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#1
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Phase splitter speculation, con't
I've been researching the pros and cons of various types of phase splitters .. . . . From a theoretical perspective, I like two approaches: the Schmidt & the Isodyne. Of those two I'm, inclined towards the Schmidt. The cons of the Schmidt are said to include: € Quality of the grid decoupling capacitor is critical; € Tubes must be paired and from a twin tube These don't look too difficult to overcome. It looks like it wouldn't be too tough to build a mu-follower version of the Schmidt. I'll probably draft one and past it on a.b.s.e. Does anyone have any advice or experience the Schmidt TUBE topology? Thanks Jon |
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#2
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"Jon Yaeger" wrote
From a theoretical perspective, I like two approaches: the Schmidt & the Isodyne. Of those two I'm, inclined towards the Schmidt. Is Schmidt a name for the LTP, or differential pair? Have you thought of using two in cascade, with one elevated, or depressed (what's the opposite of elevated?) to allow direct coupling somewhere? That seems to be the modern way. There is a fancy cross-coupled two-stage differential amp that can use mu-followers but I still can't find it. And is the Isodyne a concertina? Is this a quiz? ? Quality of the grid decoupling capacitor is critical; No more so than any other coupling capacitor, like the one on the other grid, possibly. ? Tubes must be paired and from a twin tube Why from a twin valve? For those who have built a few LTPs, do you find that in practice you always need some balance adjustment, or can you rely on matched valves and anode resistors to stay matched? It looks like it wouldn't be too tough to build a mu-follower version of the Schmidt. I'll probably draft one and past it on a.b.s.e. That would be interesting. From a recent previous post: Keep in mind that a mu-follower is constant-current, and so isn't degenerated by an unbypassed cathode resistor on the bottom. So you couldn't share a cathode resistor to help maintain balance. cheers, Ian |
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#3
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Jon Yaeger wrote:
I've been researching the pros and cons of various types of phase split= ters . . . . From a theoretical perspective, I like two approaches: the Schmidt & t= he Isodyne. Of those two I'm, inclined towards the Schmidt. The cons of the Schmidt are said to include: =80 Quality of the grid decoupling capacitor is critical; =80 Tubes must be paired and from a twin tube These don't look too difficult to overcome. It looks like it wouldn't be too tough to build a mu-follower version o= f the Schmidt. I'll probably draft one and past it on a.b.s.e. Does anyone have any advice or experience the Schmidt TUBE topology= ? Thanks Jon Fpr the real deal look at http://www.bonavolta.ch/hobby/en/audio/split.htm JLS |
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#4
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Ian Iveson wrote: deleted For those who have built a few LTPs, do you find that in practice you always need some balance adjustment, or can you rely on matched valves and anode resistors to stay matched? The inbalance of a LTP is not because of mis-match of the tubes or the plate resistors in that even if you use a _perfectly matched_ pairs, there will still be inbalance. The solution is either use a very large resistance at the tail or a pair of very high gain devices. A constant current source has a very large resistance or use a very negative supply where the other end of the tail resistor is connected and thus you can use a resistor of large value for the tail resistor. K. L. |
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#5
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"Ian Iveson" wrote in message .uk... For those who have built a few LTPs, do you find that in practice you always need some balance adjustment, or can you rely on matched valves and anode resistors to stay matched? I like to have the balance pot, sum the output on a dual beam scope, and tweak. Iain |
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#7
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"Jon Yaeger" wrote
Fpr the real deal look at http://www.bonavolta.ch/hobby/en/audio/split.htm Thanks, John. That's what I've been looking at! Aha. The isodyne is a version of the paraphase, I see. Same advice about mu-followers as last time: the main benefit would be for the first valve, which is the only one without feedback. Except where does its grid resistor go...what's going on there? Just for bias? Must have a low input impedance...235k presumably, and the input cap is necessary. You could use mu-followers instead of CFs for stacks of gain, perhaps? cheers, Ian |
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#8
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see Kimmel's paper for a differential mu-stage:
http://www7.taosnet.com/f10/mustage.html last page. cheers, Ian |
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#9
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"Jon Yaeger" wrote in message ... I've been researching the pros and cons of various types of phase splitters . . . . From a theoretical perspective, I like two approaches: the Schmidt & the Isodyne. Of those two I'm, inclined towards the Schmidt. The cons of the Schmidt are said to include: ? Quality of the grid decoupling capacitor is critical; ? Tubes must be paired and from a twin tube These don't look too difficult to overcome. It looks like it wouldn't be too tough to build a mu-follower version of the Schmidt. I'll probably draft one and past it on a.b.s.e. Does anyone have any advice or experience the Schmidt TUBE topology? Thanks Jon Well ,, I know bugger all and everybody will tell me I am wrong but who cares ... The LTP/Schmitt/paraphase/CCS tail or however you want to draw them all, splits phase and gives gain Two functions in one stage = less stages = one less time you need to turn the signal upside down = it's more stable with NFB... etc, etc ... seen all the arguments .. Sorry ..there are just too many components in there and too many variables for me to even consider this layout ... played with a test, splitter chassis for twelve months ... Result .. sorry .... A concertina is too simple to ignore and the valve is irrelevant .. any low Mu valve will do .. the OP impedances are different but that's easy to sort ... I cannot surround my valves with lots of 10% passives and expect any sense out of them .. DTN Williamson was far too clever for his own time .. The music has to go through all these bits .. the less you have ... the better it sounds .. cheers jim |
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#11
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On Thu, 24 Mar 2005 11:21:02 +1100, Patrick Turner
wrote: A differential pair amp, or LTP ( long tail pair) needs to have exactly equal RLs to each side, and a CCS "tail" from the commoned cathodes, and need only be driven at one grid, and the output voltage will be equal amplitude from each anode. The tubes can be virtually anything, 6AU6 in triode on one side, 12AU7 on the other. The only result of different devices on each side of the LTP diff amp is the rise of 2H due to the unbalanced distortion currents. No, not really *anything*. The diff pair needs to have exactly equal transconductances, too. Imbalance in Gm results in imbalance in output signal. But you know that. Chris Hornbeck |
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#12
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http://www.bonavolta.ch/hobby/en/audio/split.htm FWIW, that page is riddled with errors. Chris Hornbeck |
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#13
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Chris Hornbeck wrote: On Thu, 24 Mar 2005 11:21:02 +1100, Patrick Turner wrote: A differential pair amp, or LTP ( long tail pair) needs to have exactly equal RLs to each side, and a CCS "tail" from the commoned cathodes, and need only be driven at one grid, and the output voltage will be equal amplitude from each anode. The tubes can be virtually anything, 6AU6 in triode on one side, 12AU7 on the other. The only result of different devices on each side of the LTP diff amp is the rise of 2H due to the unbalanced distortion currents. No, not really *anything*. The diff pair needs to have exactly equal transconductances, too. Imbalance in Gm results in imbalance in output signal. But you know that. Nope. Each device for each half can have very unmatched gM, which is why I have cited trioded 6AU6 and 12AU7. The LTP with a CCS **must** have equal current *change* in either device, since the sum of the cathode currents is zero. So if you have +1mA in one side of the LTP, you must get -1mA in the other side because the combined currents at the cathode are zero, and regardless of voltage changes at the common cathode, there is no current change in the CCS tail. No currents flow in the grid circuits. So the +ve and -ve output voltages are equal if the tube current changes are equal and the RLs are equal. But the *voltage* gains of each triode might be different. Where you have a balanced voltage amp with a "short" tail, as one does in the Williamson, with a 1kohm R from commoned cathodes to 0V, you need substantial input voltage equality for output equality, and the differences in gm do matter a little, since the curcuit cannot balance itself like one can when a CCS is used. But usually the Williamson has negligible imbalance. But such a voltage amp as used in a W cannot be driven one side only; The output voltages will be very different. For balance without a CCS, as in the case of the Mullard 520 and derivatives, you need about 10k from the commoned cathodes to 0V at least, and the two RLs are quite different value for balance. As one grid is driven, about 1/2 the input voltage appears at the common cathodes, and no signal appears at the other grounded grid. The voltage across the 10k tail causes a current change which when added to the current in the second, or grounded grid triode, will equal the current change in the first or input triode. To get balance, the load of the first tube must be less than RL of the second. This makes the gains of the two tubes slightly different, and the 2H cannot perfectly cancel. But many brands use this old "cathode coupled differential pair", since its so simple and effective, and the gain of the first input grid with respect to its anode signal is about 1/2 the gain of one triode. Therefore the miller capacitance for the single input LTP is half what it otherwise would be. Patrick Turner. Chris Hornbeck |
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#14
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On Thu, 24 Mar 2005 21:12:36 +1100, Patrick Turner
wrote: Nope. Each device for each half can have very unmatched gM, which is why I have cited trioded 6AU6 and 12AU7. The LTP with a CCS **must** have equal current *change* in either device, since the sum of the cathode currents is zero. And transconductance times load equals gain, so for balanced loads, you must have balanced transconductances. Chris Hornbeck |
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#15
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"Chris Hornbeck" wrote
Slightly at cross purposes maybe. The LTP with a CCS **must** have equal current *change* in either device, since the sum of the cathode currents is zero. And transconductance times load equals gain, so for balanced loads, you must have balanced transconductances. So I can replace the valve with a resistor. As long as it has the same load resistor above it, and the cathode of the remaining valve below it. The LTP is just a folded concertina, after all. A wet carrot would work just as well. As long as the anode and output loads are perfectly matched at all frequencies and amplitudes, and the CCS has infinite resistance at all frequencies and amplitudes, and I don't need two inputs, and the ECC82 doesn't mind the non-linear behaviour of the carrot, which is why they are rarely considered fit for audio these days. cheers, Ian |
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#16
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In article , Chris Hornbeck
wrote: On Thu, 24 Mar 2005 21:12:36 +1100, Patrick Turner wrote: Nope. Each device for each half can have very unmatched gM, which is why I have cited trioded 6AU6 and 12AU7. The LTP with a CCS **must** have equal current *change* in either device, since the sum of the cathode currents is zero. And transconductance times load equals gain, so for balanced loads, you must have balanced transconductances. In addition to the transconductance and load the grid drive to each tube also figures into the equation. The flaw in your reasoning is that you are assuming that the grid drive to each tube is the same. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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#17
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On Fri, 25 Mar 2005 02:26:46 GMT, "Ian Iveson"
wrote: And transconductance times load equals gain, so for balanced loads, you must have balanced transconductances. So I can replace the valve with a resistor. As long as it has the same load resistor above it, and the cathode of the remaining valve below it. The LTP is just a folded concertina, after all. A wet carrot would work just as well. Do you know the term "transconductance"? As long as the anode and output loads are perfectly matched at all frequencies and amplitudes, and the CCS has infinite resistance at all frequencies and amplitudes, and I don't need two inputs, and the ECC82 doesn't mind the non-linear behaviour of the carrot, which is why they are rarely considered fit for audio these days. Let me rephrase my statement for the deliberately obtuse: "For equal output levels and with equal output loads, gains must be equal, so transconductances must be equal." Jeez, guys, just look it up. It 101. Chris Hornbeck |
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#18
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#19
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In article , Chris Hornbeck
wrote: On Thu, 24 Mar 2005 20:39:22 -0600, (John Byrns) wrote: And transconductance times load equals gain, so for balanced loads, you must have balanced transconductances. In addition to the transconductance and load the grid drive to each tube also figures into the equation. The flaw in your reasoning is that you are assuming that the grid drive to each tube is the same. The "grid drive to each tube" *is* the same. That's why you use a CCS. Where are you getting the notion that a CCS forces the grid drive to the two tubes to be the same? The signal voltage on the cathode adjusts to whatever is necessary to maintain the total cathode current constant, the drive to each of the grids will be different if the transconductance of the two tubes differs. This is 101, guys. Quite true, you seem to need a review. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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#21
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Chris Hornbeck wrote: On Thu, 24 Mar 2005 21:12:36 +1100, Patrick Turner wrote: Nope. Each device for each half can have very unmatched gM, which is why I have cited trioded 6AU6 and 12AU7. The LTP with a CCS **must** have equal current *change* in either device, since the sum of the cathode currents is zero. And transconductance times load equals gain, so for balanced loads, you must have balanced transconductances. I still think you are confused. for triodes, gain isn't merely Gm x RL. Gain = U x RL / ( RL + Ra ). This formula applies to all tubes, including pentodes, and where Ra is so much higher than RL that it can be ignored for approximate gain calculations, and so gain = gm x RL for a pentode. U = amplification factor = gm x Ra. One has to be aware of the 3 items involved with any tube, its gm, Ra, and U. As I said in a post just before this one, the gains of each half of the LTP with CCS and equal RL isn't the same with tubes of different U, gm, or Ra. The gain is the output voltage divided by the voltage between the grid and cathode. With two very different tube types in an LTP the vgk is very different, but the load output voltage amplitudes remain quite equal. If you still don't believe me, try making a circuit with 12AT7 in one side of the LTP and 12AU7 in the other. Use a CCS tail and equal RLs. You need to maybe place a resistor and bypass cap between the 12AU7 and CCS because the cathode bias required by the 12AU7 will be about twice that needed by the 12AT7. Once set up with equal idle current in each 1/2 of the LTP, you can measure the circuit easily with 1 kHz signal. I don't want to prove mathematically what I know is true, but I am sure you could do it for yourself if you are able to apply the above gain formula for a triode. Patrick Turner. Chris Hornbeck |
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#22
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On Fri, 25 Mar 2005 17:37:59 +1100, Patrick Turner
wrote: I still think you are confused. for triodes, gain isn't merely Gm x RL. Gain = U x RL / ( RL + Ra ). This formula applies to all tubes, including pentodes, and where Ra is so much higher than RL that it can be ignored for approximate gain calculations, and so gain = gm x RL for a pentode. U = amplification factor = gm x Ra. One has to be aware of the 3 items involved with any tube, its gm, Ra, and U. You've almost but not completely worn me out. Your case is correct with an infinite plate resistance or a zero external load resistance. Otherwise no. Please do the math. I'm tired and cranky and gone. Bye. Chris Hornbeck |
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#23
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Ian Iveson wrote: "Chris Hornbeck" wrote Slightly at cross purposes maybe. The LTP with a CCS **must** have equal current *change* in either device, since the sum of the cathode currents is zero. And transconductance times load equals gain, so for balanced loads, you must have balanced transconductances. So I can replace the valve with a resistor. As long as it has the same load resistor above it, and the cathode of the remaining valve below it. The LTP is just a folded concertina, after all. A wet carrot would work just as well. As long as the anode and output loads are perfectly matched at all frequencies and amplitudes, and the CCS has infinite resistance at all frequencies and amplitudes, and I don't need two inputs, and the ECC82 doesn't mind the non-linear behaviour of the carrot, which is why they are rarely considered fit for audio these days. cheers, Ian Unfortunately, using a resistor or wet carrot in place of a triode in an LTP with equal RL and CCS tail will NOT give equal amplitude but oppositely phased outputs. A resistor or carrot has no inverting amplifier ability, something needed to the LTP to work as an LTP. If you think you have a suitable mathematical proof to proove a resistor works, then proceed to educate me further. Patrick Turner. |
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#24
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In article , Chris Hornbeck
wrote: On Thu, 24 Mar 2005 21:48:37 -0600, (John Byrns) wrote: The "grid drive to each tube" *is* the same. That's why you use a CCS. Where are you getting the notion that a CCS forces the grid drive to the two tubes to be the same? The signal voltage on the cathode adjusts to whatever is necessary to maintain the total cathode current constant, the drive to each of the grids will be different if the transconductance of the two tubes differs. Exactly right. But gain is the product of Gm and load resistance. Load resistance is plate resistance in parallel with external loading. Are you positing a pair of devices with different Gm and identical plate resistance? No, I am positing that each of the two tubes have differing transconductance and plate resistance, I am assuming only that the load resistors for the two tubes are equal. In that case, and only in that case, equal outputs happen. In all other cases, outputs are imbalanced. A special case shouldn't be used in a sweeping statement. I'm not sure why you are calling this a special case, it seems pretty general to me? I'm sorry for my sharp tone earlier. It was uncalled for. Please forgive me. If there was something that needed forgiveness, you are forgiven. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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#25
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Patrick Turner wrote: Ian Iveson wrote: "Chris Hornbeck" wrote Slightly at cross purposes maybe. The LTP with a CCS **must** have equal current *change* in either device, since the sum of the cathode currents is zero. And transconductance times load equals gain, so for balanced loads, you must have balanced transconductances. So I can replace the valve with a resistor. As long as it has the same load resistor above it, and the cathode of the remaining valve below it. The LTP is just a folded concertina, after all. A wet carrot would work just as well. As long as the anode and output loads are perfectly matched at all frequencies and amplitudes, and the CCS has infinite resistance at all frequencies and amplitudes, and I don't need two inputs, and the ECC82 doesn't mind the non-linear behaviour of the carrot, which is why they are rarely considered fit for audio these days. cheers, Ian Unfortunately, using a resistor or wet carrot in place of a triode in an LTP with equal RL and CCS tail will NOT give equal amplitude but oppositely phased outputs. A resistor or carrot has no inverting amplifier ability, something needed to the LTP to work as an LTP. If you think you have a suitable mathematical proof to proove a resistor works, then proceed to educate me further. Patrick Turner. I need to reply to myself, since not all of what Ian said was wrong. Consider the use of a single 1/2 of an 6SN7 triode with RL = 50k, IaQ = 2 mA, and cathode taken to CCS tail with 4 mA. Then we could have a carrot with 100k resistance, and another RL of 50k to the B+, in this case let it be +300v, so with 2 mA in each RL the anode and 50k-100k junction are both sitting on +200v. If there is a -50v signal at the 50k-100k junction, there is a signal current of 1 ma in the 50k, so at the cathode of the tube, bottom of the 100k carrot, there must be a -150v signal. Then since there is no current change in the CCS, there must be a 1 mA change in the tube, and the anode voltage change is +50v. The load seen by the tube is 50k plus 150k effectively, so gain is say 19 for a 6SN7, so the Vgk needed to produce the 200v change across the tube is 10.52 v, so the whole grid input voltage to give both the +/-50v output voltages at each 50k is -160.52v. If the 100k was bypassed with a large electro, then the situation is quite different, but then similar to a concertina phase inverter, because the cathode load has been moved so it is working with a differently biased load, but a load equal to the anode load of this lone triode. So then the circuit is a true concertina phase inverter, which needs about -55.2 v of input to make +50v at the anode, and -50v at the "carrot" output. It is not in any way able to work as a differential amplifier. Patrick Turner. |
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#26
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Chris Hornbeck wrote: On Fri, 25 Mar 2005 17:37:59 +1100, Patrick Turner wrote: I still think you are confused. for triodes, gain isn't merely Gm x RL. Gain = U x RL / ( RL + Ra ). This formula applies to all tubes, including pentodes, and where Ra is so much higher than RL that it can be ignored for approximate gain calculations, and so gain = gm x RL for a pentode. U = amplification factor = gm x Ra. One has to be aware of the 3 items involved with any tube, its gm, Ra, and U. You've almost but not completely worn me out. Your case is correct with an infinite plate resistance or a zero external load resistance. Otherwise no. Please do the math. I have done the math. And I have done the practical experiment that prooves I am right. I'm tired and cranky and gone. Bye. When you return, you'll be fresh and wide awake, and i hope you find some time to do a little experimenting in your workshop. Patrick Turner. Chris Hornbeck |
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#27
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"Chris Hornbeck" wrote
Let me rephrase my statement for the deliberately obtuse: "For equal output levels and with equal output loads, gains must be equal, so transconductances must be equal." Don't pick on me, I'm just taking the ****. Sarcastic irony never really took off in the colonies... I was trying to support you in your struggle against the forces of dimwit. But it is true anyway, as far as it goes. Equal transconwotsit isn't necessary for *balance*, as long as everything else is equal. A resistor would work, as far as balance goes. In every other way it would be a poor substitute. An open circuit would give a balanced output of plus and minus zero, as would a short. Anything between would give a non-zero, balanced output. Best match would presumably be a resistor of value equal to the static resistance of the valve, to give equal static currents. An active device would also work. After all, The LTP with a CCS **must** have equal current *change* in either device, since the sum of the cathode currents is zero. So a pentagrid octode with five foreign signals would be fine, and equal to the carrot in every respect as far as balance goes. You would still get the same signal from both outputs. Just wouldn't be the one you put in, that's all. Every "*must*" begs for a reductio ad absurdam. Patrick Turner sold the worst design of LTP in the world, ever. What's missing here is that the condition of "everything else equal" can't be met, especially if you are driving an output stage. That's why it is usual to design stages with a low output impedance compared to the input of the following stage. Equal, linear loads can't be relied upon. It is also why two LTPs in cascade is fashionable I think. If one is used for driving a significant load, then it should have balanced input signals. Another used to drive the first will see an easy load, and so can be driven with unequal signals. Matching *is* important for linearity and power supply rejection. Do you know the term "transconductance"? er...I'll look it up when I check "obtuse". Jeez, guys, just look it up. It 101. What, transconthingumy, or obtuse? cheers, Ian |
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#28
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"John Byrns" wrote
...No, I am positing that each of the two tubes have differing transconductance and plate resistance, I am assuming only that the load resistors for the two tubes are equal.... No, you are also assuming that the driven loads are equal, because they are in in parallel with the load resistors. cheers, Ian |
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#29
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Ian Iveson wrote: "Chris Hornbeck" wrote Let me rephrase my statement for the deliberately obtuse: "For equal output levels and with equal output loads, gains must be equal, so transconductances must be equal." Don't pick on me, I'm just taking the ****. Sarcastic irony never really took off in the colonies... I was trying to support you in your struggle against the forces of dimwit. But it is true anyway, as far as it goes. Equal transconwotsit isn't necessary for *balance*, as long as everything else is equal. A resistor would work, as far as balance goes. In every other way it would be a poor substitute. An open circuit would give a balanced output of plus and minus zero, as would a short. Anything between would give a non-zero, balanced output. Best match would presumably be a resistor of value equal to the static resistance of the valve, to give equal static currents. An active device would also work. After all, The LTP with a CCS **must** have equal current *change* in either device, since the sum of the cathode currents is zero. So a pentagrid octode with five foreign signals would be fine, and equal to the carrot in every respect as far as balance goes. You would still get the same signal from both outputs. Just wouldn't be the one you put in, that's all. Every "*must*" begs for a reductio ad absurdam. Patrick Turner sold the worst design of LTP in the world, ever. What's missing here is that the condition of "everything else equal" can't be met, especially if you are driving an output stage. In defense of my name, and the allegedly frightful LTP associated with it, let me say that we have your alegations duly noted, and will report them to the Privvy Council and House of Lords in due course. We will be countering with a claim that on Gord Frayday, Marche, 2005, thou didst endeavour to unconscioubaly use the news group as an innappropriate selling venue for goods not approved under the Fruit and Vegetable Act, 1907, namely, that you tried to sell damp carrots, and well under the market price, and with the accompanying grossly fraudelent claim that they could be used as replacements for electronic components, thus perhaps endangering lives and contrary to the Safety Laws of Her Majestie's Realms, ie, the Bwitish Isles and Her Colonies. Patrick Turnip. That's why it is usual to design stages with a low output impedance compared to the input of the following stage. Equal, linear loads can't be relied upon. It is also why two LTPs in cascade is fashionable I think. If one is used for driving a significant load, then it should have balanced input signals. Another used to drive the first will see an easy load, and so can be driven with unequal signals. Matching *is* important for linearity and power supply rejection. Do you know the term "transconductance"? er...I'll look it up when I check "obtuse". Jeez, guys, just look it up. It 101. What, transconthingumy, or obtuse? cheers, Ian |
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#30
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For the real deal look at
http://www.bonavolta.ch/hobby/en/audio/split.htm JLS See ABSE for a differential pair implemented with dissimilar tubes. As long as the cathode return for the LTP is a real current source & the tubes are not overloaded the outputs will be equal for equal loads. Problems arise as the weaker tube runs into overload. Real problems with drift in a DC amplifier. JLS |
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#31
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In article , wrote:
For the real deal look at http://www.bonavolta.ch/hobby/en/audio/split.htm JLS See ABSE for a differential pair implemented with dissimilar tubes. As long as the cathode return for the LTP is a real current source & the tubes are not overloaded the outputs will be equal for equal loads. Problems arise as the weaker tube runs into overload. Real problems with drift in a DC amplifier. John, It might help some of the disbelievers understand how this circuit actually works if you included the signal voltage at the common cathode node so that they could observe the unequal grid drive. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ |
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#32
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"John Stewart" wrote in message ... For the real deal look at http://www.bonavolta.ch/hobby/en/audio/split.htm JLS See ABSE for a differential pair implemented with dissimilar tubes. As long as the cathode return for the LTP is a real current source & the tubes are not overloaded the outputs will be equal for equal loads. Problems arise as the weaker tube runs into overload. Real problems with drift in a DC amplifier. JLS If the current source is perfect, the 6SN7 can be replaced by a cathode-anode short and the output will still be perfectly balanced (AC wise, of course), will it not? |
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#33
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#34
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John Byrns wrote:
In article , wrote: For the real deal look at http://www.bonavolta.ch/hobby/en/audio/split.htm JLS See ABSE for a differential pair implemented with dissimilar tubes. As long as the cathode return for the LTP is a real current source & the tubes are not overloaded the outputs will be equal for equal loads. Problems arise as the weaker tube runs into overload. Real problems with drift in a DC amplifier. John, It might help some of the disbelievers understand how this circuit actually works if you included the signal voltage at the common cathode node so that they could observe the unequal grid drive. Regards, John Byrns Surf my web pages at, http://users.rcn.com/jbyrns/ Over at ABSE. I made the VM at the cathodes 100 Megs so things stayed more or less CC. Interestingly, if you switch the tube positions the gain to each plate becomes 1.508. If both triodes are 6SN7's the gain to each plate is 7.2. For both as 6SL7 the gain to each plate is 16. Cheers, John |
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#35
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BFoelsch wrote: "John Stewart" wrote in message ... For the real deal look at http://www.bonavolta.ch/hobby/en/audio/split.htm JLS See ABSE for a differential pair implemented with dissimilar tubes. As long as the cathode return for the LTP is a real current source & the tubes are not overloaded the outputs will be equal for equal loads. Problems arise as the weaker tube runs into overload. Real problems with drift in a DC amplifier. JLS If the current source is perfect, the 6SN7 can be replaced by a cathode-anode short and the output will still be perfectly balanced (AC wise, of course), will it not? An interesting concept. The current in the tubes follow the 3/2 power law, while a short (piece) of wire would be linear, I guess. That should make for more discussion. Cheers, John Stewart |
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John Stewart wrote:
BFoelsch wrote: "John Stewart" wrote in message ... For the real deal look at http://www.bonavolta.ch/hobby/en/audio/split.htm JLS See ABSE for a differential pair implemented with dissimilar tubes. As long as the cathode return for the LTP is a real current source & the tubes are not overloaded the outputs will be equal for equal loads. Problems arise as the weaker tube runs into overload. Real problems with drift in a DC amplifier. JLS If the current source is perfect, the 6SN7 can be replaced by a cathode-anode short and the output will still be perfectly balanced (AC wise, of course), will it not? An interesting concept. The current in the tubes follow the 3/2 power law, while a short (piece) of wire would be linear, I guess. That should make for more discussion. Cheers, John Stewart Thinking about that a little further, I believe there would be no gain at all. The current in the 6SL7 cathode circuit is held constant, as though connected to a very large value resistor. So I guess you get massive cathode degeneration & no output at the plate at all. JLS |
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On Fri, 25 Mar 2005 13:47:12 GMT, "Ian Iveson"
wrote: Don't pick on me, I'm just taking the ****. Sarcastic irony never really took off in the colonies... Please forgive me for any rudeness, which was certainly uncalled for. And anyways unforgivable. But please try. We're "two peoples divided by a common language." Churchill? He said most of the really cool things that W.S. was too dead for. Or was it Wilde or Noel Coward? Dead cert it wasn't Flanders and Swann, or I'd quote 'em. arguments snipped for bandwidth If I'm following you, and possibly John also, correctly your argument is based on the perfect CCS's ability to maintain a constant *sum* of the two plate currents. The CCS magic extends to proportionately varying each (active transconductance) device's actual, real, measurable grid-to-cathode-or-whatever-semiconductor-appendage-applies voltage in the magically correct amount as to send balanced, meaning equal amplitude, opposite polarity CURRENTS to the device's output. All good. And: When we think about what "the device's output" really means, in the context of a transconductance device, the shouting will die down some. I'm sorry to be so bloody dogmatic about such a trivial thing, but the newsgroup is lately overrun with terrible Through the Looking Glass ideas about all kinds of things. Fight the Wackiness! Chris Hornbeck |
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John Stewart wrote: John Stewart wrote: BFoelsch wrote: "John Stewart" wrote in message ... For the real deal look at http://www.bonavolta.ch/hobby/en/audio/split.htm JLS See ABSE for a differential pair implemented with dissimilar tubes. As long as the cathode return for the LTP is a real current source & the tubes are not overloaded the outputs will be equal for equal loads. Problems arise as the weaker tube runs into overload. Real problems with drift in a DC amplifier. JLS If the current source is perfect, the 6SN7 can be replaced by a cathode-anode short and the output will still be perfectly balanced (AC wise, of course), will it not? An interesting concept. The current in the tubes follow the 3/2 power law, while a short (piece) of wire would be linear, I guess. That should make for more discussion. Cheers, John Stewart Thinking about that a little further, I believe there would be no gain at all. The current in the 6SL7 cathode circuit is held constant, as though connected to a very large value resistor. So I guess you get massive cathode degeneration & no output at the plate at all. JLS Actually, with only 5 ma in the CC source, all of that would run thru the 27K that was in the 6SN7 plate circuit & pull the 6SL7 cathode up so that the 6SL7 would be cut off. To get the 6SL7 turned on you would need something like 11 ma from the CC source to get the 6SL7 biased on. That would get it's cathode down to +3 volts so conduction could begin. JLS |
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"Chris Hornbeck" wrote
...To hammer the issue into the ground, gain is the product of transconductance and effective load resistance. There is no tap-dancing around this fundamental fact.... No-one has said the gain is equal. It doesn't have to be, for balance. The valves are both in closed-loop feedback circuits. What happens to the valve in order to lead to a balanced output into equal loads is neither here nor there. Ohms law trumps all: equal currents into equal loads results in equal output voltages. Full stop. So you have to fit what you know about transconductance around that fact, not vice versa. Ohms law is not tap dancing. At least one John has unravelled the conundrum, potentially. The CCS works by varying its output voltage in order to achieve the correct constant current. So the voltage at the cathodes varies until both valves pass the same current. This is possible because the valve receiving the signal at its grid perceives the consequent cathode voltage as negative feedback. One thing we have missed out, BTW, is the capacitance to the grids. Current lost to or gained from the grids from or to anodes and cathodes is not accounted for so far. cheers, Ian |
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On Sat, 26 Mar 2005 06:01:31 GMT, "Ian Iveson"
wrote: All OK up to he So the voltage at the cathodes varies until both valves pass the same current. This is possible because the valve receiving the signal at its grid perceives the consequent cathode voltage as negative feedback. You're either yanking my chain, or your model is flawed. Although inclined to give you the benefit of the doubt and believe the former, I'll respond to the latter. Don't overthink this. It's not anything new and it's not even interesting. The division of voltages across the diff-pair of transconductance devices is purely algebraic and inversely proportional to their transconductances. Period. No magic, no feedback. It's a simple voltage divider. A voltage appears across a pair of series'd transconductance devices' inputs. The voltage is divided between the pair in inverse linear proportion to their transconductances. Each device's share of the voltage generates an output current that is the product of its share of the voltage and its transconductance. The gain of each device is the simple linear product of its transconductance and its load impedance. Guys, there just isn't any more to it than this. No handwaving is necessary or allowed. Chris Hornbeck "Excuse me, since when is getting paid for it not USING the property?" -Bob Olhsson |
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