Looks like it can be scaled back to 100 mW.
This is the exciter i recently bought and described about.
Please bear in mind that this exciter and the asmax-1 use linear final stages, so the efficiency will be low, so 100mw dc input of the final stage is likely to be in the order of 25mw rf out.
Paul.
The nice thing is that it's 50 ohms. So...you could get a QRP Power/SWR meter and tweak it for desired power.
Rich: ENTER COMMENTS HERE
Would be better to have this in a finished cabinet.
Mark
The nice thing is that it's 50 ohms. So...you could get a QRP Power/SWR meter and tweak it for desired power.
For now I'll keep trying to support the Part 15 community.
1) A transmitter designed to deliver its maximum output power into a 50 +j 0 ohm load does NOT have an output impedance of 50 +j 0 ohms. If it did, only 1/2 of its output power would be dissipated by an antenna system having a 50 +j 0 ohm load impedance -- and not all of that load power necessarily would produce radiation useful for "broadcasting."
Note that a dummy load can provide a 50 +j 0 ohm load impedance to a transmitter, while producing essentially no useful e-m radiation.
2) A typical "QRP Power/SWR meter" designed for use in systems driving a 50-ohm r-f load is accurate (to its specification) only when the load impedance it sees at its output port is exactly 50 +j 0 ohms.
3) Relatively few antenna systems used by Part 15 AM stations have a feedpoint impedance of 50 +j 0 ohms.
4) Such inexpensive meters most likely are not sensitive enough or frequency-selective enough to respond accurately to the power levels needed to ~prove compliance with Part 15 AM -- even when the details about its antenna system are accurately known.
So what is the output impdance of transmitter which is advertised as 50 ohms?
Most of the commercially made amateur equipment specifies 50 ohms.
My Nautel LPFM transmitter specifies 50 ohms.
I know if there is a mismatch between transmitter/feedline/antenna standing waves occur which reduce power to the antenna and can result in damage to the transmitter from reflected power.
I also know that 50% efficiency is the best you can get in order to transfer maximum power to a load.
Here is a link to info which relates to the idea of maximum power transfer with regard to transmitters and other.
ELECTRONIC DESIGN publication
By the way, that's why they call it a "dummy load". It not supposed to radiate.
So what is the output imp(e)dance of transmitter which is advertised as 50 ohms? Most of the commercially made amateur equipment specifies 50 ohms. My Nautel LPFM transmitter specifies 50 ohms. ...
_________
Such specs show the LOAD IMPEDANCE into which those transmitters safely can deliver their rated power, for a given set of conditions.
They do not specify the source impedances of such transmitters.
not 100mW TPO. in reality will anyone care if it's 100mW TPO vs input to final? probably not, but technically per 19.219 it is 100mW input and if you get a persnickity FCC agent who goes by a strict dry reading of 15.219 and doesn't give any leeway you may find yourself in hot water with this Tx, but for intermittant testing and properly setup legal carrier current this is an excellent transmitter, not so much so for 24/7 15.219.
i am an advocate of C-QuAM and wish somoen would make an FCC certified 100mW AM Stereo Tx that can load into either 50 ohms or a short 3m vertical.
basically an AMt3000 only with the addition of C-QuAM.
any takers on this chanllenge? Phil B./SSTRan paying attention? Hamilton? Chez Radio? Radio Systems?
Clip from page 559 of the NAB Engineering Handbook (8th Edition) --
Power Amplifier Output Source Impedance
At the milliwatt levels used in RF test equipment, it is customary to use 50 Ω as the standard impedance. Consequently, test equipment is typically designed to provide both 50 Ω source and load impedances to systems under test. This approach minimizes any reflections to systems under test since both the transmitter (source) and the termination (load) will absorb reflected energy. A 50 Ω source impedance is usually provided by placing a 50 Ω build out resistor in series with a low impedance voltage source (Thévenin equivalent). The closed circuit voltage with this configuration
is exactly 1/2 of the open circuit voltage, meaning that half of the total available RF power is dissipated in the source resistance. The best possible efficiency for this system is 50% assuming the voltage source is 100% efficient without the source resistance.
It becomes obvious that while an FM transmitter is designed to drive a 50 Ω load, it does not itself have an output source impedance of 50 Ω. In order to achieve high efficiency, the transmitter must have a very low output source impedance so that nearly all of the power is delivered to the load. The plate dissipation indirectly represents some of the power lost within the low source
resistance. Since the low source impedance of the transmitter provides a mismatch to reflected power from the load, this power is almost totally reflected back from the transmitter output stage toward the load again.
Part 15 Engineer Said
i am an advocate of C-QuAM and wish somoen would make an FCC certified 100mW AM Stereo Tx that can load into either 50 ohms or a short 3m vertical.
basically an AMt3000 only with the addition of C-QuAM.
MrBruce Said:
You might want to follow and participate in my topic regarding the Chris Cuff C-Quam AM transmitter. That topic is also open to discussions regarding other brand named C-Quam Stereo Transmitters and the curcuitry involved to make them work.
Load Impedance has been brought up in that topic. Please have a read.
Bruce.
There is an argument to be made that power does not get reflected as a result of a mismatch.