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#81
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
On Wed, 20 Aug 2008 22:09:04 -0500, flipper wrote:
On Wed, 20 Aug 2008 17:33:34 -0700, The Phantom wrote: On Tue, 05 Aug 2008 13:04:52 -0500, John Byrns wrote: SNIP Could you elaborate on what the goof was? The goof was not a typo or an error in a formula, his idea of placing a build out resistor in the cathode of a concertina phase inverter to equalize the source impedances of the plate and cathode circuits was simply a goofy idea. It was a "bright" idea intended to fix an imagined problem that didn't actually exist, that instead created a real problem. I saw the magazine article I'm posting the page from the magazine article where he shows a schematic "...featuring--possibly for the first time--cathode build-out resistor in the driver stage.", and several letters that followed. It's over on ABSE. Interesting. And interesting to compare McFadden's letter to my original post because I said almost exactly the same thing He wrote "The circuit behaves as if the output resistance at both ports is the same {as the source resistance of a cathode follower...]" And I wrote "...the thing acts as if the source impedances are identical. I.E. there is no difference in HF roll off (with equal value grid stoppers)." I'm want to put forth the notion that it's not necessary to use phrases like "...behaves as if..." and "...acts as if...". The phrases suggest that the circuit "behaves as if" some condition were true, but which isn't really true. What is in fact true, is that as used in audio, the Concertina has a differential load (or call it a balanced load; it's the same thing). We need not say that it "...behaves as if it had a differential load."; it *does* have a differential (balanced) load. The outputs, P and K, of the Concertina are exactly equal in amplitude and 180 out of phase; v(P) = -v(K) as long as the loads are identical. This means that if a load of Z is applied from each of P and K to ground, we may lift the grounded ends of each Z and connect them together. The junction of the two Z's will have no voltage present because when -v(t) is present at P and v(t) is present at K, they cancel at the junction of the two Z's. That means that the circuit behavior is the same, whether that junction is grounded or not, and if the junction is *not* grounded, then we have a purely differential load of 2*Z. This means that connecting identical loads of Z (one end grounded, of course) to each of P and K is *exactly* the same as connecting a load of 2*Z *between* P and K; differentially in other words. Z may have a capacitive part, and this doesn't alter the previous argument. If a purely differential load 2*Z is applied, it may be treated as separate loads of Z at each of P and K, and by the well-known explanation (which I mention below can be found in RDH4), the current through whatever load is applied to P will be the same as the current through that same load applied to K. And therefore the high frequency rolloffs at P and K will be identical. Because the Concertina has a vacuum tube as one of its components, its admittance matrix is not bilateral; the transfer impedance from node i to node j is not the same as in the reverse direction. This causes the driving point impedance at P and at K to be different than the half differential output impedance. Since the intended load on the Concertina is a differential (balanced) load, if we want to calculate things like decrease in output voltage due to loading, or high frequency rolloff due to loading by capacitance, we must use the differential output impedance, because the load is differential. However, we may sometimes wish to calculate the effect of unequal capacitive loads, perhaps due to strays. In that case, the use of the driving point impedances at P and K is called for. The differential output impedance could be measured in the real world by connecting a small LCR meter between P and K: http://www.mcmconnect.com/tenma/prod...tance%20Meters This meter should be able to measure resistance with an AC stimulus (1 kHz typically) instead of a DC stimulus. The meter will inject a test current from one of its leads while simultaneously withdrawing the same current from the other lead; in other words, it injects a current of +i(t) from one lead and a current of -i(t) from the other lead. This allows a measurement *between* nodes P and K; a *differential* measurement. This same method can be used to find the differential impedance between a pair of nodes in a circuit when doing a mathematical analysis. We simply inject a current of +1 at one node while simultaneously injecting a current of -1 at the other node. Using a current of magnitude 1 results in a voltage appearing at a node, due to the injection of such a current, that is numerically equal to the impedance at that node. If we (mathematically) inject a current of +1 into P and simultaneously inject -1 into K and calculate the voltage appearing at each node due to those currents, those voltages will be (numerically) the impedances at the nodes. The differential impedance will be the sum of the individual impedances appearing at P and K while injecting the +1/-1 differential currents. The impedance at the node where a current of -1 is injected is the negative of the voltage appearing there. That differential impedance can treated as one quantity, or it can be considered split into a P component and a K component. When the individual components are used, then capacitive loading can also be split into separate components. In other words, a 100 pF differential (balanced) capacitive load can be considered to be separate 50 pF capacitors applied simultaneously to P and K for the purpose of determining high frequency behavior. But the effect of the two 50 pF capacitive loads is identical to the effect of a single 100 pF differential capacitive load. As it happens, when we inject +1/-1 currents at P and K we find that the individual impedances at P and K are identical and their sum is the differential impedance. So, we might say that the impedance at each node, P and K, which we should use to compute the effect of further (differential) loading, is the "half differential output impedance". And, when we are careful to apply any additional loads to P and K simultaneously, and in identical value, we should realize that we are really applying a *differential* load. I recommend using the phrases "differential output impedance" and "driving point (output) impedance". If we only use "output impedance", we have one person meaning one thing and another person meaning another thing and acrimony results. There's no need for this. The circuit can be analyzed exactly and without resort to waffle-words like "...behaves as though...". We need not show that the circuit "...behaves as though..."; we can show how the circuit behaves in fact. We can do fairly simple mathematical analyses which will show how the circuit behaves for loads applied to one output, P or K, at a time or for loads applied to both outputs simultaneously (a differential, or balanced, load). I'm going to post over on ABSE, analysis of the circuit using the admittance technique of Jacob Shekel. His paper has been posted over there already. I also made the same 'intuitive' argument with "If, however, you accept that the concertina is balanced as long as the two loads are equal then it's an 'of course' the roll off is the same because the loads are equal, they just vary (equally) with frequency." (Balanced as long as the loads are equal because anode and cathode current through the one tube must be one and the same.) I wonder when this explanation first began to be used. I looked in my RDH3 (printed in 1940) and on page 10 there is a schematic and short description of the circuit we now call the Concertina. There is no discussion of impedances at all. By the time of RDH4 (1952), there is discussion of impedances. RDH4 says, on page 330: "The effective output resistance is different for the two output channels, since P operates with current feedback and K with voltage feedback." They then give expressions for the "effective output resistance" (what they are referring to is what nowadays would be called the "driving point impedance") in formulas 30 and 31, but I'm sure that formula 30 is incorrect. They continue: "...but this does not affect the balance at either low or high frequencies when the total effective impedance of channel P is equal to that of channel K. The same signal plate current which flows through one impedance Zp also flows through the other impedance Zk, and if Zp=Zk then the two output voltages are equal. As I've shown, when you apply identical load impedances Z to each of P and K, you are in effect applying a differential load of 2*Z, and the behavior of the circuit can be analyzed that way. The expression given in formula 34a in RDH4 is what I am now calling the "half differential output impedance". This is the impedance that can be attributed to each of P and K for the purpose of determining circuit behavior when identical loads are simultaneously applied to P and K. About the only difference is I went on to explain that the 'classic unequal output impedance model works when you consider different generator voltages with "If you keep the 'different impedance' analysis the thing is as cathode impedance drops anode gain increases in exact proportion to the drop across the 'larger' anode impedance." And I have been attacked and insulted by Byrns for it. Btw, I agree with your implied criticism (at least that's how I interpreted it) that Jones tries to 'hedge' his 'build out' resistor with the 'problem' of grid drive. He continues to analyze this as an 'output impedance' issue but, IMO, that only serves to muddy one's understanding because the actual behavior is different than a simple 'output impedance' model suggests. I.E., it isn't that the signal on the 'high impedance' side is (uniformly) 'reduced', as one might think with a simple impedance model, it has a large negative spike when the cathode tries to drive +ve, in addition, of course, to the positive clipping. Due, of course, to the impedances but not in the manner one would expect for 'Ro into a load'. I *have* seen the 'problem' he refers to but I tend to view the issue as an 'upset' condition with the huge anode (and some on the cathode in reverse) swing introducing spurious HF components that cause feedback instability issues, depending on the OPT characteristics and amount of GNFB. I mentioned that in an earlier post about grid stoppers on the output tubes noting that it only takes 'one', on the cathode side, but that I always put two, equal, grid stoppers. and the first edition of his book where he presented this goofy idea, I have never seen the second edition of his book to see how he extricated himself from the predicament he created for himself, although I have been told by people that have seen the second edition that he did somehow extricate himself. |
#82
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
In article ,
flipper wrote: On Tue, 19 Aug 2008 18:37:40 -0500, John Byrns wrote: Can you give a clear definition of what makes something a "--Thevenin--- 'output impedance'"? Yes, as I have repeated so many times I've lost count, a derived impedance you can plug into RC, or any other simple impedance calculation, and get a meaningful result with. As in, if I slap a capacitor on the output of an arbitrary circuit how will the output of that arbitrary circuit behave? Simple, take the --Thevenin--- output impedance and use it for the R in the RC calculation. That's what a --Thevenin--- output impedance *is* and why Thevenin bothered to develop the model The values you calculate in the 'unequal output impedance' model don't work that way. The value derived with the equal output impedance model does. Hi Flipper, Thanks for making a start at defining what a Thevenin model is. Please pardon my trimming this discussion back to a single item, I would like to try and keep things more understandable, and less confusing, by discussing only one issue at a time, starting with the Thevenin model, before moving on to the next issue. I am clearly a little foggy as to what the restrictions and constraints on a proper Thevenin model are? IIRC you said that my model, as defined by the equations I posted, with the typos corrected, is not a Thevenin model, is that correct? It is my understanding that the Thevenin model consists of a voltage generator and an impedance, clearly there are some restrictions and constraints on these that my model fails to meet? Can you more completely explain the restrictions and constraints on the Thevenin voltage generator and impedance, and how my model fails to meet the requirements? You speak to "the --Thevenin--- output impedance" above, saying that it is a "simple impedance". I believe the Thevenin model also includes a " --Thevenin--- voltage generator", can you explain the requirements for the Thevenin voltage generator? Finally, I have snipped off much of your post in my attempt to keep things clear and simple, in the part that I snipped off you mentioned your model for the "CPI" at one point. Could you post the equations representing "the --Thevenin--- output impedance" and "the --Thevenin--- voltage generator" used in your model, as an example of a proper Thevenin model? -- Regards, John Byrns Surf my web pages at, http://fmamradios.com/ |
#83
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
"The Phantom" wrote in message
... I'm want to put forth the notion that it's not necessary to use phrases like "...behaves as if..." and "...acts as if...". The phrases suggest that the circuit "behaves as if" some condition were true, but which isn't really true. I disagree. Everything you say about the circuit is really a "...behaves as if..." statement if you back off a level or two and consider all of the abstractions that go into traditional circuit analysis. So, it it really a question of choosing the appropriate level of abstraction for the situation. You, like Byrns, seem to be a pretty pedantic fellow. And that's fine, but I have a different focus. And it's not for lack of the wherewithal to be pedantic if I wanted. There is a whole other aspect to the so-called "controversy" on this board that has nothing at all to do with circuits and everything to do with abusive personalities. Whatever. -Henry |
#84
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
In article ,
The Phantom wrote: On Wed, 20 Aug 2008 22:09:04 -0500, flipper wrote: On Wed, 20 Aug 2008 17:33:34 -0700, The Phantom wrote: On Tue, 05 Aug 2008 13:04:52 -0500, John Byrns wrote: SNIP Could you elaborate on what the goof was? The goof was not a typo or an error in a formula, his idea of placing a build out resistor in the cathode of a concertina phase inverter to equalize the source impedances of the plate and cathode circuits was simply a goofy idea. It was a "bright" idea intended to fix an imagined problem that didn't actually exist, that instead created a real problem. I saw the magazine article I'm posting the page from the magazine article where he shows a schematic "...featuring--possibly for the first time--cathode build-out resistor in the driver stage.", and several letters that followed. It's over on ABSE. Interesting. And interesting to compare McFadden's letter to my original post because I said almost exactly the same thing He wrote "The circuit behaves as if the output resistance at both ports is the same {as the source resistance of a cathode follower...]" And I wrote "...the thing acts as if the source impedances are identical. I.E. there is no difference in HF roll off (with equal value grid stoppers)." I'm want to put forth the notion that it's not necessary to use phrases like "...behaves as if..." and "...acts as if...". The phrases suggest that the circuit "behaves as if" some condition were true, but which isn't really true. What is in fact true, is that as used in audio, the Concertina has a differential load (or call it a balanced load; it's the same thing). We need not say that it "...behaves as if it had a differential load."; it *does* have a differential (balanced) load. The outputs, P and K, of the Concertina are exactly equal in amplitude and 180 out of phase; v(P) = -v(K) as long as the loads are identical. This means that if a load of Z is applied from each of P and K to ground, we may lift the grounded ends of each Z and connect them together. The junction of the two Z's will have no voltage present because when -v(t) is present at P and v(t) is present at K, they cancel at the junction of the two Z's. That means that the circuit behavior is the same, whether that junction is grounded or not, and if the junction is *not* grounded, then we have a purely differential load of 2*Z. This means that connecting identical loads of Z (one end grounded, of course) to each of P and K is *exactly* the same as connecting a load of 2*Z *between* P and K; differentially in other words. Z may have a capacitive part, and this doesn't alter the previous argument. If a purely differential load 2*Z is applied, it may be treated as separate loads of Z at each of P and K, and by the well-known explanation (which I mention below can be found in RDH4), the current through whatever load is applied to P will be the same as the current through that same load applied to K. And therefore the high frequency rolloffs at P and K will be identical. Because the Concertina has a vacuum tube as one of its components, its admittance matrix is not bilateral; the transfer impedance from node i to node j is not the same as in the reverse direction. This causes the driving point impedance at P and at K to be different than the half differential output impedance. Since the intended load on the Concertina is a differential (balanced) load, if we want to calculate things like decrease in output voltage due to loading, or high frequency rolloff due to loading by capacitance, we must use the differential output impedance, because the load is differential. However, we may sometimes wish to calculate the effect of unequal capacitive loads, perhaps due to strays. In that case, the use of the driving point impedances at P and K is called for. The differential output impedance could be measured in the real world by connecting a small LCR meter between P and K: http://www.mcmconnect.com/tenma/prod...tance%20Meters This meter should be able to measure resistance with an AC stimulus (1 kHz typically) instead of a DC stimulus. The meter will inject a test current from one of its leads while simultaneously withdrawing the same current from the other lead; in other words, it injects a current of +i(t) from one lead and a current of -i(t) from the other lead. This allows a measurement *between* nodes P and K; a *differential* measurement. This same method can be used to find the differential impedance between a pair of nodes in a circuit when doing a mathematical analysis. We simply inject a current of +1 at one node while simultaneously injecting a current of -1 at the other node. Using a current of magnitude 1 results in a voltage appearing at a node, due to the injection of such a current, that is numerically equal to the impedance at that node. If we (mathematically) inject a current of +1 into P and simultaneously inject -1 into K and calculate the voltage appearing at each node due to those currents, those voltages will be (numerically) the impedances at the nodes. The differential impedance will be the sum of the individual impedances appearing at P and K while injecting the +1/-1 differential currents. The impedance at the node where a current of -1 is injected is the negative of the voltage appearing there. That differential impedance can treated as one quantity, or it can be considered split into a P component and a K component. When the individual components are used, then capacitive loading can also be split into separate components. In other words, a 100 pF differential (balanced) capacitive load can be considered to be separate 50 pF capacitors applied simultaneously to P and K for the purpose of determining high frequency behavior. But the effect of the two 50 pF capacitive loads is identical to the effect of a single 100 pF differential capacitive load. Hi Rodger, Wouldn't a 100 pF differential capacitive load be the equivalent of separate 200 pF capacitive loads applied simultaneously to P and K, not 50 pF loads? As it happens, when we inject +1/-1 currents at P and K we find that the individual impedances at P and K are identical and their sum is the differential impedance. So, we might say that the impedance at each node, P and K, which we should use to compute the effect of further (differential) loading, is the "half differential output impedance". And, when we are careful to apply any additional loads to P and K simultaneously, and in identical value, we should realize that we are really applying a *differential* load. I recommend using the phrases "differential output impedance" and "driving point (output) impedance". If we only use "output impedance", we have one person meaning one thing and another person meaning another thing and acrimony results. There's no need for this. The circuit can be analyzed exactly and without resort to waffle-words like "...behaves as though...". We need not show that the circuit "...behaves as though..."; we can show how the circuit behaves in fact. We can do fairly simple mathematical analyses which will show how the circuit behaves for loads applied to one output, P or K, at a time or for loads applied to both outputs simultaneously (a differential, or balanced, load). I'm going to post over on ABSE, analysis of the circuit using the admittance technique of Jacob Shekel. His paper has been posted over there already. I also made the same 'intuitive' argument with "If, however, you accept that the concertina is balanced as long as the two loads are equal then it's an 'of course' the roll off is the same because the loads are equal, they just vary (equally) with frequency." (Balanced as long as the loads are equal because anode and cathode current through the one tube must be one and the same.) I wonder when this explanation first began to be used. I looked in my RDH3 (printed in 1940) and on page 10 there is a schematic and short description of the circuit we now call the Concertina. There is no discussion of impedances at all. By the time of RDH4 (1952), there is discussion of impedances. RDH4 says, on page 330: "The effective output resistance is different for the two output channels, since P operates with current feedback and K with voltage feedback." They then give expressions for the "effective output resistance" (what they are referring to is what nowadays would be called the "driving point impedance") in formulas 30 and 31, but I'm sure that formula 30 is incorrect. They continue: "...but this does not affect the balance at either low or high frequencies when the total effective impedance of channel P is equal to that of channel K. The same signal plate current which flows through one impedance Zp also flows through the other impedance Zk, and if Zp=Zk then the two output voltages are equal. As I've shown, when you apply identical load impedances Z to each of P and K, you are in effect applying a differential load of 2*Z, and the behavior of the circuit can be analyzed that way. The expression given in formula 34a in RDH4 is what I am now calling the "half differential output impedance". This is the impedance that can be attributed to each of P and K for the purpose of determining circuit behavior when identical loads are simultaneously applied to P and K. About the only difference is I went on to explain that the 'classic unequal output impedance model works when you consider different generator voltages with "If you keep the 'different impedance' analysis the thing is as cathode impedance drops anode gain increases in exact proportion to the drop across the 'larger' anode impedance." And I have been attacked and insulted by Byrns for it. Btw, I agree with your implied criticism (at least that's how I interpreted it) that Jones tries to 'hedge' his 'build out' resistor with the 'problem' of grid drive. He continues to analyze this as an 'output impedance' issue but, IMO, that only serves to muddy one's understanding because the actual behavior is different than a simple 'output impedance' model suggests. I.E., it isn't that the signal on the 'high impedance' side is (uniformly) 'reduced', as one might think with a simple impedance model, it has a large negative spike when the cathode tries to drive +ve, in addition, of course, to the positive clipping. Due, of course, to the impedances but not in the manner one would expect for 'Ro into a load'. I *have* seen the 'problem' he refers to but I tend to view the issue as an 'upset' condition with the huge anode (and some on the cathode in reverse) swing introducing spurious HF components that cause feedback instability issues, depending on the OPT characteristics and amount of GNFB. I mentioned that in an earlier post about grid stoppers on the output tubes noting that it only takes 'one', on the cathode side, but that I always put two, equal, grid stoppers. and the first edition of his book where he presented this goofy idea, I have never seen the second edition of his book to see how he extricated himself from the predicament he created for himself, although I have been told by people that have seen the second edition that he did somehow extricate himself. -- Regards, John Byrns Surf my web pages at, http://fmamradios.com/ |
#85
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
On Thu, 21 Aug 2008 18:38:59 -0500, John Byrns
wrote: SNIP Hi Rodger, Wouldn't a 100 pF differential capacitive load be the equivalent of separate 200 pF capacitive loads applied simultaneously to P and K, not 50 pF loads? Yes, of course, capacitors in series and all that. It's hard to do such a long post without a boo-boo or two creeping in. As we all know. :-) |
#86
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
On Thu, 21 Aug 2008 19:38:24 -0400, "Henry Pasternack" wrote:
"The Phantom" wrote in message .. . I'm want to put forth the notion that it's not necessary to use phrases like "...behaves as if..." and "...acts as if...". The phrases suggest that the circuit "behaves as if" some condition were true, but which isn't really true. I disagree. Everything you say about the circuit is really a "...behaves as if..." statement if you back off a level or two and consider all of the abstractions that go into traditional circuit analysis. So, it it really a question of choosing the appropriate level of abstraction for the situation. You, like Byrns, seem to be a pretty pedantic fellow. And that's fine, but I have a different focus. What is that focus? And it's not for lack of the wherewithal to be pedantic if I wanted. There is a whole other aspect to the so-called "controversy" on this board that has nothing at all to do with circuits and everything to do with abusive personalities. It seems to me that the first step toward that kind of abuse is bringing personalities into the discussion in the first place. I don't see what personalities have to do with the working of a Concertina phase splitter, so I don't mention them. Whatever. -Henry |
#87
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
By the time of RDH4 (1952), there is discussion of impedances. RDH4 says, on page 330: "The effective output resistance is different for the two output channels, since P operates with current feedback and K with voltage feedback." They then give expressions for the "effective output resistance" (what they are referring to is what nowadays would be called the "driving point impedance") in formulas 30 and 31, but I'm sure that formula 30 is incorrect. Equation 30 is what effective Rp' of the tube is at the anode. Its quite correct. The R looking into the anode of the triode is Rp' and with an unbypassed Rk, is exactly what they say, Rp + ( [µ + 1]Rk. The Rout from the anode at the anode if tested separately would be the above Rpp' figure in parallel with whatever anode RL you have. I gave a summary which everyone has ignored in a post...... Subject: Output resistance of the CPI or "concertina phase inverter" Date: Mon, 18 Aug 2008 20:47:31 +1000 From: Patrick Turner Newsgroups: rec.audio.tubes Much has been said about the CPI last week, and many must be wondering what the F... does all that hot air mean. When let's consider a 1/2 6SN7 used in a typical CPI case in an amp. Let us assume that at the chosen Ia for the triode, µ = 20, and Ra = 10k. Let us examine the gains and output resistance of the two outputs assuming that the load at anode and cathode will always remain equal up to 32 kHz, so we don't have to worry about capaictance effects. Suppose total RL for the triode = 40k, so cathode RL = 20k = anode RL = 20k. Internal open loop gain OLG of the tube, A, = 20 x 40k / ( 10k + 40k ) = 16.0. So if +1V appears across Vgk, there will be -8V at anode, and +8V at cathode, and to get this you must apply +8V + +1V to the grid, so total closed loop gain, CLG = A' = -16/+9 = -1.778 total, or -0.889 if the gain is between grid and anode only, or +0.889 if measured between grid and cathode. So for +9V VOLTAGE CHANGE applied to the grid, you get -8V at a, and +8 at k and Ia change = 8V / 20k = 0.4mA. Now let's say we change the load conditions so IaQ and EaQ remains the same, but RL total = 20k, so RL at a and k each equal 10k. CLG = 20 x 20 / (10 + 20) = 13.333. So for +1V change across Vgk, there must be -/+ 6.666V change at anode and cathode respectively. A' between grid and anode = -6.666 / ( 6.666 + 1 ) = -6.666 / 7.666 = -0.8695. And +0.8695 gain between grid and cathode. So let's apply the same +9V change to the grid as we did when RL = 20k + 20k, and we'll get Va = -7.82V, and Vk = +7.82V. Current flow = 7.82V / 10k = 0.782mA So, what is the output resistance of the two outputs in our circuit? For one thing, while ever RLa = RLk, the two remain equal. But we can work out from the two load states what is the Rout for each with 9V applied to either load state of 20k or 10k at a and k. With 20k+20k, Io = 0.4mA, 10k+10k Io = 0.782mA, so current change between the two loadings = the difference = 0.382mA. The Va or Vk change is from 8.000V to 7.82V, which gives a V change = 0.174 V. Rout = Vchange / Ichange = 0.174V / 0.382mA = 0.455k = 455 ohms. What else can we say? We could say Rout measured between anone and cathode = 2 x 455 ohms because there are two oppositely moving voltage changes so Ra-k = 910 ohms. Now what if the same tube under identical Ia and Ea conditions was set up as a cathode follower? Asume there was a CCS cathode current sink, Rout = 1/gm, and gm = µ / Ra so Rout of the CF = Ra / µ = 10k / 20 = 500 ohms. The CPI can be said to have Rout at each of the two terminals slightly below that of a pure CF. So a CF has an Rout slightly higher than either Rout of th CPI with resistance loading on which it depends to work. But the CPI Rout expressed as between a and k is 910 ohms, and well above the CF case. There is of course an easier way to calculate the Rout a to k for the CPI. If the loads always remain the same for a and k, its as if the CPI had transformer coupled loadings between a to 0V and k and 0V, and then it is said to operate with voltage NFB instead of current NFB. The effective Ra, or Rout, with the FB where 1/2 the total Vo a to k is fed back becomes simply Ra' = Ra / ( 1 + [µ x ß] ) . In the case of the 1/2 6SN7, we get Ra' = 10k / ( 1 + [ 20 x 0.5 ] ), = 10k / 11 = 909 ohms, which remarkably enough, is very close to what I so laboriously calculated above. Measurements should confirm all the above, providing the µ and Ra of the tube are exactly as I stated above for all tests. One could have two transformer windings, one for a and the other for k, to accomodate the difference of a and k operating Vdc, and have a tertiary winding to alter the loading. Rout will be the same as calculated as long as the tranny has a large amount of Lp. Because of the variations of Ra and gm due to Ia variations, and thus slight changes to µ as a result of µ = gm x Ra, there isn't any need to measure such things as long as the circuit works as required in an amplifier. We don't need to count photons during a sunrise, or calculate what allowance must be made for mist or clouds, if its a beautiful sunrise, all is well. Clear enough for the nit pickers here? Patrick Turner. |
#88
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
"The Phantom" wrote in message
... You, like Byrns, seem to be a pretty pedantic fellow. And that's fine, but I have a different focus. What is that focus? I wrote a lengthy respose, but decided not to post it. If you would like to see what I had to say, you are welcome to email me privately. -Henry moc.ncr@kcanretsaph |
#89
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
On Aug 22, 4:12*am, "Henry Pasternack" wrote:
"The Phantom" wrote in message ... You, like Byrns, seem to be a pretty pedantic fellow. *And that's fine, but I have a different focus. What is that focus? I wrote a lengthy respose, but decided not to post it. *If you would like to see what I had to say, you are welcome to email me privately. -Henry moc.ncr@kcanretsaph Yup. Plodnick lost the argument on its merits, so now he whines that the whole world is against him, he calls them names, and at the same time offers to conspire behind their backs with anyone stupid enough to enter into private correspondence with such a discredited and incompetent plotter. Par for the course. Plodnick is a slow learner. What else is new. Andre Jute Observer |
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Audio Cyclopedia - A highly recommended book
On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner
wrote: By the time of RDH4 (1952), there is discussion of impedances. RDH4 says, on page 330: "The effective output resistance is different for the two output channels, since P operates with current feedback and K with voltage feedback." They then give expressions for the "effective output resistance" (what they are referring to is what nowadays would be called the "driving point impedance") in formulas 30 and 31, but I'm sure that formula 30 is incorrect. Equation 30 is what effective Rp' of the tube is at the anode. Its quite correct. The R looking into the anode of the triode is Rp' and with an unbypassed Rk, is exactly what they say, Rp + ( [µ + 1]Rk. The Rout from the anode at the anode if tested separately would be the above Rpp' figure in parallel with whatever anode RL you have. In my copy of RDH4, on page 330 they have two lines like this: "Channel P: rp' = (u - 1)Rk where Rk = RL (30) (from equation 25a, Sect 1.)" But on page 313, equation 25a says: "rp' = rp + (u +1)R3 (25a)" How do they get equation 30 from equation 25a? |
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Audio Cyclopedia - A highly recommended book
On Wed, 13 Aug 2008 10:50:58 GMT, Patrick Turner
wrote: flipper wrote: On Tue, 12 Aug 2008 16:22:18 GMT, Patrick Turner wrote: However the most satisfying solution would be to abandon the model entirely, along with the questions it raises, and simply build the physical circuit and measure the resulting source impedances. Unfortunately it appears that would just take us back to square one, again raising the crucial question of how to properly measure the source impedances. Just building something, observing it, measuring it and deciding what is there is trifle to complex and real for some around here. They hafta get orf their bums, and go into their workshop, and spend time actually doing something. You get yourself into trouble with those incessant knee jerk snide remarks of yours because this matter came up as a result of my doing exactly that. I BUILT the damn thing and measured it. So, with all due respect to you and your bums, sit on it. OK, Point taken, but some here are arm chair wanabes. But ya don't need to all that. Blind Freddy can see that if you measure the Rout at the anode by adjusting the anode load and only the anode load, the Rout = Vchange / Ichange, which is how you measure the Rout of anything. Not when equal loads is inherent to the circuit and a required condition. But when one talks about the Rout at one point in a circuit, what is usually meant is that its taken as is at that point, and without any load adjustments elsewhere. What you are describing here is the driving point impedance; what flipper is describing is the differential impedance. The Vchange / I change can be used to measure Rout easily for any amp output. In the case of the CPI, it can be done if you specifiy that the load change to anode and cathode will be equal. Then the because the same current flows through cathode load, tube, and anode load, the Ichange is equal, and because the TWO loads are kept equal, V change will also be equal, so you'd think Rout is the same from both terminals. |
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Audio Cyclopedia - A highly recommended book
1 ___ | ___ +-+. ,+-+-|___|--|___|-GND )|( 50 10k )|( ___ -. ,+-+-|___|-GND )|( 1k )|( ___ ___ +-+' '+-+-|___|.-|___|-GND 50 | 10k 2 Imagine that you had a circuit like this, with the transformer turns ratio being 1 to 1, and the primary driven with a 1 kHz source having zero ohms output impedance. 1 ___ | +-+. ,----|___|- )|( 50 )|( ___ )|(----|___|-GND )|( 10k )|( ___ +-+' '----|___|- 50 | 2 Now suppose you measure the impedance at node 1 with respect to ground. It will be something like 10k ohms. This is a driving point impedance. Now measure the impedance *between" nodes 1 and 2; it will be something like 100 ohms. This is a differential impedance. The differential output impedance is not the same as the driving point output impedance, as can be easily seen. It doesn't seem to me that there should be controversy about what the circuit does if we choose our terminology correctly. It's important to distinguish between driving point impedance and differential impedance. When making the differential measurement, half of the differential impedance is at node 1 and half at node 2. If we had a circuit like this: 1 ___ | +-+. ,----|___|- )|( 60 )|( ___ )|(----|___|-GND )|( 10k )|( ___ +-+' '----|___|- 40 | 2 we would still have a differential output impedance of about 100 ohms, but we would no longer have half the differential impedance at each node. The Concertina is like the first circuit. The differential output impedance is split, half and half, between P and K (which is rather fortuitous). |
#93
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Audio Cyclopedia - A highly recommended book
The Phantom wrote: On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner wrote: By the time of RDH4 (1952), there is discussion of impedances. RDH4 says, on page 330: "The effective output resistance is different for the two output channels, since P operates with current feedback and K with voltage feedback." They then give expressions for the "effective output resistance" (what they are referring to is what nowadays would be called the "driving point impedance") in formulas 30 and 31, but I'm sure that formula 30 is incorrect. Equation 30 is what effective Rp' of the tube is at the anode. Its quite correct. The R looking into the anode of the triode is Rp' and with an unbypassed Rk, is exactly what they say, Rp + ( [µ + 1]Rk. The Rout from the anode at the anode if tested separately would be the above Rpp' figure in parallel with whatever anode RL you have. In my copy of RDH4, on page 330 they have two lines like this: "Channel P: rp' = (u - 1)Rk where Rk = RL (30) (from equation 25a, Sect 1.)" In my RDH4, Pg 330, equation 30 reads... Channel P rp' = rp + ( µ + 1 )Rk where Rk = RL. ( from eqtn 25A, Sect 1 ). You have misquoted Equation 30. But on page 313, equation 25a says: "rp' = rp + (u +1)R3 (25a)" How do they get equation 30 from equation 25a? There is no inconsistency. R3 in eqtn 25a is Rk in eqtn 30. RDH4 gives what is exactly right for the Rout at the anode for the CPI if measured as a separate outlet to power some following stage. Ditto for the cathode Rout as spelled out in eqtn (31). Rout at Channel K : rp' = ( rp + RL ) / ( µ + 1 ) If you follow my treatise in a recent post, you'll see what Rout of K and P will be where always RL = RK as in a real amp using CPI.... |
#94
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Audio Cyclopedia - A highly recommended book
The Phantom wrote: On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner wrote: By the time of RDH4 (1952), there is discussion of impedances. RDH4 says, on page 330: "The effective output resistance is different for the two output channels, since P operates with current feedback and K with voltage feedback." They then give expressions for the "effective output resistance" (what they are referring to is what nowadays would be called the "driving point impedance") in formulas 30 and 31, but I'm sure that formula 30 is incorrect. Equation 30 is what effective Rp' of the tube is at the anode. Its quite correct. The R looking into the anode of the triode is Rp' and with an unbypassed Rk, is exactly what they say, Rp + ( [µ + 1]Rk. The Rout from the anode at the anode if tested separately would be the above Rpp' figure in parallel with whatever anode RL you have. In my copy of RDH4, on page 330 they have two lines like this: "Channel P: rp' = (u - 1)Rk where Rk = RL (30) (from equation 25a, Sect 1.)" In my RDH4, Pg 330, equation 30 reads... Channel P rp' = rp + ( µ + 1 )Rk where Rk = RL. ( from eqtn 25A, Sect 1 ). You have misquoted Equation 30. But on page 313, equation 25a says: "rp' = rp + (u +1)R3 (25a)" How do they get equation 30 from equation 25a? There is no inconsistency. R3 in eqtn 25a is Rk in eqtn 30. RDH4 gives what is exactly right for the Rout at the anode for the CPI if measured as a separate outlet to power some following stage. Ditto for the cathode Rout as spelled out in eqtn (31). Rout at Channel K : rp' = ( rp + RL ) / ( µ + 1 ) If you follow my treatise in a recent post, you'll see what Rout of K and P will be where always RL = RK as in a real amp using CPI.... Sorry I forgot to add the post and sign it; Not one person at r.a.t has disagreed with anything in the post below. Either its dead right and correct or nobody bothered to read it. ------------------------------------------------------------------------------------- Subject: Output resistance of the CPI or "concertina phase inverter" Much has been said about the CPI last week, and many must be wondering what the F... does all that hot air mean. When let's consider a 1/2 6SN7 used in a typical CPI case in an amp. Let us assume that at the chosen Ia for the triode, µ = 20, and Ra = 10k. Let us examine the gains and output resistance of the two outputs assuming that the load at anode and cathode will always remain equal up to 32 kHz, so we don't have to worry about capaictance effects. Suppose total RL for the triode = 40k, so cathode RL = 20k = anode RL = 20k. Internal open loop gain OLG of the tube, A, = 20 x 40k / ( 10k + 40k ) = 16.0. So if +1V appears across Vgk, there will be -8V at anode, and +8V at cathode, and to get this you must apply +8V + +1V to the grid, so total closed loop gain, CLG = A' = -16/+9 = -1.778 total, or -0.889 if the gain is between grid and anode only, or +0.889 if measured between grid and cathode. So for +9V VOLTAGE CHANGE applied to the grid, you get -8V at a, and +8 at k and Ia change = 8V / 20k = 0.4mA. Now let's say we change the load conditions so IaQ and EaQ remains the same, but RL total = 20k, so RL at a and k each equal 10k. CLG = 20 x 20 / (10 + 20) = 13.333. So for +1V change across Vgk, there must be -/+ 6.666V change at anode and cathode respectively. A' between grid and anode = -6.666 / ( 6.666 + 1 ) = -6.666 / 7.666 = -0.8695. And +0.8695 gain between grid and cathode. So let's apply the same +9V change to the grid as we did when RL = 20k + 20k, and we'll get Va = -7.82V, and Vk = +7.82V. Current flow = 7.82V / 10k = 0.782mA So, what is the output resistance of the two outputs in our circuit? For one thing, while ever RLa = RLk, the two remain equal. But we can work out from the two load states what is the Rout for each with 9V applied to either load state of 20k or 10k at a and k. With 20k+20k, Io = 0.4mA, 10k+10k Io = 0.782mA, so current change between the two loadings = the difference = 0.382mA. The Va or Vk change is from 8.000V to 7.82V, which gives a V change = 0.174 V. Rout = Vchange / Ichange = 0.174V / 0.382mA = 0.455k = 455 ohms. What else can we say? We could say Rout measured between anode and cathode = 2 x 455 ohms because there are two oppositely moving voltage changes so Ra-k = 910 ohms. Now what if the same tube under identical Ia and Ea conditions was set up as a cathode follower? Asume there was a CCS cathode current sink, Rout = 1/gm, and gm = µ / Ra so Rout of the CF = Ra / µ = 10k / 20 = 500 ohms. The CPI can be said to have Rout at each of the two terminals slightly below that of a pure CF. So a CF has an Rout slightly higher than either Rout of th CPI with resistance loading on which it depends to work. But the CPI Rout expressed as between a and k is 910 ohms, and well above the CF case. There is of course an easier way to calculate the Rout a to k for the CPI. If the loads always remain the same for a and k, its as if the CPI had transformer coupled loadings between a to 0V and k and 0V, and then it is said to operate with voltage NFB instead of current NFB. The effective Ra, or Rout, with the FB where 1/2 the total Vo a to k is fed back becomes simply Ra' = Ra / ( 1 + [µ x ß] ) . In the case of the 1/2 6SN7, we get Ra' = 10k / ( 1 + [ 20 x 0.5 ] ), = 10k / 11 = 909 ohms, which remarkably enough, is very close to what I so laboriously calculated above. Measurements should confirm all the above, providing the µ and Ra of the tube are exactly as I stated above for all tests. One could have two transformer windings, one for a and the other for k, to accomodate the difference of a and k operating Vdc, and have a tertiary winding to alter the loading. Rout will be the same as calculated as long as the tranny has a large amount of Lp. Because of the variations of Ra and gm due to Ia variations, and thus slight changes to µ as a result of µ = gm x Ra, there isn't any need to measure such things as long as the circuit works as required in an amplifier. We don't need to count photons during a sunrise, or calculate what allowance must be made for mist or clouds, if its a beautiful sunrise, all is well. Clear enough for the nit pickers here? Patrick Turner. |
#95
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Audio Cyclopedia - A highly recommended book
In article ,
The Phantom wrote: On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner wrote: By the time of RDH4 (1952), there is discussion of impedances. RDH4 says, on page 330: "The effective output resistance is different for the two output channels, since P operates with current feedback and K with voltage feedback." They then give expressions for the "effective output resistance" (what they are referring to is what nowadays would be called the "driving point impedance") in formulas 30 and 31, but I'm sure that formula 30 is incorrect. Equation 30 is what effective Rp' of the tube is at the anode. Its quite correct. The R looking into the anode of the triode is Rp' and with an unbypassed Rk, is exactly what they say, Rp + ( [µ + 1]Rk. The Rout from the anode at the anode if tested separately would be the above Rpp' figure in parallel with whatever anode RL you have. In my copy of RDH4, on page 330 they have two lines like this: "Channel P: rp' = (u - 1)Rk where Rk = RL (30) (from equation 25a, Sect 1.)" But on page 313, equation 25a says: "rp' = rp + (u +1)R3 (25a)" How do they get equation 30 from equation 25a? Seems straight forward to me, on page 330 of my copy of the RDH equation 30 for "Channel P" is shown as: rp' = rp + u * Rk Equation 25a on page 313 of my copy of the RDH is shown as: rp' = rp + u * R3 These two equations appear virtually identical to me. The problem I see is that while they are the same, they differ from equation 5 in the Byrns report which I believe to be the correct version, and is as follows: rp' = rp + (u + 1) * Zk -- Regards, John Byrns Surf my web pages at, http://fmamradios.com/ |
#96
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Audio Cyclopedia - A highly recommended book
The reason for wanting to know the Concertina "output impedance" is to be
able calculate the effect of further loading on the frequency response. If "equal output impedance" model is used, then how will the effect of unequal loading be calculated? This is a very real concern, because the stray capacitances at the plate and cathode will not likely be identical. And, of course, a model which only deals with the unequal driving point impedances will not properly give the effect of the balanced loading which will dominate the loads (we hope). It's not necessary to use only one of these. Better to do a detailed analysis using Shekel's method such as the one I posted a week or so ago. From this analysis, expressions for the gain from grid to plate and to cathode are derived. These expressions can then be used to calculate the frequency responses from grid to plate and cathode regardless of whether the loads are balanced or not; the correct result will be obtained for all loads. Using the following symbols: ra = plate resistance of the tube gm = transconductance of the tube Zp = additional load applied to plate; include all loads in this, including such things as stray capacitances. Zk = additional load applied to cathode. Then the grid to plate gain is given by: ra*gm*Zp - ------------------- (ra*gm+1)*Zk+Zp+ra The grid to cathode gain is given by: ra*gm*Zk ------------------- (ra*gm+1)*Zk+Zp+ra If Zp = Zk, then these expressions are identical, which suggests that the impedances at plate and cathode are equal before additional identical loads are applied. This is quite true, because as I've explained elsewhere in this thread, applying identical loads simultaneously to plate and cathode is exactly equivalent to applying a differential (balanced) load to the Concertina, and we would expect that would not upset the equality of gains at plate and cathode. Knowing the output impedances allows us to calculate frequency response by treating the outputs as sources with some impedance and then using a voltage divider formula to calculate the effect of loads. If the loads are perfectly balanced, then the appropriate impedance to use for these calculations is what I've called the "half differential output impedance". But what if the loads aren't perfectly balanced? Then I suppose we could separate the load into a balanced component and an unbalanced component, solve separately and use superposition to get the final result. It's much easier to just use the gain expressions above and not worry about output impedances. You get a guaranteed correct result for frequency response no matter whether the loads are perfectly balanced, or if there is a slight unbalanced component. I've posted an analysis and some plots of the 3 Concertina output impedances (2 driving point and 1 differential) versus frequency. There's also a plot of plate and cathode frequency response with a slightly unbalanced capacitive load. |
#97
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Audio Cyclopedia - A highly recommended book
On Wed, 27 Aug 2008 13:15:55 -0500, John Byrns
wrote: In article , The Phantom wrote: On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner wrote: By the time of RDH4 (1952), there is discussion of impedances. RDH4 says, on page 330: "The effective output resistance is different for the two output channels, since P operates with current feedback and K with voltage feedback." They then give expressions for the "effective output resistance" (what they are referring to is what nowadays would be called the "driving point impedance") in formulas 30 and 31, but I'm sure that formula 30 is incorrect. Equation 30 is what effective Rp' of the tube is at the anode. Its quite correct. The R looking into the anode of the triode is Rp' and with an unbypassed Rk, is exactly what they say, Rp + ( [µ + 1]Rk. The Rout from the anode at the anode if tested separately would be the above Rpp' figure in parallel with whatever anode RL you have. In my copy of RDH4, on page 330 they have two lines like this: "Channel P: rp' = (u - 1)Rk where Rk = RL (30) (from equation 25a, Sect 1.)" But on page 313, equation 25a says: "rp' = rp + (u +1)R3 (25a)" How do they get equation 30 from equation 25a? Seems straight forward to me, on page 330 of my copy of the RDH equation 30 for "Channel P" is shown as: rp' = rp + u * Rk My copy of RDH4 has: rp' = (u - 1)Rk for equation 30 Is it really the case that your RDH is different from mine? There is a downloadable copy available on the web, and it is the same as my printed copy. What copyright/printing date is your copy? Equation 25a on page 313 of my copy of the RDH is shown as: rp' = rp + u * R3 My copy of RDH4 on page 313 for equation 25a has: rp' = rp + (u + 1)R3 but you are saying that your copy has (and I've cut and pasted from just above): rp' = rp + u * R3 They're not the same. How can this be? These two equations appear virtually identical to me. The problem I see is that while they are the same, they differ from equation 5 in the Byrns report which I believe to be the correct version, and is as follows: rp' = rp + (u + 1) * Zk |
#98
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Audio Cyclopedia - A highly recommended book
On Wed, 27 Aug 2008 20:48:59 -0500, flipper wrote:
On Tue, 26 Aug 2008 11:37:26 -0700, The Phantom wrote: On Wed, 13 Aug 2008 10:50:58 GMT, Patrick Turner wrote: flipper wrote: On Tue, 12 Aug 2008 16:22:18 GMT, Patrick Turner wrote: However the most satisfying solution would be to abandon the model entirely, along with the questions it raises, and simply build the physical circuit and measure the resulting source impedances. Unfortunately it appears that would just take us back to square one, again raising the crucial question of how to properly measure the source impedances. Just building something, observing it, measuring it and deciding what is there is trifle to complex and real for some around here. They hafta get orf their bums, and go into their workshop, and spend time actually doing something. You get yourself into trouble with those incessant knee jerk snide remarks of yours because this matter came up as a result of my doing exactly that. I BUILT the damn thing and measured it. So, with all due respect to you and your bums, sit on it. OK, Point taken, but some here are arm chair wanabes. But ya don't need to all that. Blind Freddy can see that if you measure the Rout at the anode by adjusting the anode load and only the anode load, the Rout = Vchange / Ichange, which is how you measure the Rout of anything. Not when equal loads is inherent to the circuit and a required condition. But when one talks about the Rout at one point in a circuit, what is usually meant is that its taken as is at that point, and without any load adjustments elsewhere. What you are describing here is the driving point impedance; what flipper is describing is the differential impedance. No, what I described is a Thevenin style 'two' output impedance model. Do you mean that there are two output ports, each having the same output impedance, and the output voltage at each port is the same magnitude as the other, but the opposite polarity? The Vchange / I change can be used to measure Rout easily for any amp output. In the case of the CPI, it can be done if you specifiy that the load change to anode and cathode will be equal. Then the because the same current flows through cathode load, tube, and anode load, the Ichange is equal, and because the TWO loads are kept equal, V change will also be equal, so you'd think Rout is the same from both terminals. |
#99
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Audio Cyclopedia - A highly recommended book
The Phantom wrote: The reason for wanting to know the Concertina "output impedance" is to be able calculate the effect of further loading on the frequency response. If "equal output impedance" model is used, then how will the effect of unequal loading be calculated? This is a very real concern, because the stray capacitances at the plate and cathode will not likely be identical. And, of course, a model which only deals with the unequal driving point impedances will not properly give the effect of the balanced loading which will dominate the loads (we hope). It's not necessary to use only one of these. Better to do a detailed analysis using Shekel's method such as the one I posted a week or so ago. From this analysis, expressions for the gain from grid to plate and to cathode are derived. These expressions can then be used to calculate the frequency responses from grid to plate and cathode regardless of whether the loads are balanced or not; the correct result will be obtained for all loads. Using the following symbols: ra = plate resistance of the tube gm = transconductance of the tube Zp = additional load applied to plate; include all loads in this, including such things as stray capacitances. Zk = additional load applied to cathode. Then the grid to plate gain is given by: ra*gm*Zp - ------------------- (ra*gm+1)*Zk+Zp+ra The grid to cathode gain is given by: ra*gm*Zk ------------------- (ra*gm+1)*Zk+Zp+ra If Zp = Zk, then these expressions are identical, which suggests that the impedances at plate and cathode are equal before additional identical loads are applied. This is quite true, because as I've explained elsewhere in this thread, applying identical loads simultaneously to plate and cathode is exactly equivalent to applying a differential (balanced) load to the Concertina, and we would expect that would not upset the equality of gains at plate and cathode. Knowing the output impedances allows us to calculate frequency response by treating the outputs as sources with some impedance and then using a voltage divider formula to calculate the effect of loads. If the loads are perfectly balanced, then the appropriate impedance to use for these calculations is what I've called the "half differential output impedance". But what if the loads aren't perfectly balanced? Then I suppose we could separate the load into a balanced component and an unbalanced component, solve separately and use superposition to get the final result. It's much easier to just use the gain expressions above and not worry about output impedances. You get a guaranteed correct result for frequency response no matter whether the loads are perfectly balanced, or if there is a slight unbalanced component. I've posted an analysis and some plots of the 3 Concertina output impedances (2 driving point and 1 differential) versus frequency. There's also a plot of plate and cathode frequency response with a slightly unbalanced capacitive load. I guess all this deserves some sort of Concertina Medal for the most serious attempt to mathematically define what happens above say 20kHz to a CPI as a result of capacitance effects due to strays, Miller, and loading from following tube Miller C etc. I just observe the roll offs one gets, after getting the CPI to work as well as i want it to at 1 kHz. Whatever will be will be. If a CPI drives an output stage, as in the case of a Dynaco ST70, then the Miller C "seen by" the CPI varies with output load because gain changes with output load, and the load change is actually dynamically changing all the time because F changes, and speaker Z changes with F. Miller C reduces with a C load at the output, so tubes should drive ESL speakers well. Because speaker loads are not so consistent and simple as a pure resistance dummy load, I know I could work it all out mathematically if I tried, but the real situation does not make me want to try. In a Williamson, the balanced drive amp buffers the CPI from both output tube Miller C and the onset of grid I of the output tubes. Meanwhile the CPI buffers the VI input triode from the Miller C of the balanced amp. Last time I built a Williamson with 6CG7 throughout, and measured F pole at the output tube grids the pole was above 250kHz, and above anything else I have ever tried. The Williamson line up is a fast circuit. Somewhat too fast perhaps, because Willy amps tend to be HF unstable with a pure 0.1uF to 0.47uF load, so HF gain has to be reduced with a shelving network to reduce HF gain above 30kHz. The CPI output resistance below 20kHz is a simple thing to understand if you make one assumption, and that is that RL at a and k will always be equal, and all tests to measure Rout at a or k are to be made with RL at a and k always remaining equal, so hence Rout at a = Rout at k. Patrick Turner. |
#100
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Audio Cyclopedia - A highly recommended book
In article ,
The Phantom wrote: On Wed, 27 Aug 2008 13:15:55 -0500, John Byrns wrote: In article , The Phantom wrote: On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner wrote: By the time of RDH4 (1952), there is discussion of impedances. RDH4 says, on page 330: "The effective output resistance is different for the two output channels, since P operates with current feedback and K with voltage feedback." They then give expressions for the "effective output resistance" (what they are referring to is what nowadays would be called the "driving point impedance") in formulas 30 and 31, but I'm sure that formula 30 is incorrect. Equation 30 is what effective Rp' of the tube is at the anode. Its quite correct. The R looking into the anode of the triode is Rp' and with an unbypassed Rk, is exactly what they say, Rp + ( [µ + 1]Rk. The Rout from the anode at the anode if tested separately would be the above Rpp' figure in parallel with whatever anode RL you have. In my copy of RDH4, on page 330 they have two lines like this: "Channel P: rp' = (u - 1)Rk where Rk = RL (30) (from equation 25a, Sect 1.)" But on page 313, equation 25a says: "rp' = rp + (u +1)R3 (25a)" How do they get equation 30 from equation 25a? Seems straight forward to me, on page 330 of my copy of the RDH equation 30 for "Channel P" is shown as: rp' = rp + u * Rk My copy of RDH4 has: rp' = (u - 1)Rk for equation 30 Is it really the case that your RDH is different from mine? Yes, it is really true if your copy doesn't show equation 30 on page 330 as: "rp' = rp + uRk" There is a downloadable copy available on the web, and it is the same as my printed copy. What copyright/printing date is your copy? The copyright notice in my copy does not specify a date, various dates are mentioned elsewhere including 1952, 1953, and February 1954. The latter being the date it was reproduced in the USA for RCA Victor. I can find none of the usual numbers indicating what printing my copy might be. Equation 25a on page 313 of my copy of the RDH is shown as: rp' = rp + u * R3 My copy of RDH4 on page 313 for equation 25a has: rp' = rp + (u + 1)R3 but you are saying that your copy has (and I've cut and pasted from just above): rp' = rp + u * R3 It appears exactly as: "rp' = rp + uR3" They're not the same. How can this be? I guess it was one of the wonders of modern printing technology circa 1954. -- Regards, John Byrns Surf my web pages at, http://fmamradios.com/ |
#101
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Audio Cyclopedia - A highly recommended book
There is a downloadable RDH4 at:
http://headfonz.rutgers.edu/RDH4/ I checked chapter 7, and it appears to be exactly like my printed copy. If you want a version to compare, there it is. If I look carefully, it's apparent that the typeface of equation 30 in my copy is different than that of equation 31. Changes were made somewhere along the way. Does your copy have an equation 34a? It appears to have been derived from the paper by George Jones that is available at http://www.diybanter.com/attachment....2&d=1213179423 |
#102
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Audio Cyclopedia - A highly recommended book
On Wed, 27 Aug 2008 01:37:29 GMT, Patrick Turner
wrote: The Phantom wrote: On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner wrote: By the time of RDH4 (1952), there is discussion of impedances. RDH4 says, on page 330: "The effective output resistance is different for the two output channels, since P operates with current feedback and K with voltage feedback." They then give expressions for the "effective output resistance" (what they are referring to is what nowadays would be called the "driving point impedance") in formulas 30 and 31, but I'm sure that formula 30 is incorrect. Equation 30 is what effective Rp' of the tube is at the anode. Its quite correct. The R looking into the anode of the triode is Rp' and with an unbypassed Rk, is exactly what they say, Rp + ( [µ + 1]Rk. The Rout from the anode at the anode if tested separately would be the above Rpp' figure in parallel with whatever anode RL you have. In my copy of RDH4, on page 330 they have two lines like this: "Channel P: rp' = (u - 1)Rk where Rk = RL (30) (from equation 25a, Sect 1.)" In my RDH4, Pg 330, equation 30 reads... Channel P rp' = rp + ( µ + 1 )Rk where Rk = RL. ( from eqtn 25A, Sect 1 ). You have misquoted Equation 30. In another post, John Byrns says that his copy has: rp' = rp + u * Rk for equation 30. My copy of RDH4 has: rp' = (u - 1)Rk where Rk = RL which is the same as the downloadable copy of RDH4 at http://headfonz.rutgers.edu/RDH4/ has. and you (Patrick) say that yours has: rp' = rp + ( µ + 1 )Rk where Rk = RL It would seem that there are at least 3 different expressions for equation 30 in various editions/printings of RDH4. I wonder what other differences there are? But on page 313, equation 25a says: "rp' = rp + (u +1)R3 (25a)" How do they get equation 30 from equation 25a? There is no inconsistency. R3 in eqtn 25a is Rk in eqtn 30. RDH4 gives what is exactly right for the Rout at the anode for the CPI if measured as a separate outlet to power some following stage. Ditto for the cathode Rout as spelled out in eqtn (31). Rout at Channel K : rp' = ( rp + RL ) / ( µ + 1 ) If you follow my treatise in a recent post, you'll see what Rout of K and P will be where always RL = RK as in a real amp using CPI.... |
#103
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
In article ,
The Phantom wrote: On Wed, 27 Aug 2008 01:37:29 GMT, Patrick Turner wrote: The Phantom wrote: On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner wrote: By the time of RDH4 (1952), there is discussion of impedances. RDH4 says, on page 330: "The effective output resistance is different for the two output channels, since P operates with current feedback and K with voltage feedback." They then give expressions for the "effective output resistance" (what they are referring to is what nowadays would be called the "driving point impedance") in formulas 30 and 31, but I'm sure that formula 30 is incorrect. Equation 30 is what effective Rp' of the tube is at the anode. Its quite correct. The R looking into the anode of the triode is Rp' and with an unbypassed Rk, is exactly what they say, Rp + ( [µ + 1]Rk. The Rout from the anode at the anode if tested separately would be the above Rpp' figure in parallel with whatever anode RL you have. In my copy of RDH4, on page 330 they have two lines like this: "Channel P: rp' = (u - 1)Rk where Rk = RL (30) (from equation 25a, Sect 1.)" In my RDH4, Pg 330, equation 30 reads... Channel P rp' = rp + ( µ + 1 )Rk where Rk = RL. ( from eqtn 25A, Sect 1 ). You have misquoted Equation 30. In another post, John Byrns says that his copy has: rp' = rp + u * Rk for equation 30. My copy of RDH4 has: rp' = (u - 1)Rk where Rk = RL which is the same as the downloadable copy of RDH4 at http://headfonz.rutgers.edu/RDH4/ has. and you (Patrick) say that yours has: rp' = rp + ( µ + 1 )Rk where Rk = RL It would seem that there are at least 3 different expressions for equation 30 in various editions/printings of RDH4. I wonder what other differences there are? That was exactly the thought that came to my mind too. -- Regards, John Byrns Surf my web pages at, http://fmamradios.com/ |
#104
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
Henry,
Good to see you are alive and kicking. Pop me and email. Spike Henry Pasternack wrote: "The Phantom" wrote in message ... You, like Byrns, seem to be a pretty pedantic fellow. And that's fine, but I have a different focus. What is that focus? I wrote a lengthy respose, but decided not to post it. If you would like to see what I had to say, you are welcome to email me privately. -Henry moc.ncr@kcanretsaph |
#105
Posted to rec.audio.tubes
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Audio Cyclopedia - A highly recommended book
Testing 1, 2 3.
Spike wrote: Henry, Good to see you are alive and kicking. Pop me and email. Spike Henry Pasternack wrote: "The Phantom" wrote in message ... You, like Byrns, seem to be a pretty pedantic fellow. And that's fine, but I have a different focus. What is that focus? I wrote a lengthy respose, but decided not to post it. If you would like to see what I had to say, you are welcome to email me privately. -Henry moc.ncr@kcanretsaph |
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