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#1
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Class of Operation
Dear Rodents,
Just when you think you're getting a grasp on this tube business, something comes along to humble you. For example, years ago I owned a Threshold S-500e. I bought it because of Nelson Pass' unique and patented circuit called Stassis. It produced a significant amount of class A and during heavy demands, slid over into class AB. Now some rats like Patrick can actually design an amp, such as his 12 KT88s to produce 100W in class A and then move over to class AB or AB1. I know that the biasing has a lot to do with the class of operation, but how do you know when it's reached its class limit? Is that like Mr. Pass' Stassis design? If not an answer (I know this can't be answered with a few words) then perhaps a direction to a good book that will address this issue. Here's a good one that may be related to the aforementioned topic.... Is there a discernable difference between a 6550 or EL-34 in AB1, Triode mode and the same tubes in Class A, Pentode? Thanks in advance gents. Cordially, west |
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#2
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"west" ha scritto nel messaggio ... Dear Rodents, Just when you think you're getting a grasp on this tube business, something comes along to humble you. For example, years ago I owned a Threshold S-500e. I bought it because of Nelson Pass' unique and patented circuit called Stassis. It produced a significant amount of class A and during heavy demands, slid over into class AB. Now some rats like Patrick can actually design an amp, such as his 12 KT88s to produce 100W in class A and then move over to class AB or AB1. I know that the biasing has a lot to do with the class of operation, but how do you know when it's reached its class limit? Ideally, reducing the load line to a simple straight one, it can be seen from anode curves in the usual way. Locate idle point, pick a ruler, draw the loadline. Then the amount of power produced in Class A is - by definition - the one corresponding to the voltage swing from the intersection of the loadline with the 0 mA ("X") axis to the symmetrical grid voltage with respect to the idle value. Example: EL34 triode, V anode 450V, idle point @ 30 mA, grid bias -53V, output impedance 5k. At V grid=-95V the valve reaches the cut-off, ie, after a negative voltage swing of 95-53=42V. 53-42=11 = the amount of Class A power corresponds to the swing between the cutoff and the point at Vgrid= -11V (appr. 180V, 90mA). Voltage at cutoff is 600V, then delta V = 600, delta I = 90. Power is the product of the RMS I and V values, which in turn are given by the "peak to peak" values divided by 2*(sqrt2) = power = delta V * delta I / 8 = 6.75W per each tube. It is to be pointed out that this corresponds to 10% 2nd harm. distortion. Is that like Mr. Pass' Stassis design? If not an answer (I know this can't be answered with a few words) then perhaps a direction to a good book that will address this issue. Here's a good one that may be related to the aforementioned topic.... Is there a discernable difference between a 6550 or EL-34 in AB1, Triode mode Triodes distortion is mainly 2nd harmonic, and is cancelled by PP operation, therefore a PP output in triode or UL is intrinsically "correct" and can produce a pleasing sound with no or little feedback. Pentodes produce most odd-order distortion, which is much more disturbing (it is not harmonically related with the base tone, therefore giving a "fingernails on the blackboard" effect). Odd-order harmonics are not cancelled by PP operation, and the nulling of the 2nd h. makes them even more annoying. This is why pentode output stages are usually provided with LOTS of feedback. IMHO pentodes are a viable solution only if You want LOTS of watts and do not want to have a power station of Your own. A couple of LS50 (GU50 in CCCP form) in Class B can deliver some 100WPC with only 70W of total dissipation, but the circuit shall be very well engineered and in any case I'll suggest it only for "loud" music where tonal nuances are not strictly required. BTW, I don't see any point in Class A pentodes: the scope of a pentode is to be efficient, therefore it's made for Class B. Ciao Fabio and the same tubes in Class A, Pentode? Thanks in advance gents. Cordially, west |
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#3
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west wrote: Dear Rodents, Just when you think you're getting a grasp on this tube business, something comes along to humble you. For example, years ago I owned a Threshold S-500e. I bought it because of Nelson Pass' unique and patented circuit called Stassis. It produced a significant amount of class A and during heavy demands, slid over into class AB. Now some rats like Patrick can actually design an amp, such as his 12 KT88s to produce 100W in class A and then move over to class AB or AB1. I know that the biasing has a lot to do with the class of operation, but how do you know when it's reached its class limit? Is that like Mr. Pass' Stassis design? If not an answer (I know this can't be answered with a few words) then perhaps a direction to a good book that will address this issue. Here's a good one that may be related to the aforementioned topic.... Is there a discernable difference between a 6550 or EL-34 in AB1, Triode mode and the same tubes in Class A, Pentode? Thanks in advance gents. Cordially, west Say uou have a pentode PP amp with a pair of EL34 set up with Ea = 400v, Eg2 = 380v, fixed bias and Ia at idle = 40 mA per tube. Pd at idle is 16 watts each. You can work out what is called the anode to anode load but squaring the turn ratio of the OPT and multiplying by the secondary load, and adding 10% to cover the winding resistances of P and reflected S seen by the tubes. OK, so we have OPT ratio of 20:1, so ZR = 400:1, so if there is 8 ohms at the sec we get RL anode to anode = 3,200 ohms plus the Rw, so actually about 3,200 + 320 ohms = 3,520 ohms total. While in class A for the first few watts each tube sees 1/2 x RLa-a, or 1,760 ohms. But when one tube cuts off for class AB action, the load of the other tube turning on further is 1/4 x RLa-a = 880 ohms. At the transition point between class A and AB, its as if a cheeky dude pulled one tube out of it's socket; its cut off due to the -ve going Vg1. With only one tube connected for the "B" part of the cycle, the OPT ratio changes. While in class A the full primary is being used to deliver voltage change x current change to the load. But with only one tube driving, the OPT turn ratio is effectively halved from 20 to 10:1, so the ZR is reduced 4 times from 3520:8 to 880:8. Now for a pentode with a sharp cut off characteristic, the peak swing downwards of anode voltage at the point of class A cessation is = 2 x Ia at idle x 0.25 RLa-a. So since in this example 0.25 RLa-a = 880 ohms, the class A peak swing down is 880 x 2 x 0.04 = 70.4 peak volts. Each tube cuts off at the same voltage swing, thus the anode to anode voltage swing from cut off to cut off is 70.4 x 2 peak volts, = 140.8 pkv. All the anode to anode voltage swing within this 140.8 peak volts calculated is the class A swing, and the swing outside it is class B. So 140.8 peak v = 99.54 vrms and it works into RL a-a and the class A power produced by the tubes is ( 99.54 x 99.54 ) / 3,520 = 2.81 watts. Since there are winding losses, subtract about 10%, so at the sec we see a lousy 2.53 watts only of class A power, and a bit distorted compared to the situation if we had Ea lower, Ia higher for this value of load. We can draw a load line of 880 ohms ( 1/4 RLa-a ) on the anode curces for a single tube, but starting from Ea on the axis, or at 400v in this case, and up to where it inersects the vertical Ia axis at 400 / 880 = 0.454 amps. The plate resistance line where Eg1 = 0V will intersect the load line in the knee of the curve and at Ia = approx 0.36 amps and Ea = 90v. Thus the anode swing is from 400 down to 90, or 310 peak volts. The same thing occurs in both tubes, and regardless of class A content. So we have 620 pk volts anode to anode = 438 vrms so we get a total AB1 power output of ( 438 x 438 ) / 3,520 = 54 watts, less winding losses of say 10%, so about 49 watts. ( but power supply sag with an increasing DC flow into the CT for the total class AB power will reduce the Ea at maximum AB power by perhaps 10%, so the swing is actually reduced, and you'll be lucky to get 42 watts at clipping with the wind behind it. ) Its easiest to think about what goes on in one output tube at a time, rather than thinking about the composite load for a pair of tubes. So for the first 2.81 watts at the anodes, each tube sees a 1760 ohm class A load, and this load is just like a single ended tube loaded with 1,760 ohms. In practice, since the two tubes are never perfectly matched, the loads seen by each differ slightly according to the differences in gm. But beyound the 2.81 watts and after one tube cuts off, the load then changes to 880 ohms, so in afct the class A load line is a bent line for AB pentodes. Triodes have the most gradual cut off of all tubes so the bent load line is more a curve from RLa-a / 2 to RLa-a / 4, and since cut off is more gradual, the class AB triode transition does not produce as many of high numbered odd order switching harmonics that occurs in a pentode amp. With a pure class A pentode amp the doubling of idle current will nearly still occur in each tube when loaded by whatever class A load it has. where we can assume the peak downwards class A swing on eac anode = 0.95 x Ea at isle, the this voltage max swing is 380v. Since the idle current of 40 mA is almost doubled during this swing, there is 40 mA I change for a 380v voltage change and RL must be = E / I = 380 / 0.04 = 9,500 ohms. This is the class A load required for *pure* class A for 1 tube. Since there is the same swing on the other anode, but in the opposite direction, there is 760 pk v a-a = 537.3 vrms a-a, and RLa-a = 2 x 9,500 = 19,000 ohms, so pure class A power is approx = ( 537.3 x 537.3 ) / 19,000 = 15.2 watts. The power input does not change with class A PP. So the input power here is 2 x 400 x 0.04 = 32 watts, and plate efficiency is an expected % = 100 x 15.2 / 32 = 47.5%, although this ignores screen current wastage. With pure class A, the tubes may be operated with a higher Ia, lower Eg1 bias voltage, and cathode bias since the DC energy flow does not change much. So with B+ set up for 400v, and about 27 volts across the cathode RC circuits, Ea = 373v, Eg2 would be 350v, and if Pd was allowed to be 24 watts, then Ia would be 24 / 373 = 64 mA. Power input would be 48 watts, not including wastage on cathode circuits and screen currents. So if efficiency was 45%, a practical figure to achieve, then the class A power would be 21.6 watts. Now from all what I have said, and to test if you are all still awake or not, What RL a-a is required to achieve 21.6 watts of pure class A under the conditions in the preceding paragraph? Hint :- it will be less than 19,000 ohms. With the tube curves for pentodes, 43% UL and triodes in your minds, what RL a-a is used for the above conditions and to get pure class A for each mode of operation? Patrick Turner. |
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#4
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"Fabio Berutti" wrote in message ... [...] Triodes distortion is mainly 2nd harmonic, [...] Why? [...] Pentodes produce most odd-order distortion, [...] Why? [...] Odd-order harmonics are not cancelled by PP operation, [...] Why not? Buona sera Fabio, TIA for your answer, ciao, Jan. |
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#5
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On Tue, 26 Apr 2005 22:38:56 +0200, "J.Koning"
wrote: "Fabio Berutti" wrote in message ... [...] Triodes distortion is mainly 2nd harmonic, [...] Why? Because it is. The full answer is highly technical. Does it matter? [...] Pentodes produce most odd-order distortion, [...] Why? Because they do. The full answer is highly technical. Does it matter? [...] Odd-order harmonics are not cancelled by PP operation, [...] Why not? Because P-P operation can only cancel symmetrical artifacts. Odd-order distortion is by nature asymmetrical. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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#6
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"J.Koning" wrote: "Fabio Berutti" wrote in message ... [...] Triodes distortion is mainly 2nd harmonic, [...] Why? [...] Pentodes produce most odd-order distortion, [...] Why? [...] Odd-order harmonics are not cancelled by PP operation, [...] Why not? A good read of RDH4 should provide some understanding. Also read anything you can find written by Professor Child in the 1930s. Patrick Turner. Buona sera Fabio, TIA for your answer, ciao, Jan. |
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#7
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Stewart Pinkerton wrote: On Tue, 26 Apr 2005 22:38:56 +0200, "J.Koning" wrote: "Fabio Berutti" wrote in message ... [...] Triodes distortion is mainly 2nd harmonic, [...] Why? Because it is. The full answer is highly technical. Does it matter? [...] Pentodes produce most odd-order distortion, [...] Why? Because they do. The full answer is highly technical. Does it matter? [...] Odd-order harmonics are not cancelled by PP operation, [...] Why not? Because P-P operation can only cancel symmetrical artifacts. Odd-order distortion is by nature asymmetrical. You are wrong yet again. Even order harmonics are asymmetrical. There is a different shape of the wave either side of the zero crossing point. Looking at 2H distortion, this is obvious. Odd order harmonics produce the same wave shape each side of zero volts, and are symmetrical. The pointing or flattening of a sine wave's crests are the same for 3H and for +ve and -ve crests. Only the even order harmonics can be cancelled by PP action. Beware the Oinkerton, for he brings BS to our group. Patrick Turner. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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#8
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On Wed, 27 Apr 2005 00:25:51 GMT, Patrick Turner
wrote: [...] Odd-order harmonics are not cancelled by PP operation, [...] Why not? Because P-P operation can only cancel symmetrical artifacts. Odd-order distortion is by nature asymmetrical. You are wrong yet again. Yes, I know, got my symmetrical and asymmetrical the wrong way round. The 'DOH!' tends to come about a millisecond after you hit the 'send' key........ :-( Even order harmonics are asymmetrical. There is a different shape of the wave either side of the zero crossing point. Looking at 2H distortion, this is obvious. Quite so. Odd order harmonics produce the same wave shape each side of zero volts, and are symmetrical. Indeed they are. The pointing or flattening of a sine wave's crests are the same for 3H and for +ve and -ve crests. Yup. Only the even order harmonics can be cancelled by PP action. At least we agree on that one! :-) Beware the Oinkerton, for he brings BS to our group. Beware the Turner, for he is a bitter old ****. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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#9
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"Stewart Pinkerton" wrote in message ... Beware the Turner, for he is a bitter old ****. I think you may find that, like most of us, he is actually younger than yourself, SP. Pots and kettles? PT is one of the most informative posters on this group. Take a look at his detailed explanatory post elsewhere in this thread. Good, solid, useful stuff. That's the kind of information that makes RAT what it is. As a "precision analogue engineer" (your term) when can we expect posts of such quality from your goodself? Cordially, Iain |
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#10
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"Stewart Pinkerton" wrote in message ... [...] Triodes distortion is mainly 2nd harmonic, [...] Why? Because it is. The full answer is highly technical. Does it matter? I'm not afraid to do some reading, so, throw-up. [...] Pentodes produce most odd-order distortion, [...] Why? Because they do. The full answer is highly technical. Does it matter? I'm not afraid to do some reading, so, throw-up. Rgds, Jan. |
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#11
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"Patrick Turner" wrote in message ... [...] Triodes distortion is mainly 2nd harmonic, [...] Why? [...] Pentodes produce most odd-order distortion, [...] Why? [...] Odd-order harmonics are not cancelled by PP operation, [...] Why not? A good read of RDH4 should provide some understanding. Also read anything you can find written by Professor Child in the 1930s. Hi Patrick, I've been turning RDH4 pages but could'nt find anything relevant. Must be dissonant harmonic brainwaves. You got a page hint? TIA, Jan. |
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#12
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In article , "Iain M Churches"
wrote: "Stewart Pinkerton" wrote in message ... Beware the Turner, for he is a bitter old ****. I think you may find that, like most of us, he is actually younger than yourself, SP. Pots and kettles? PT is one of the most informative posters on this group. Take a look at his detailed explanatory post elsewhere in this thread. Good, solid, useful stuff. That's the kind of information that makes RAT what it is. As a "precision analogue engineer" (your term) when can we expect posts of such quality from your goodself? Don't hold your breath, SP tends to speak in generalities, without understanding what's behind them, or how they apply in a particular case. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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#13
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"J.Koning" wrote: "Patrick Turner" wrote in message ... [...] Triodes distortion is mainly 2nd harmonic, [...] Why? [...] Pentodes produce most odd-order distortion, [...] Why? [...] Odd-order harmonics are not cancelled by PP operation, [...] Why not? A good read of RDH4 should provide some understanding. Also read anything you can find written by Professor Child in the 1930s. Hi Patrick, I've been turning RDH4 pages but could'nt find anything relevant. Must be dissonant harmonic brainwaves. You got a page hint? RDH4 has an enormous amount on distortion in triodes and pentodes and beam tetrodes. Many pages cover the subject. There isn't one particular page which sums it all up. Professor Child wrote about triodes in Terman's 1937 book, Radio Engineering, which is a very good read for anyone wanting to get detail not so easily found on the web. He explains the basics fairly well, but the math is somewhat heavy. Try googling using 'history of triodes' as the search words. You might spend days reading. Better still, build a bread board amp and make a single stage preamp using a common 6AU6, and then a triode such as 12AX7. Then you get a pc based analyser program like SpecLab and analyse the thd to heart's content. Trying different tube topologies is a great way to enjoy wet sundays. You will need a sound card, and some way of keeping excessive plate DC and signal voltages out of the sound card input. One dude I know has alligator clips and wires all over the loungeroom floor, he just listens, and never uses pentodes, so he doesn't have to measure... He's trying to get a horn speaker system sounding OK. He has a basic understanding of the concepts, and its all he needs. Patrick Turner. TIA, Jan. |
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#14
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On Wed, 27 Apr 2005 11:50:57 +0300, "Iain M Churches"
wrote: "Stewart Pinkerton" wrote in message .. . Beware the Turner, for he is a bitter old ****. I think you may find that, like most of us, he is actually younger than yourself, SP. Pots and kettles? He's 57, and qualifies on all three counts for most of our gentle readers. Well, maybe not on RAT, which is a kind of eventide home for the terminally bewildered...... :-) PT is one of the most informative posters on this group. Take a look at his detailed explanatory post elsewhere in this thread. Good, solid, useful stuff. Indeed, he does know his technical stuff, which is why his weird opinions about tube amps are so much the stranger. That's the kind of information that makes RAT what it is. As a "precision analogue engineer" (your term) when can we expect posts of such quality from your goodself? I make them all the time, but rarely on RAT, and it's fair to say that the audio newgroups are pretty moribund at the moment. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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#16
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On Wed, 27 Apr 2005 13:16:02 +0200, "J.Koning"
wrote: "Stewart Pinkerton" wrote in message .. . [...] Triodes distortion is mainly 2nd harmonic, [...] Why? Because it is. The full answer is highly technical. Does it matter? I'm not afraid to do some reading, so, throw-up. [...] Pentodes produce most odd-order distortion, [...] Why? Because they do. The full answer is highly technical. Does it matter? I'm not afraid to do some reading, so, throw-up. Get hold of RDH4, it's available online to download from various sites. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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#17
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"Stewart Pinkerton" wrote in message ... On Wed, 27 Apr 2005 11:50:57 +0300, "Iain M Churches" wrote: "Stewart Pinkerton" wrote in message . .. Beware the Turner, for he is a bitter old ****. I think you may find that, like most of us, he is actually younger than yourself, SP. Pots and kettles? He's 57, and qualifies on all three counts for most of our gentle readers. Well, maybe not on RAT, which is a kind of eventide home for the terminally bewildered...... :-) So that means you won't be staying? :-)) PT is one of the most informative posters on this group. Take a look at his detailed explanatory post elsewhere in this thread. Good, solid, useful stuff. Indeed, he does know his technical stuff, which is why his weird opinions about tube amps are so much the stranger. Oddly enough, his opinions about tube amps mirror exactly those of several well informed authors on the subject, and others who actually have tube amp building experience. There is nothing like actually doing it. I am sure that you too would find it more satisfying than jiust criticising other people's efforts. OK. OK. I know what you are going to say. Your stock reply "Been there, done that!" That's the kind of information that makes RAT what it is. As a "precision analogue engineer" (your term) when can we expect posts of such quality from your goodself? I make them all the time, but rarely on RAT, ..... (snip) All the time? I must complain to my ISP he must be filtering 90% of your posts:-))) One thing I have noticed is that you seem to be at loggerheads with many members of every group to which you subscribe. (Thank God there are good groups to which you have no access:-)) It has also been pointed out to me that many have left UKRA due to your presence there. That's a loss to the group. I was hoping that the question of high voltage DC on psu and amplifier interconnects would have been a subject on which the "Man from Marconi" could have helped us out. A mil spec reference may be just what we are looking for. Cordially, Iain |
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#18
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On Wed, 27 Apr 2005 21:42:30 +0300, "Iain M Churches"
wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 27 Apr 2005 11:50:57 +0300, "Iain M Churches" wrote: "Stewart Pinkerton" wrote in message ... Beware the Turner, for he is a bitter old ****. I think you may find that, like most of us, he is actually younger than yourself, SP. Pots and kettles? He's 57, and qualifies on all three counts for most of our gentle readers. Well, maybe not on RAT, which is a kind of eventide home for the terminally bewildered...... :-) So that means you won't be staying? :-)) Longer than you, probably. You don't strike me as someone with any staying power, after your self-promotional bull**** is debunked. Not to worry though, I'm sure that there are many more dodgy Scandinavian videos awaiting your recording talents........ PT is one of the most informative posters on this group. Take a look at his detailed explanatory post elsewhere in this thread. Good, solid, useful stuff. Indeed, he does know his technical stuff, which is why his weird opinions about tube amps are so much the stranger. Oddly enough, his opinions about tube amps mirror exactly those of several well informed authors on the subject, and others who actually have tube amp building experience. It's always possible to find *someone* who will agree with you. This does not alter the wider reality. There is nothing like actually doing it. I am sure that you too would find it more satisfying than jiust criticising other people's efforts. OK. OK. I know what you are going to say. Your stock reply "Been there, done that!" Correct, I have indeed been there and done that, for even longer than you have. Interesting that you will dismiss *my* extensive experience of valve amps simply because I'm a professional engineer, and have moved on. Would you say the same about someone brought up with steam engines, who prefers internal combustion? That's the kind of information that makes RAT what it is. As a "precision analogue engineer" (your term) when can we expect posts of such quality from your goodself? I make them all the time, but rarely on RAT, ..... (snip) All the time? I must complain to my ISP he must be filtering 90% of your posts:-))) None so blind as those who will not read................ One thing I have noticed is that you seem to be at loggerheads with many members of every group to which you subscribe. (Thank God there are good groups to which you have no access:-)) OTOH, I am generally in agreement with the *rational* members of those groups. And of course, I have little interest in mutual admiration societies who repel opposing viewpoints by censorship. In my younger days, we called them Nazis..................................... Interesting that you do not promote your valvecentric bull**** on rec.audio.high-end, a *moderated* newsgroup where I am a regular poster. Do you not consider valve amplifiers to be 'high-end'? It has also been pointed out to me that many have left UKRA due to your presence there. That's a loss to the group. No, it hasn't, that's just another of your many pathetic lies. Have you considered returning to the UK and joining New Labour? I was hoping that the question of high voltage DC on psu and amplifier interconnects would have been a subject on which the "Man from Marconi" could have helped us out. A mil spec reference may be just what we are looking for. I already gave you one - MIL 5015, although that would be more 'the man from Hughes Aircraft' than 'the man from Marconi SDS'. However, your problem is not the availability of excellent HT connectors and cables, which has always been a given, but the willingness of your insurer to evaluate the risk. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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#19
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J.Koning" wrote
Triodes distortion is mainly 2nd harmonic, [...] Why? [...] Pentodes produce most odd-order distortion, [...] Why? [...] Odd-order harmonics are not cancelled by PP operation, [...] Why not? A good read of RDH4 should provide some understanding. Also read anything you can find written by Professor Child in the 1930s. Hi Patrick, I've been turning RDH4 pages but could'nt find anything relevant. Must be dissonant harmonic brainwaves. You got a page hint? Basically, a single triode amplifies the positive-going part of the input more than the negative: it compresses in one direction and expands in the other. A typically loaded pentode compresses in both directions, although more in one than the other. Consider a sine-wave. If you add even harmonics, then peaks are emphasised in one direction, and flattened in the other. This is similar to triode distortion. Odd harmonics tend to flatten in both directions, producing a symmetrical distortion. This is more like the typically-loaded pentode. Somewhere in RDH you will find a distortion ruler. No kidding. You place it on the anode characteristics of the valve, along the load line. You can estimate 2nd and 3rd harmonics with it AFAIR. Anyway, now consider a PP stage. Although the input signal is inverted, the harmonics produced by the stage itself are not. With even harmonics, one valve is expanding at the same rate as the other is compressing at any time, so the two effects cancel. With odd harmonics, they are both compressing at the same time, so the effects don't cancel. Incidentally, you should be able to see that odd harmonics tend to be associated with IMD. The maths of valves isn't simple in detail, but the general rules expressed in RDH are good enough for most purposes. The "why" is tricky, but the "what" is fairly straightforward. Is it possible to do a word-search in Acrobat reader? I could find references quick if someone can tell me how to search. cheers, Ian |
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#20
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Ian Iveson wrote:
Is it possible to do a word-search in Acrobat reader? I could find references quick if someone can tell me how to search. cheers, Ian Yes. In the oldest version of Acrobat Reader on my Mac, v5.x, there is a pair of binoculars in the picon bar across the top. You should also be able to find a "Find" command in a pull-down menu. That would be minimum functionality. Better versions of Reader, in something called Acrobat, offers greater opportunities to bugger around with PDFs, which is weird since one of the original attractions of PDFs, next in desirability to cross-platform WYSIWIG, was that nobody could bugger around with your text and presentation... If you still can't do it, let us know and I will install a PC copy of Adobe Reader into Windows XP which I have running under Virtual PC in partition, and find the commands for you. Andre Jute Ian Iveson wrote: J.Koning" wrote Triodes distortion is mainly 2nd harmonic, [...] Why? [...] Pentodes produce most odd-order distortion, [...] Why? [...] Odd-order harmonics are not cancelled by PP operation, [...] Why not? A good read of RDH4 should provide some understanding. Also read anything you can find written by Professor Child in the 1930s. Hi Patrick, I've been turning RDH4 pages but could'nt find anything relevant. Must be dissonant harmonic brainwaves. You got a page hint? Basically, a single triode amplifies the positive-going part of the input more than the negative: it compresses in one direction and expands in the other. A typically loaded pentode compresses in both directions, although more in one than the other. Consider a sine-wave. If you add even harmonics, then peaks are emphasised in one direction, and flattened in the other. This is similar to triode distortion. Odd harmonics tend to flatten in both directions, producing a symmetrical distortion. This is more like the typically-loaded pentode. Somewhere in RDH you will find a distortion ruler. No kidding. You place it on the anode characteristics of the valve, along the load line. You can estimate 2nd and 3rd harmonics with it AFAIR. Anyway, now consider a PP stage. Although the input signal is inverted, the harmonics produced by the stage itself are not. With even harmonics, one valve is expanding at the same rate as the other is compressing at any time, so the two effects cancel. With odd harmonics, they are both compressing at the same time, so the effects don't cancel. Incidentally, you should be able to see that odd harmonics tend to be associated with IMD. The maths of valves isn't simple in detail, but the general rules expressed in RDH are good enough for most purposes. The "why" is tricky, but the "what" is fairly straightforward. Is it possible to do a word-search in Acrobat reader? I could find references quick if someone can tell me how to search. cheers, Ian |
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#21
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In article , "Ian Iveson"
wrote: Basically, a single triode amplifies the positive-going part of the input more than the negative: it compresses in one direction and expands in the other. A typically loaded pentode compresses in both directions, although more in one than the other. With a pentode the loading is critical, with the loading just so there is no 2nd harmonic distortion, and by a stroke of luck, or unluck depending on how you look at it, this loading is close to the optimum loading for maximum power output. change the load one way and the pentode compresses like a triode, change it the other way and the opposite happens. Consider a sine-wave. If you add even harmonics, then peaks are emphasised in one direction, and flattened in the other. This is similar to triode distortion. Odd harmonics tend to flatten in both directions, producing a symmetrical distortion. This is more like the typically-loaded pentode. Somewhere in RDH you will find a distortion ruler. No kidding. You place it on the anode characteristics of the valve, along the load line. You can estimate 2nd and 3rd harmonics with it AFAIR. Anyway, now consider a PP stage. Although the input signal is inverted, the harmonics produced by the stage itself are not. With even harmonics, one valve is expanding at the same rate as the other is compressing at any time, so the two effects cancel. With odd harmonics, they are both compressing at the same time, so the effects don't cancel. Incidentally, you should be able to see that odd harmonics tend to be associated with IMD. Actually both even and odd harmonics are associated with IMD, the distortion products are just located differently with respect the original tones. If you have two input tones close together in frequency, even order distortion will produce difference tones far removed from the original tones, while odd order distortion produces difference tones close in frequency to the original input tones. Hope I got that the right way around. The maths of valves isn't simple in detail, but the general rules expressed in RDH are good enough for most purposes. The "why" is tricky, but the "what" is fairly straightforward. Actually the maths are relatively trivial. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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In article .com,
" wrote: Ian Iveson wrote: Is it possible to do a word-search in Acrobat reader? I could find references quick if someone can tell me how to search. cheers, Ian Yes. In the oldest version of Acrobat Reader on my Mac, v5.x, there is a pair of binoculars in the picon bar across the top. You should also be able to find a "Find" command in a pull-down menu. That would be minimum functionality. Andre, how come you aren't using "Adobe Reader" v7.0 on your Mac? Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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"John Byrns" wrote
Actually both even and odd harmonics are associated with IMD, the distortion products are just located differently with respect the original tones. If you have two input tones close together in frequency, even order distortion will produce difference tones far removed from the original tones, while odd order distortion produces difference tones close in frequency to the original input tones. Hope I got that the right way around. I didn't intend to give the impression that IMD is associated *only* with 3H, just that it is the most obvious illustration of how IMD works. IMD is associated with 3H in RDH I believe. It has been repeated so many times I have taken it as read. You must be right...*any* amplitude distortion will produce IMD of one sort or another. Presumably the modulation frequency is f for even H, and 2f for odd. It would be interesting to see some quantitative account. Also an accepted definition of IMD would be handy. Actually the maths are relatively trivial. "Relatively trivial" could mean anything, since you don't say relative to what. No, the maths is not trivial. No simple equation has been found to represent a pentode, and even the common triode function is a distant approximation at the extremes of operation. If you have a "trivial" function, perhaps you could share it? "not simple in detail" and "trivial" are not mutually exclusive, BTW. It is possible to be complex without being complicated. cheers, Ian |
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"Ian Iveson" wrote in message news:u3Wbe.2605 Hi Ian, Basically, a single triode amplifies the positive-going part of the input more than the negative: it compresses in one direction and expands in the other. I see. A typically loaded pentode compresses in both directions, although more in one than the other. I see that too. Sort of sliding along the loadline and measuring/comparing distance of -Vg intersections. For triodes the mechanism is clear. For penthodes I just did'nt see it. Your remarks and some fiddling with a ruler on an actual Va/Ia characteristic on paper did it. Consider a sine-wave. If you add even harmonics, then peaks are emphasised in one direction, and flattened in the other. This is similar to triode distortion. That's clear. Odd harmonics tend to flatten in both directions, producing a symmetrical distortion. This is more like the typically-loaded pentode. That's clear too. Somewhere in RDH you will find a distortion ruler. No kidding. You place it on the anode characteristics of the valve, along the load line. You can estimate 2nd and 3rd harmonics with it AFAIR. That's right, it's on page 550. Unfortunatly it works for 2H only, not for 3H. Accordingly the Espley and Farren method works. Anyway, now consider a PP stage. Although the input signal is inverted, the harmonics produced by the stage itself are not. With even harmonics, one valve is expanding at the same rate as the other is compressing at any time, so the two effects cancel. With odd harmonics, they are both compressing at the same time, so the effects don't cancel. Yeaahhhh. Incidentally, you should be able to see that odd harmonics tend to be associated with IMD. Not yet, but one day I'll get it. Another bunch of remarks might make that day tomorrow. The maths of valves isn't simple in detail, but the general rules expressed in RDH are good enough for most purposes. The "why" is tricky, but the "what" is fairly straightforward. For me, insight originates from basic understanding, not from math. Your remarks triggered basic understanding! Is it possible to do a word-search in Acrobat reader? I could find references quick if someone can tell me how to search. As others pointed out, use the binocular icon. I used v.5, at present v.6. Both support word-search. But watch out! It does'nt work with scanned documents AFAIK. Thanks, Jan. |
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"Stewart Pinkerton" wrote in message ... On Wed, 27 Apr 2005 21:42:30 +0300, "Iain M Churches" wrote: "Stewart Pinkerton" wrote in message . .. On Wed, 27 Apr 2005 11:50:57 +0300, "Iain M Churches" wrote: "Stewart Pinkerton" wrote in message m... Beware the Turner, for he is a bitter old ****. I think you may find that, like most of us, he is actually younger than yourself, SP. Pots and kettles? He's 57, and qualifies on all three counts for most of our gentle readers. Well, maybe not on RAT, which is a kind of eventide home for the terminally bewildered...... :-) So that means you won't be staying? :-)) Longer than you, probably. That's very bad news indeed for RAT:-(( Not to worry though, I'm sure that there are many more dodgy Scandinavian videos awaiting your recording talents........ I am off to Prague again in a few days. to continue the Dvorak project. Why not come and listen and learn? No, on the other hand don't. You would probably find it difficult to keep your mouth shut for the duration of a complete take. Your presence would be an embarrassment to us all. One thing I have noticed is that you seem to be at loggerheads with many members of every group to which you subscribe. (Thank God there are good groups to which you have no access:-)) OTOH, I am generally in agreement with the *rational* members of those groups. It certainly doesn't look like that. Your unpleasant profile has earned you a reputation across three continents:-)) A local group refer to you affectionately as "Apinanperse" Google anyone? Interesting that you do not promote your valvecentric bull**** on rec.audio.high-end, a *moderated* newsgroup where I am a regular poster. Do you not consider valve amplifiers to be 'high-end'? I don't have time for any more NGs. I make time for RAT and the closed groups (to which I am pleased to note you are denied access) It has also been pointed out to me that many have left UKRA due to your presence there. That's a loss to the group. No, it hasn't, that's just another of your many pathetic lies. A former member of UKRA is compiling a list. I was hoping that the question of high voltage DC on psu and amplifier interconnects would have been a subject on which the "Man from Marconi" could have helped us out. A mil spec reference may be just what we are looking for. I already gave you one - MIL 5015, although that would be more 'the man from Hughes Aircraft' than 'the man from Marconi SDS'. But, surprise, surprise, the man from Marconi and Hughes Aircraft was talking through his hat:-(( I telephoned to Farnell technical support this morning to order some connectors. MIL 5015 is approved to 50V AC/DC only. Even your information on a straightforward matter like HV DC connectors is not to be relied upon. You had better keep to your stamp licking in the mail room. Safer for all of use:-) As mentioned on UKRA, I shall not be reading or responding to your posts in the future. Cordially, Iain |
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wrote
Yes. In the oldest version of Acrobat Reader on my Mac, v5.x, there is a pair of binoculars in the picon bar across the top. You should also be able to find a "Find" command in a pull-down menu. That would be minimum functionality. Better versions of Reader, in something called Acrobat, offers greater opportunities to bugger around with PDFs, which is weird since one of the original attractions of PDFs, next in desirability to cross-platform WYSIWIG, was that nobody could bugger around with your text and presentation... If you still can't do it, let us know and I will install a PC copy of Adobe Reader into Windows XP which I have running under Virtual PC in partition, and find the commands for you. Edit/Find. Thanks, and yes, what's a reader got an edit menu for. I have the binoculars too. Searching for ruler got nowhere coz they call it a rule. 13.2.(i) fig 13.4 shows a "power output rule" or "5% distortion rule", and fig 13.5 shows a rule for graphical determination of other percentages of 2H. Can't find one for 3H but I'm sure I've seen it somewhere. I wonder if anyone has ever actually used one of these? cheers, Ian |
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"J.Koning" wrote
[below] Cool. The relationship between intermodulation and harmonic distortion is discussed in RDH 14.3. Basically, if the amplification is varied by one frequency component of the signal, then a second component will be amplitude-modulated by the first. Easiest to imagine with a large low-frequency sine wave combined with one of much higher frequency and much lower amplitude. The combination will appear on a scope as the second riding on the first. If the lower frequency is subject to 3H, for example, then amplification will reduce at its peaks. The reduction will equally apply to the higher frequency component, which will thus be amplitude-modulated, in this case at twice the frequency of the lower. The relationship between this amplitude modulation and additional frequency components is not so easy to see. Arguing backwards, if you pluck two guitar strings of close but not identical pitch, then you will hear a low, "beat" frequency: what sounds like a single note varying in amplitude. This is because the two frequencies slip in and out of phase, alternately reinforcing and cancelling each other. That demonstrates a relationship between amplitude modulation and the summing of frequencies. In reverse, if you have an amplitude-modulated frequency, then you should be able to find a sum of constant-amplitude frequencies that would be equivalent. Thinking one stage further, the sum of frequencies would need to have the same frequency as the original, and this can only be done by adding one frequency above, and one below. Hence "side bands", which are symmetrical. This is about as far as my mental picture goes...this paragraph might be rubbish. Soon someone will tell me. cheers, Ian Basically, a single triode amplifies the positive-going part of the input more than the negative: it compresses in one direction and expands in the other. I see. A typically loaded pentode compresses in both directions, although more in one than the other. I see that too. Sort of sliding along the loadline and measuring/comparing distance of -Vg intersections. For triodes the mechanism is clear. For penthodes I just did'nt see it. Your remarks and some fiddling with a ruler on an actual Va/Ia characteristic on paper did it. Consider a sine-wave. If you add even harmonics, then peaks are emphasised in one direction, and flattened in the other. This is similar to triode distortion. That's clear. Odd harmonics tend to flatten in both directions, producing a symmetrical distortion. This is more like the typically-loaded pentode. That's clear too. Somewhere in RDH you will find a distortion ruler. No kidding. You place it on the anode characteristics of the valve, along the load line. You can estimate 2nd and 3rd harmonics with it AFAIR. That's right, it's on page 550. Unfortunatly it works for 2H only, not for 3H. Accordingly the Espley and Farren method works. Anyway, now consider a PP stage. Although the input signal is inverted, the harmonics produced by the stage itself are not. With even harmonics, one valve is expanding at the same rate as the other is compressing at any time, so the two effects cancel. With odd harmonics, they are both compressing at the same time, so the effects don't cancel. Yeaahhhh. Incidentally, you should be able to see that odd harmonics tend to be associated with IMD. Not yet, but one day I'll get it. Another bunch of remarks might make that day tomorrow. The maths of valves isn't simple in detail, but the general rules expressed in RDH are good enough for most purposes. The "why" is tricky, but the "what" is fairly straightforward. For me, insight originates from basic understanding, not from math. Your remarks triggered basic understanding! Is it possible to do a word-search in Acrobat reader? I could find references quick if someone can tell me how to search. As others pointed out, use the binocular icon. I used v.5, at present v.6. Both support word-search. But watch out! It does'nt work with scanned documents AFAIK. Thanks, Jan. |
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Ian Iveson wrote: "J.Koning" wrote [below] Cool. The relationship between intermodulation and harmonic distortion is discussed in RDH 14.3. Basically, if the amplification is varied by one frequency component of the signal, then a second component will be amplitude-modulated by the first. Easiest to imagine with a large low-frequency sine wave combined with one of much higher frequency and much lower amplitude. The combination will appear on a scope as the second riding on the first. If the lower frequency is subject to 3H, for example, then amplification will reduce at its peaks. The reduction will equally apply to the higher frequency component, which will thus be amplitude-modulated, in this case at twice the frequency of the lower. The relationship between this amplitude modulation and additional frequency components is not so easy to see. Arguing backwards, if you pluck two guitar strings of close but not identical pitch, then you will hear a low, "beat" frequency: what sounds like a single note varying in amplitude. This is because the two frequencies slip in and out of phase, alternately reinforcing and cancelling each other. That demonstrates a relationship between amplitude modulation and the summing of frequencies. In reverse, if you have an amplitude-modulated frequency, then you should be able to find a sum of constant-amplitude frequencies that would be equivalent. Thinking one stage further, the sum of frequencies would need to have the same frequency as the original, and this can only be done by adding one frequency above, and one below. Hence "side bands", which are symmetrical. This is about as far as my mental picture goes...this paragraph might be rubbish. Soon someone will tell me. cheers, Ian Basically, imho, you have a fair grasp of what is happening. The traditional way to measure IMD of an amp is to apply two frequencies not harmonically related, so a 4:1 voltage ratio exists where there is a higher voltage of say 100 Hz and lower voltage of say 5 kHz. The level is adjusted so that the combined output signal is 1 dB short of clipping anywhere. This simulates an amp with the usual high bass content compared to a usually lower amount of treble content, but content to which our ears are very sensitive. At the test equipment we can filter out all the 100 Hz signal quite easily. This will leave just the 5 kHz. Its amplitude will appear to vary acording to the transfer function of the amp. Usually the HF signal will appear on an CRO like an amplitude modulated signal. The HF signal can be passed through a diode + RC detector, and the shape of the variations in amplitude easily seen as the modulation signal. Often it looks like the shape of the LF signal. The amplitude of the modulating signal can be easily measured, and this amplitude expressed as a fraction of the original HF signal, and the fraction is multiplied by 100 to bring it to a percentage. So it may be that we measure 2 vrms of 5 kHz, and the amplitude modulation signal is say 0.05 vrms, so the % IMD = ( 0.05 / 2 ) x 100 = 2.5 %, and that's on the high side for a hi-fi amp but is low for a guitar amp, or anything with no NFB at all. If we had sufficiently high Q bandpass filters, we may find that if we filtered the 5 kHz amplitude modulated signal, we'd find at least 3 different frequencies present, 4,900, 5,000, and 5,100 Hz. If the transfer nature of the amp was more uneven or if there were a complex set of harmonic spectra present, then you'd be able to filter out a lot more artifacts each side of the 5 kHz centre frequency. Note that 4,900, 5,000, and 5,100 Hz are not musically related, so they tend to sound not good. But music has hundreds of fundemental tones all intermodulating each other in any amplifier. If one could xtract all the products of intermodulation and play it through the speakers without the original music it sounds like a grungy noise rising and falling with the level of the music. RDH4 gives the amount of IMD one could expect with a given amount of THD, and IMD using the 4:1 LF:HF voltages is usually around 3 to 5 times the THD measurement where the % is the IMD of the HF signal. PP amps usually have predominantly 3H in the THD. Its phase in relation to the fundemental F is either subtractive or additive to the peaks of the fundemental. Often a peaked wave has 5H present as well, and anyone who has tried to use 6DQ6 line output tetrodes for a beam tetrode PP amp will understand what I am saying. The tendency for an amp to at least innitially produce a triangular wave form sounds the worst imho. Most class AB pentode or beam tubes suffer a reduction of tube gain as the plate current increases despite the increase in gm with the higher plate current. Its because the load changes fron 1/2 RLa-a to 1/4 RLa-a during each large signal swing beyound the cut off points for each output tube. Most guitar amps are set up to have a small amount of class A, lots of class B, so the thd is somewhat higher than a class AB amp made for hi-fi. With higher thd there is higher imd. The reasons we see ripples on the amplitude of a HF signal when a large LF signal is also present is due to the change of output tube gain during the LF cycle. With PP, the change of gain occurs at each +ve and -ve peak. With SE tubes, there is more gain on one side of zero than the other. SE Triodes usually have more gain with increasing plate current with all load values, and the thd is predominatly 2H at first but at high levels there are many other harmonics produced. SE pentode or beam tubes have both odd and even numbered harmonics in the THD at all levels, and the phase of the harmonics can vary with load, as well as the mix. Pentodes and beam tubes are regarded as poor performers for hi-fi, but wonderful for guitar amps, since the harmonic richness adds much life to music from a guitar, or an organ. Anyone who has heard a Hammond organ with a full tube amp and an accompanying Leslie speaker with tube amp knows that transistor amps just are quite inadequate. Usually if you have a hi-fi amp tube amp which measures 0.2% thd at 50 watts and with the first 15 watts in class A and you only use an average of a watt like most ppl then you have an amp which will sound fairly blameless for hi-fi, but maybe a bit dead as an instrument amp. There is a school of thought who say that the best way to avoid the pitfalls of too much thd and imd is to use a single SE triode for the output stage and use only a small % of the power available. SET thd/imd is about the least blameable of all types of thd and imd. Hence the interest in 5 watt SET amps which are best utilised when the speaker sensitivity is over 96 dB/W/M, which includes most horn loaded speakers. In this case the use of NFB loops is forbidden because the amount applied needed to reduce Ro and improove Ro is usually less than 10 dB, and this much NFB causes the thd of mainly 2H to be reduced, but it also creates some 3H because the 2H fed back into the amp from the output is intermodulated and a 3H product is formed. In practice I don't myself think it matters, since the extra 3H will be at a very low level if the 2H was less than 1% to start with. Distortion products sound worse as their F rises where they are the products of signal below about 5 kHz. Subjective horribleness = N squared / 4, where N is the harmonic number, and the reference is an amount of 2H. So if you had 0.1% of 5H, its as bad as ( 5 x 5 / 4 ) % of 2H, ie 0.1% of 5H = 0.625% of 2H. Opinions vary on the subjective effects of small quantities of thd and imd. But most class A triode amps, either PP or SE types used well away from clipping will produce music second to none if you do it right. I know one "Dark Metal" guitarist who makes music which sounds like a continual series of jumbo jets crashing into the house. He also prefers PP triodes. So each unto their own. Basically, a single triode amplifies the positive-going part of the input more than the negative: it compresses in one direction and expands in the other. I see. A typically loaded pentode compresses in both directions, although more in one than the other. I see that too. Sort of sliding along the loadline and measuring/comparing distance of -Vg intersections. For triodes the mechanism is clear. For penthodes I just did'nt see it. Your remarks and some fiddling with a ruler on an actual Va/Ia characteristic on paper did it. Consider a sine-wave. If you add even harmonics, then peaks are emphasised in one direction, and flattened in the other. This is similar to triode distortion. That's clear. Odd harmonics tend to flatten in both directions, producing a symmetrical distortion. This is more like the typically-loaded pentode. That's clear too. Somewhere in RDH you will find a distortion ruler. No kidding. You place it on the anode characteristics of the valve, along the load line. You can estimate 2nd and 3rd harmonics with it AFAIR. That's right, it's on page 550. Unfortunatly it works for 2H only, not for 3H. Accordingly the Espley and Farren method works. Anyway, now consider a PP stage. Although the input signal is inverted, the harmonics produced by the stage itself are not. With even harmonics, one valve is expanding at the same rate as the other is compressing at any time, so the two effects cancel. With odd harmonics, they are both compressing at the same time, so the effects don't cancel. Yeaahhhh. Incidentally, you should be able to see that odd harmonics tend to be associated with IMD. Not yet, but one day I'll get it. Another bunch of remarks might make that day tomorrow. The maths of valves isn't simple in detail, but the general rules expressed in RDH are good enough for most purposes. The "why" is tricky, but the "what" is fairly straightforward. For me, insight originates from basic understanding, not from math. Your remarks triggered basic understanding! Is it possible to do a word-search in Acrobat reader? I could find references quick if someone can tell me how to search. As others pointed out, use the binocular icon. I used v.5, at present v.6. Both support word-search. But watch out! It does'nt work with scanned documents AFAIK. Thanks, Jan. |
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In article , "Ian
Iveson" wrote: "John Byrns" wrote Actually both even and odd harmonics are associated with IMD, the distortion products are just located differently with respect the original tones. If you have two input tones close together in frequency, even order distortion will produce difference tones far removed from the original tones, while odd order distortion produces difference tones close in frequency to the original input tones. Hope I got that the right way around. I didn't intend to give the impression that IMD is associated *only* with 3H, just that it is the most obvious illustration of how IMD works. IMD is associated with 3H in RDH I believe. It has been repeated so many times I have taken it as read. You must be right...*any* amplitude distortion will produce IMD of one sort or another. Presumably the modulation frequency is f for even H, and 2f for odd. It would be interesting to see some quantitative account. Also an accepted definition of IMD would be handy. Actually the maths are relatively trivial. "Relatively trivial" could mean anything, since you don't say relative to what. No, the maths is not trivial. No simple equation has been found to represent a pentode, and even the common triode function is a distant approximation at the extremes of operation. If you have a "trivial" function, perhaps you could share it? Sorry, I misunderstood your statement about the "maths". I thought you were speaking of the "maths" relating harmonic and IM distortion given a transfer function, I see now you were actually speaking of the "maths" for the transfer function of a triode, which unfortunately I don't have to share. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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In article , Patrick Turner
wrote: Ian Iveson wrote: "J.Koning" wrote [below] Cool. The relationship between intermodulation and harmonic distortion is discussed in RDH 14.3. Basically, if the amplification is varied by one frequency component of the signal, then a second component will be amplitude-modulated by the first. Easiest to imagine with a large low-frequency sine wave combined with one of much higher frequency and much lower amplitude. The combination will appear on a scope as the second riding on the first. If the lower frequency is subject to 3H, for example, then amplification will reduce at its peaks. The reduction will equally apply to the higher frequency component, which will thus be amplitude-modulated, in this case at twice the frequency of the lower. The relationship between this amplitude modulation and additional frequency components is not so easy to see. Arguing backwards, if you pluck two guitar strings of close but not identical pitch, then you will hear a low, "beat" frequency: what sounds like a single note varying in amplitude. This is because the two frequencies slip in and out of phase, alternately reinforcing and cancelling each other. That demonstrates a relationship between amplitude modulation and the summing of frequencies. In reverse, if you have an amplitude-modulated frequency, then you should be able to find a sum of constant-amplitude frequencies that would be equivalent. Thinking one stage further, the sum of frequencies would need to have the same frequency as the original, and this can only be done by adding one frequency above, and one below. Hence "side bands", which are symmetrical. This is about as far as my mental picture goes...this paragraph might be rubbish. Soon someone will tell me. cheers, Ian Basically, imho, you have a fair grasp of what is happening. The traditional way to measure IMD of an amp is to apply two frequencies not harmonically related, so a 4:1 voltage ratio exists where there is a higher voltage of say 100 Hz and lower voltage of say 5 kHz. I don't have the exact details close to hand, but I thought the traditional way of measuring IMD was to apply two signals close in frequency, say 12 kHz and 13 kHz, in a ratio of 1:1, and measure the amplitude of the difference frequency produced at 1 kHz as a measure of the amount of IM distortion? Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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On Thu, 28 Apr 2005 13:51:43 +0300, "Iain M Churches"
wrote: "Stewart Pinkerton" wrote in message .. . On Wed, 27 Apr 2005 21:42:30 +0300, "Iain M Churches" wrote: "Stewart Pinkerton" wrote in message ... On Wed, 27 Apr 2005 11:50:57 +0300, "Iain M Churches" wrote: "Stewart Pinkerton" wrote in message om... Beware the Turner, for he is a bitter old ****. I think you may find that, like most of us, he is actually younger than yourself, SP. Pots and kettles? He's 57, and qualifies on all three counts for most of our gentle readers. Well, maybe not on RAT, which is a kind of eventide home for the terminally bewildered...... :-) So that means you won't be staying? :-)) Longer than you, probably. That's very bad news indeed for RAT:-(( Only in your opinion, and we all know what that's worth..... Not to worry though, I'm sure that there are many more dodgy Scandinavian videos awaiting your recording talents........ I am off to Prague again in a few days. to continue the Dvorak project. Why not come and listen and learn? No, on the other hand don't. You would probably find it difficult to keep your mouth shut for the duration of a complete take. Your presence would be an embarrassment to us all. Oh, I'm sure you're quite capable of embarrassing yourself without my help, you've done it often enough in these newsgroups! One thing I have noticed is that you seem to be at loggerheads with many members of every group to which you subscribe. (Thank God there are good groups to which you have no access:-)) OTOH, I am generally in agreement with the *rational* members of those groups. It certainly doesn't look like that. Your unpleasant profile has earned you a reputation across three continents:-)) A local group refer to you affectionately as "Apinanperse" Google anyone? Your constant lies have earned you global contempt. Interesting that you do not promote your valvecentric bull**** on rec.audio.high-end, a *moderated* newsgroup where I am a regular poster. Do you not consider valve amplifiers to be 'high-end'? I don't have time for any more NGs. I make time for RAT and the closed groups (to which I am pleased to note you are denied access) I am happy to avoid your corruption wherever possible. It has also been pointed out to me that many have left UKRA due to your presence there. That's a loss to the group. No, it hasn't, that's just another of your many pathetic lies. A former member of UKRA is compiling a list. Yeah, right, of course he is. I was hoping that the question of high voltage DC on psu and amplifier interconnects would have been a subject on which the "Man from Marconi" could have helped us out. A mil spec reference may be just what we are looking for. I already gave you one - MIL 5015, although that would be more 'the man from Hughes Aircraft' than 'the man from Marconi SDS'. But, surprise, surprise, the man from Marconi and Hughes Aircraft was talking through his hat:-(( I telephoned to Farnell technical support this morning to order some connectors. MIL 5015 is approved to 50V AC/DC only. You really are an ass. Depending on the style of inserts used, MIL-C-5015 connectors can be rated up to 4200 V DC. Check out http://www.in2connect.co.uk/site/Array%205015.pdf for some useful information. However, if you want to go really bananas, try: http://www.reynoldsindustries.com/ for the real deal in serious HT connectors! Even your information on a straightforward matter like HV DC connectors is not to be relied upon. You had better keep to your stamp licking in the mail room. Safer for all of use:-) You had better keep to recording soundtracks for dodgy Finnish videos. Much more pleasant for all of us than your pathetic pontificating and self-promotion on RAT. As mentioned on UKRA, I shall not be reading or responding to your posts in the future. Good. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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John Byrns wrote:
In article , Patrick Turner wrote: Ian Iveson wrote: "J.Koning" wrote [below] Cool. The relationship between intermodulation and harmonic distortion is discussed in RDH 14.3. Basically, if the amplification is varied by one frequency component of the signal, then a second component will be amplitude-modulated by the first. Easiest to imagine with a large low-frequency sine wave combined with one of much higher frequency and much lower amplitude. The combination will appear on a scope as the second riding on the first. If the lower frequency is subject to 3H, for example, then amplification will reduce at its peaks. The reduction will equally apply to the higher frequency component, which will thus be amplitude-modulated, in this case at twice the frequency of the lower. The relationship between this amplitude modulation and additional frequency components is not so easy to see. Arguing backwards, if you pluck two guitar strings of close but not identical pitch, then you will hear a low, "beat" frequency: what sounds like a single note varying in amplitude. This is because the two frequencies slip in and out of phase, alternately reinforcing and cancelling each other. That demonstrates a relationship between amplitude modulation and the summing of frequencies. In reverse, if you have an amplitude-modulated frequency, then you should be able to find a sum of constant-amplitude frequencies that would be equivalent. Thinking one stage further, the sum of frequencies would need to have the same frequency as the original, and this can only be done by adding one frequency above, and one below. Hence "side bands", which are symmetrical. This is about as far as my mental picture goes...this paragraph might be rubbish. Soon someone will tell me. cheers, Ian Basically, imho, you have a fair grasp of what is happening. The traditional way to measure IMD of an amp is to apply two frequencies not harmonically related, so a 4:1 voltage ratio exists where there is a higher voltage of say 100 Hz and lower voltage of say 5 kHz. I don't have the exact details close to hand, but I thought the traditional way of measuring IMD was to apply two signals close in frequency, say 12 kHz and 13 kHz, in a ratio of 1:1, and measure the amplitude of the difference frequency produced at 1 kHz as a measure of the amount of IM distortion? Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ John- Go to http://www.tvhandbook.com/support/pd...hapter13_3.pdf where you will find an excellant paper certified free of Bull****. About 19 pages in pdf. Enjoy, John Stewart |
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John Byrns wrote:
In article .com, " wrote: In the oldest version of Acrobat Reader on my Mac, v5.x, t...... ....That would be minimum functionality. Andre, how come you aren't using "Adobe Reader" v7.0 on your Mac? Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ I have Adobe Reader v6 and v7 as well as v5. But I don't see the point of using the latest software just because it is the latest; I upgrade only when I consistently find an advantage in requiring some facility the earlier version doesn't have. Version 5 loads faster than the newer versions. My requirement for reading PDFs is intermittent and unpredictable and usually arises as a side issue to whatever I am doing so starting the programme causes an interruption, therefore it matters to me how long it takes to start up. Contrast QuarkXPress, which takes a relatively long time to launch (say 20-30s, relatively long to the others I use), but which I use consistently for hours for the main task when I do use it, so I don't mind the launch time. I have a gigabyte of fast RAM allocated for the programmes, and more to operate the video, and a very smooth OS to allocate RAM between multiple programmes, all operating on a 1.2GHz bus, so I can start up every programme I use regularly (MSWord, QuarkXPress, Illustrator, Photoshop) at startup without a problem*, but none of these programmes cause such a delay that it is worth the effort to set it up. I have my comms and some utilities, about a dozen in all that I use all the time (a multiple virtual screen switcher, for instance, which saves me having several large screens on my desk), set to launch at startup. Generally, I am not a fashion victim. I don't care what is new, I ask first, 'What will it do for me?' I always operate several OS updates and at least one hardware generation** behind Apple; I am nobody's unpaid beta-tester! Andre Jute *At the end of the day I often have twenty to thirty programmes, half of them major memory hogs handling graphics, running side by side without a problem. I recently looked into buying a faster machine and calculated I would save about two minutes in a sixteen hour day. It just isn't worth a 5-6K expenditure for another maxed out machine. I am at the very margin of speed for my sort of work, though if I ever again do much video (as in designing film or music videos) the position may change. As an example, I often simultaneously do 3D CAD with my keyboard and mouse, dictate text for instant transcription into via ViaVoice into Word, write with my other hand on a digital pad more text for instant handwriting recognition and insertion into another Word document I am editing, lay out pages as I edit them in another window, watch a movie in a small window, handle several strands of comms at once (internet, newsgroup, mail, video conference), perform batch operations on sets of graphics for the layout in another window, and several other semi-deliberate functions not requiring my immediate attention, plus of course all the background operations. In all this the machinery keeps up; at last computers have bumped up against human capabilities in manual dexterity and mental multitasking. It is in this context that waiting for a newer version of Acrobat Reader to open becomes an intolerable interruption of the workflow. It is also in this context that a Mac scores so heavily over the PC and justifies its price: it does all this seamlessly, whereas on an Microsoft operating system (and programmes too, because they take their cue from MS) the interruptions dictated by its crudeness and lack of ergonomics would at the end of the day mount up to two hours lost and frustration wrecking at least some of the done work. That's what makes a Mac worth the money, that at the end of the day you survey a big chunk of work done right yet feel pretty relaxed; but you want deliberately to protect that advantage by not becoming a fashion victim/beta tester for Apple, Adobe or Microsoft. (Nobody can help becoming a beta-tester for the wretched makers of QuarkXPress.) **Though my present Mac (a G4 iMac which when maxed out at the time I bought it had a better spec than the then-current tower such as I always used before) is a most pleasing machine, probably the relatively best (to its period, what else was available, capabilities, longevity, eventual value of money) Mac I ever owned was an 840AV Quadra which I kept for a decade and whose breadth of video capabilities has still not been exceeded, though it has long been overtaken in speed. I bought that 840AV on the day Apple declared it obsolete and a senior VP said "good riddance; we lost money on every single one we sold". I paid a premium because suddenly every top graphic designer in the world, who before thought it outrageously overpriced, decided there was nothing else that would the job... |
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Patrick Turner wrote:
I know one "Dark Metal" guitarist who makes music which sounds like a continual series of jumbo jets crashing into the house. He also prefers PP triodes. Wouldn't that be Black Metal then? I can't explain why I like it, but I do  Except for the fact that most Black Metal bands actually suck, but thefew ones that have the magic are great... but I guess I digress So each unto their own. Indeed. Regards, Kimjand |
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#35
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John Byrns wrote: In article , Patrick Turner wrote: Ian Iveson wrote: "J.Koning" wrote [below] Cool. The relationship between intermodulation and harmonic distortion is discussed in RDH 14.3. Basically, if the amplification is varied by one frequency component of the signal, then a second component will be amplitude-modulated by the first. Easiest to imagine with a large low-frequency sine wave combined with one of much higher frequency and much lower amplitude. The combination will appear on a scope as the second riding on the first. If the lower frequency is subject to 3H, for example, then amplification will reduce at its peaks. The reduction will equally apply to the higher frequency component, which will thus be amplitude-modulated, in this case at twice the frequency of the lower. The relationship between this amplitude modulation and additional frequency components is not so easy to see. Arguing backwards, if you pluck two guitar strings of close but not identical pitch, then you will hear a low, "beat" frequency: what sounds like a single note varying in amplitude. This is because the two frequencies slip in and out of phase, alternately reinforcing and cancelling each other. That demonstrates a relationship between amplitude modulation and the summing of frequencies. In reverse, if you have an amplitude-modulated frequency, then you should be able to find a sum of constant-amplitude frequencies that would be equivalent. Thinking one stage further, the sum of frequencies would need to have the same frequency as the original, and this can only be done by adding one frequency above, and one below. Hence "side bands", which are symmetrical. This is about as far as my mental picture goes...this paragraph might be rubbish. Soon someone will tell me. cheers, Ian Basically, imho, you have a fair grasp of what is happening. The traditional way to measure IMD of an amp is to apply two frequencies not harmonically related, so a 4:1 voltage ratio exists where there is a higher voltage of say 100 Hz and lower voltage of say 5 kHz. I don't have the exact details close to hand, but I thought the traditional way of measuring IMD was to apply two signals close in frequency, say 12 kHz and 13 kHz, in a ratio of 1:1, and measure the amplitude of the difference frequency produced at 1 kHz as a measure of the amount of IM distortion? Afaik, and I don't know everything, RDH4 has little on testing IMD with a pair of same amplititude HF tones. It was always thought that the worst and most likely imd could most easily be provoked by a large amplitude LF tone's effect on a smaller amplitude HF tone. One reason is that if one uses a frequency near the cut off of the OPT, the saturation effects would also add to a degradation of the transfer make the imd worse. So if an amp uses 25 Hz at LF then this usually affects the outcome more than using 100 Hz. If not, then you have a very good tube amp indeed. Some say that if you have equal tones of 5 kHz and 12 kHz, then its easy to filter out the 7 kHz and 17 kHz IMD products. This takes two accurate stable oscillators, and two accurate bandpass filters. The use of any pair of F which are 1 kHz apart would be ok. Patrick Turner. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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Patrick Turner wrote:
John Byrns wrote: In article , Patrick Turner wrote: Ian Iveson wrote: "J.Koning" wrote [below] Cool. The relationship between intermodulation and harmonic distortion is discussed in RDH 14.3. Basically, if the amplification is varied by one frequency component of the signal, then a second component will be amplitude-modulated by the first. Easiest to imagine with a large low-frequency sine wave combined with one of much higher frequency and much lower amplitude. The combination will appear on a scope as the second riding on the first. If the lower frequency is subject to 3H, for example, then amplification will reduce at its peaks. The reduction will equally apply to the higher frequency component, which will thus be amplitude-modulated, in this case at twice the frequency of the lower. The relationship between this amplitude modulation and additional frequency components is not so easy to see. Arguing backwards, if you pluck two guitar strings of close but not identical pitch, then you will hear a low, "beat" frequency: what sounds like a single note varying in amplitude. This is because the two frequencies slip in and out of phase, alternately reinforcing and cancelling each other. That demonstrates a relationship between amplitude modulation and the summing of frequencies. In reverse, if you have an amplitude-modulated frequency, then you should be able to find a sum of constant-amplitude frequencies that would be equivalent. Thinking one stage further, the sum of frequencies would need to have the same frequency as the original, and this can only be done by adding one frequency above, and one below. Hence "side bands", which are symmetrical. This is about as far as my mental picture goes...this paragraph might be rubbish. Soon someone will tell me. cheers, Ian Basically, imho, you have a fair grasp of what is happening. The traditional way to measure IMD of an amp is to apply two frequencies not harmonically related, so a 4:1 voltage ratio exists where there is a higher voltage of say 100 Hz and lower voltage of say 5 kHz. I don't have the exact details close to hand, but I thought the traditional way of measuring IMD was to apply two signals close in frequency, say 12 kHz and 13 kHz, in a ratio of 1:1, and measure the amplitude of the difference frequency produced at 1 kHz as a measure of the amount of IM distortion? Afaik, and I don't know everything, RDH4 has little on testing IMD with a pair of same amplititude HF tones. It was always thought that the worst and most likely imd could most easily be provoked by a large amplitude LF tone's effect on a smaller amplitude HF tone. One reason is that if one uses a frequency near the cut off of the OPT, the saturation effects would also add to a degradation of the transfer make the imd worse. So if an amp uses 25 Hz at LF then this usually affects the outcome more than using 100 Hz. If not, then you have a very good tube amp indeed. Some say that if you have equal tones of 5 kHz and 12 kHz, then its easy to filter out the 7 kHz and 17 kHz IMD products. This takes two accurate stable oscillators, and two accurate bandpass filters. The use of any pair of F which are 1 kHz apart would be ok. Not really. For example, if you were to take one & two KHZ as the test tones many of the IMD's would be buried under the resulting THD's. Need to pick a pair of frequencies having no simple harmonic relationship, within the band of interest. I frequently use equal amplitude tones of one & 1.6 KHZ in a five KHZ window for a quick look. It is very revealing. I do that with a Pico Tech ADC-216, so can look down 96 db. JLS Patrick Turner. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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John Stewart wrote: (snip), John- Go to http://www.tvhandbook.com/support/pd...hapter13_3.pdf where you will find an excellant paper certified free of Bull****. About 19 pages in pdf. Thanks JH. Its a pity one cannot save the document. The first part concerning thd and imd is straightforward, but described far more longwindedly than I have in a previous post. Patrick Turner. Enjoy, John Stewart |
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Kim Johan Andersson wrote: Patrick Turner wrote: I know one "Dark Metal" guitarist who makes music which sounds like a continual series of jumbo jets crashing into the house. He also prefers PP triodes. Wouldn't that be Black Metal then? Nah, when planes crash, lotsa silver coloured puddles of aluminium laying around, a few charred bones and fried plastics. "Dark" is a favourite trendy word used to describe anything mysterious and new and unsettling. So there is dark music, dark movie plots, dark art. I can't explain why I like it, but I do Except for the fact that most Black Metal bands actually suck, but thefew ones that have the magic are great... but I guess I digress Upon what to they suck prey tell? I'd rather read the news paper than listen to any modern idea of dark music, unless its a top notch band. I like Louis Armstrong, and Sammy Davis, and many dark artists..... Patrick Turner. So each unto their own. Indeed. Regards, Kimjand |
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John Stewart wrote: Patrick Turner wrote: The use of any pair of F which are 1 kHz apart would be ok. Not really. For example, if you were to take one & two KHZ as the test tones many of the IMD's would be buried under the resulting THD's. Need to pick a pair of frequencies having no simple harmonic relationship, within the band of interest. I frequently use equal amplitude tones of one & 1.6 KHZ in a five KHZ window for a quick look. It is very revealing. I do that with a Pico Tech ADC-216, so can look down 96 db. I would remove the word "any" in my sentence is misleading. Obviously one wouldn't choose an F where the thd would coincide with the expected imd product. 800 Hz and 1.8 kHz would be ok. Patrick Turner. JLS Patrick Turner. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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In article , Patrick Turner
wrote: John Stewart wrote: (snip), John- Go to http://www.tvhandbook.com/support/pd...hapter13_3.pdf where you will find an excellant paper certified free of Bull****. About 19 pages in pdf. Thanks JH. Its a pity one cannot save the document. What do you mean "Its a pity one cannot save the document"? I had no trouble saving the document, although I am not sure why one might want to save it for any length of time, it really isn't a very inspiring document. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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