|
Hi OM,
This goes into the intricacies of how forced propositions do not yield a forceful argument. On Tue, 01 Mar 2005 18:06:18 GMT, gwhite wrote: You don't know the output impedance because you don't have a way of determining it by swinging the output full-scale. This is more properly an admission from you, than a projected inability upon us. You may not know how, but this does not prevent me from expressing a value that is suitably accurate. Now, within the field of measurement, no statement is accurate without an expression of its range of error. However, in this regard accuracy is still a remote issue as you offer nothing of practical consideration and have failed to respond to a simple example to provide context. Richard Harrison, , KB5WZI, has in this sense already done the heavy lifting with: From the specifications page also, the power reguirement is TX: 18A 13.8V DC. It`s a linear amplifier. Only 40% efficiency. The designer probably was more interested in low harmonics than efficiency. The final by itself only takes part of the 18A ao its efficiency is more than 40%. continuing.... Even for class A, large signals will/can have rail to rail swing. This marks an artificial imposition not required to respond to the spirit of the topic. Such swings are not necessary. The device will not be linear for large swings: sinusoidal input swing will not result in a sinusoidal output swing. This is immaterial to impedance and is a set-up of another artificial imposition: the Thevenin Model (which was specifically dismissed). Hence we are into a cascade of impositions. But "impedance" is a sinusoidal (s-domain) concept. This is baloney cut thick. S Domains (?) are at best a modern contrivance to model well behaved small signal devices. Their utility follow theory, they do not drive theory. So how can you define an impedance--a sinusoidal concept--when the waveform is not sinusoidal for an inputted sine wave? There are no sine waves in nature, so by this contortion of logic from above there are no s-domains (?). Why are there no sine waves in nature? Because nature is bounded by the Big Bang (a discontinuity) at one end, and has yet to fulfill its infinite extent. In other words, tedious appeals to artificial impositions of purity fail at the gate for their sheer collapse of internal logic. This kind of stuff appeals to arm-chair theorists who find themselves impotent to perform. The point is that the output impedance is time dependent ("causes" the non-sinusoid output for sinusoid drive), which rather makes the concept questionable. As I wrote earlier, one might decide to consider a time averaged impedance, but I'm not clear on what the utility would be. Classic performance anxiety. Engineers learn to live with limitation and to express results and sources of error so that others can judge merit. Priests are better suited with mulling over these issues of ambiguity. There is no "presumption." Linear parameters and theorems totally ignore practical limitations--this is a fact and you can look it up in just about any text on circuit analysis. Knowledge limited. There are many suitable texts that offer a wider spectrum of discussion that are fully capable of answering these issues. However, it is made worse that most of this stuff is derivable from first principles and no recourse to vaster libraries is actually needed. The simple linear model is perfectly okay for small signal devices. It isn't okay for large signal devices. And yet there is no substantive illustration to prove this ambiguous point. What constitutes small, and what demarcates large? Such nebulous thinking clouds the obvious observation that the full range of devices themselves operate on only one principle. What is limited is the human component of their perception, not the physical reality of their operation. The faulty choice of models (S Parameters) is not the fault of either Physics or the devices when they diverge from the crutch of calculation against the wrong mathematical expression. In any case, load pull equipment does not make the pretense of defining output impedance of an active large signal device. It does say what the load needs to be to acquire maximum power out of the device. This is simply the statement from a lack of experience. Thevenins and conjugate matching (for maximum power transfer) are explicitly linear small signal device models. Their use in RF PA output design is a misapplication. These statements are drawn from thin air. So to return to a common question that seems to defy 2 out of 3 analysis (and many demurred along the way) - A simple test of a practical situation with a practical Amateur grade transistor model 100W transmitter commonly available for more than 20-30 years now: 1. Presuming CW mode into a "matched load" (any definition will do); Any definition won't do, and for this discussion the specific "won't do" is using conjugate matching which is a small signal (linear) model. Given the failure to provide any discussion for either or any form of matching suggests a lack fluency in any of them. *You* brought up Thevenins and armchair philosophy regarding it, not me. I rejected it as an unnecessary filigree, but I notice in the quotes above that you readily embraced it as a necessary imposition. I said Thevenins was irrelevent, and now you appear to agree with me. Ken effectively brought up conjugate matching, not me. This compounded with the denial of Thevenin is quickly closing the available matching mechanisms. If it is not about Thevenin, and it is not about Conjugation, then I am willing to wait to hear what it IS about. ....But not really. I have little faith that the difference is appreciated nor how many ways a match may be accomplished or for what ends. The original comment I was challenging was: "...the antenna works as an impedance mathcing network that matches the output stages impedance to the radiation resistance." I am always suspicious of how a quoted claim is couched by the rebutter (cut and paste from the original is always available and citing the link to the complete contextual post is hardly Herculean). However, responding to the bald statement, I find nothing objectionable about it. I simply wanted to make it clear that the "matching" done was not an issue of "output impedance" per se. It is an issue of how the transistor is to be loaded to extract maximum ouput power. Again, a presumption not brought to the table. It may follow as a consequence, but it is not a necessary condition. Our questioner who started this thread is undoubtedly interested in the outcome in terms of maximum radiation for a limited power - it is a chain of causality that is a forced step matching issue from the battery to the æther. This is a first principle of successful production engineering. 73's Richard Clark, KB7QHC |
Asimov wrote:
In fact the power difference is always zero, so there is no reason for the device to cool. Here's the reason the device cools under load. Plate loss = Eb*Ib - Ip^2*RL With no signal, the plate loss is Eb*Ib. When the signal is increased from zero, any power delivered to the load is subtracted from Eb*Ib, thus causing the device to cool. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Richard Clark wrote:
There are no sine waves in nature, so by this contortion of logic from above there are no s-domains (?). Why are there no sine waves in nature? Because nature is bounded by the Big Bang (a discontinuity) at one end, and has yet to fulfill its infinite extent. One would think that a 12 billion year windowing would be close enough. :-) -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Asimov wrote:
harmonic generation... Why do the instructions on my stereo amp warn against running the amp with no speakers attached? -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
I read in sci.electronics.design that Cecil Moore
wrote (in ) about '1/4 vs 1/2 wavelength antenna', on Wed, 2 Mar 2005: Richard Clark wrote: There are no sine waves in nature, so by this contortion of logic from above there are no s-domains (?). Why are there no sine waves in nature? Because nature is bounded by the Big Bang (a discontinuity) at one end, and has yet to fulfill its infinite extent. One would think that a 12 billion year windowing would be close enough. :-) Not only that, but since by definition the Universe started at T=0, any 'sine wave' that starts at a positive zero-crossing is at any later time indistinguishable from a real one that started at T=0. -- 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 |
Asimov wrote:
"Let`s look at it from the dynamic point of view (loss in a Class A amplifier)." The no-signal loss of a Class A amplifier is 100%. It equals volts x amps and appears in the amplifier. Now feed a signal to the ideal amplifier set just below the clipping level. Average d-c power is unchanged from the unloaded and no-signal conditions. Connect a matched load resistor to the amplifier output. If physically small, the resistor may become warm with heat that were it not for the load would be otherwise dissipated in the amplifier. Input power to the Class A amplifier is unchanging. Finding the internal resistance theoretically is simple. It is simply the open-circuit output voltage divided by the short-circuit current. Open-circuit voltage at full output and short-circuit current may be severe. Internal resistance can be found under less stressful conditions. Internal resistance will drop the voltage to any load reasistance. Use the voltage-divider formula to calculate the internal resistance. With pure resistances, half the open-circuit volts are dropped by the internal resistance when the load is a match. A power amplifier`s internal impedance can be determined. Power output from a Class A amplifier cools it. Best regards, Richard Harrison, KB5WZI |
"Cecil Moore" bravely wrote to "All" (02 Mar 05 07:55:04)
--- on the heady topic of " Say what you mean." CM From: Cecil Moore CM Xref: aeinews rec.radio.amateur.antenna:26244 CM Asimov wrote: harmonic generation... CM Why do the instructions on my stereo amp warn against CM running the amp with no speakers attached? Because then the screen tries to carry the plate signal, the reactance in the output transformer is not damped, and because a tube is sensitive to voltage, this quickly leads to a molten hole in the side? Watts to plasma. A*s*i*m*o*v |
Ken Smith wrote:
The strongest argument for dropping the impedance matching concept is PA efficiency, and therefore maximum signal swing. Obtaining maximum swing is a load line issue. What do you mean by "maximum signal swing" in this context. I can get a bigger swing by leaving the output completely unloaded and hence causing the actual efficiency to be zero. LOL. Sure, the purpose of a power amp is to actually extract power. This is a good start. Perhaps a simplistic (and of course idealized) class A example would help. And I want to remind that this is a simplification of the first order design cut. The first assumption/idealization for the class A example would be to demarcate between strong and weak non-linearity. This demarcation is basically the boundary of clipping, both positive and negative. That is, we want our class A amp to swing to the rails but not go beyond. In our class A design we will accept weak non-linearity but not strong non-linearity. Lets say we have selected a class A device/amplifier for which we can statically dissipate 10W. The drain DC circuit is simply an RF choke (see, it is a simple example!). Let's say the quiescent values are a 10 V supply and 1 A of current (10 W). The question is: how do we load the device to extract maximum power given the "no clipping" (strong non-linearity) constraint? Say we load it with 20 ohms, what happens? The max positive swing before clipping is Id*rL = 1*20 = 20 V. The max negative swing is, of course, Vd = 10 V. Since the 10 V is the lesser of the two swings, our non-clipping design constraint limits us to 20 Vp-p. So we can deliver (Vp)^2/(2*rL) = (10)^2/(2*20) = 2.5 W. Say we load it with 5 ohms, what happens? The max positive swing before clipping is Id*rL = 1*5 = 5 V. The max negative swing is, of course, Vd = 10 V. Since the 5 V is the lesser of the two swings, our non-clipping design constraint limits us to 10 Vp-p. So we can deliver (Vp)^2/(2*rL) = (5)^2/(2*5) = 2.5 W. Say we load it with 10 ohms, what happens? The max positive swing before clipping is Id*rL = 1*10 = 10 V. The max negative swing is, of course, Vd = 10 V. Since they are equal, we get 20 V of p-p swing. So we can deliver (Vp)^2/(2*rL) = (10)^2/(2*10) = 5 W. Our circuit loaded with 10 ohms delivers twice as much power as with the lesser 5 ohms or greater 20 ohms. That is, extracted output power is peaking at some finite non-zero value. This is also easily seen to be most efficient point for this simplistic example. In no way was the ouput-Z of the amplifier considered in deciding how to load it for the purpose of extracting maximum power from the circuit. The output-R is completely irrelevent. This example is intended to be illustrative rather than exact. The reactive component issue is still there too. Reactive loads cause increased currents in the output stage without delivering any power to the load so they still need to be reduced as much as practical. Yes, I already noted that for that portion of the impedance, it should be tuned out *as best* possible. "...(to be fair, the time-averaged reactive output component is tuned out as best possible)." |
Richard Clark wrote:
On Tue, 01 Mar 2005 18:06:18 GMT, gwhite wrote: It is about DC to RF efficiency, Put a number to it. as I've been pointing out since my first post, and which you initially commented was "nonsense" Hi OM, And so it remains with additional elaborations not quoted here. but now seem to agree with. Seeming is a rather insubstantial thing to hang your theories on. Well they are apparently your's too! Your own example of testing your own PA said absolutely zip about output-Z. The most you could say is how the circuit is loaded and its RF/DC efficiency. You're agreeing with me and can't even seem to recognize it. "Impedance matching" meant in the normal sense of conjugate matching for maximum transfer of power And this reveals the error of "Seeming" because the so-called meaning you ascribe is this same nonsense. Here's the original quote [Ken]: "When the correct matching is done, the antenna works as an impedance mathcing network that matches the output stages impedance to the radiation resistance." He brought up "matching to the output impedance" (of the device), not me. There is no "misinterpretation of meaning" when it comes to making statements about matching output impedance to a load impedance. The meaning is well-understood and precise. It means conjugate matching for maximum power transfer, and this is explicitly sourced from small signal theory. Small signal theory is oblivious to practical factors like supply rails and efficiency. These practical factors are paramount in PA design. Thus to apply a theory that ignores paramount factors is to beg a design which will likely be non-optimal. Pay more attention to reading instead of writing. I'm paying attention, you agree with me but don't have the background to understand it. It has been pointed out more than once, and by several, that Matching comes under many headings. The most frequent violation is the mixing of concepts and specifications (your text is littered with such clashes). No, you still don't get it. I don't have a problem with saying it is "matched." For example, I said a PA needs to be "load-line matched." This has a specific meaning, and that meaning explicitly isn't "impedanced matched," which means something else. If you don't bother to know what the words mean, I might as well speak Swahili. is a misapplied small signal concept/model. I think that is all I've really been saying. And I preserved this clash quoted above as an example. If there is any misapplication, you brought it to the table with this forced presumption. There is no forced presumption. The words have explicit definitions. If you don't know the language, you have no way of communicating. The misapplication of S parameters to a large signal amplifier is one thing, to project this error backwards into the fictive theory that there is some difference between large and small signal BEHAVIOR (not modeling) is tailoring the argument to suit a poorly framed thesis. The models come from behavior and/or device physics, and were developed for the express purpose of efficient design methodology. Small signal models can (and do) conveniently ignore large signal concerns such as efficiency and supply rails, because such concerns are irrelevent in the small signal milieu. To apply a model to a milieu for which the model is not suited begs a non-optimal design. The output-impedance concept itself is quite dubious for large signal amplifiers. None of your dissertation reveals any practical substantiation, hence it falls into the realm of armchair theory. We get plenty of that embroidered with photonic wave theory that is far more amusing. You are off track. |
I read in sci.electronics.design that gwhite wrote
(in ) about '1/4 vs 1/2 wavelength antenna', on Wed, 2 Mar 2005: I might as well speak Swahili. Good idea! Furahini mkaimbe. The wrangling is getting tiresome. -- 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 |
On Wed, 02 Mar 2005 19:06:32 GMT, gwhite wrote:
but now seem to agree with. Seeming is a rather insubstantial thing to hang your theories on. Well they are apparently your's too! Hi OM, From seeming to appearances - leaps of faith are better suited for debate at the Vatican. The remainder, unquoted due to repetition of the same basic errors, has already been commented upon in another posting. Oh, except the more entertaining jousts: Pay more attention to reading instead of writing. I'm paying attention, you agree with me but don't have the background to understand it. Mmm-hmm :-) To be so eagerly embraced as a fellow fool! Something of the chess equivalent of the sacrificial queen gambit. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
Hi OM, This goes into the intricacies of how forced propositions do not yield a forceful argument. LOL. On Tue, 01 Mar 2005 18:06:18 GMT, gwhite wrote: You don't know the output impedance because you don't have a way of determining it by swinging the output full-scale. This is more properly an admission from you, than a projected inability upon us. You may not know how, but this does not prevent me from expressing a value that is suitably accurate. Now, within the field of measurement, no statement is accurate without an expression of its range of error. However, in this regard accuracy is still a remote issue as you offer nothing of practical consideration and have failed to respond to a simple example to provide context. Sheesh! Richard Harrison, , KB5WZI, has in this sense already done the heavy lifting with: From the specifications page also, the power reguirement is TX: 18A 13.8V DC. It`s a linear amplifier. Only 40% efficiency. The designer probably was more interested in low harmonics than efficiency. The final by itself only takes part of the 18A ao its efficiency is more than 40%. Efficiency seems to be important enough to mention. continuing.... Even for class A, large signals will/can have rail to rail swing. This marks an artificial imposition not required to respond to the spirit of the topic. Such swings are not necessary. No one said they "are necessary." But not driving "as hard as possible" simply means you are wasting power and paying for a bigger device than you need to. The device will not be linear for large swings: sinusoidal input swing will not result in a sinusoidal output swing. This is immaterial to impedance,... Oh? The definition of impedance is: Z = V/I V and I are sinusoid (phasors), *by definition*. It is as if you don't know the definition of impedance. and is a set-up of another artificial imposition: the Thevenin Model (which was specifically dismissed). Hence we are into a cascade of impositions. But "impedance" is a sinusoidal (s-domain) concept. This is baloney cut thick. S Domains (?) are at best a modern contrivance to model well behaved small signal devices. S-domain *is* linear circuit theory. Their utility follow theory, they do not drive theory. It *is* linear circuit theory. The theory was developed for its utility. http://www.amazon.com/exec/obidos/tg.../-/0801869099/ So how can you define an impedance--a sinusoidal concept--when the waveform is not sinusoidal for an inputted sine wave? There are no sine waves in nature, so by this contortion of logic from above there are no s-domains (?). What are you talking about? No circuit is perfectly linear, and no one I knows claims such. That does not invalidate linear theory, nor denigrate its utility properly applied. Many circuits are "sufficiently linear," and "care" little about supply rails and efficiency. Why are there no sine waves in nature? Because nature is bounded by the Big Bang (a discontinuity) at one end, and has yet to fulfill its infinite extent. I'm not religious, but you beg me. Ohmigod! In other words, tedious appeals to artificial impositions of purity fail at the gate for their sheer collapse of internal logic. This kind of stuff appeals to arm-chair theorists who find themselves impotent to perform. Suit yourself. Go ahead and apply theory to that for which it was not designed to handle. In fact, you don't do it -- your own example about testing your PA stated absolutely nothing about linear theory, or output impedance of the device. I use (apply) linear theory a good share of the time. That doesn't mean I don't recognize its limitations as a theory (a model). The point is that the output impedance is time dependent ("causes" the non-sinusoid output for sinusoid drive), which rather makes the concept questionable. As I wrote earlier, one might decide to consider a time averaged impedance, but I'm not clear on what the utility would be. Classic performance anxiety. Engineers learn to live with limitation and to express results and sources of error so that others can judge merit. Priests are better suited with mulling over these issues of ambiguity. Wow. More importantly, engineers select appropriate models for the design task. They don't bother with ones that have no application to the task at hand. There is no "presumption." Linear parameters and theorems totally ignore practical limitations--this is a fact and you can look it up in just about any text on circuit analysis. Knowledge limited. There are many suitable texts that offer a wider spectrum of discussion that are fully capable of answering these issues. Yeah, like for example: http://www.amazon.com/exec/obidos/tg.../-/0890069891/ However, it is made worse that most of this stuff is derivable from first principles and no recourse to vaster libraries is actually needed. Yes, load line matching is certainly a first principle. The simple linear model is perfectly okay for small signal devices. It isn't okay for large signal devices. And yet there is no substantive illustration to prove this ambiguous point. What constitutes small, and what demarcates large? Maybe you didn't read those first principles quite closely enough. Nor have you read this thread well. Large signal amplifiers -- i.e. power amplifiers -- "care" about DC to RF efficiency and supply rails. Small signal amplifiers don't "care" about that. Such nebulous thinking clouds the obvious observation that the full range of devices themselves operate on only one principle. Quite afraid to ask, but being brave, I ask: what "one principle" is it "that the full range of devices themselves operate" upon? What is limited is the human component of their perception, not the physical reality of their operation. And you critiqued me for nonsense. The faulty choice of models (S Parameters) is not the fault of either Physics or the devices when they diverge from the crutch of calculation against the wrong mathematical expression. And no one said so. In any case, load pull equipment does not make the pretense of defining output impedance of an active large signal device. It does say what the load needs to be to acquire maximum power out of the device. This is simply the statement from a lack of experience. No, it is a fact of the matter. You don't know what the equipment does. Thevenins and conjugate matching (for maximum power transfer) are explicitly linear small signal device models. Their use in RF PA output design is a misapplication. These statements are drawn from thin air. No, for PA design, the thevenin impedance of the output source never enters "the equation." Thus pretending that it "is there" is an unfounded assertion. You asserted thevenins to PA design, now prove it. You can't. So to return to a common question that seems to defy 2 out of 3 analysis (and many demurred along the way) - A simple test of a practical situation with a practical Amateur grade transistor model 100W transmitter commonly available for more than 20-30 years now: 1. Presuming CW mode into a "matched load" (any definition will do); Any definition won't do, and for this discussion the specific "won't do" is using conjugate matching which is a small signal (linear) model. Given the failure to provide any discussion for either or any form of matching suggests a lack fluency in any of them. What utter ignorance of what has actually been written. In my very first post I described the first order cut of matching technique. *You* brought up Thevenins and armchair philosophy regarding it, not me. I rejected it as an unnecessary filigree,... Exactly. It is not necessary. But you brought it up, and Ken implied a simile with "impedance matching." You might wonder why it is not necessary. You might even ask the question wondering if the reason it never shows up is because it would be a misapplication of the concept. ... but I notice in the quotes above that you readily embraced it as a necessary imposition. I said Thevenins was irrelevent, and now you appear to agree with me. Ken effectively brought up conjugate matching, not me. This compounded with the denial of Thevenin is quickly closing the available matching mechanisms. If it is not about Thevenin, and it is not about Conjugation, then I am willing to wait to hear what it IS about. Ah, at last a relevent question/statement. See my first post in this thread. ...But not really. I have little faith that the difference is appreciated,... You don't appreciate it because you don't understand it. That's not my problem. nor how many ways a match may be accomplished or for what ends. If you don't know what the end is for an RF PA, how could you hope to scratch a meaningful and optimal solution? The original comment I was challenging was: "...the antenna works as an impedance mathcing network that matches the output stages impedance to the radiation resistance." I am always suspicious of how a quoted claim is couched by the rebutter (cut and paste from the original is always available and citing the link to the complete contextual post is hardly Herculean). LOL. I guess you don't appreciate convenience. However, responding to the bald statement, I find nothing objectionable about it. That's because you don't understand the difference between impedance matching and ac load line matching. I simply wanted to make it clear that the "matching" done was not an issue of "output impedance" per se. It is an issue of how the transistor is to be loaded to extract maximum ouput power. Again, a presumption not brought to the table. It was brought to the table in my first post to this thread. It may follow as a consequence, but it is not a necessary condition. Our questioner who started this thread is undoubtedly interested in the outcome in terms of maximum radiation for a limited power - it is a chain of causality that is a forced step matching issue from the battery to the æther. This is a first principle of successful production engineering. How would you know about first principles of production engineering and what does it have to do with this thread? |
John Woodgate wrote:
I read in sci.electronics.design that Cecil Moore wrote (in ) about '1/4 vs 1/2 wavelength antenna', on Wed, 2 Mar 2005: Richard Clark wrote: There are no sine waves in nature, so by this contortion of logic from above there are no s-domains (?). Why are there no sine waves in nature? Because nature is bounded by the Big Bang (a discontinuity) at one end, and has yet to fulfill its infinite extent. One would think that a 12 billion year windowing would be close enough. :-) Not only that, but since by definition the Universe started at T=0, any 'sine wave' that starts at a positive zero-crossing is at any later time indistinguishable from a real one that started at T=0. Not if we were there the moment the later wave turned on. I heard that amateur operators hate splatter. RC appears to be an exception, however. |
I read in sci.electronics.design that gwhite wrote
(in ) about '1/4 vs 1/2 wavelength antenna', on Wed, 2 Mar 2005: John Woodgate wrote: Not only that, but since by definition the Universe started at T=0, any 'sine wave' that starts at a positive zero-crossing is at any later time indistinguishable from a real one that started at T=0. Not if we were there the moment the later wave turned on. I heard that amateur operators hate splatter. RC appears to be an exception, however. See 'at any later time' in my text. -- 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 |
John Woodgate wrote:
I read in sci.electronics.design that gwhite wrote (in ) about '1/4 vs 1/2 wavelength antenna', on Wed, 2 Mar 2005: John Woodgate wrote: Not only that, but since by definition the Universe started at T=0, any 'sine wave' that starts at a positive zero-crossing is at any later time indistinguishable from a real one that started at T=0. Not if we were there the moment the later wave turned on. I heard that amateur operators hate splatter. RC appears to be an exception, however. See 'at any later time' in my text. Oh yeah. |
On Wed, 02 Mar 2005 20:30:45 GMT, gwhite wrote:
I said Thevenins was irrelevent, and now you appear to agree with me. Ken effectively brought up conjugate matching, not me. This compounded with the denial of Thevenin is quickly closing the available matching mechanisms. If it is not about Thevenin, and it is not about Conjugation, then I am willing to wait to hear what it IS about. Ah, at last a relevent question/statement. See my first post in this thread. Mmm-Hmm On Wed, 23 Feb 2005 19:08:20 GMT, gwhite wrote: RF transmitters are not impedance matched to antennae in the sense of maximum transfer of power. Hi OM, As I've noted in the past, you can fill a library with negative assertions without ever offering an answer, eg.: RF transmitters are not Nuclear resonated to antennae in the sense of maximum transfer of power. RF transmitters are not impedance matched to antennae in the sense of maximum balance of payments. RF transmitters are not cosmically matched to antennae in the sense of maximum psychrotropic power. The list could go on, be completely accurate, and yet never actually mean anything in the end much as the nonsense you offered from the start. You sighed with content at being offered a "relevent question/statement" Your re-iterative response contains the same (how could it be otherwise?) slack of precision that started this. Want to try again? You could have as easily expressed what sense they ARE matched, but instead this time offer what Basis of Matching you are attempting to describe. This is the more rigorous approach that eliminates vague descriptions and uses standard terms. If you have to query about what "Basis" means (used by professionals - namely metrologists who can quantify Output Z of all sources) - then we can skip it as a topic out of the reach of amateur discussion. Note: Again, RF PA's should be load-line matched. Does not qualify as a Basis. It is suggestive of one, but because you indiscriminately mix several Basis within your discussions, it is your responsibility to be precise. If you can accomplish this, then we can proceed to review how little it all matters. Barring resolving any of these issues of precise language, I notice that you rather enjoy fruitless jousting with them than challenging my support of Ken's (supposed) statement that you say is your focus: However, responding to the bald statement, I find nothing objectionable about it. That's because you don't understand the difference between impedance matching and ac load line matching. We will leave that as another dead-end. 73's Richard Clark, KB7QHC |
Ken Smith wrote:
In article , gwhite wrote: [...] You entirely missed the point. You don't know the output impedance because you don't have a way of determining it by swinging the output full-scale. You don't have to swing the output full-scale to measure the impedance. 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.) Any change in the load, no matter how small, will cause a change in the output voltage and the output current. 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. From these you can calculate the output impedance at the current operating point. When a transistor is operating under large signal conditions into a tuned load, there is still an output impedance and this impedance still discribes what will happen for small changes in the load. 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. Ropt = Vmax/Imax where Ropt Rgen, if not (Rgen + Ropt)/(Rgen*Ropt) = Vmax/Imax So even looking into the PA output in the small signal sense (or tweaking the impedance as you suggest), we won't likely see Ropt = Rgen, because we are dealing with some practical design limitations not accounted for in linear theory. Perhaps a couple of quotes from Cripps would be nice: http://www.amazon.com/exec/obidos/tg.../-/0890069891/ "The load-line match is a real-world compromise that is necessary to extract the maximum power from RF transistors and at the same time keep the RF voltage swing within specified limits and/or the available DC supply." p13 "A final note here concerns the nebulous and highly questionable concept of large signal impedance. The reason for the load-line match is to accommodate the maximum allowable current and voltage swings at the transistor output. That says nothing about the impedance of the device, which remains the same throughout the linear range. Once a device starts to operate in a significantly nonlinear fashion, the apparent value of the impedances will change, but the whole concept of impedance starts to break down as well, because the wave forms no longer are sinusoidal." p14 |
In article , gwhite wrote:
Ken Smith wrote: The strongest argument for dropping the impedance matching concept is PA efficiency, and therefore maximum signal swing. Obtaining maximum swing is a load line issue. What do you mean by "maximum signal swing" in this context. I can get a bigger swing by leaving the output completely unloaded and hence causing the actual efficiency to be zero. LOL. Sure, the purpose of a power amp is to actually extract power. This is a good start. No, the purpose of the power amp is to deliver power, not extract it. Perhaps a simplistic (and of course idealized) class A example would help. And I want to remind that this is a simplification of the first order design cut. 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. [...] Our circuit loaded with 10 ohms delivers twice as much power as with the lesser 5 ohms or greater 20 ohms. That is, extracted output power is peaking at some finite non-zero value. This is also easily seen to be most efficient point for this simplistic example. 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. -- -- forging knowledge |
In article , gwhite wrote:
[...] Here's the original quote [Ken]: "When the correct matching is done, the antenna works as an impedance mathcing network that matches the output stages impedance to the radiation resistance." 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. -- -- forging knowledge |
In article ,
John Woodgate wrote: [...] The point is that if you want to talk/write about one of these impedances, you need, to prevent misunderstanding, use a precise term, such as 'incremental output source impedance' and define it. 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. -- -- forging knowledge |
Ken Smith wrote:
In article , gwhite wrote: Ken Smith wrote: The strongest argument for dropping the impedance matching concept is PA efficiency, and therefore maximum signal swing. Obtaining maximum swing is a load line issue. What do you mean by "maximum signal swing" in this context. I can get a bigger swing by leaving the output completely unloaded and hence causing the actual efficiency to be zero. LOL. Sure, the purpose of a power amp is to actually extract power. This is a good start. No, the purpose of the power amp is to deliver power, not extract it. 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. Perhaps a simplistic (and of course idealized) class A example would help. And I want to remind that this is a simplification of the first order design cut. 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. 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. Our circuit loaded with 10 ohms delivers twice as much power as with the lesser 5 ohms or greater 20 ohms. That is, extracted output power is peaking at some finite non-zero value. This is also easily seen to be most efficient point for this simplistic example. 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. 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. |
Ken Smith wrote:
In article , gwhite wrote: [...] Here's the original quote [Ken]: "When the correct matching is done, the antenna works as an impedance mathcing network that matches the output stages impedance to the radiation resistance." 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. 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. 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. |
G. White wrote:
"This example is intended to be illustrative rather than exact." Nobody contradicts that maximum power from an amplifier requires an impedance match, so far. Nobody contradicts that efficiency in an ideal Class A amplifier can never exceed 50%, so far. Some texts show waveform distortion typical of a single-ended Class A amplifier and note that "maximum undistorted power" requires a load impedance 2 to 3 times that of the amplifying device. G. White gives a peak to peak voltage of 20 and a power out of 5 watts. RMS volts are 7.07 in this case. The load then is the square of the volts divided by the power. This is 50/5 or 10 ohms for the load. If this matches his amplifier, 5 watts is the maximum output. G. White said that he did not consider the output-Z of the amplifier in its loading. Ignoring an amplifier`s impedance does not make it go away. Ignoring an amplifier`s impedance does not revoke the maximum power transfer theorem, either. Best regards, Richard Harrison, KB5WZI |
Richard, your method of measuring internal impedance of an amplifier sounds
interesting. (o/c volts divided by s/c current) At present I have no facilities to make such measurements. Perhaps a few curious readers, who do have facilities, might make crude measurements on their HF rigs and report approximate impedance values on one or two bands on this newsgroup. Some sort of average could be obtained. ---- Reg, G4FGQ =================================== "Richard Harrison" wrote in message ... Asimov wrote: "Let`s look at it from the dynamic point of view (loss in a Class A amplifier)." The no-signal loss of a Class A amplifier is 100%. It equals volts x amps and appears in the amplifier. Now feed a signal to the ideal amplifier set just below the clipping level. Average d-c power is unchanged from the unloaded and no-signal conditions. Connect a matched load resistor to the amplifier output. If physically small, the resistor may become warm with heat that were it not for the load would be otherwise dissipated in the amplifier. Input power to the Class A amplifier is unchanging. Finding the internal resistance theoretically is simple. It is simply the open-circuit output voltage divided by the short-circuit current. Open-circuit voltage at full output and short-circuit current may be severe. Internal resistance can be found under less stressful conditions. Internal resistance will drop the voltage to any load reasistance. Use the voltage-divider formula to calculate the internal resistance. With pure resistances, half the open-circuit volts are dropped by the internal resistance when the load is a match. A power amplifier`s internal impedance can be determined. Power output from a Class A amplifier cools it. Best regards, Richard Harrison, KB5WZI |
Asimov wrote:
"Cecil Moore" bravely wrote: CM Why do the instructions on my stereo amp warn against CM running the amp with no speakers attached? Because then the screen tries to carry the plate signal, the reactance in the output transformer is not damped, and because a tube is sensitive to voltage, this quickly leads to a molten hole in the side? Watts to plasma. Sorry, it's solid state - no screen. Let's face it - unloaded amplifiers tend to burn up no matter what class they are running. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Reg Edwards wrote:
Richard, your method of measuring internal impedance of an amplifier sounds interesting. (o/c volts divided by s/c current) Reg, I tried that one time in college. I got as far as torching two ARC-5 1625's driving an open circuit so I don't know what would have happened with a short circuit. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Cecil Moore wrote:
Reg Edwards wrote: Richard, your method of measuring internal impedance of an amplifier sounds interesting. (o/c volts divided by s/c current) Reg, I tried that one time in college. I got as far as torching two ARC-5 1625's driving an open circuit so I don't know what would have happened with a short circuit. Some of Motorola's devices might actually be able to stand Reg's little test. The MRF150, for instance is advertised as being able to withstand a 30:1 VSWR at all phase angles. I think Reg should offer to pay for any damage sustained as the result of this test. 73, Tom Donaly, KA6RUH |
"Rich Grise" bravely wrote to "All" (03 Mar 05 23:00:40)
--- on the heady topic of " 1/4 vs 1/2 wavelength antenna" RG From: Rich Grise RG Xref: aeinews rec.radio.amateur.antenna:26329 RG sci.electronics.design:1386 RG On Thu, 03 Mar 2005 20:53:48 +0000, John Woodgate wrote: I read in sci.electronics.design that gwhite wrote ... 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!(;-) RG Yeah - isn't that why the TOOB amps had those taps? So you could get RG that rich, full-bodied TOOB sound? ;-) No, the taps are there to maximize the power output. If one uses the 8 ohm tap with a 4 ohm speaker the power is reduced and the same if one uses a 16 ohm speaker. Audio output matching has little to do with plate resistance in a beam tube or pentode. It has everything to do with plate voltage swing and maximum plate current. The less plate resistance the more power can get to the load. Beam tube curves look a lot like a transistor's collector saturation curves. However, some tube amplifiers had a current feedback control which would increase the output impedance to equal the speaker's. Needless to say it severely reduced the speaker damping resulting in exagerated frequency response artifacts. A*s*i*m*o*v .... Without ignorance, knowledge is powerless. |
Reg Edwards wrote:
Richard, your method of measuring internal impedance of an amplifier sounds interesting. (o/c volts divided by s/c current) Reg, I tried that one time in college. I got as far as torching two ARC-5 1625's driving an open circuit so I don't know what would have happened with a short circuit. -- 73, Cecil ============================== Cecil, Presumably you hadn't heard of Ohms Law. The internal resistance of a generator is independent of its internal voltage. Just reduce the drive level to some small, no particular value, and stop making excuses. ( Somebody will say it IS dependent. But we are interested only in ball-park accuracy. And in any case the operating point will be within the normal range of operation between no-drive and full drive, or between no modulation and full modulation.) ---- Reg, G4FGQ |
Reg Edwards wrote:
The internal resistance of a generator is independent of its internal voltage. Just reduce the drive level to some small, no particular value, and stop making excuses. ( Somebody will say it IS dependent. But we are interested only in ball-park accuracy. And in any case the operating point will be within the normal range of operation between no-drive and full drive, or between no modulation and full modulation.) That might work better for Class-AB than Class-C. If one reduces drive level on a Class-C, one is reducing the 'on' time of the device, thus changing the internal impedance probably somewhat inversely proportional to the 'on' time. If I remember correctly, my IC-706 would not fold back under any load condition when run at minimum power. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
"Cecil Moore" bravely wrote to "All" (02 Mar 05 20:05:49)
--- on the heady topic of " Say what you mean." CM From: Cecil Moore CM Xref: aeinews rec.radio.amateur.antenna:26289 CM Asimov wrote: Watts to plasma. CM Sorry, it's solid state - no screen. Let's face it - unloaded CM amplifiers tend to burn up no matter what class they are running. Why don't you repeat that to the folk who think their amp cools when it outputs watts? A*s*i*m*o*v .... New computer? But I like my vacuum tubes... They keep me warm. |
Asimov wrote:
"I was confusing it with Class B which does warm up the heatsink" True. Class B amplifiers can be single-ended for R-F if used with the right tank circuit to supply the missing 1/2-cycle. More often they are push-pull to reduce harmonics. SSB final amplifiers are biased to near current cut-off. During modulation gaps, current falls to nearly zero. When current is zero, so is the dissipation. Reg mentioned the blistering heat of the 6N7. Thai`s my recollection too when getting nearly 10 watts of audio from the small twin-triode octal-based metal tube. Reduce the audio volume and the heat is reduced simultaneously. Best regards, Richard Harrison, KB5WZI |
Richard Clark wrote:
As I've noted in the past, you can fill a library with negative assertions... 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. The list could go on,... LOL. Given your pattern, I am sure it will. You sighed with content at being offered a "relevent question/statement" Your re-iterative response contains the same (how could it be otherwise?) slack of precision that started this. Want to try again? 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. You could have as easily expressed what sense they ARE matched, For what seems like the billionth time now: they are load-line matched. ...but instead this time offer what Basis of Matching you are attempting to describe. 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. This is the more rigorous approach that eliminates vague descriptions and uses standard terms. If you have to query about what "Basis" means (used by professionals - namely metrologists who can quantify Output Z of all sources) - then we can skip it as a topic out of the reach of amateur discussion. 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. Note: Again, RF PA's should be load-line matched. Does not qualify as a Basis. 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! It is suggestive of one, but because you indiscriminately mix several Basis within your discussions, it is your responsibility to be precise. 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 can accomplish this, then we can proceed to review how little it all matters. 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. Barring resolving any of these issues of precise language,...] 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. I notice that you rather enjoy... 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. ..fruitless jousting with them than challenging my support of Ken's (supposed) statement that you say is your focus: However, responding to the bald statement, I find nothing objectionable about it. That's because you don't understand the difference between impedance matching and ac load line matching. We will leave that as another dead-end. 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: 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 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!(;-) -- 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 |
On Thu, 03 Mar 2005 20:25:16 GMT, gwhite wrote:
You sighed with content at being offered a "relevent question/statement" Your re-iterative response contains the same (how could it be otherwise?) slack of precision that started this. Want to try again? Not really. .... I notice that you rather enjoy fruitless jousting with them than challenging my support of Ken's (supposed) statement that you say is your focus: However, responding to the bald statement, I find nothing objectionable about it. That's because you don't understand the difference between impedance matching and ac load line matching. We will leave that as another dead-end. I suspect you will. Hi OM, 224 line postings to produce this little qualitative information? :-) 73's Richard Clark, KB7QHC |
On Thu, 3 Mar 2005 20:53:48 +0000, John Woodgate
wrote: 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. Hi John, I hope that was a joke. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
224 line postings to produce this little qualitative information? :-) I see one line here with no content. You don't have any argument because you have zero understanding. No one can cure that but you. But can you? "Stupid is as stupid does." -- Forrest Gump |
John Woodgate wrote:
I read in sci.electronics.design that gwhite wrote (in ) about '1/4 vs 1/2 wavelength antenna', on Thu, 3 Mar 2005: 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 And the power dissipated in the device is also 0.445 W. 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. 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. 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%). 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!(;-) Nice one. |
On Thu, 03 Mar 2005 20:53:48 +0000, John Woodgate wrote:
I read in sci.electronics.design that gwhite wrote .... 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!(;-) Yeah - isn't that why the TOOB amps had those taps? So you could get that rich, full-bodied TOOB sound? ;-) Cheers! Rich |
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