The usual design path for RF power amplifiers is to find the load resistance that dissipates the target power level with Vpp a little under twice the power supply voltage. An impedance 'matching' section is then inserted that transforms the feedline and antenna impedance to that resistance.
Looking back into the power amplifier output, the impedance seen is then not at all equal to the feedline impedance.
Look up Helge Granberg's RF power amplifier design application notes, if you don't believe me.
On a sunny day (Sun, 12 Oct 2014 14:25:05 +0200) it happened jeroen Belleman wrote in :
That is silly, the impedance matching network, be it in form a a Pi filter of transformer is always part of the transmitter (design), look at this nice 2.4 GHz board:
formatting link
I'd say you are way of your rocker if you claimn the output matching circuit is not part of the design. I have done high and both low power RF amps, and the output coupling is ALWAYS part of the design.
That also goes for any audio amp, the tubes and some transistor amps used transformers, and sure the transfomer is part of the design.
No, it doesn't. A transmitter can have a proper match (defined through various engineering tradeoffs such as maximum power output, efficiency, compression point, SWR tolerance, minimum distortion, etc.), AND have s22 values wildly different from 1.0 + 0j.
The non-RF statement being, consider an audio amp with Zo > R_L. These achieve their goals just fine, despite disobeying the power transfer theorem by a huge margin.
The power transfer theorem is nonetheless true in these examples, as long as the conditions for the theorem remain valid. You'll draw a damn sight more power from a low-Z audio power amplifier at, say, 0.03 ohms, than you will at 8 ohms, as long as the Thevenin equivalent source voltage (or Norton current) remains constant. It's when other engineering considerations cut in (like peak current capacity, or RMS ratings of the amplifier) that the theorem goes out the window, because that part people keep forgetting: it's explicitly for linear systems.
If the transmitter output impedance is the same as the antenna and line (say 50R) then sure it is *matched* but the efficiency is at most 50%. The existence of class C or class E output stages clearly shows DC input to RF output efficiencies above 50% are possible. It is better to say the transmitter is designed such it operates at optimum when loaded by (say) 50R but its output impedance is not necessarily anything near 50R.
On a sunny day (Sun, 12 Oct 2014 17:26:17 +0100) it happened piglet wrote in :
No that is not correct, transmission lines with some impedance can be lossless, you still need to match to that impedance to prevent reflections.
In an other way, and I think the whole confusion comes from the low Zout audio amps that people say work better because of better damping of the speakers, in a radio transmitter you need to drive an antenna, the antenna has a certain impedance, it needs to be coupled to a transmitetr with a cable of the same impeddance, or any other impedance via a transformer / balun what not. Do you can leave out the cable and say drive antenna directly, via any transformer you like, but that transformer would then have to be such that the output stage of the transmitter works in a range where it can actually provide some power at minimum loss.
That comes from the output topologie as you already noticed.
For a simple single stage you have supply voltage and current swing -> for zout of the output device (transistor, MOSFET) being very high (say penthode like0, these 2 things, voltage and current, gives you the impedance. Now you know how to design your transformer turns ratio or Pi filter values to match the impedance you drive. Its not like an audio complemantary emitter follower with lots of feedback to make it look like zero ohm (it of course is not, connect a zero ohm load and see the smoke).
It is more like that good old 8 Ohm speaker via an output transformer in the anode of a EL84 or whatever have you 'merricans for standard output tube, that turns ratio is simple math, and the transformer part of the amplifier (design). Even then feedback is sometimes used from that transformer secundary.. All part of the design.
Yes, of course, but its purpose is *not* to make the transmitter output look like a Z0 source, but to make the Z0 load look like whatever is needed to satisfy the constraints imposed by the required output power, the active device used and its power supply.
Again, look up Helge Granberg's notes, for example.
On a sunny day (Sun, 12 Oct 2014 14:25:05 +0200) it happened jeroen Belleman wrote in :
Yes, but it appears that there is confusion on just what "matching" means.
"Match" strictly means "The same". Does the colour of the jacket match the trousers?
"Match" in electronics, often means, "Getting some optimum pair of values"
The fundamental principle of "matching" a tube amp to its speaker load is "What is the maximum voltage and current ratings of the output device" ? Period.
Matching has absolutely noting to do with frequency response or maximum power transfer.
Say, a tube has an voltage rating of 500V at 50ma. Its supply voltage would then be set to 250V, so that it would swing up to 500V and down to 0V (ideal for simplicity), in a transformer output stage.
To get 50ma into the load it would be 250V/50ma = 5k ohms. Thus a transformer is required to transform 5k to the standard, 4, 8 and 16 ohm taps.
The fact that there is some nominal output resistance of the tubes, is quite irrelevant.
A 2N3055 at 60V/15A is 4 ohms, hence why no transformer is required for a bipolar amplifier.
I don't know where you are getting that idea from, but I believe you're off the mark just a bit, especially with the class C amp.
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.
I raised in the old school of things! TUBES with PI"S! :)
On a sunny day (Sun, 12 Oct 2014 08:50:37 -0500) it happened "Tim Williams" wrote in :
If a zero source impedance drives a transmission line loaded by its characteristic impedance, the system will have a pure flat frequency response, with no reflections. Period. There is no theoretical need to have the source matched to the line. However...
A very low source impedance, say from an ASIC, invariably will have drive inductance, say 2nH. At 5 GHz, this is 62 ohms. This will reduce the voltage going across say, a 50 ohm load by quite a bit. So, stick a capacitor in series and cancel its reactance out. If capacitance is attenuating a 50 ohm source, stick a parallel inductor across it. This is called "matching" in RF, but it isn't really. Match, strictly, means "the same".
Essentially, in a nutshell, all PA "matching" resolves down to eliminating the effects of capacitance s/c the signal, or inductance o/c the signal, by cancelling them with its opposite, and getting the optimum ratio of V/I consistent with device ratings. No one in their right mind is going to deliberately add a real, power using resistance to "match" a radiation antenna resistance, and thereby waste half the power down the toilet. Indeed, there is a significant effort going on in producing 40MHz BW switching regulators to use in envelope tracked RF PAs to improve their efficiency. Unfortunately, much of RF waffle, hides the true nature of what the big picture view of the plot actually is.
On a sunny day (Sun, 12 Oct 2014 18:48:26 +0100) it happened "Kevin Aylward" wrote in :
Kevin, (once?) 100% agree. that is why I mentioned 'penthode' as its output impedance is (for reasonbale values of it) infinite.
But if you needed to drive say 100 Ohms, you would need one. This was about RF output amps, it obviously does not deliver much power to drive a 300 Ohm antenna with a [transformed to] 4 Ohm output amp as the voltage will be too low, lower then when you had transformed it 'up'. its logical.
Can anyone explain what Jan is going on about "penthode"? I can't find this anywhere in Google searches. Every page they link to is actually about pentodes. I even tried translating it and nothing shows up except pentodes. I had asked him about this once before and he claimed he was talking about penthode and not pentode. Is he just using his own spelling but is unable to explain that?
Makes no sense. The antenna should be largely ohmic at the radiation frequency, to accept RF from the transmission line and launch it into space. If it does reflect some power back up the transmission line, back into the transmitter, we'd *want* the transmitter to re-reflect it, not absorb it, to let the antanna have another whack at radiating it.
If you want reasonable efficiency, any generator, DC to RF, should be lower impedance than the load that it drives. Conjugate-matched impedance implies < 50% efficiency. You don't want that in a power station or any non-trivial transmitter.
--
John Larkin Highland Technology, Inc
jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com
Exactly the point I tried to make earlier. The fact that all the time we have transmitters exceed 50% efficiency means they cannot have a matching 50R/75R whatever resistive output impedance. After all it is the load that creates a reflection if the line is mismatched. And nearly all the time the transmitter Z is not the same as lines Zo. It is the load-to-line mismatch that first creates the return.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.