Playing along with the idea that there is some meaningful fixed Z of the device for large swings, yes you would have to do so to prove the concept. You would need to prove that output Z was the same for driving 1 W into the output as for driving 100 W into the output. I also predict that even the small signal output Z of the power amp will not be that conjugate impedance you think it is for a properly designed PA. (I am not making a claim that it would *never* be so for any PA.)
Likewise, a change in the output Z would do the same thing. Since you're presuming linearity, we can include gain linearity. I.e., the gain with "-10 dB" of drive is the same as the gain with "0 dB" drive. I'll define the 0 dB gain as associated with the 1 db compression point. Since the gain is defined as linear (really fixed regardless of drive), and the load is fixed, something must have "caused" the compression. A way to *model* the compression is a changed output Z as a function of drive. While I realize this is an unconventional view of output compression modeling, I believe it is fair, since you are making the linear presumption. I think this is fair also because the impedance concept is a linear/sinusoid one. Under that presumption, you've given me license to disregard distortion.
Let's do another example.
Say the device we've selected has an Imax rating of 1 amp and a generator resistance of 100 ohms. Per standard linear theory, we do our norton model of Igen in parallel with the 100 ohms. Under standard conjugate matching theory, we should load it with 100 ohms. Now with the 100 ohm load, we get a 50 V peak for Imax = 1 amp. But what if both our DC supply and device breakdown won't allow this? We have a practical limiting Vmax not at all included in linear theory. Due to breakdown or supply rail concerns, we'll see our Imax quite short of the 1 amp we expect when the device is loaded with 100 ohms. We won't be getting all the power out of it we "expect" because of practical limitations not built into linear conjugate matching theory.
How do we select the best load, since conjugate loading clearly does not use the device to its full potential? We seek Ropt, or what is commonly referred to as the load line match.
No, the purpose of the power amp is to deliver power, not extract it.
Don't bother with the over simplified Class A case. RF power amplification is rarely done class and and it is a digression from the actual topic.
[...]
At some point as you decrease the resistance, the output will drop to zero as the amplifier fails or it will start to decrease in some more controlled manner as the protection circuits take control. If we assume the latter case, it is easy to see that the power reaches a maximum value and then decreases as the resistance is lowered. The point at which the power is at the maximum is the point at which the load is matched. If you make a small change in the load and observe the voltage and current when that small change is made, you will see that that is indeed the output impedance of the amplifier. I think this is the part you are not grasping.
Yes, I stand by and have just in another part of the thread once again explained that indeed the impedance is matched. ie: If you make a small change in the impedance in any direction the power decreases. Increasing the resistance is the obvious one. The other three are because the protection circuits act. The OP had a completed transmitter he was connecting to a length of wire.
You are right. I really needed to be more clear in the first posting I did. That bridge has now been crossed and this is getting tiresome. If the OP doesn't come in with more questions, I'm out of here.
Well really, once the device and supply have been determined, we can indeed view it as extraction. We'll load it to extract the most. If you want to mince words and call it "deliver," that is fine.
And
Well, class A is certainly done. Two cases are where the extra little bit of linearity is desired and at high frequencies, were PAE starts to take a bite as the gain drops below 10 dB.
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No, this is exactly where I'm saying you are incorrect. You are not getting the practical limitations and are mistakenly applying linear concepts. It doesn't work if you want to extract maximum power from the DC supply through a real device, converting the DC power into RF power.
Driven to max swing, this is true. But it is because of asymmetrical clipping, not because of conjugate mismatch. For lower drives, what you say won't necessarily be true *unless* you've mis-designed according to conjugate match ideals. Your argument is circular.
If you design for conjugate match, you're right. I'm saying: don't do that. If I design for load line match and you design for conjugate max (both pf us using the same device and supply), I will get a higher peak power than you will. However, you'll get to be right about how your amp acts regarding diverging from conjugate load. But it is irrelevent: you made a fundamental mistake.
The troublesome assertion is not the negative one. It is that RF PA's are conjugate matched. Neither you nor Ken has provided a single example of such a design that also extracts the maximum amount of "linear" power from a device and essentially its power supply (after all, that is what it is: a _power_ amp). Your example said nothing about output-Z, which suggests you have no clue, since you didn't even remotely address the issue.
For Ken's part, he recently obfuscated by dismissing an example that was primarily intended to be illustrative, but yet holding the salient points. He completely ignored (or didn't understand) the clipping issue. Further obfuscation was provided by talking about "protection circuitry," which may or may not exist in a circuit, but adds zero to a discussion regarding how the PA is to be loaded. "Protection" is a non-stater because the PA is either off or impaired.
Ken's argument is circular. He say's that if a design is done for conjugate match, then it will behave as if it is conjugately matched. Well of course (or at least sort of under specific test conditions and circuits)! It is self-fullfilling prophecy but it unfortunately makes no statement regarding obtaining the maximum power out of the circuit in the sense of turning DC power into RF power (yes, *extracting* power from the DC supply and transformed to RF). This is paramount to PA design. To use the device to maximum efficacy, as Cripps puts it, a load-line match is needed. Ken's "conjugate match" design won't do that, and that's why PA's aren't designed that way.
The bottom line is that if I design an amp via load line techniques using the same device and power supply as Ken (him using conj-match), my amp will deliver higher unclipped PEP than his. That is the factual result you resist. Now if you want to pay for extra power and big devices, that's your business--go ahead and attempt to conj-match your amp--but engineers who design PA's don't do that.
Another idealized and hypothetical example to elucidate the load-line principle is offered.
Let's say we have a 10 W FET we'll build into a class A circuit. An RF choke is used to supply drain current. We DC bias it to Vd = 10 V and Id = 1 A. Just for argument sake, let's say it has a constant internal resistance of 110 ohms and the device will break down at 25 V. According to the most idealized and standard load-line theory, we should load it to rL = Vd/Id = 10 Ohms. This idealization includes the definition of positive and negative clipping -- whichever comes "first" -- of being the operational limit for output voltage swing. Clipping is associated with severe distortion.
Since we need rL to be 10 ohms, and Ri = 110 ohms, we need to make the actual load resistor equal to: RL = 11 Ohms. Let's check that result and see if it meets the clipping constraint for maximum available power.
positive swing = Id*rL = 1*10 = 10 V negative swing = Vd = 10 V Power delivered to RL: Pload = 10^2/(2*11) = 4.55 W The efficiency is a little under 50% because of the internal resistance. Note the Load resistance is decidely not the conjugate of the internal resistance.
Let's spot check the load to see if it at least appears to be the peak available power, by testing two loads "immediately" on either side of our optimum 11 ohms.
Let RL = 10 ohms positive swing = Id*rL = 1*9.17 = 9.17 V negative swing = Vd = 10 V Since we positive clip at 9.17 V, we are limited by our design clipping constraint to only driving the PA such that 2*9.17 V is the maximum available voltage swing. Power delivered to RL: Pload = 9.17^2/(2*10) = 4.20 W
Let RL = 12 ohms positive swing = Id*rL = 1*10.82 = 10.82 V negative swing = Vd = 10 V Since we negative clip at 10 V, we are limited by our design clipping constraint to driving the PA such that 2*10 V is the maximum available voltage swing. Power delivered to RL: Pload = 10^2/(2*12) = 4.17 W
Sure enough, the power peaked at a load of 11 ohms, just like load-line theory says it will. Now let's see what the available power hit of conjugate matching is.
By definition, conj-match insists RL = Ri = 110 ohms. Again we are limited in our clipping constraint by static drain current, and supply voltage, specifically 10 V.
Our negative swing limit is, as ever, 10 V (the drain voltage).
positive swing = Id*rL = 1*55 = 55 V
This would breakdown the device, but the lower negative swing will force us to back down the drive to meet the design defined clipping constraint.
Pload = 10^2/(2*110) = 0.455 W
Conjugate matching resulted in a 10*log(0.455/4.55) = 10 dB available power hit. Power amplifiers are not designed with conjugate matching in mind. You don't need to re-invent the wheel. Just follow well established principles when doing cookie cutter PA design.
LOL. Given your pattern, I am sure it will.
Not really. The problem isn't precision, it is you can't, or refuse, to comprehend what is being said, which I presume is why you instead write with the most bizarre terms and phrasology that has nothing of import to the topic at hand.
For what seems like the billionth time now: they are load-line matched.
I've given a didactic example (actually a couple), you just don't--or more likely won't--get it. If you don't like my example, you can refer to Cripps, who is considered one of the preeminant RF PA experts in the world.
Even more simplistic is Malvino's discussion on pp177-185 of the first edition ((c) 1968) of "Transistor Circuit Approximations." It is basically a technician level description, so perhaps it is well-suited to you. In academics, load-line theory is presented down to tech level courses and up across to engineering. That some engineers and techs aren't clear on the load-line concept for PA's (or
*any* circuit needing a wide symmetrical swing) is notwithstanding.
I see you still don't know what impedance is. In any case, it doesn't mean that looking into a properly designed PA output with a network analyzer confirms the conj-match precept, it doesn't.
Impedance is a *linear* conception, a portion of linear theory, and again by definition:
Z = V/I
V and I are sinusoids (phasors). But with power amps, substantial non-linearity exists (destroying the linearity assumption of impedance), thus applying a linearly defined concept to a non-linear milieu is a misapplication. You are attempting, as is Ken, to stuff a square peg down a round hole. Why?
The concept is even questionable for the most linear of the power amps: class A. In any case, given real devices with real supplies, the conj-match ideal is next to worthless. While I could agree that the borderline may be fuzzy regarding where and when to drop the impedance notion, it still stands that the concept is not useful in determining how to optimally load an RF PA.
At this point you own the conj-match assertion as much as Ken. Prove it! You can't because it is fundamentally incorrect.
Load-line matching is such a basic electronic concept it is unbelievable how oblivious you are to the concept. Read a basic book. Don't rely on me: look it up and do your own design!
You just like to hear yourself talk. I've been explicit and precise. You just don't know anything about the elementary electronics principle of load line matching. I presume this is why your comments have zero substantive responsiveness.
If you keep ignoring what I've written, and that which is written in elementary electronics texts, you can remain happily ignorant of understanding the simple-basic-fundamental concept presented. Your choice.
The guy ignorant of the definition of impedance and that s-domain theory *is* linear circuit theory (and more goodies) is talking about "precise language." Amusing.
No, I don't enjoy it at all. Your lack of electronic understanding is dismal, especially given your tone. It would have been a lot easier for me if Ken hadn't made the erroneous statement in the first place and made a correct one instead. That would have been my preferance.
I suspect you will. I already understand it -- you're the one who doesn't.
"One of the principal differences between linear RF amplifier design and PA design is that, for optimum power, the output of the device is not presented with the impedance required for a linear conjugate match. That causes much consternation and has been the subject of extensive controversy about the meaning and nature of conjugate matching. It is necessary, therefore, to swallow that apparently unpalatable result as early as possible (Section 1.5), before going on to give it more extended interpretation and analysis (Chapter 2)." -- Cripps, p1
The quote is on Page 1. Swallow it now. Learn something for a change.
I read in sci.electronics.design that gwhite wrote (in ) about '1/4 vs 1/2 wavelength antenna', on Thu, 3 Mar 2005:
And the power dissipated in the device is also 0.445 W. Matching according to the 'maximum power theorem' or conjugate matching, results in equal power in the PA and load. That's why it isn't useful for power amplifiers.
Doesn't everyone know that an audio amplifier that id designed to feed an 8 ohm load MUST have an output source impedance of 0.0000001 ohms or less. An output source impedance of 8 ohms would dramatically decrease the electromagnetic damping on the loudspeaker voice-coil - by the huge factor of .... two!(;-)
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Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
I think he just meant that damping factor is important in an audio amp.
At least I hope that's what he meant.
He forgot to mention that for that output impedance to be relevant, you need superconducting wire to the speakers as well as superconducting voice coils.
Thanx, your two bits were worth more than the academic plug nickel. This is something that our original poster should hearken to as his needs were obviously production oriented. Bench experience will trump cut-and-paste theory in a heart-beat.
However, triple stub is pretty aggressive. How long did it take you to flatten response?
I think it is 1A*10V - 0.455 W = 9.545 W ^^^^^^ ^^^^^^^ DC input Power Power delivered to RL
The resistance dissipated in the "internal AC resistance" is equal to RL in the conj-match condition. Of course, we're ignoring input power here, which is "small" when the gain is > +20 dB.
Amusingly for my hypothetical class A conj-match example, the "equal power dissipation" isn't such a big deal, since it is class A and the fractional power dissipated in either the internal AC resistance or the external load resistance is rather small compared to DC dissipation (less than 10%).
Wow this is a long thread. Don't really know where I should put my two bits in, but here it goes.
I have designed several RF PA sections in the past. 500MHz at about 50W. Pretty easy stuff if you have the right tools and know how to use them. The tools I like using for matching the power output FET is two triple stub tuners. One on the input of the FET and one on the output. So it goes...pre-amp (50 ohm output) -> stub tuner -> FET -> stub tuner -> 50 ohm dummy pad -> spectrum analyzer. Then just tune the stubs for the performance you desire, these include: efficiency (thermal issues), harmonic content, spurious emissions, load VSWR considerations, cold start, ect. Then remove the FET and look into the triple stub tuners with the network analyzer. Model and duplicate the network out of discrete components that can handle the voltage/power, send the design off to the enviro test lab, and head home early for the day.
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