ratio. It is not a power ratio. There exists a PSWR but it is defined as the square of the VSWR but doesn't relate to the power in the wavefronts. Your example of a shaken rope is good and, in fact, I have used it in the classroom but the height of the waves moving along the rope are amplitudes analogous to voltage.
Post 19:
You are certainly correct that the SWR depends on many factors and you listed many such valid points, but, to focus on my question, suppose a perfect 50 ohm antenna is fed by a 50 ohm line and thus the SWR at the transmitter end of the feedline is 1:1. What makes a transmitter "designed to drive a 50 ohm load" actually designed to drive a 50 ohm load? In such a case are there any reflections and if so where?
You might guess that I am leading somewhere with this so help me think this through.
Neil
A perfect 50 ohm antenna w/ a perfect 50 ohm feedline connected to a perfect 50 ohm transmitter would have a perfect 1:1 SWR. How do we know this? Because that is what we described with those components. It cannot be anything else.
A transmitter is designed, using the proper component values, to couple its internally generated RF power to an external load. That load is comprised of the feedline and antenna. If the output impedance of the transmitter is not 50 ohms, then there is a mismatch and SWR would occur.
The measurement (symptom) of that mismatch may be measured in a number of ways (one does not need an SWR meter). One obvious indicator is when the final stage transistor melts or pops because of the excessive reverse current flow. A more meaningful measurement could be made by observing the waveforms on the output components using an oscilloscope. If the observed waveforms do not correspond to the expected waveforms per the design and the math used to select components, then there is a problem.
I'm not sure how else to explain this........
Mike
Mike, you are straight on with what you posted and I am not trying to argue with anything you wrote, rather I would like to focus now on the output stage of the transmitter. It took quite a while for me to wrap my brain around this and I would be interested in what you and others think.
If the output Z of the transmitter is 50 ohms and it drives a 50 ohm load then half the power input to the final will be lost in the final as heat so this is not what we want to do. In fact, for best efficiency the output Z should be zero.
You correctly stated that the output Z of a common emitter transistor is set by the collector resistor and this is where half the power is lost when the load equals this resistance. What if we could make the antenna the only resistive load in the collector circuit? Then all the power will go to the antenna.
Here's what we can do: eliminate the collector resistor and replace it with the Z of the load transformed by a matching LC network to be appropriate to the V and I needs of the final transistor. This way almost all of the power is delivered to the load seen by the collector which turns out to be the antenna. In other words, the antenna load is the collector "resistor".
This is the approach I used when I designed my "high efficiency" part 15 AM transmitter and it works to give an efficiency of 85%. The needed collector load turned out to be around 120 ohms when operating at 100 mW. The losses in my design are primarily in the toroids used in the matching network and the choke used to supply collector current...
This approach also worked for some VHF power amplifiers which I designed and built yielding efficiencies around 70%. In these amps, the required collector Z is in the tenths of ohms and the high currents cause losses in the matching coils.
It might be something to pursue in getting your SSTRAN to work with a 50 ohm load by supplying the collector current through a choke and transforming the 50 ohm load to the Z needed at the collector for correct operation. The calculation to estimate this collector equivalent resistance is simple but it needs to be changed according to the particular circuitry used.
Going ahead with your planned antenna with the toroid you added is probably the best next step but maybe later some more changes could be made.
What do you think?
Neil
Neil,
Your appraoch to loading the final stage of an RF amplifier is intriguing. I had thought of something similar when making my modifications to the AMT3000. The choke L8 on the collector of Q5 could act as the amplifier output impedance if R18 (820 ohm) is removed. In fact, I believe that is the case when the SSTRAN instructions for the base loaded antenna are applied.
However, I reconsidered after calculating the inductive reactance of L8 at 1600 KHz; it's about 10K ohms. With R18 removed, that increases the voltage drop across L8 but also limits the current flow considerably. I'll need to go back through the calculations to determine if 100 milliwatts is still achievable.
As for my center-loaded antenna; I will keep this forum posted with updates as a build and deploy the antenna (weather permitting).
Mike
Center loading is nice. I made mine by warppin a 4" PVC pipe with aluminum flashing for bandwidth.
Below is a NEC4.2 analysis of the performance difference between base loading and center loading a 3-m Part 15 AM monopole.
The plots show that even though center loading produces somewhat higher SWR and somewhat narrower SWR bandwidth than base loading, the center-loaded system produces a somewhat higher groundwave field at 1 km than the base-loaded system.
The reason for this is related to the greater average value of antenna currrent along the length of the monopole (for the same applied power), when using center loading.
Neil,
I came across this webpage recently. It appears to support your theory of a choke-only output stage.
http://electronbunker.ca/eb/AntennaMatch.html
Mike
The comments sections in the NEC graphic show the loading coil in the wrong units of measure.
The numerical values are correct, but the units are µH, not pF.
Below is a corrected version...
Rich,
Nice work. That looks like a gain advantage of about 2 dB for the center-loaded antenna. I assume the NEC calculations are with a nominal 50 ohm feedpoint impedance? Would you be able to recalculate with the SWR of both antenna models adjusted to 1.1:1 for a better comparison?
Thank you
Mike
Mike - Once the loading coil has offset the capacitive reactance of the antenna system then the feedpoint Z consists only of the ohmic sum of the radiation resistance, loading coil r-f resistance, and the r-f resistance of the earth ground in use
For these NEC studies those are shown as 84.5 ohms for the base-loaded version, and 103 ohms for the center-loaded version.
The SWR values shown in the graphics are based on a 50-ohm source (transmitter) driving those systems, as you are wanting/working to provide for an SSTRAN transmitter.
Matching a 50 ohm source to the 84.5 ohm and 103 ohm antenna system feedpoint impedances would take a suitable network (other than a loading coil), and/or changing the length of the monopole.
Mike, thanks for putting up the link on antenna matching. The writeup does support making the antenna system the only resistive load for the output stage so the final stage losses are minimum. It should be noted that the idea of a "50 ohm output imedance" for a transmitter is not valid. It is correct to say that it is designed to drive a 50 ohm load or whatever other load impedance is used. The key is to match the high impedance at the collector or plate to the load impedance which the author has done with his Pi network.
Also, note that he chose 1MHz as the transmit frequency which requires much higner inductance than would be needed at the top of the MW broadcast band. Reducing the L means less I squared R loss in the coils.
Designs such as are needed for high efficiency do not happen in an afternoon but the time and effort spent to get it right is worth the effort. Please keep us posted on the progress of your project, even if there are dead ends for this is a good way to learn for all of us.
Neil