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Output classes A and AB
In article ,
Eeyore wrote: Patrick Turner wrote: One cannot have distortion cancelling by one tube cancelling that in another when one is cut off. THANK YOU ! Basics do matter. Indeed they do, but neither Patrick, myself, or anyone else is correct on every issue. In this case Patrick has vigorously asserted that this view, which he holds in common with you, is true, but he has failed to even attempt an argument that might demonstrate its truth. Patrick is an extremely skilled and talented fellow in the practical aspects of tube amp design and construction, but he has a very limited understanding of what is going on behind the scenes in the theory of tube amp operation. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
On Oct 26, 7:09 am, Eeyore
wrote: Andre Jute wrote: Eeyore wrote: John Byrns wrote: Patrick Turner wrote: Cancelation of even order harmonics occurs in amps working in class AB during that part of the wave forms which are in class A, ie, the bits either side of the zero crossing. But once each tube moves into cut off, nothing is cancelled. Patrick, I'm surprised to hear you say this. What are you trying to tell us, that the even order harmonics are only cancelled during those parts of the cycle when both tubes are conducting, but that the even order distortion components reappear during those parts of the cycle when only one tube is conducting? If you actually believe that you should go back to the books and study the theory of harmonic distortion more carefully. I hope you didn't get this notion from the RDH4, I haven't read the RDH4's harmonic distortion explanation, but if this is what it says I have just lost any respect I had for the book. In a perfectly balanced PP amplifier the even order harmonic distortion is completely cancelled even when the tubes are cut off for parts of the cycle. I'd love to know how that happens. There's no cancellation of ANYTHING once one side has ceased conducting ! Graham Holy ****! Did I say yet that Poopie is ignorant and incompetent? Nah, nobody can be that stupid and uninformed about tube basics. Poopie must be cracking a joke. For the first time in his life. Good on yer, cobber! If you can't be smart and informed, at least you can try to be a clown, give people a giggle. As an alleged 'wordsmith' you of all people ought to understand what cancellation means. Apparently it went right over your head though. Only to be expected from an ignorant non-technical ****wit. Graham Yeah, Poopie, you're the man when it comes to cancellations: your definition of Class A, okay until then, cancelled out when you added the superfluous words "under any signal condition". Unsigned out of contempt |
Output classes A and AB
On Oct 26, 7:21 am, Eeyore
wrote: Andre Jute wrote: Poopie Stevenson aka Eeyore wrote: Andre Jute wrote: Eeyore wrote: Andre Jute wrote: All of that follows logically from Poopie's absurd redefinition of Class A as a Class in which "the output device(s)never cease conducting *under any signal condition*," (emphasis added). It's ludicrous. It's actually the only accurate definition. I've already demonstrated several times that your words "under any signal condition" make your definition grossly inaccurate. But you're an ignorant **** and what you say is a load of ********. Even when I'm right? You're NOT right. Your ignorance is simply confusing you. HOWEVER, to keep you happy I am happy to modify to modify my definition for clarity. I already posted this once but I suppose you like to argue more than anything. "the output device(s)never cease conducting *under any signal condition* within the rated specification". To be honest, to have to explicitly state "within the rated specification" is really a case of pandering to fools, which certainly describes YOU, Jootikins. Graham Make up your mind, Poopie. Either Class A operation is possible only with limited signal or it is possible, as you erroneously claim, "under any signal condition". One or the other, not both. There are about 500 messages across two threads in which you shilly- shally about this gross error you, Pearce and Krueger have made. Which is it? Andre Jute ....who knows his own mind |
Output classes A and AB
On Oct 26, 9:36 am, Eeyore
wrote: John Byrns wrote: Eeyore wrote: John Byrns wrote: (Don Pearce) wrote: Eeyore wrote: I'd love to know how that happens. There's no cancellation of ANYTHING once one side has ceased conducting ! Because if you add an even harmonic to a signal, you have to make it asymmetrical. You always get a peak coinciding with a trough on one half cycle, followed by a peak coinciding with a peak on the next. If you modify the signal to remove any asymmetry, you must by definition remove the even harmonics. Finally a man who understands the theory! But it's not by ** CANCELLATION ** in the case of AB operation beyond the crossover point. That's my issue with the description. It does have that effect but the use of the word *cancellation* is wong IMHO. There should be another way to describe it. Cancelation is the right word, ********. the two tubes, even when they, "operate beyond the crossover point", When one tube has ceased conducting, there's NOTHING TO CANCEL, you IGNORANT ****WITTED ****. CANCELLATION IS THE *** WRONG WORD ***. In fact it's ADDITION of waveforms, not cancellation. Graham But d'y'see, dear old Poopster, the total harmonic distortion is less, so something has been subtracted from the result. The net sense is the same as cancellation. A little algebra, if you can handle it, would help. Or a little, a very little sensitivity to the English language would help you arrive at the same conclusion just from reading the explanations from Patrick and Pearcey and even Arny getting it right for once. Andre Jute The anti-pedant |
Output classes A and AB
On Oct 26, 10:06 am, Eeyore
wrote: Patrick Turner wrote: One cannot have distortion cancelling by one tube cancelling that in another when one is cut off. THANK YOU ! Basics do matter. Graham Here's a basic you should learn off by heart, Poopie: Class A operation requires a limited signal so that the device(s) are never driven to any point where they can stop conducting. Andre Jute Always ready to help a newbie |
Output classes A and AB
John Byrns wrote: In article , Patrick Turner wrote: John Byrns wrote: In article , Patrick Turner wrote: This also means that once the Ia travels below 10% of the idle value, the gm of the tube cutting off has diminished to such a low value the other tube turning on harder is providing virtually all the Ichange x Vchange across the available load, and is the only device coupled through only 1/2 the OPT primary to the load, so the RL seen by this tube turning on hard has reduced to 1/2 its class A load, or 1/4 of the nominal RL a-a, and in this case its 1.25k. The load is the same as that for a class B amp. Isn't 1.25k too low a load for getting maximum power from a KT88 in triode mode, even in class B? No. If the RL a-a = 5k, then the class B load is 1.25k, and if Ea = 500V, then max Ia at grid current is about 220mA. If you run AB2, you get a heck of a lot more Ia up to around 350mA. KT88 ca easily make 500mA, depending on loads etc. One can get 140W from a pair in AB2 in tetrode. But I was asking about the best load for a class B triode amp, is 1.25k too low for a KT-88? I guess I will have to see if I can find the triode plate curves for the KT-88, or maybe I can substitute the 6550 curves. There is no best load for a class B triode amp. Class B is a horrid way to build any amp. Maybe you meant low bias class AB. Do the load line analysis, or have a look at my website pages and print out a set of curves for 6550 which are virtually the same as KT88. http://www.turneraudio.com.au/loadma...p-triodes.html What is all this talk about ARC's anyone else's rules that would keep you from handing out free copies of their abominable concoctious junk, assuming you drew the schematic your self? You mentioned this same issue in connection with the ManleyLabs amplifier you modified, my understanding is that they only have protection for schematics they have drawn, if you draw your own schematic of the same circuit, they have no rights with regard to it. Any Lawyers out there care to comment? I fell OK about just letting folks know what they could do to rebuild a Manley or ARC or start from scratch and use the schematic I will be posting at my site. There is nothing I can gain by posting a copy of the original schematic these companies use. I was not talking about the original schematic drawn by these companies, I was talking about a schematic of the same circuit that you or anyone else may have drawn, it is my understanding that there is nothing to prevent you from legally posting such a schematic, illustrating the same circuit as the company circuit, you just can't post the schematic drawn by the company. I have zero reason or time available to re-draw anyone else's original schematic, and have no wish to disturb the minds of the ppl who work in prestigious US audio companies any more than I may have. But I doubt they are aware of my existance. What I have to say is aimed really at those with a really keen interest in such matters AND who understand such things AND who can read a schematic AND understand the effects of layouts, AND who have time to use a soldering iron. Maybe only 2.69 people in the world are actually interested.... Many companies do NOT like ppl posting copies of their schematics on the web, and I have no intention of offending them by doing so. I am not suggesting that you should do it, but it is my understanding that they have no say in your posting a schematic you drew of their circuit. It would be ungentlemanly for me to copy out and post a schematic of theirs without their consent especially if basically I was doing it to tell everyone what a POS it was. Its better for them, me, and the public if I merely leave out POS descriptions, and say "Here's an alternative that works better than the original..." Then anyone really keen can focus in on it, and maybe do the same thing for themeselves, or hire me or somebody else to do similar. I am only trying to get other maker's gear to stop smoking and sing better. I have now accumulated several schematics used in a range of PP amps and they are worth publishing at my website when I have time because they are fine tested designs for anyone to try. Of course you are going to offend them by doing that, and they may retaliate by denying you access to replacement parts. I am free to post alternative schematics used in the cases of their amps though, You are also free to post your rendition of the schematic for their original circuit. The schematic I have come up with for some of these amps is totally mine, and to use my design instad of the original meant removal of 80% of the parts and tracks on the board and starting all over again. As I understand the situation they only have rights to and control over their drawing of the original circuit, you are free to create and distribute copies of a new drawing of the circuit that was drawn by you. I may have that right as you suggest, but I don't feel its right to copy their schematic out slightly differently in appearance and post it. The intellectual content IS THEIRS, and remains theirs even after I have drawn it up myself. So if anyone wants to see really what I am on about, they have to find their own copy of the original schematic. To get that you have to own one of their amps and be able to quote a serial number. I am not in the mood to be seen to publically question all these companies might do. I need only say what I have done in response to being presented with samples of their amps that had bad smoking habits. In general, its my personal opinion that major US companies have forgotten how to build simple fine amplifiers, and have drifted to complexity, weight, size, high cost, and lots of do-dahs and bells and whistles that do nothing for the sound. Meanwhile, in general, there is an appalling lack of respect for good biasing methods of output tubes. Their engineers seem to have misplaced optimism about reliability in power amps. There is never any active protection. But ****e happens anyway..... I shouldn't ever have to be telephoned by someone saying to me "My nice new brand XXX tube amp was "fixed" elsewhere, but pharqued up again a fortnight later and the sound went really bad, and it blows fuses..." But most of my last 12 months work was with ppl having to cope with results of so called engineers. I respect engineers in general, but so often its a dumb apprentice who is used to design the amp in way too little time. Sales are down, and engineers cost serious money. Engineers are professionals, and unlike tradesmen like myself they put an extra faerking zero on the prices they charge. Companies only hire them if the cost can be justified by the sales figures. And sales by US majors are probably falling as ppl turn to chinese crap imports. I make no apology for my cynicism. Patrick Turner. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
Eeyore wrote: Patrick Turner wrote: One cannot have distortion cancelling by one tube cancelling that in another when one is cut off. THANK YOU ! Basics do matter. But Graham, once the tubes stop cancelling each other's even order distortions when they move from class A to AB, their non linear current behaviour is utterly attrocious, and each tube only 1/2 amplicates the signal. Yet the VOLTAGE outcome across the OPT primary is substantially linear. So some would say that by means of the SUMMING action of the OPT, there is cancelling going on. I am simply saying the summing action merely obstructs the gross non linearity of currents from being current in the OPT secondary. Its Mysterious, this whole simple business. But any AB amp can be made to be a class A amp if the load value is simply raised high enough to prevent cut off occuring, which I define as being the reduction in Ia to less than 1/10 of the idle current for each tube. Cut off could also be described as being where distortion in current waves in each tube exceeds 5% to 10%. Patrick Turner. Graham |
Output classes A and AB
Eeyore wrote: Patrick Turner wrote: But the SUM of the joint action of each tube in class AB with very non linear currents gives a linear voltage outcome. And should not be confused with genuine CANCELLATION of distortion by Class A push-pull operation. Agreed, and see my last post. Patrick Turner. Graham |
Output classes A and AB
John Byrns wrote: In article , Eeyore wrote: Patrick Turner wrote: One cannot have distortion cancelling by one tube cancelling that in another when one is cut off. THANK YOU ! Basics do matter. Indeed they do, but neither Patrick, myself, or anyone else is correct on every issue. In this case Patrick has vigorously asserted that this view, which he holds in common with you, is true, but he has failed to even attempt an argument that might demonstrate its truth. Patrick is an extremely skilled and talented fellow in the practical aspects of tube amp design and construction, but he has a very limited understanding of what is going on behind the scenes in the theory of tube amp operation. I think my website might indicate that your are not quite right about my levels of understanding. I should not mention whether or not I should be worried about your levels of understanding. I just let ppl decide for themseleves. But I do know what is going on in each output tube of an AB pair, milisecond by milisecond, electron by electron. Patrick Turner. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
In article ,
Patrick Turner wrote: John Byrns wrote: In article , Patrick Turner wrote: John Byrns wrote: In article , Patrick Turner wrote: This also means that once the Ia travels below 10% of the idle value, the gm of the tube cutting off has diminished to such a low value the other tube turning on harder is providing virtually all the Ichange x Vchange across the available load, and is the only device coupled through only 1/2 the OPT primary to the load, so the RL seen by this tube turning on hard has reduced to 1/2 its class A load, or 1/4 of the nominal RL a-a, and in this case its 1.25k. The load is the same as that for a class B amp. Isn't 1.25k too low a load for getting maximum power from a KT88 in triode mode, even in class B? No. If the RL a-a = 5k, then the class B load is 1.25k, and if Ea = 500V, then max Ia at grid current is about 220mA. If you run AB2, you get a heck of a lot more Ia up to around 350mA. KT88 ca easily make 500mA, depending on loads etc. One can get 140W from a pair in AB2 in tetrode. But I was asking about the best load for a class B triode amp, is 1.25k too low for a KT-88? I guess I will have to see if I can find the triode plate curves for the KT-88, or maybe I can substitute the 6550 curves. There is no best load for a class B triode amp. Class B is a horrid way to build any amp. That is certainly a broad brush stroke, McIntosh did a nice business in what were essentially class B amplifiers. Many of the older readers here enjoyed Rock & Roll music during their teen years delivered via the class B amplifier at the local AM radio station, without "horrid" results. Maybe you meant low bias class AB. No, I actually meant class B. Do the load line analysis, or have a look at my website pages and print out a set of curves for 6550 which are virtually the same as KT88. http://www.turneraudio.com.au/loadma...p-triodes.html I did the analysis as I earlier had said that I would. 1.25k does seem to be a reasonable class B load for the KT-88/6550 just as you said. I asked the question because I have not built any amps with this tube and am not familiar with it beyond the fact that the Quadraplex VTRs at the Television Station where I worked as a youth had a couple dozen 6550s in each VTR. The class B load of 1.25k seemed low to me relative to your 2.5k class A load, but I made two erroneous assumptions in asking that question. First I didn't realize that ra for the KT-88 is as low as it is, and second I didn't take into consideration that your class A load line is dissipation limited rather than voltage limited as with the class B load line. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
In article ,
Patrick Turner wrote: John Byrns wrote: In article , Eeyore wrote: Patrick Turner wrote: One cannot have distortion cancelling by one tube cancelling that in another when one is cut off. THANK YOU ! Basics do matter. Indeed they do, but neither Patrick, myself, or anyone else is correct on every issue. In this case Patrick has vigorously asserted that this view, which he holds in common with you, is true, but he has failed to even attempt an argument that might demonstrate its truth. Patrick is an extremely skilled and talented fellow in the practical aspects of tube amp design and construction, but he has a very limited understanding of what is going on behind the scenes in the theory of tube amp operation. I think my website might indicate that your are not quite right about my levels of understanding. Can you point out where on your website it explains why "One cannot have distortion cancelling by one tube cancelling that in another when one is cut off"? I admit I do not understand this notion that even order distortion can't be canceled when one tube is cut off, and I would really appreciate an explanation from someone that understands why cancellation should fail to occur? Why I don't understand this is because even order distortion is canceled in class B amplifiers, where one or the other tube is cutoff over almost the entire cycle, yet cancellation still occurs! I am having trouble reconciling this with your claim that cancelation can't occur. Also the related observation that there is no even order distortion in a sine wave which has had both the tops and bottoms of the sine wave symmetrically clipped off. Please explain. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
"John Byrns" wrote in message
... In article , Patrick Turner wrote: John Byrns wrote: In article , Eeyore wrote: Patrick Turner wrote: One cannot have distortion cancelling by one tube cancelling that in another when one is cut off. THANK YOU ! Basics do matter. Indeed they do, but neither Patrick, myself, or anyone else is correct on every issue. In this case Patrick has vigorously asserted that this view, which he holds in common with you, is true, but he has failed to even attempt an argument that might demonstrate its truth. Patrick is an extremely skilled and talented fellow in the practical aspects of tube amp design and construction, but he has a very limited understanding of what is going on behind the scenes in the theory of tube amp operation. I think my website might indicate that your are not quite right about my levels of understanding. Can you point out where on your website it explains why "One cannot have distortion cancelling by one tube cancelling that in another when one is cut off"? I admit I do not understand this notion that even order distortion can't be canceled when one tube is cut off, and I would really appreciate an explanation from someone that understands why cancellation should fail to occur? Why I don't understand this is because even order distortion is canceled in class B amplifiers, where one or the other tube is cutoff over almost the entire cycle, yet cancellation still occurs! I am having trouble reconciling this with your claim that cancelation can't occur. Also the related observation that there is no even order distortion in a sine wave which has had both the tops and bottoms of the sine wave symmetrically clipped off. Please explain. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ John: I think that the problem that Patrick may be having with this concept is more one of semantics than anything else. It's hard to intuitively view "cancellation" of even harmonic terms if both tubes are not conducting at the same time. Let's view it with a very small amount of math thrown in. Sometimes intuition does not take one far enough. Take a pure Class B amplifier. Each tube or transistor conducts for 180 degrees (forget about any crossover effects). The current in each half of the OPT primary, neglecting any magnetizing current is a half sine. Do a Fourier expansion on this half sine and it's full of harmonics...all even ones. But as long as the two tubes are well matched, the two half primary currents are of identical shape and add in the OPT secondary to produce a full sine, symmetrical about its zero voltage axis. The even harmonics are gone. Is this cancellation? Well I guess you might call it that. Basically any waveform symmetrical about its zero voltage axis will contain no even harmonics...even a symmetrical square wave. Similarly, if we symmetrically clip the tops off both the positive and negative half cycles of our sine we do not see any even harmonic content either for the same reason. The ideal push-pull output stage using matched devices is not going to produce even order distortion regardless of its operating class, whether it be Class A, Class AB or Class B. If in any doubt, see pages 300-301 of RDH4, or any university level math text covering Fourier analysis. Best Regards : Doug Bannard, P. Eng. |
Output classes A and AB
On Oct 28, 2:29 am, "Doug Bannard" wrote:
"John Byrns" wrote in message ... In article , Patrick Turner wrote: John Byrns wrote: In article , Eeyore wrote: Patrick Turner wrote: One cannot have distortion cancelling by one tube cancelling that in another when one is cut off. THANK YOU ! Basics do matter. Indeed they do, but neither Patrick, myself, or anyone else is correct on every issue. In this case Patrick has vigorously asserted that this view, which he holds in common with you, is true, but he has failed to even attempt an argument that might demonstrate its truth. Patrick is an extremely skilled and talented fellow in the practical aspects of tube amp design and construction, but he has a very limited understanding of what is going on behind the scenes in the theory of tube amp operation. I think my website might indicate that your are not quite right about my levels of understanding. Can you point out where on your website it explains why "One cannot have distortion cancelling by one tube cancelling that in another when one is cut off"? I admit I do not understand this notion that even order distortion can't be canceled when one tube is cut off, and I would really appreciate an explanation from someone that understands why cancellation should fail to occur? Why I don't understand this is because even order distortion is canceled in class B amplifiers, where one or the other tube is cutoff over almost the entire cycle, yet cancellation still occurs! I am having trouble reconciling this with your claim that cancelation can't occur. Also the related observation that there is no even order distortion in a sine wave which has had both the tops and bottoms of the sine wave symmetrically clipped off. Please explain. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ John: I think that the problem that Patrick may be having with this concept is more one of semantics than anything else. It's hard to intuitively view "cancellation" of even harmonic terms if both tubes are not conducting at the same time. Let's view it with a very small amount of math thrown in. Sometimes intuition does not take one far enough. Take a pure Class B amplifier. Each tube or transistor conducts for 180 degrees (forget about any crossover effects). The current in each half of the OPT primary, neglecting any magnetizing current is a half sine. Do a Fourier expansion on this half sine and it's full of harmonics...all even ones. But as long as the two tubes are well matched, the two half primary currents are of identical shape and add in the OPT secondary to produce a full sine, symmetrical about its zero voltage axis. The even harmonics are gone. Is this cancellation? Well I guess you might call it that. Basically any waveform symmetrical about its zero voltage axis will contain no even harmonics...even a symmetrical square wave. Similarly, if we symmetrically clip the tops off both the positive and negative half cycles of our sine we do not see any even harmonic content either for the same reason. The ideal push-pull output stage using matched devices is not going to produce even order distortion regardless of its operating class, whether it be Class A, Class AB or Class B. If in any doubt, see pages 300-301 of RDH4, or any university level math text covering Fourier analysis. Best Regards : Doug Bannard, P. Eng.- Hide quoted text - - Show quoted text - That looks like not created v. cancelled from here...:) cheers, Douglas |
Output classes A and AB
On Oct 27, 4:29 pm, "Doug Bannard" wrote:
"John Byrns" wrote in message {snip] I admit I do not understand this notion that even order distortion can't be canceled when one tube is cut off, and I would really appreciate an explanation from someone that understands why cancellation should fail to occur? Why I don't understand this is because even order distortion is canceled in class B amplifiers, where one or the other tube is cutoff over almost the entire cycle, yet cancellation still occurs! I am having trouble reconciling this with your claim that cancelation can't occur. Also the related observation that there is no even order distortion in a sine wave which has had both the tops and bottoms of the sine wave symmetrically clipped off. Please explain. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ John: I think that the problem that Patrick may be having with this concept is more one of semantics than anything else. It's hard to intuitively view "cancellation" of even harmonic terms if both tubes are not conducting at the same time. Let's view it with a very small amount of math thrown in. Sometimes intuition does not take one far enough. Take a pure Class B amplifier. Each tube or transistor conducts for 180 degrees (forget about any crossover effects). The current in each half of the OPT primary, neglecting any magnetizing current is a half sine. Do a Fourier expansion on this half sine and it's full of harmonics...all even ones. But as long as the two tubes are well matched, the two half primary currents are of identical shape and add in the OPT secondary to produce a full sine, symmetrical about its zero voltage axis. The even harmonics are gone. Is this cancellation? Well I guess you might call it that. Well, I'd call it cancellation. I didn't want to give my take on this and get the subthread invaded by the usual fartcatchers who follow me around before John received a sound answer, and yours is a good and complete answer; thanks Doug from me too. I don't see how anyone but a pedant can object to the "removal result" of the combination of a positive and a negative half-cycle being called a cancellation; it is an "addition" only in a mathematical sense (and it goes against the grain!). But Patrick, clinging so tightly to the formal definition, had me wondering whether there was something else that made my commonsense view wrong. Basically any waveform symmetrical about its zero voltage axis will contain no even harmonics...even a symmetrical square wave. Similarly, if we symmetrically clip the tops off both the positive and negative half cycles of our sine we do not see any even harmonic content either for the same reason. The ideal push-pull output stage using matched devices is not going to produce even order distortion regardless of its operating class, whether it be Class A, Class AB or Class B. If in any doubt, see pages 300-301 of RDH4, or any university level math text covering Fourier analysis. Nah. We don't need to resort to such desperate measures while we have you and Patrick to explain things to us. Best Regards : Doug Bannard, P. Eng. Thanks again for restoring my faith in the simple things... Andre Jute There are more things in heaven and earth, Horatio -- Will the Shake Your first take on a problem is usually the right one -- John Z DeLorean, when we were young |
Output classes A and AB
John Byrns wrote: In article , Patrick Turner wrote: John Byrns wrote: In article , Eeyore wrote: Patrick Turner wrote: One cannot have distortion cancelling by one tube cancelling that in another when one is cut off. THANK YOU ! Basics do matter. Indeed they do, but neither Patrick, myself, or anyone else is correct on every issue. In this case Patrick has vigorously asserted that this view, which he holds in common with you, is true, but he has failed to even attempt an argument that might demonstrate its truth. Patrick is an extremely skilled and talented fellow in the practical aspects of tube amp design and construction, but he has a very limited understanding of what is going on behind the scenes in the theory of tube amp operation. I think my website might indicate that your are not quite right about my levels of understanding. Can you point out where on your website it explains why "One cannot have distortion cancelling by one tube cancelling that in another when one is cut off"? No, because I endorse what many books tell us such as RDH4. I refuse to spoon feed ppl about the basics. But simple use of the brain about what happens inside each tube of a PP pair in class A, AB or B or C etc reveals to most minds what happens. I admit I do not understand this notion that even order distortion can't be canceled when one tube is cut off, and I would really appreciate an explanation from someone that understands why cancellation should fail to occur? Once one tube is well and truly cut right off, the other is conducting delivery of all the power through only 1/2 the OPT primary, and its as if the other cut off tube is unplugged for that part of the wave. Each output tube takes a turn at current delivery by means of turn on harder to produce voltage change which then SUM but without cancelation of current. The cancelation thinge is when the tubes are in class A and the 2H is cancelled, similarly to in any LTP. The Class A Wiulliamson has a common cathode resistance which is unbypassed to assist the cancellation, but ideally, a CCS should be used, or even a choke for the cathode resistance. One may argue whether having each cathode with separate RC bypass networks is better than a shared cathode impedance/resistance. But the individual cathode RC or fixed bias does allow class AB operation, wheras the common cathode R/Z does not. Work out the 2H current in each tube while in class A of each tube and see how such currents are applied across the primary. The reason 2H is low in class A is that the 2H currents are the same phase at each end of the OPT primary which cannot produce voltage in the load if the current is applied in common mode to both ends of the load. Why I don't understand this is because even order distortion is canceled in class B amplifiers, where one or the other tube is cutoff over almost the entire cycle, yet cancellation still occurs! Class B and C amps have a summing action of their turn on currents which pull the voltage in oposite directio. There is huge distortion in the Class B device current, but the total action produces a linear voltage outcome. a pair of complementary pnp and npn SS devices do exactly the same thing. But its not unusual that the amount of V swing achieved in each direction by each class B device is not equal so some 2H will allways appear in the output. But usually, the 3H dominates H distortion products. I am having trouble reconciling this with your claim that cancelation can't occur. You have to see the distinction between each tube in class A "sharing the load" and the summing actions once cut off has occurred. One could say correctly that the severe current distortions of each device in class B are cancelled by means of the summing. In effect they are. In class A, what one tube does with the load current affect the other tube. If you have one tube with high gm and the other with a low gm, then the amount of class A power produced in each tube varies. This in effect is because the load seen by each tube working as an SE tube varies, and where you have RLa-a = 10k, then the class A load of each tube = 5k in theory. Tubes ain't any more perfect than I am, and you will find that perhaps one tube "sees" 4.5k, and the other sees 5.5k. Careful measurements with 10 ohm current sensing R in each anode or cathode circuit will tell you about the current flow in each tube. Also the related observation that there is no even order distortion in a sine wave which has had both the tops and bottoms of the sine wave symmetrically clipped off. A sine wave symetrically clipped usually has a pile of odd numbered H, and not much 2H. Some 2H is usually there though. Try measuring one of your AB amps at 1W, 3.2W, 10W, 32W and as clipping progesses from mild to severe. Patrick Turner. Please explain. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
In article . com,
Multi-grid wrote: On Oct 28, 2:29 am, "Doug Bannard" wrote: "John Byrns" wrote in message ... In article , Patrick Turner wrote: John Byrns wrote: In article , Eeyore wrote: Patrick Turner wrote: One cannot have distortion cancelling by one tube cancelling that in another when one is cut off. THANK YOU ! Basics do matter. Indeed they do, but neither Patrick, myself, or anyone else is correct on every issue. In this case Patrick has vigorously asserted that this view, which he holds in common with you, is true, but he has failed to even attempt an argument that might demonstrate its truth. Patrick is an extremely skilled and talented fellow in the practical aspects of tube amp design and construction, but he has a very limited understanding of what is going on behind the scenes in the theory of tube amp operation. I think my website might indicate that your are not quite right about my levels of understanding. Can you point out where on your website it explains why "One cannot have distortion cancelling by one tube cancelling that in another when one is cut off"? I admit I do not understand this notion that even order distortion can't be canceled when one tube is cut off, and I would really appreciate an explanation from someone that understands why cancellation should fail to occur? Why I don't understand this is because even order distortion is canceled in class B amplifiers, where one or the other tube is cutoff over almost the entire cycle, yet cancellation still occurs! I am having trouble reconciling this with your claim that cancelation can't occur. Also the related observation that there is no even order distortion in a sine wave which has had both the tops and bottoms of the sine wave symmetrically clipped off. Please explain. John: I think that the problem that Patrick may be having with this concept is more one of semantics than anything else. It's hard to intuitively view "cancellation" of even harmonic terms if both tubes are not conducting at the same time. Let's view it with a very small amount of math thrown in. Sometimes intuition does not take one far enough. Take a pure Class B amplifier. Each tube or transistor conducts for 180 degrees (forget about any crossover effects). The current in each half of the OPT primary, neglecting any magnetizing current is a half sine. Do a Fourier expansion on this half sine and it's full of harmonics...all even ones. But as long as the two tubes are well matched, the two half primary currents are of identical shape and add in the OPT secondary to produce a full sine, symmetrical about its zero voltage axis. The even harmonics are gone. Is this cancellation? Well I guess you might call it that. Basically any waveform symmetrical about its zero voltage axis will contain no even harmonics...even a symmetrical square wave. Similarly, if we symmetrically clip the tops off both the positive and negative half cycles of our sine we do not see any even harmonic content either for the same reason. The ideal push-pull output stage using matched devices is not going to produce even order distortion regardless of its operating class, whether it be Class A, Class AB or Class B. If in any doubt, see pages 300-301 of RDH4, or any university level math text covering Fourier analysis. Best Regards : Doug Bannard, P. Eng. That looks like not created v. cancelled from here...:) It wouldn't look like "not created" if you used a current probe to observe the anode current waveforms of each of the two PP output tubes. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
In article ,
Patrick Turner wrote: John Byrns wrote: In article , Patrick Turner wrote: John Byrns wrote: In article , Eeyore wrote: Patrick Turner wrote: One cannot have distortion cancelling by one tube cancelling that in another when one is cut off. THANK YOU ! Basics do matter. Indeed they do, but neither Patrick, myself, or anyone else is correct on every issue. In this case Patrick has vigorously asserted that this view, which he holds in common with you, is true, but he has failed to even attempt an argument that might demonstrate its truth. Patrick is an extremely skilled and talented fellow in the practical aspects of tube amp design and construction, but he has a very limited understanding of what is going on behind the scenes in the theory of tube amp operation. I think my website might indicate that your are not quite right about my levels of understanding. Can you point out where on your website it explains why "One cannot have distortion cancelling by one tube cancelling that in another when one is cut off"? No, because I endorse what many books tell us such as RDH4. Yes, and if we read, understand, and apply what the RDH4 tells us, we understand that even harmonic cancellation does occur in class AB amplifiers, even when the tubes are cutoff for part of the cycle. That is what this discussion is all about, Multi-grid, at least I think Multi-grid was the first, made the claim that even harmonic cancellation doesn't occur when the tubes in a class AB amplifier are cut off, then both you and Eeyore jumped in to back up his claim. Have you read pages 300 & 301 of the RDH4 as Doug Bannard, not to be confused with the Multi-grid Doug, suggested? There is not a lot there, but it is a start and hopefully might lead you to follow up by pursuing some more complete references on Fourier analysis and the theory behind it. I refuse to spoon feed ppl about the basics. But simple use of the brain about what happens inside each tube of a PP pair in class A, AB or B or C etc reveals to most minds what happens. Simple use of the brain reveals that the cathode current of each tube of a class A, AB or B amplifier contains even order harmonic currents, among others. What simple use of the brain apparently doesn't reveal to some is the reason these even harmonic currents are canceled in the output of a PP amplifier. If the two PP sides of the amplifier are identical, then the cancellation will be complete, if the to sides are not identical, for any one of a multitude of reasons, including but not limited to, mismatched tubes then the cancellation will only be partial. I admit I do not understand this notion that even order distortion can't be canceled when one tube is cut off, and I would really appreciate an explanation from someone that understands why cancellation should fail to occur? Once one tube is well and truly cut right off, the other is conducting delivery of all the power through only 1/2 the OPT primary, and its as if the other cut off tube is unplugged for that part of the wave. Yes, that is obvious, but not especially relevant. Each output tube takes a turn at current delivery by means of turn on harder to produce voltage change which then SUM but without cancelation of current. It is not necessary for each tube to be continuously delivering current for the cancellation of the even order harmonic currents from the two tubes to cancel, it is only necessary that the two currents exhibit the required symmetry with respect to one another. The harmonic currents flow even when the tube is cutoff, it's just that all the currents, including the 2H current, happen to add to zero for the period the tube is cutoff. The cancelation thinge is when the tubes are in class A and the 2H is cancelled, similarly to in any LTP. "The cancellation thingy" is also when the tubes are in class AB or even class B, where the 2H is also canceled. The Class A Wiulliamson has a common cathode resistance which is unbypassed to assist the cancellation, but ideally, a CCS should be used, or even a choke for the cathode resistance. One may argue whether having each cathode with separate RC bypass networks is better than a shared cathode impedance/resistance. But the individual cathode RC or fixed bias does allow class AB operation, wheras the common cathode R/Z does not. This is a diversion into a side issue but let me ask, why does a common cathode RC not allow class AB? Granted fixed bias provides far better class AB operation than does cathode bias, but given that we are using cathode bias I fail to see how a common cathode RC would prevent class AB operation anymore than separate cathode RCs do? What separate cathode RCs do is reduce to some extent the effect of mismatched tubes in the two sides of the PP circuit, but given well matched tubes a common RC does not prevent class AB operation. Work out the 2H current in each tube while in class A of each tube and see how such currents are applied across the primary. The reason 2H is low in class A is that the 2H currents are the same phase at each end of the OPT primary which cannot produce voltage in the load if the current is applied in common mode to both ends of the load. The same is also true of the even order harmonic currents in class AB and class B amplifiers! I am beginning to get a glimmer of what your problem is here, it appears that you are not correctly identifying the even order harmonic currents that are applied to the output transformer primary in common mode. You are looking at the total current waveform from each tube, which is causing you to become confused. When the tube is cut off, the even order harmonic currents continue to flow through the tube, its just that the DC, fundamental, and all the harmonics together sum to zero during the time the tube is cutoff, however this does not imply that the current of any particular harmonic is aero during the time of cutoff, hence cancellation goes on as if the tube had not cutoff. Why I don't understand this is because even order distortion is canceled in class B amplifiers, where one or the other tube is cutoff over almost the entire cycle, yet cancellation still occurs! Class B and C amps have a summing action of their turn on currents which pull the voltage in oposite directio. It sounds like you are describing the push pull action of any class of amplifier? There is huge distortion in the Class B device current, but the total action produces a linear voltage outcome. And that is exactly the point, which you persist in trying to deny. Although odd order non-linearity does remain in class A, AB, and B amplifiers. a pair of complementary pnp and npn SS devices do exactly the same thing. But its not unusual that the amount of V swing achieved in each direction by each class B device is not equal so some 2H will allways appear in the output. But usually, the 3H dominates H distortion products. You are simply stating that in the real world, devices, drive signals and other parameters of the two sides of a PP amplifier may not be perfectly matched, and as a result the even harmonic cancellation will be incomplete, but partial cancellation still occurs reducing the even order harmonic distortion, even when one or the other tube is cutoff. I am having trouble reconciling this with your claim that cancelation can't occur. You have to see the distinction between each tube in class A "sharing the load" and the summing actions once cut off has occurred. One could say correctly that the severe current distortions of each device in class B are cancelled by means of the summing. In effect they are. In class A, what one tube does with the load current affect the other tube. If you have one tube with high gm and the other with a low gm, then the amount of class A power produced in each tube varies. This in effect is because the load seen by each tube working as an SE tube varies, and where you have RLa-a = 10k, then the class A load of each tube = 5k in theory. Tubes ain't any more perfect than I am, and you will find that perhaps one tube "sees" 4.5k, and the other sees 5.5k. Careful measurements with 10 ohm current sensing R in each anode or cathode circuit will tell you about the current flow in each tube. Not only will the load seen by each of the two tubes in a class A amplifier vary depending on the relative gm of the two tubes, but even with two perfectly matched tubes the relative load each sees will vary over the complete audio cycle due to the dynamic changes in the tubes characteristics over the cycle. Also the related observation that there is no even order distortion in a sine wave which has had both the tops and bottoms of the sine wave symmetrically clipped off. A sine wave symetrically clipped usually has a pile of odd numbered H, and not much 2H. Some 2H is usually there though. There is no 2H if the clipping is symmetrical. Try measuring one of your AB amps at 1W, 3.2W, 10W, 32W and as clipping progesses from mild to severe. I'm not sure what you think this experiment proves? All it proves is that a voltage amplifier or driver stage is producing some 2H at high levels of drive, or that the output stage begins clipping asymmetrically at high output levels due to some circuit feature or imperfection, I don't see what it has to do with the issue of even order harmonic cancellation? If one wanted to go to the trouble I have little doubt that given some effort one could design a class AB amplifier that wouldn't exhibit the problems you are alluding to. You are simply dragging a Red Herring across the path with this issue which is unrelated to the issue of even order harmonic cancellation in class AB amplifiers when the tubes cutoff. And these sorts of problems aren't restricted to class AB amplifiers, there are class A PP amplifiers that show a rise in even order, including 2H, distortion even before clipping sets in. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
In article ,
flipper wrote: On Sun, 28 Oct 2007 13:12:57 GMT, Patrick Turner wrote: In class A, what one tube does with the load current affect the other tube. Not with pentodes (or any SS device) it doesn't. (Ideal) Pentodes are essentially voltage controlled current sources and will source that current regardless of what the 'other' is doing. Yes, but what the other tube is doing will affect load seen by the first tube, which will affect the voltage on its anode, if not its anode current. I think all this is just going to further confuse poor old Patrick, we should probably let him alone so he can get back to his shed to wind some output transformers and catch up with the backlog of Big US amps awaiting modifications. While he working on that, his sub conscious can work on the cutoff problem and perhaps an Epiphany will eventually come. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
In article ,
flipper wrote: On Sun, 28 Oct 2007 21:40:26 GMT, John Byrns wrote: In article , flipper wrote: On Sun, 28 Oct 2007 13:12:57 GMT, Patrick Turner wrote: In class A, what one tube does with the load current affect the other tube. Not with pentodes (or any SS device) it doesn't. (Ideal) Pentodes are essentially voltage controlled current sources and will source that current regardless of what the 'other' is doing. Yes, but what the other tube is doing will affect load seen by the first tube, which will affect the voltage on its anode, if not its anode current. Of course, but Patrick was speaking of what happens 'in the tube' and (ideal) pentode conduction is independent of anode voltage The load change will affect voltage but as long as everything is 'identical' (which is never the 'real world' case, of course) both sides will see and do the same thing so, again, even harmonics are canceled. Odd harmonics are another matter. The point here, though, was that 'tube interaction', and 'mutual' conduction, isn't the 'canceling' mechanism. It's the OPT summing two equal but opposite waveforms and it's inherent to what even harmonics 'are'. With both halves 'equal but opposite' they just plain don't exist no matter 'how' you got there and how many you started with, because they 'cancel'. Even if it's just equal but opposite 1ns blips, no even order harmonics. I think all this is just going to further confuse poor old Patrick, we should probably let him alone so he can get back to his shed to wind some output transformers and catch up with the backlog of Big US amps awaiting modifications. While he working on that, his sub conscious can work on the cutoff problem and perhaps an Epiphany will eventually come. Well, let's not get too smug about it because there could/should be 'some' interaction with triodes so his description has intuitive merit in that case. But it's a special case due to plate feedback in triodes (which could/should affect all harmonics to some extent) and not PP, per see. I may not have been very clear, but my second sentence was referring to the issue of whether or not cutoff makes even order cancelation impossible. I consider the interaction question I referred to in the first sentence to be only a side issue to the original question of cutoff defeating even order cancelation. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
John Byrns wrote: In article , Patrick Turner wrote: John Byrns wrote: In article , Patrick Turner wrote: John Byrns wrote: In article , Eeyore wrote: Patrick Turner wrote: One cannot have distortion cancelling by one tube cancelling that in another when one is cut off. THANK YOU ! Basics do matter. Indeed they do, but neither Patrick, myself, or anyone else is correct on every issue. In this case Patrick has vigorously asserted that this view, which he holds in common with you, is true, but he has failed to even attempt an argument that might demonstrate its truth. Patrick is an extremely skilled and talented fellow in the practical aspects of tube amp design and construction, but he has a very limited understanding of what is going on behind the scenes in the theory of tube amp operation. I think my website might indicate that your are not quite right about my levels of understanding. Can you point out where on your website it explains why "One cannot have distortion cancelling by one tube cancelling that in another when one is cut off"? No, because I endorse what many books tell us such as RDH4. Yes, and if we read, understand, and apply what the RDH4 tells us, we understand that even harmonic cancellation does occur in class AB amplifiers, even when the tubes are cutoff for part of the cycle. That is what this discussion is all about, Multi-grid, at least I think Multi-grid was the first, made the claim that even harmonic cancellation doesn't occur when the tubes in a class AB amplifier are cut off, then both you and Eeyore jumped in to back up his claim. I have lost track of whatever it was that Multigrid might have said. But once one tube cuts off, the part of the cycle handled by the other tube alone generates a substantially linear output voltage when summed with the other tube's efforts. IMHO, the summing action of tubes or other devices in class B does not include the cancelation of harmonics because of the common mode application of the near identical even harmonics of the same phase of both tubes across the whole primary. The class B action means the current wave of the device is a series of 1/2 sine waves with a flat part and many harmonics are present if you ever wanted to filter them out and quantify them. But these harmonics don't appear in the load. Fact. There isn't mutual cancellation like in class A. But there is crossover distortion as one device turns off, and the other turns on. Class B is like the two guys sawing the log with the long bush saw, and each guy only pulls the saw from centre position his way and pushes it back to centre then lets go. The mens' current production is severely distorted, and jerky, stop'start in fact, but they manage to saw the log with the saw travelling one way and the other like an output voltage due to the summed applications of the efforts, ie, currents. Once one device cuts off, no current flows, so there are no distortion currents produced by that device. So the only relevant issue is what each tube does when it turns on, and has the load power under its control. Have you read pages 300 & 301 of the RDH4 as Doug Bannard, not to be confused with the Multi-grid Doug, suggested? Page 299 through to pg 303 is about Fourier Series and Harmonics. There is so much high falooting math I don't understand very much, and i don't need to. The pages do not contain a large range of down to earth simple current wave forms of PP amps with the list and typical amplitudes of the harmonics present. There is not a lot there, but it is a start and hopefully might lead you to follow up by pursuing some more complete references on Fourier analysis and the theory behind it. I do not need to become fully familiar with Fourier analysis. I am not formally university educated, and don'y undertsand the math. I make amops and do whatever relevant hands on measuring and analysis that I need to do without putting on the garb of the white coated boffin. I do know that were I to measure the harmonics of a typical AB PP amps current waves, I would find a shirtfull and bootfull of harmonics when the amp is well into class AB. But a very small fraction of them find their way into the secondary of an OPT, or into a direct connected load of complementary pnp and npn devices. I refuse to spoon feed ppl about the basics. But simple use of the brain about what happens inside each tube of a PP pair in class A, AB or B or C etc reveals to most minds what happens. Simple use of the brain reveals that the cathode current of each tube of a class A, AB or B amplifier contains even order harmonic currents, among others. What simple use of the brain apparently doesn't reveal to some is the reason these even harmonic currents are canceled in the output of a PP amplifier. In a class B amp, the output current produced by each output tube is substantially linear with input voltage for each 1/2 wave of the sine wave. Its tuned off for the other 1/2. Its mate does the same substantially linear job on the other 1/2 of the wave not handled by the first. Its a simple case of each device substantially linearly amplifying each 1/2 wave. The Fourier Analysis might describe things with math, and any given repeating waveform can be expressed in terms of various amplitudes of sine wave harmonics with amplitudes and phase relationships required to give the original un-analysed non sine wave. The quantified harmonics are simply not relevant in the class B case. It could be argued cancelation does take place, but I see things simply as a arched 1/2 wave being a linear effort with ZERO effort made upon the other 1/2 of the wave. Now were you to analyse and quantify and draw the graphs for the fundememental and harmonics of EACH class B device, you could congratulate yourself on your wizardry, but just getting a PC spectral analysis program to do it all for you could be less prone to errors after days of work doing the RDH4 math "by hand". After doing the graphs of the current waves displaying them so they represent the currents flowing in/out of each end of the OPT primary, perhaps you'd find that lo and behold, when the harmonic waves are all viewed, it is seen they applied in common mode to each end of the load and thus cannot appear as load voltage current since the amplitude and phase of the harmonics is the same at each end of the load, ie, each end of the OPT. But I see no need to do this labourious and unecessary math and graphing task which may satisfy the lofty smartarse acedemics. RDH4 does not have too many full descriptions to simply illustrate what I have been saying. So what use are the smartarse acedemics? have they earned their dinner? I don't see they have unless they make easy for lesser mortals to understand the concepts. So the Fourier analysis is merely confusing. I see the tubes simply turning on and off to do various parts of the wave, and while 'on', they have a linear control of output current, but while off, they have no control. In some SS amps, there are two crossover regions as the output voltage lifts its own rail voltages as in class H, all very complicated in Fourier terms but all you need to know is the timing sequence of what happens during wave cycles. The Soundcraftsmen amp was class H. If the two PP sides of the amplifier are identical, then the cancellation will be complete, if the to sides are not identical, for any one of a multitude of reasons, including but not limited to, mismatched tubes then the cancellation will only be partial. The perfectly complementary action of two devices in class B never ever occurs. Crossover distortion of some kind spoils the attempt at perfection. Where you have EL34 on one side and 6L6 on the other of a PP AB amp, expect a large amplitude DIFFERENCE in the applied commonality of harmonic currents applied to each end of the OPT, so you'd get considerably more resultant THD than if you had a matched pair of the same tube type number. But in common sense terms, each of the pair of tubes are just switching on and off, and while off, each tube is oblivious to its Fourier productions. Its bloody turned off, to there ain't nothin going on. If a tube could talk, it'd tell you I am NOT making any harmonics while I is turned off, OK. You might answer the cheeky brat of a tube by saying but all your harmonics add to a flat line of zero current while you are turned off, and don't argue, because you can't argue with that guy Fourier. So all these many harmonics are there, but appear not to be there for 1/2 a wave.... I admit I do not understand this notion that even order distortion can't be canceled when one tube is cut off, and I would really appreciate an explanation from someone that understands why cancellation should fail to occur? Once one tube is well and truly cut right off, the other is conducting delivery of all the power through only 1/2 the OPT primary, and its as if the other cut off tube is unplugged for that part of the wave. Yes, that is obvious, but not especially relevant. We cannot remove tubes and plug them back in fast enough to demonstrate the common sense notion... Each output tube takes a turn at current delivery by means of turn on harder to produce voltage change which then SUM but without cancelation of current. It is not necessary for each tube to be continuously delivering current for the cancellation of the even order harmonic currents from the two tubes to cancel, it is only necessary that the two currents exhibit the required symmetry with respect to one another. The harmonic currents flow even when the tube is cutoff, it's just that all the currents, including the 2H current, happen to add to zero for the period the tube is cutoff. See the above comments. The cancelation thinge is when the tubes are in class A and the 2H is cancelled, similarly to in any LTP. "The cancellation thingy" is also when the tubes are in class AB or even class B, where the 2H is also canceled. One could prove almost anything with Fourier math.... Its easier to see the mutually cancelling 2H of each tube in a class A amp. But while tubes are cut right off, summed distortion currents in that tube = zero. I have to say this does not mean the distortion currents are NOT PRESENT. They must be, but all summ to zero amps for the flat line of the tube when cut off. If you filter out harmonics of the current wave of a single tube of the clas B amp, you'll see the plethera of harmonics, and these are all sine wave signals which flow continuously, and without cutting off, but which when added will give you zero current for 1/2 the cycle wave at fundememtal F. The Class A Wiulliamson has a common cathode resistance which is unbypassed to assist the cancellation, but ideally, a CCS should be used, or even a choke for the cathode resistance. One may argue whether having each cathode with separate RC bypass networks is better than a shared cathode impedance/resistance. But the individual cathode RC or fixed bias does allow class AB operation, wheras the common cathode R/Z does not. This is a diversion into a side issue but let me ask, why does a common cathode RC not allow class AB? Because after one tube cuts off, the common Rk is the Rk for one tube only, and the Ra of the tube rises acording to (µ + 1) x Rk, and severe 3H distortion results. Granted fixed bias provides far better class AB operation than does cathode bias, but given that we are using cathode bias I fail to see how a common cathode RC would prevent class AB operation anymore than separate cathode RCs do? Well, the C bypassing the common Rk does improve AB working considerably. Quad-II has 180 ohms plus 40uF between cathode FB winding and 0V. So for music at well below clipping the biasing stays fixed. But with a sine wave the biasing voltage Ek rises dramatically in AB and is as bad as separate RC to each cathode. What separate cathode RCs do is reduce to some extent the effect of mismatched tubes in the two sides of the PP circuit, but given well matched tubes a common RC does not prevent class AB operation. True. Its the low ZC that counts. If the C is ommitted, class AB is horrid. The separate RC give far better self regulation of bias than a common RC does. Work out the 2H current in each tube while in class A of each tube and see how such currents are applied across the primary. The reason 2H is low in class A is that the 2H currents are the same phase at each end of the OPT primary which cannot produce voltage in the load if the current is applied in common mode to both ends of the load. The same is also true of the even order harmonic currents in class AB and class B amplifiers! Well yes, but they are less easy to see, as i explained above. In effect we agree then, but practically speaking a tube cut off has zero harmonic distortion currents for the time its cut off. I am beginning to get a glimmer of what your problem is here, it appears that you are not correctly identifying the even order harmonic currents that are applied to the output transformer primary in common mode. You are looking at the total current waveform from each tube, which is causing you to become confused. When the tube is cut off, the even order harmonic currents continue to flow through the tube, its just that the DC, fundamental, and all the harmonics together sum to zero during the time the tube is cutoff, however this does not imply that the current of any particular harmonic is aero during the time of cutoff, hence cancellation goes on as if the tube had not cutoff. I explained this above. The humble man in the street can be forgiven if he doubts any currents of any kind flow when the DC and harmonic currents are ZERO when the tube is cut off. Why I don't understand this is because even order distortion is canceled in class B amplifiers, where one or the other tube is cutoff over almost the entire cycle, yet cancellation still occurs! Class B and C amps have a summing action of their turn on currents which pull the voltage in oposite directio. It sounds like you are describing the push pull action of any class of amplifier? There is huge distortion in the Class B device current, but the total action produces a linear voltage outcome. And that is exactly the point, which you persist in trying to deny. Although odd order non-linearity does remain in class A, AB, and B amplifiers. a pair of complementary pnp and npn SS devices do exactly the same thing. But its not unusual that the amount of V swing achieved in each direction by each class B device is not equal so some 2H will allways appear in the output. But usually, the 3H dominates H distortion products. You are simply stating that in the real world, devices, drive signals and other parameters of the two sides of a PP amplifier may not be perfectly matched, and as a result the even harmonic cancellation will be incomplete, but partial cancellation still occurs reducing the even order harmonic distortion, even when one or the other tube is cutoff. The even order artifacts in PP are generated when the two halves of the sine wave fundemental are amplified to a different amplitude; ie, the transfer curve of each tube isn't the same... I am having trouble reconciling this with your claim that cancelation can't occur. You have to see the distinction between each tube in class A "sharing the load" and the summing actions once cut off has occurred. One could say correctly that the severe current distortions of each device in class B are cancelled by means of the summing. In effect they are. In class A, what one tube does with the load current affect the other tube. If you have one tube with high gm and the other with a low gm, then the amount of class A power produced in each tube varies. This in effect is because the load seen by each tube working as an SE tube varies, and where you have RLa-a = 10k, then the class A load of each tube = 5k in theory. Tubes ain't any more perfect than I am, and you will find that perhaps one tube "sees" 4.5k, and the other sees 5.5k. Careful measurements with 10 ohm current sensing R in each anode or cathode circuit will tell you about the current flow in each tube. Not only will the load seen by each of the two tubes in a class A amplifier vary depending on the relative gm of the two tubes, but even with two perfectly matched tubes the relative load each sees will vary over the complete audio cycle due to the dynamic changes in the tubes characteristics over the cycle. And you will get resultant intermodulations.... But class A PP remains basically more linear process than SE... Also the related observation that there is no even order distortion in a sine wave which has had both the tops and bottoms of the sine wave symmetrically clipped off. A sine wave symetrically clipped usually has a pile of odd numbered H, and not much 2H. Some 2H is usually there though. There is no 2H if the clipping is symmetrical. But only if the clipping is symetrical. In many amps it is not perfectly symetrical, and 2H remains substantial. Try measuring one of your AB amps at 1W, 3.2W, 10W, 32W and as clipping progesses from mild to severe. I'm not sure what you think this experiment proves? Depends what you measure. If its the amplitude of each harmonic one will get some surprising results showing varying levels of each harmonic as outpt voltage is increased. For any given amp, its very difficult to predict and or calculate the rate of rise in harmonics with output voltage. Better you merely assume you'll get THD, and then measure it to know how much. Some empiricle assumptions can be made, but to know, you must measure. All it proves is that a voltage amplifier or driver stage is producing some 2H at high levels of drive, or that the output stage begins clipping asymmetrically at high output levels due to some circuit feature or imperfection, I don't see what it has to do with the issue of even order harmonic cancellation? If one wanted to go to the trouble I have little doubt that given some effort one could design a class AB amplifier that wouldn't exhibit the problems you are alluding to. You are simply dragging a Red Herring across the path with this issue which is unrelated to the issue of even order harmonic cancellation in class AB amplifiers when the tubes cutoff. And these sorts of problems aren't restricted to class AB amplifiers, there are class A PP amplifiers that show a rise in even order, including 2H, distortion even before clipping sets in. Don't worry, be happy. Patrick Turner. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
flipper wrote: On Sun, 28 Oct 2007 13:12:57 GMT, Patrick Turner wrote: snip. I mentioned... In class A, what one tube does with the load current affect the other tube. You said... Not with pentodes (or any SS device) it doesn't. (Ideal) Pentodes are essentially voltage controlled current sources and will source that current regardless of what the 'other' is doing. But the pentode (or tetrode) gm could be different, and thus if so, current turn on/off is different despite the same amplitude of grid input signals and total summed output voltage being the same for each tube. So the load each pentode sees is also different. It does not follow that just because you have PP pentodes instead of some other devices such as triodes, bjts, or mosfets, that in PP class A the pentodes will always be experiencing the same load as each other. But even with un-matched pentodes, the amount of class A 2H cancelling is so substantial that the 3H is still usually much more than any 2H left due to poor tube matching. That one PP output tube never sees an exactly equal load to the other is of negligible consequence because the THD in a couple of watts from a pair of tubes capable of 30 watts of class A will be inherently low. Patrick Turner. |
Output classes A and AB
In article ,
Patrick Turner wrote: John Byrns wrote: In article , Patrick Turner wrote: No, because I endorse what many books tell us such as RDH4. Yes, and if we read, understand, and apply what the RDH4 tells us, we understand that even harmonic cancellation does occur in class AB amplifiers, even when the tubes are cutoff for part of the cycle. That is what this discussion is all about, Multi-grid, at least I think Multi-grid was the first, made the claim that even harmonic cancellation doesn't occur when the tubes in a class AB amplifier are cut off, then both you and Eeyore jumped in to back up his claim. I have lost track of whatever it was that Multigrid might have said. Multi-grid said something to the effect that 2H cancellation can't occur in a class AB amplifier when one one or the other of the tubes is cutoff, the same as you are, or at least were previously saying. But once one tube cuts off, the part of the cycle handled by the other tube alone generates a substantially linear output voltage when summed with the other tube's efforts. IMHO, the summing action of tubes or other devices in class B does not include the cancelation of harmonics because of the common mode application of the near identical even harmonics of the same phase of both tubes across the whole primary. But that is just saying in different words that 2H cancellation does occur even while one or the other tube is cutoff, "common mode application of the near identical even harmonics of the same phase of both tubes across the whole primary" is simply the mechanism for accomplishing the 2H cancellation. We are saying the same thing in different words. The class B action means the current wave of the device is a series of 1/2 sine waves with a flat part and many harmonics are present if you ever wanted to filter them out and quantify them. But these harmonics don't appear in the load. Fact. Exactly, that is the whole point of 2H and even harmonic cancellation in a PP stage, the odd harmonics do appear in the load however because they appear across the output transformer as a differential signal, not a common mode signal. There isn't mutual cancellation like in class A. But there is crossover distortion as one device turns off, and the other turns on. I don't see the distinction between class A and class B operation, the 2H and even harmonic cancellation mechanism is the same for both classes of operation, the only difference is the amplitude of the harmonics generated. In both cases the 2H and even harmonics appear at the output transformer as a common mode signal and are canceled in the load by the action of the transformer. Once one device cuts off, no current flows, so there are no distortion currents produced by that device. That does not follow, all that follows is that the sum of the DC current, the fundamental current, and the harmonic distortion currents sum to zero during the interval that the tube is cut off. That is not the same thing as saying there are no distortion currents flowing while the tube is cut off. Mr. Fourier explains this for us. [Big Snip] But I see no need to do this labourious and unecessary math and graphing task which may satisfy the lofty smartarse acedemics. Exactly, there is no need for you to do it because the academics have all ready done it for you, so you already know that the 2H and other even order harmonics will cancel in a PP stage, be it class A, class B, or anywhere in between, or even if it is PP class C. RDH4 does not have too many full descriptions to simply illustrate what I have been saying. So what use are the smartarse acedemics? have they earned their dinner? Certainly they have, you are beginning to sound a little like Ian. The cancelation thinge is when the tubes are in class A and the 2H is cancelled, similarly to in any LTP. "The cancellation thingy" is also when the tubes are in class AB or even class B, where the 2H is also canceled. One could prove almost anything with Fourier math.... Its easier to see the mutually cancelling 2H of each tube in a class A amp. To me it is easier to see the mutually canceling 2H of each tube in a class B amp because the amplitudes of the harmonics are larger and as a result are easier to observe. But while tubes are cut right off, summed distortion currents in that tube = zero. I have to say this does not mean the distortion currents are NOT PRESENT. They must be, but all summ to zero amps for the flat line of the tube when cut off. If you filter out harmonics of the current wave of a single tube of the clas B amp, you'll see the plethera of harmonics, and these are all sine wave signals which flow continuously, and without cutting off, but which when added will give you zero current for 1/2 the cycle wave at fundememtal F. Exactly, I think you have got it now, the even order distortion currents cancel even while the tubes are cut off in either class AB or class B. Work out the 2H current in each tube while in class A of each tube and see how such currents are applied across the primary. The reason 2H is low in class A is that the 2H currents are the same phase at each end of the OPT primary which cannot produce voltage in the load if the current is applied in common mode to both ends of the load. The same is also true of the even order harmonic currents in class AB and class B amplifiers! Well yes, but they are less easy to see, as i explained above. Your reasoning on this seems backwards, I think we both agree that the harmonic currents in a class AB stage are larger than those in a class A stage, I have always found it easier to see large things, within reason, than to see small things. In effect we agree then, but practically speaking a tube cut off has zero harmonic distortion currents for the time its cut off. No, a tube cutoff does not have zero harmonic distortion currents for the time it is cut off, what happens is that the harmonic distortion currents, along with the fundamental current, and the DC bias current sum to zero over that interval, which is a completely different thing than saying the distortion currents are zero during that interval. You even said this yourself a few paragraphs up from here where you said "I have to say this does not mean the distortion currents are NOT PRESENT. They must be, but all summ to zero amps for the flat line of the tube when cut off." Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
On Oct 29, 2:14 pm, Patrick Turner wrote:
I have lost track of whatever it was that Multigrid might have said. I expect so has Cuddles Multi-grid. He's a troll. He says whatever comes into his head; it doesn't have to make sense, it just has to stir up someone. Andre Jute "You don't need global feedback to build a good-sounding amplifier." -- Henry Pasternack |
If it cancels like a cancellation...
If it walks like a cancellation, if it talks like a cancellation, if
it sounds like a cancellation, if it causes less total secind harmonic like a cancellation, if it *cancels* like a cancellation, it *is* a cancellation. I don't care who calls it an addition. That's just algebraic sleight of hand performed inside the halves of a transformer: a cancellation by any other name. Quack. Eeeh. Deeeh. Andre Jute Visit Jute on Amps at http://members.lycos.co.uk/fiultra/ "wonderfully well written and reasoned information for the tube audio constructor" John Broskie TubeCAD & GlassWare "an unbelievably comprehensive web site containing vital gems of wisdom" Stuart Perry Hi-Fi News & Record Review |
Output classes A and AB
I don't see the distinction between class A and class B operation, the 2H and even harmonic cancellation mechanism is the same for both classes of operation, the only difference is the amplitude of the harmonics generated. In both cases the 2H and even harmonics appear at the output transformer as a common mode signal and are canceled in the load by the action of the transformer. Fair enough... Once one device cuts off, no current flows, so there are no distortion currents produced by that device. That does not follow, all that follows is that the sum of the DC current, the fundamental current, and the harmonic distortion currents sum to zero during the interval that the tube is cut off. That is not the same thing as saying there are no distortion currents flowing while the tube is cut off. Mr. Fourier explains this for us. But for ordinary ppl its difficult for them to see that a number of harmonics flowing continually can sum to give a straight line signal with zero current at all. [Big Snip] But I see no need to do this labourious and unecessary math and graphing task which may satisfy the lofty smartarse acedemics. Exactly, there is no need for you to do it because the academics have all ready done it for you, so you already know that the 2H and other even order harmonics will cancel in a PP stage, be it class A, class B, or anywhere in between, or even if it is PP class C. RDH4 does not have too many full descriptions to simply illustrate what I have been saying. So what use are the smartarse acedemics? have they earned their dinner? Certainly they have, you are beginning to sound a little like Ian. Hmm, some might say that the ordinary man would only feed them porridge for dinner.... Me?, like Ian? heaven forbid! The cancelation thinge is when the tubes are in class A and the 2H is cancelled, similarly to in any LTP. "The cancellation thingy" is also when the tubes are in class AB or even class B, where the 2H is also canceled. One could prove almost anything with Fourier math.... Its easier to see the mutually cancelling 2H of each tube in a class A amp. To me it is easier to see the mutually canceling 2H of each tube in a class B amp because the amplitudes of the harmonics are larger and as a result are easier to observe. Well they would be. I built a tunable filter with Q = 50 for all H between 1kHz and 10kHz, maybe I will connect to a cathode one day when an typical class AB signal is passing. But while tubes are cut right off, summed distortion currents in that tube = zero. I have to say this does not mean the distortion currents are NOT PRESENT. They must be, but all summ to zero amps for the flat line of the tube when cut off. If you filter out harmonics of the current wave of a single tube of the clas B amp, you'll see the plethera of harmonics, and these are all sine wave signals which flow continuously, and without cutting off, but which when added will give you zero current for 1/2 the cycle wave at fundememtal F. Exactly, I think you have got it now, the even order distortion currents cancel even while the tubes are cut off in either class AB or class B. Its not obvious to most though. A cut off tube has no currents, period. Work out the 2H current in each tube while in class A of each tube and see how such currents are applied across the primary. The reason 2H is low in class A is that the 2H currents are the same phase at each end of the OPT primary which cannot produce voltage in the load if the current is applied in common mode to both ends of the load. The same is also true of the even order harmonic currents in class AB and class B amplifiers! Well yes, but they are less easy to see, as i explained above. Your reasoning on this seems backwards, I think we both agree that the harmonic currents in a class AB stage are larger than those in a class A stage, I have always found it easier to see large things, within reason, than to see small things. In effect we agree then, but practically speaking a tube cut off has zero harmonic distortion currents for the time its cut off. No, a tube cutoff does not have zero harmonic distortion currents for the time it is cut off, what happens is that the harmonic distortion currents, along with the fundamental current, and the DC bias current sum to zero over that interval, which is a completely different thing than saying the distortion currents are zero during that interval. You even said this yourself a few paragraphs up from here where you said "I have to say this does not mean the distortion currents are NOT PRESENT. They must be, but all summ to zero amps for the flat line of the tube when cut off." OK, I am just showing the difference between commonsense perceptions and theoretical perceptions involving Fourier analysis. Complementary pairs of transistors which operate close to class B and spend all their lives each only half amplifying music create the same perception dilemnas. One should be able to filter then measure the many harmonic currents in one of the 0.22 ohm emitter resistors and get a string of harmonics in a filter which will add up to zero current for 1/2 the wave. Then when you measure the other 0.22 ohm R there should be a similar set of harmonic currents present which must shunt those of the other bjt, giving a net current applied to the SINGLE output terminal which must sum substantially close to zero, leaving only crossover distortions. Its never explained this way in the books though. There are never any graphs pf all the harmonics shown in correct to scale amplitudes and phase. Patrick Turner Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
There is no best load for a class B triode amp. Class B is a horrid way to build any amp. That is certainly a broad brush stroke, McIntosh did a nice business in what were essentially class B amplifiers. Many of the older readers here enjoyed Rock & Roll music during their teen years delivered via the class B amplifier at the local AM radio station, without "horrid" results. The crock and troll was so full of distortions and screaming gits that using low distortion radio gear was wasted on them. Maybe you meant low bias class AB. No, I actually meant class B. Well, class B is where corssover distortion is not pretty.... Do the load line analysis, or have a look at my website pages and print out a set of curves for 6550 which are virtually the same as KT88. http://www.turneraudio.com.au/loadma...p-triodes.html I did the analysis as I earlier had said that I would. 1.25k does seem to be a reasonable class B load for the KT-88/6550 just as you said. But only where Ea isn't too high. But KT88 were run with up to Ea = 800V, in beam mode, and with fixed bias and for 140 watts and working class AB2, and large PO is available. TT21 were ideal at this. I forget the load value used. I asked the question because I have not built any amps with this tube and am not familiar with it beyond the fact that the Quadraplex VTRs at the Television Station where I worked as a youth had a couple dozen 6550s in each VTR. The class B load of 1.25k seemed low to me relative to your 2.5k class A load, but I made two erroneous assumptions in asking that question. First I didn't realize that ra for the KT-88 is as low as it is, and second I didn't take into consideration that your class A load line is dissipation limited rather than voltage limited as with the class B load line. 5k is a typical RLa-a for a pair of KT88. I have seen many with 3.2k. When wound up to full power with a sine wave they might smoke. Even 4k was used with Quad-II for a pair of KT66. So when the amp works AB, the B part of the operation means each tube sees only 1k on alternate wave crests. I quite like Ea = 500V and RLa-a = 8k for KT88/6550. This means the B load is 2k min, and distortion in UL or CFB isn't too much. The change in load from the class A load of 1/2 RLa-a to class B load of 1/4 RLa-a means that there is a dynamic gain reduction on each wave peak leading to 3H and other odd H, 5H especially. Class AB beam and pentode amps are worst, triode AB amps are better, and make the best AB amps. UL is between the two with regard to harmonics generated by the cut off and load changes. Patrick Turner. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
If it cancels like a cancellation...
Andre Jute wrote: If it walks like a cancellation, if it talks like a cancellation, if it sounds like a cancellation, if it causes less total secind harmonic like a cancellation, if it *cancels* like a cancellation, it *is* a cancellation. I don't care who calls it an addition. That's just algebraic sleight of hand performed inside the halves of a transformer: a cancellation by any other name. Politicians can sum and cancel almost any economic strategems, mostly leaving behind them an odd order of broken promises, and gaining no realization they were ever wrong. Of some engineers, similar could be said, but for the man in the street unused to Fourier, a cut off tube is as silent, harmless, dead, and as distortion current free as a dead politician. Even better than a dead cat. We have a Federal Election in a few weeks and the BS is really flowing well, and all sorts of arguments happening. The Liberal pollies can't see that reducing wages and putting up interest rates really hurts ppl all the while saying they are not doing either. The pollies would tell us that income and expenditure currents of many types flow and sum together to cancel flows and cut offs in the next person along to give a single wonderful income and fundemental wellbeing to all. A Mr Furious dreamed up the theory of frequent political expedients. I heard that a candidate Blue Party candidate, Mr Phark Yutu is looking to get a large vote. Anyway, back at the amplifier, too much switching and stop start amplifying is dreadful for the music. The less we rely on Fourrier, the better. Patrick Turner. Quack. Eeeh. Deeeh. Andre Jute Visit Jute on Amps at http://members.lycos.co.uk/fiultra/ "wonderfully well written and reasoned information for the tube audio constructor" John Broskie TubeCAD & GlassWare "an unbelievably comprehensive web site containing vital gems of wisdom" Stuart Perry Hi-Fi News & Record Review |
Output classes A and AB
"John Byrns" wrote in message ... : In article .com, : Andre Jute wrote: : : On Oct 25, 7:40 am, Eeyore : wrote: : : I'd love to know how that happens. There's no cancellation of ANYTHING once : one : side has ceased conducting ! : : Graham : : Holy ****! Did I say yet that Poopie is ignorant and incompetent? : : Nah, nobody can be that stupid and uninformed about tube basics. : : There are a lot of "stupid and uninformed" people around, there are at : least three people involved in this discussion that have expressed this : same belief as Eeyore, they are Multi-grid, Patrick Turner, and Eeyore. : : : Regards, : : John Byrns it seems the error made is in using the wrong mental model, that is seeing the transformer primary as a resistance. sure, PP operating on a pure R load would cease to cancel when one side would be cutoff, as it would no longer partake in the transfer curve for PP ! with a transformer, however, the transfer curve is a function of the tight coupling of the primary halves, so always PP ;-) (not the current, but the core flux is what matters) Rudy |
Output classes A and AB
Rudy wrote: "John Byrns" wrote in message ... : In article .com, : Andre Jute wrote: : : On Oct 25, 7:40 am, Eeyore : wrote: : : I'd love to know how that happens. There's no cancellation of ANYTHING once : one : side has ceased conducting ! : : Graham : : Holy ****! Did I say yet that Poopie is ignorant and incompetent? : : Nah, nobody can be that stupid and uninformed about tube basics. : : There are a lot of "stupid and uninformed" people around, there are at : least three people involved in this discussion that have expressed this : same belief as Eeyore, they are Multi-grid, Patrick Turner, and Eeyore. : : : Regards, : : John Byrns it seems the error made is in using the wrong mental model, that is seeing the transformer primary as a resistance. To the vacuum tubes, RLa-a IS a resistance. One R set up set up with a CT so any signal voltage applied between the end and the CT is applied in opposite phase to the other end, courtesy of the action of the transformer. Its RL a-a. This transformed single R appears to the tubes somewhat variably. In class B, while a tube is conducting, the load on the tube is 1/4 RLa-a, and when both conduct in class A the load on each tube is 1/2RLa-a. The secondary R load is always reflected to the two tubes as RLa-a, with a CT. The Ra of the tubes seen at the secondary varies. Say you have Ra = 1.2k for EL34 in triode, and 6k:6 ohms OPT. ZR = 1,000:1 In class A, Ra-a = 2.4k and is transformed to 2.4ohms by the OPT. In class B, only one tube is connected via 1/2 the primary, so only one Ra = 1.2k and OPT Z ratio has changed to 1.5k:6, because 1/2 the OPT primary isn't operating because one tube is cut off, so turn ratio is halved, Z ratio Ra in class B = 1.5k/6 = 250, so Ra at the output = 1.5k/250 = 6 ohms. The Rout of the amplifier at the sec is Ra transformed, and notice the change in Rout between A and AB. Its responsible for major 3H, 5H and other distortions. sure, PP operating on a pure R load would cease to cancel when one side would be cutoff, as it would no longer partake in the transfer curve for PP ! with a transformer, however, the transfer curve is a function of the tight coupling of the primary halves, so always PP ;-) (not the current, but the core flux is what matters) If one runs a class B PP amp with only ONE output tube, you will get a badly distorted signal like the one you'd get with a pure un-magnetically coupled R between B+ and the anode, only at each end of the OPT pri there would be two phases of the same distorted signal. The load with one tube and OPT is the same as with a single R taken to a suitably higher B+ than used with an OPT. Fourier could describe what harmonics flow all the time despite cut off. Furious arguemnts have followed about him. Patrick Turner. Rudy |
Output classes A and AB
"Patrick Turner" wrote in message ... : : : Rudy wrote: : : "John Byrns" wrote in message : ... : : In article .com, : : Andre Jute wrote: : : : : On Oct 25, 7:40 am, Eeyore : : wrote: : : : : I'd love to know how that happens. There's no cancellation of ANYTHING : once : : one : : side has ceased conducting ! : : : : Graham : : : : Holy ****! Did I say yet that Poopie is ignorant and incompetent? : : : : Nah, nobody can be that stupid and uninformed about tube basics. : : : : There are a lot of "stupid and uninformed" people around, there are at : : least three people involved in this discussion that have expressed this : : same belief as Eeyore, they are Multi-grid, Patrick Turner, and Eeyore. : : : : : : Regards, : : : : John Byrns : : it seems the error made is in using the wrong mental model, that is : seeing the transformer primary as a resistance. : : To the vacuum tubes, RLa-a IS a resistance. for the purpose of loadline analyses, that is a fair approximation. for analyzing the actual working mechanism, it lacks accounting for the element of time, as the transformer actually stores energy comparable to a fast rotating heavy object storing rotational impulse so in case a tube is cutoff, it's plate will happily follow to wherever that winding will take it. (well, not so happy when certain V's are exceeded, perhaps:) the output result will just be more or less symmetrically compressed, 3H, 5H, etc. this is distinctly different from a transistor PP, mostly apparent when you think of asymmetrical signals (music) driving it ;-) the load each PP tube sees is not suddenly changed at cutoff, it is in fact only balanced somewhat is a small region around the bias point, as with larger swing the dynamic plate resistance will vary more between the tubes, 3H will start to appear a measure of 'true class A' power of an AB pp, meaning the 3H would be low without nfb, would be the region where current is below twice the bias and cutoff AND dynamic plate R of the output tubes doesn't change more than "X" percent. .... to be discussed, heh, Rudy : : One R set up set up with a CT so any signal voltage applied between the : end and the CT : is applied in opposite phase to the other end, courtesy of the action of : the transformer. : : Its RL a-a. : : This transformed single R appears to the tubes somewhat variably. : In class B, while a tube is conducting, the load on the tube is 1/4 : RLa-a, : and when both conduct in class A the load on each tube is 1/2RLa-a. : : The secondary R load is always reflected to the two tubes as RLa-a, with : a CT. : The Ra of the tubes seen at the secondary varies. : : Say you have Ra = 1.2k for EL34 in triode, and 6k:6 ohms OPT. ZR = : 1,000:1 : In class A, Ra-a = 2.4k and is transformed to 2.4ohms by the OPT. : In class B, only one tube is connected via 1/2 the primary, so only one : Ra = 1.2k : and OPT Z ratio has changed to 1.5k:6, because 1/2 the OPT primary isn't : operating : because one tube is cut off, so turn ratio is halved, Z ratio Ra in : class B = 1.5k/6 = 250, : so Ra at the output = 1.5k/250 = 6 ohms. : : The Rout of the amplifier at the sec is Ra transformed, and notice the : change in Rout : between A and AB. Its responsible for major 3H, 5H and other : distortions. : : : : sure, PP operating on a pure R load would cease to : cancel when one side would be cutoff, as it would no longer : partake in the transfer curve for PP ! : with a transformer, however, the transfer curve is a function : of the tight coupling of the primary halves, so always PP ;-) : (not the current, but the core flux is what matters) : : If one runs a class B PP amp with only ONE output tube, : you will get a badly distorted signal like the one you'd get : with a pure un-magnetically coupled R between B+ and the anode, : only at each end of the OPT pri there would be two phases of the same : distorted signal. : The load with one tube and OPT is the same as with a single R taken to a : suitably higher B+ : than used with an OPT. : : Fourier could describe what harmonics flow all the time despite cut off. : : Furious arguemnts have followed about him. : : Patrick Turner. : : : Rudy |
Output classes A and AB
In article ,
"Rudy" wrote: for the purpose of loadline analyses, that is a fair approximation. for analyzing the actual working mechanism, it lacks accounting for the element of time, as the transformer actually stores energy comparable to a fast rotating heavy object storing rotational impulse so in case a tube is cutoff, it's plate will happily follow to wherever that winding will take it. (well, not so happy when certain V's are exceeded, perhaps:) Huh, wouldn't the transformer store energy more like a spring does, after all the schematic symbols are very similar? If "the transformer actually stores energy comparable to a fast rotating heavy object" wouldn't that cause a 6 dB/octave roll-off, resulting in a serious loss of high frequencies? Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
"John Byrns" wrote in message ... : In article , : "Rudy" wrote: : : for the purpose of loadline analyses, that is a fair approximation. : for analyzing the actual working mechanism, it lacks accounting for : the element of time, as the transformer actually stores energy : comparable to a fast rotating heavy object storing rotational impulse : so in case a tube is cutoff, it's plate will happily follow to wherever : that winding will take it. : (well, not so happy when certain V's are exceeded, perhaps:) : : Huh, wouldn't the transformer store energy more like a spring does, : after all the schematic symbols are very similar? : : If "the transformer actually stores energy comparable to a fast rotating : heavy object" wouldn't that cause a 6 dB/octave roll-off, resulting in a : serious loss of high frequencies? : : : Regards, : : John Byrns : err, yes, i just wanted to emphasize the 'flux doesn't stop on a dime' aspect. :-) Rudy |
Output classes A and AB
On Oct 31, 4:46 pm, John Byrns wrote:
In article , "Rudy" wrote: for the purpose of loadline analyses, that is a fair approximation. for analyzing the actual working mechanism, it lacks accounting for the element of time, as the transformer actually stores energy comparable to a fast rotating heavy object storing rotational impulse so in case a tube is cutoff, it's plate will happily follow to wherever that winding will take it. (well, not so happy when certain V's are exceeded, perhaps:) Huh, wouldn't the transformer store energy more like a spring does, after all the schematic symbols are very similar? If "the transformer actually stores energy comparable to a fast rotating heavy object" wouldn't that cause a 6 dB/octave roll-off, resulting in a serious loss of high frequencies? Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ Yes it does. This is from the leakage inductance. Things come to a complete stop when leakage L and capacitance resonate...:) cheers, Douglas |
Output classes A and AB
In article .com,
Multi-grid wrote: On Oct 31, 4:46 pm, John Byrns wrote: In article , "Rudy" wrote: for the purpose of loadline analyses, that is a fair approximation. for analyzing the actual working mechanism, it lacks accounting for the element of time, as the transformer actually stores energy comparable to a fast rotating heavy object storing rotational impulse so in case a tube is cutoff, it's plate will happily follow to wherever that winding will take it. (well, not so happy when certain V's are exceeded, perhaps:) Huh, wouldn't the transformer store energy more like a spring does, after all the schematic symbols are very similar? If "the transformer actually stores energy comparable to a fast rotating heavy object" wouldn't that cause a 6 dB/octave roll-off, resulting in a serious loss of high frequencies? Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ Yes it does. This is from the leakage inductance. Things come to a complete stop when leakage L and capacitance resonate...:) We specify a well designed and wound transformer with low leakage inductance so that it doesn't cause high frequency roll-off problems until well above the audio range. The impact of excess leakage inductance is also minimized by using a pentode output tube, which given your name I am sure you know all about. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
Rudy wrote: "Patrick Turner" wrote in message ... : : : Rudy wrote: : : "John Byrns" wrote in message : ... : : In article .com, : : Andre Jute wrote: : : : : On Oct 25, 7:40 am, Eeyore : : wrote: : : : : I'd love to know how that happens. There's no cancellation of ANYTHING : once : : one : : side has ceased conducting ! : : : : Graham : : : : Holy ****! Did I say yet that Poopie is ignorant and incompetent? : : : : Nah, nobody can be that stupid and uninformed about tube basics. : : : : There are a lot of "stupid and uninformed" people around, there are at : : least three people involved in this discussion that have expressed this : : same belief as Eeyore, they are Multi-grid, Patrick Turner, and Eeyore. : : : : : : Regards, : : : : John Byrns : : it seems the error made is in using the wrong mental model, that is : seeing the transformer primary as a resistance. : : To the vacuum tubes, RLa-a IS a resistance. for the purpose of loadline analyses, that is a fair approximation. for analyzing the actual working mechanism, it lacks accounting for the element of time, as the transformer actually stores energy comparable to a fast rotating heavy object storing rotational impulse so in case a tube is cutoff, it's plate will happily follow to wherever that winding will take it. (well, not so happy when certain V's are exceeded, perhaps:) The analogy is fine, as long as you remember there is no such thing as momentum in electronics. the output result will just be more or less symmetrically compressed, 3H, 5H, etc. this is distinctly different from a transistor PP, mostly apparent when you think of asymmetrical signals (music) driving it ;-) Asymetrical music signals will still have a dc content of zero, and there's very nearly always equal energy each side of zero. In a bjt output stage or a tube stage where the load is taken from the junction of the two output bjts or tubes, the Fourier waves are shunting one another, rather than being applied common mode as John B is so adament about. If one device is omitted, the class B waves are like you get with an SE tube biased on the brink of cut off. the load each PP tube sees is not suddenly changed at cutoff, it is in fact only balanced somewhat is a small region around the bias point, as with larger swing the dynamic plate resistance will vary more between the tubes, 3H will start to appear a measure of 'true class A' power of an AB pp, meaning the 3H would be low without nfb, would be the region where current is below twice the bias and cutoff AND dynamic plate R of the output tubes doesn't change more than "X" percent. Hmm, a little hard to follow you, but while a tube is cut off there is *no load* connected to it, unless in other words we say the Fourier currents all add to zero current for 1/2 a wave and are flowing in tube and 1/2 the OPT, but are conveniently summing to 0.0amps dc or ac for that 1/2 cycle. Usually 3H is HIGH witout NFB and unacceptably high when tubes are biased for class B at low levels, then as PO increases the % reduces, then inceases again near clipping. This was a problem with early SS amps; the early designers didn't know how to deal with the SS distortions at low levels where tube amps at the same level were much freer from distortions, and spectra was benign. The early designers could not afford to have hot idling class AB stages because germanium transistors went phut all too easily when they got hot. So they were forced int making all their crummy creations using class B working, and with token idle current. Bean counters made damn well sure OPTs were damn well eliminated from the SS amps during the transition from tubes to SS. For awhile, input transformers on SS amps were used because of the appallingly low base input resistance of the early power transistors, and besides, there were lots of out of work OPT winders available to wind simpler cheap IPT for SS to fill them in until permanent retrenchment, retirement or the sack. The appalling design ideas carried over into the days of early silicon bjts which turned out to be much more rugged than the germanium types. But SS heralded the use of class B, and the use of ever increasing amounts of NFB, so that we have amps with 40dB of local series voltage NFB in the form of emitter follower connection, and another 60dB of global NFB, to crossover distortion is difficult to identify as a spurious artifact amoungst the macro 3H and 2H which dominate the THD spectra just below clipping, but which rarely exceed 0.005% if the designer knows his stuff. In effct, many VAS stages in SS amps are merely a gain bjt with CC collector load to power a relaetively high impedance base input impedance of darligton output stage device arrays. The Single Ended VAS stage often has gain = 5,000x, and THD of this stage is say 5% at say 25Vrms, mainly 2H and 3H, very much like a pentode's spectra. The VAS stage THD well exceeds the output stage THD because of the emitter follower NFB. The 60dB of global NFB reduces the 5% of VAS thd to 0.005%. Because tubes have lower gain when used in audio circuits than available with bjts, and because of stability reasons, a large amount of NFB cannot be applied, with a total of 40 dB being about the limit. Patrick Turner. ... to be discussed, heh, Rudy : : One R set up set up with a CT so any signal voltage applied between the : end and the CT : is applied in opposite phase to the other end, courtesy of the action of : the transformer. : : Its RL a-a. : : This transformed single R appears to the tubes somewhat variably. : In class B, while a tube is conducting, the load on the tube is 1/4 : RLa-a, : and when both conduct in class A the load on each tube is 1/2RLa-a. : : The secondary R load is always reflected to the two tubes as RLa-a, with : a CT. : The Ra of the tubes seen at the secondary varies. : : Say you have Ra = 1.2k for EL34 in triode, and 6k:6 ohms OPT. ZR = : 1,000:1 : In class A, Ra-a = 2.4k and is transformed to 2.4ohms by the OPT. : In class B, only one tube is connected via 1/2 the primary, so only one : Ra = 1.2k : and OPT Z ratio has changed to 1.5k:6, because 1/2 the OPT primary isn't : operating : because one tube is cut off, so turn ratio is halved, Z ratio Ra in : class B = 1.5k/6 = 250, : so Ra at the output = 1.5k/250 = 6 ohms. : : The Rout of the amplifier at the sec is Ra transformed, and notice the : change in Rout : between A and AB. Its responsible for major 3H, 5H and other : distortions. : : : : sure, PP operating on a pure R load would cease to : cancel when one side would be cutoff, as it would no longer : partake in the transfer curve for PP ! : with a transformer, however, the transfer curve is a function : of the tight coupling of the primary halves, so always PP ;-) : (not the current, but the core flux is what matters) : : If one runs a class B PP amp with only ONE output tube, : you will get a badly distorted signal like the one you'd get : with a pure un-magnetically coupled R between B+ and the anode, : only at each end of the OPT pri there would be two phases of the same : distorted signal. : The load with one tube and OPT is the same as with a single R taken to a : suitably higher B+ : than used with an OPT. : : Fourier could describe what harmonics flow all the time despite cut off. : : Furious arguemnts have followed about him. : : Patrick Turner. : : : Rudy |
Output classes A and AB
John Byrns wrote:
Rudy wrote: for the purpose of loadline analyses, that is a fair approximation. for analyzing the actual working mechanism, it lacks accounting for the element of time, as the transformer actually stores energy comparable to a fast rotating heavy object storing rotational impulse so in case a tube is cutoff, it's plate will happily follow to wherever that winding will take it. (well, not so happy when certain V's are exceeded, perhaps:) : Huh, wouldn't the transformer store energy more like a spring does, : after all the schematic symbols are very similar? A capacitor is analogous to a spring. Don't be fooled by the symbols. A resistor is analogous to a damper, incidentally. : If "the transformer actually stores energy comparable to a fast rotating : heavy object" wouldn't that cause a 6 dB/octave roll-off, resulting in a : serious loss of high frequencies? Low frequencies, considering the mass is in shunt. Yes, it does. The concept of momentum in electronics is a bit interesting. Is LI concerved? If I have a perfect inductor passing current in a loop, and switch a second, equal, inductor into the circuit in series with the first, will the current be halved? If instead I double the inductance by inserting a core into the first inductor, will the current be also halved? In both cases large voltages would accompany the change, analogous to collision forces. Ian |
Output classes A and AB
In article ,
"Ian Iveson" wrote: John Byrns wrote: Rudy wrote: for the purpose of loadline analyses, that is a fair approximation. for analyzing the actual working mechanism, it lacks accounting for the element of time, as the transformer actually stores energy comparable to a fast rotating heavy object storing rotational impulse so in case a tube is cutoff, it's plate will happily follow to wherever that winding will take it. (well, not so happy when certain V's are exceeded, perhaps:) : Huh, wouldn't the transformer store energy more like a spring does, : after all the schematic symbols are very similar? A capacitor is analogous to a spring. Don't be fooled by the symbols. A resistor is analogous to a damper, incidentally. It depends on which dual you are using, an inductor can also be analogous to a spring and a capacitor to a mass. Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Output classes A and AB
On Thu, 01 Nov 2007 12:26:01 -0500, John Byrns
wrote: In article , "Ian Iveson" wrote: John Byrns wrote: Rudy wrote: for the purpose of loadline analyses, that is a fair approximation. for analyzing the actual working mechanism, it lacks accounting for the element of time, as the transformer actually stores energy comparable to a fast rotating heavy object storing rotational impulse so in case a tube is cutoff, it's plate will happily follow to wherever that winding will take it. (well, not so happy when certain V's are exceeded, perhaps:) : Huh, wouldn't the transformer store energy more like a spring does, : after all the schematic symbols are very similar? A capacitor is analogous to a spring. Don't be fooled by the symbols. A resistor is analogous to a damper, incidentally. It depends on which dual you are using, an inductor can also be analogous to a spring and a capacitor to a mass. Regards, John Byrns I haven't given this any thought, but my intuition says that when you give an inductor a kick with a voltage, it is reluctant to get moving (current flow), which makes it more like the mass. How would you intuit that the other way? d -- Pearce Consulting http://www.pearce.uk.com |
Output classes A and AB
On Thu, 01 Nov 2007 13:40:47 -0500, John Byrns
wrote: In article , (Don Pearce) wrote: On Thu, 01 Nov 2007 12:26:01 -0500, John Byrns wrote: In article , "Ian Iveson" wrote: John Byrns wrote: Rudy wrote: for the purpose of loadline analyses, that is a fair approximation. for analyzing the actual working mechanism, it lacks accounting for the element of time, as the transformer actually stores energy comparable to a fast rotating heavy object storing rotational impulse so in case a tube is cutoff, it's plate will happily follow to wherever that winding will take it. (well, not so happy when certain V's are exceeded, perhaps:) : Huh, wouldn't the transformer store energy more like a spring does, : after all the schematic symbols are very similar? A capacitor is analogous to a spring. Don't be fooled by the symbols. A resistor is analogous to a damper, incidentally. It depends on which dual you are using, an inductor can also be analogous to a spring and a capacitor to a mass. Regards, John Byrns I haven't given this any thought, but my intuition says that when you give an inductor a kick with a voltage, it is reluctant to get moving (current flow), which makes it more like the mass. How would you intuit that the other way? It all depends on whether you are using the Force-voltage analogy or the Force-current analogy. Similarly you can redraw a purely electrical circuit so that all the capacitors become inductors, the inductors become capacitors, the resistors become conductances, voltages become currents, and currents become voltages. Thought you might say that! No, I've never bought into the notion of current causing voltage, it is always the other way round in my head. You apply a voltage, and a current results. d -- Pearce Consulting http://www.pearce.uk.com |
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