And it would only take one reflection or to the distort the intended transmit waveform. Not so good. More reflections is worse.
Actually impedance match matters primarily in the area of (RF) transmitters and not so much anywhere else. Theory only types totally misunderstand this. In transmitters it is a matter of power direction control. The most important reason for antenna matchers(couplers) is preventing the power reflections. You do not any any transmit power being wasted by reflection back into the transmitter output where it will have to be lost (dissipated).
The same way we get high efficiency switching power supplies. Very little time spent in the linear region, and energy storage to get the proper waveforms. Think it through.
On a sunny day (Sun, 12 Oct 2014 12:27:42 -0700) it happened John Larkin wrote in :
Wel, see it however you like. You mentioned antenna tuner', do you know what that is?
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An antenna tuner makes a complex impedance antenna (for example too short or too long of a part of the wavelenth) into a real impedance (ohmic in your words). It has nothing whatsoever to do with the transmitter design.
The transmitetr wants to see a real impedance, and normally one that matches the cable connecting to the (now ohmic) antenna. So your jive about antenna tunah is not releated to this issue we were talking about.
You still do not get it.
Ub | __ coupling C L (_____ | |------------------ | ia anode
cathode | ///
The voltage at the LC circuit will at the best (theoreticaly) swing from 0V to 2 x Ub. The tube can provide a current ia that varies from zero to whatever the specs say. In case of a high Zi of the tube (penthode) that gives you the impedance. That probably is a rather high value in this case, and should be matched to a 50 Ohm cable, or any other ohm cable, that will then go to the antenna. The old way was a simple transformer cuoplig (as drawn) but that leaves a lot of harmonics (say distortion) into the antenna. The modern way is a Pi filter, basically a transformer and low pass in one (the C ratio is now the transformation factor).
In case tube is a triode, those have a very low Zi, and that should be thought in parallel to the output. We are lucky these days with transistors and MOSFETS that mainly have an I versus U that goes flat, so a high impedance. This is the way it is, the way I didid, the way people doit, and probably the way people will keep doingit.
Some transceivers have the antenna tuner build in, some antena tuners look like pi filters,
The small signal output impedance of a device is totally unrelated to the maximum voltage and current that the device can handle, or swing that it provides.
Consider an EL34 tube with 400V anode/cathode voltage at a grid voltage of -10V resulting in 150 ma anode current. This means that its DC resistance is 2.67k. At Vgrid of 0V, the current is 300 ma, or a DC resistance of 1.3k. However, its dynamic output resistance is around 10k in both cases.
It is not clear to me what you are saying.
The load needs to be matched to a number that has the dimensions of resistance, but has nothing to do with resistance. The load should be "matched" to the ratio of the maximum voltage rating and current rating of the device.
I agree that you don't want reflections from the load, but I would say that what is more important then wasted power, is preventing the excessive voltage that can result from a mismatched load. This can breakdown the drive transistor.
On a sunny day (Sun, 12 Oct 2014 17:26:17 +0100) it happened piglet wrote in :
It was correct. The loss is in the transmitter output resistance. No one has claimed that there is loss in the transmission line.
There are no reflection when driving from a zero source impedance, when the transmission line has a load matched to its characteristic impedance. Go and Spice it.
No. The confusion comes from not understanding the system.
The are low resistance output because driving a speaker with a voltage source gives the flattest frequency response. If current driven, the grequency response will look something like a loudspeaker impedance plot, which is all over the place.
Yes.
Yes.
Yes.
No, it is not zout of the output device. The voltage and current swing says nothing about either the large signal or small signal output impedance of an amplifier.
VLoad = Vin . RLoad/(Rload+Ro)
For any arbitrary value of Ro, a Vin can be chosen to give the same voltage and current at the load.
The maximum voltage divided by maximum current is a number that does not physical exist. This number is the the number required to determine the transformer turns ratio to drive say, a 4 ohm load.
On a sunny day (Mon, 13 Oct 2014 18:40:57 +0100) it happened "Kevin Aylward" wrote in :
Yes I agree, but that turns ratio is what this discussion (on my part) was all about, that transformer is part of the amplifier design. It is silly to say it is not part of the design, as without it the design simply does not work.
I even showed a picture of a board where you can see that (2.4 GHz) transformer.. And it is the same for the old penthode output transformer to drive that 4 or 8 Ohm speaker, or the early 2 or 6 transistoe radios where a push pull output stage with a driver - AND an output transformer was used. Those guys knew how to get the maximum _power_ amplification out of a transistor. Modern designs just throw in 100 k not matched on chip transistors.
On a sunny day (Mon, 13 Oct 2014 18:40:57 +0100) it happened "Kevin Aylward" wrote in :
Yes I agree, but that turns ratio is what this discussion (on my part) was all about, that transformer is part of the amplifier design. It is silly to say it is not part of the design, as without it the design simply does not work.
Yes, the 6 transistor superhet radio I had when I was a lad. An OC71 rated at 20V and 10ma. On a 9V supply, the ratio would be 9/.01 = 90 ohms, or maybe 200 ohms allowing for some current margin, so a problem without a transformer. However, today cheap small signal transistors with an 1A, means not a problem now.
A source can have almost any reasonable "output impedance". IEEE has a definition of a lossless as well as a normal output resistance. The main reason you don't want high mismatch at the load end is that it not only causes some additional line loss, but in television broadcasting, the reflections can cause ghosting (before digital, that is... I don't know about now).
The dipole in radio amateur operation is very rarely 70+j0 ohms apart on some point frequencies. Usually it is something else and often very reactive on some frequencies.
The open wire 300 to 600 ohm feeder have a very low losses on HF compared e.g. with coaxial cable.
Since there are a missmatch between the feed line and the antenna, some power is reflected back from the antenna towards the transmitter. The aim is to get the generated RF power to finally radiate to the surrounding space, so the reflected power needs to be reflected back at the low end (transmitter/ATU) back to the line and again some port will radiate from the antenna.
In old tube transmitters, the pi (CLC) network at the transmitter (needed to convert the high anode impedance to some usable levels) could also be set to provide a conjugate match, re-reflecting the reflected wave back towards the antenna.
In transistor equipment designed for 50+j0 ohm output impedance use an external Antenna Tuning Unit (ATU) often a T circuit to make sure the reflected wave is reflected back to the antenna and not enter the transmitter, where it would be dissipated into heat and in the worst case destroy the transmitter.
In this configuration, some power is reflected back and forth in the feed line, but eventually radiate into the space. Since the wave moves several times through the feedline, some RF power is dissipated into heat in each pass. If the transmission line losses at the operating frequency is minimal (as in the case of the open wire line), this is not an issue.
However, using a lossy line (such as coaxial cable at higher frequencies) will attenuate the reflection in each pass and hence the final radiated power will be severely limited. This is the reason that this kind of tricks are not used on VHF+.
One other way of looking the ATU+line+antenna is to think about it as a big resonant circuit and the ATU simply feeds power into this resonant circuit. At the antenna end, some of the resonant circuit "leaks" into the open space through the antenna :-).
"Everything from the output device, tube or transistor, is suppose to pass through a series of networks that will allow for proper matching of the transmission line, and that thing attached at the end, matches the transmission line."
That does not happen using a open wire feeder to a dipole. I don't care whether it is 35+j50, 100+j3, etc., a 300 to 600 ohm balanced feed line will not match to a half wave antenna. So, in a low loss line, you accept the mismatch between the line and "that thing attached at the end" and compensate (if needed) as necessary at the transmitter.
Who cares? My statement stands. In order to have the load match the transmission line there must be a tuner or equivalent located at the antenna.
I know. Again, so what? My statement is that "that thing" at the other end does not have to match the transmission line. It's nice if the transmission line is low loss, of course.
Yes, that will happen.
I'm too tired to comment on the rest of your stuff.
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