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#41
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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 Ian Iverson made the assertion that one tube could be replaced by a resistor equal to the static resistance of one triode. He suggested a damp carrot be used, and still you'd get equal inverted output, but gain of about 0.35 from input to each anode.. I did the math on the circuit arrangement on the follow up post. If the "damp carrot" resistance is well bypassed by a an electro cap, the circuit works as a normal CPI, with the same gain. I cannot think of any use for such a creature as a Turnip-Carrot PI. Patrick Turnip. |
#42
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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. Here is a copy of my reply to Ian :- ---------------------------- 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. ----------------------------------- JLS |
#43
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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 Something to ponder. Yet another phase splitter at ABSE. JLS |
#44
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Not of any interest on RAT of course, but the early McIntosh SS power
amplifier, the MC2100, used this technique. A high-compliance current source (3K6 with 80 V) in the collector circuit of a BJT. You can argue all you want about gain and transconductance, KCL is hard to overcome. "John Stewart" wrote in message ... 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 Something to ponder. Yet another phase splitter at ABSE. JLS |
#45
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"Chris Hornbeck" wrote in message ... 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. So, if the division of voltages is inversely proportional to transconductance, and gain is directly proportional to transconductance, then the transconductance drops out, no? |
#46
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"Chris Hornbeck" wrote
All OK up to he Cool. So you agree that unmatched valves produce equal output voltages in a differential pair with matched loads and a CCS at its tail? 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. Whatever takes your fancy. But if you put these two paragraphs together... 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. ....it amounts to... So the voltage at the cathodes varies until both valves pass the same current. That is, the voltage is divided such that currents are equal. What I said explains how that happens. 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. Now you are wandering off track again. Perhaps you could explain, for example, how come there is no feedback? The gain of each device is the simple linear product of its transconductance and its load impedance. Open loop gain, yes. But they are not open loop. Guys, there just isn't any more to it than this. No handwaving is necessary or allowed. Er, you are the one waving hands. You have made no reference to the specific circuit we are discussing, but just repeat fundamentals without explaining how they apply in this case. If you do try and explain in terms of the actual circuit, you may find that we agree. But not if your fundamentalism denies the simple truth about balance. cheers, Ian |
#47
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On Sat, 26 Mar 2005 10:26:06 -0500, "BFoelsch"
wrote: So, if the division of voltages is inversely proportional to transconductance, and gain is directly proportional to transconductance, then the transconductance drops out, no? Yes, for the current outputs of the transconductance devices. In special cases, including matched output loading, infinite internal impedance and zero external impedance, mismatched transconductances can give matched voltage outputs. But these are special cases, not a general rule. Chris Hornbeck "Excuse me, since when is getting paid for it not USING the property?" -Bob Olhsson |
#48
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On Sat, 26 Mar 2005 15:39:43 GMT, "Ian Iveson"
wrote: Now you are wandering off track again. Perhaps you could explain, for example, how come there is no feedback? What feedback? You *are* yanking my chain, aren't you? Chris Hornbeck "Excuse me, since when is getting paid for it not USING the property?" -Bob Olhsson |
#49
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"Patrick Turner" wrote
Ian Iverson made the assertion that one tube could be replaced by a resistor equal to the static resistance of one triode. He suggested a damp carrot be used, and still you'd get equal inverted output, but gain of about 0.35 from input to each anode.. I did the math on the circuit arrangement on the follow up post. If the "damp carrot" resistance is well bypassed by a an electro cap, the circuit works as a normal CPI, with the same gain. I cannot think of any use for such a creature as a Turnip-Carrot PI. I V E S O N Two syllables, please. http://www.ivesonaudio.pwp.blueyonder.co.uk/carrot.GIF cheers, Ian |
#50
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ps figures are static values, graphs are both with 1V pk signal at
1k. zoom to actual size and don't complain you can't read it |
#51
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"Chris Hornbeck" wrote
What feedback? You *are* yanking my chain, aren't you? Well, at least I know what you're missing now. Signal input to triode is Vgk. Vk varies in the case of the LTP. cheers, Ian |
#52
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pps carrots are actually quite dangerous in HV circuits so please
don't try this at home. Morrisons carrots, with slight trace of broccoli, have poor conductance but they might be dead because one of them is a bit furry. I bought them because they were on special offer with the broccoli, which I ate. Difference in cooking time would have meant cutting the carrots into little strips and I couldn't be arsed so they were abandoned in the bottom of the fridge. Anyway, 95k required a slice 1cm diameter 2mm thick sandwiched between two 1p pieces, previously cleaned with brown sauce and rinsed. Carrot is not a well-behaved load, taking a second or two to settle, like a big cap and resistor in series. Possibly characteristics are similar to a cap with polarising dielectric. Maybe it's a semiconductor. Perhaps a grid could be threaded between the anode and cathode? The other problem is long-term stability. A bonsai carrot might be OK. Wouldn't need replacing, just an occasional trim and watering. Multi-terminal carrots can be grown if you put some stones in the soil. cheers, Ian |
#53
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On Sat, 26 Mar 2005 20:26:08 GMT, "Ian Iveson"
wrote: "Chris Hornbeck" wrote What feedback? You *are* yanking my chain, aren't you? Well, at least I know what you're missing now. Signal input to triode is Vgk. Vk varies in the case of the LTP. A valve's cathode looks to the outside world like a resistor equal to the reciprocal of its transconductance. It has all the characteristics of a real resistor, both from the viewpoint of an imaginary idealized tranconductance "deamon" inside the valve and from the viewpoint of the external world. It even has the noise characteristics of a real resistor of appropriate temperature. IOW, each valve sees each other's cathode as a resistor, and neither sees the (idealized) CCS. It's too big. The resistors make a simple voltage divider. There is *no* feedback involved. To put it another way, if the grounded grid valve were replaced by a resistor equal to 1/Gm the signal voltage appearing at the cathodes/CCS junction would not change. There is *no* feedback involved. And there is no such thing as "internal feedback in triodes" either. That's whack. Chris Hornbeck |
#54
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So you have to fit what you know about transconductance around that fact, not vice versa. Ohms law is not tap dancing. Isn't the key to the scheme is that the current from the constant current source (sink actually) is split between the two triodes? One triode has the input signal applied to its grid, the other triode's grid connected to a non varying bias voltage (say ground). Adjust the input signal on the first triode and make it draw say 1ma more current from the CCS. But the only way that can happen is if the other triode decreases its current by 1ma. Use the same value resistors in both plate circuits, and you'd see inversion from one plate to the other. 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. A few picofarads. Not of significance at audio frequencies. |
#55
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"robert casey" wrote
Isn't the key to the scheme is that the current from the constant current source (sink actually) is split between the two triodes? One triode has the input signal applied to its grid, the other triode's grid connected to a non varying bias voltage (say ground). Adjust the input signal on the first triode and make it draw say 1ma more current from the CCS. But the only way that can happen is if the other triode decreases its current by 1ma. Use the same value resistors in both plate circuits, and you'd see inversion from one plate to the other. Yup, we are all but one agreed on this. That's why I said ohm's law isn't tap dancing. ...Not of significance at audio frequencies. Famous last words! You are probably right. Even so, in the presence of current source/sink there is a need to be careful because of high impedance. A few tens of pF can become significant. cheers, Ian |
#56
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"Chris Hornbeck" wrote
[below in full] You are not applying yourself to the matter in hand, still. You are boxing with strange shadows, setting up cardboard villains and chopping them down. I don't like to watch. The feedback halves the gain. I've posted a picture. Others have posted circuits. Either make your refutation clear and specific or please stop pestering. cheers, Ian in message ... On Sat, 26 Mar 2005 20:26:08 GMT, "Ian Iveson" wrote: "Chris Hornbeck" wrote What feedback? You *are* yanking my chain, aren't you? Well, at least I know what you're missing now. Signal input to triode is Vgk. Vk varies in the case of the LTP. A valve's cathode looks to the outside world like a resistor equal to the reciprocal of its transconductance. It has all the characteristics of a real resistor, both from the viewpoint of an imaginary idealized tranconductance "deamon" inside the valve and from the viewpoint of the external world. It even has the noise characteristics of a real resistor of appropriate temperature. IOW, each valve sees each other's cathode as a resistor, and neither sees the (idealized) CCS. It's too big. The resistors make a simple voltage divider. There is *no* feedback involved. To put it another way, if the grounded grid valve were replaced by a resistor equal to 1/Gm the signal voltage appearing at the cathodes/CCS junction would not change. There is *no* feedback involved. And there is no such thing as "internal feedback in triodes" either. That's whack. Chris Hornbeck |
#57
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On Sun, 27 Mar 2005 03:06:33 GMT, "Ian Iveson"
wrote: ...Not of significance at audio frequencies. Famous last words! You are probably right. Even so, in the presence of current source/sink there is a need to be careful because of high impedance. A few tens of pF can become significant. The impedance at the cathode of a valve is the reciprocal of its transconductance, (typically a coupla K ohms -ish). Until you decide to believe this, I'll shut up. Thanks, Chris Hornbeck |
#58
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Ian Iveson wrote: "Patrick Turner" wrote Ian Iverson made the assertion that one tube could be replaced by a resistor equal to the static resistance of one triode. He suggested a damp carrot be used, and still you'd get equal inverted output, but gain of about 0.35 from input to each anode.. I did the math on the circuit arrangement on the follow up post. If the "damp carrot" resistance is well bypassed by a an electro cap, the circuit works as a normal CPI, with the same gain. I cannot think of any use for such a creature as a Turnip-Carrot PI. I V E S O N Sory. Seems I put in an 'r' where it wasn't needed Don't let it wurie you too much. Two syllables, please. http://www.ivesonaudio.pwp.blueyonder.co.uk/carrot.GIF So this circuit also confirms that if CCS and equal RL are used, and gM, Ra and u are both quite differnt, you get equual VO? Patrick Turner. cheers, Ian |
#59
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Chris Hornbeck wrote: On Sat, 26 Mar 2005 20:26:08 GMT, "Ian Iveson" wrote: "Chris Hornbeck" wrote What feedback? You *are* yanking my chain, aren't you? Well, at least I know what you're missing now. Signal input to triode is Vgk. Vk varies in the case of the LTP. A valve's cathode looks to the outside world like a resistor equal to the reciprocal of its transconductance. Yes true, but only for the case of a CF, where Rkin = 1/gm For where there is a load between the anode and B+, Rkin = RL/A, where A is the gain of the tube with that RL. And if there is a cathode resistor, values of Rkin have to be adjusted for the parallel R. It has all the characteristics of a real resistor, both from the viewpoint of an imaginary idealized tranconductance "deamon" inside the valve and from the viewpoint of the external world. It even has the noise characteristics of a real resistor of appropriate temperature. IOW, each valve sees each other's cathode as a resistor, and neither sees the (idealized) CCS. It's too big. The resistors make a simple voltage divider. There is *no* feedback involved. Hmm, but in an LTP, each tube has mutual negative current, since the cathode of one tube is connected to the Rkin of the other. So if Dn occurs in one tube it also must occur inverted at its cathode, and be reduced by series current NFB. The output impedance from one anode of the LTP is less than the output resistance of the same tube in a one tube common cathode circuit with full cathode bypassing. But where the loads imposed on both the LTP anodes are the same, they act as if the Ro was the same as a single tube in common cathode mode. To put it another way, if the grounded grid valve were replaced by a resistor equal to 1/Gm the signal voltage appearing at the cathodes/CCS junction would not change. There is *no* feedback involved. And there is no such thing as "internal feedback in triodes" either. That's whack. NFB in triodes was defined by professor Child in Terman's Radio Engineering in 1937. I wouldn't dare say you are wrong, and would only ever wonder what you would think about what the Professor Child wrote, all those years ago. Patrick Turner. Chris Hornbeck |
#60
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"Patrick Turner" wrote
I V E S O N Sory. Seems I put in an 'r' where it wasn't needed Don't let it wurie you too much. There has been a tendency for the runts of the family to add a syllable. Several epochs past, Ive the French butcher became entangled with a Scandinavian girl about 30 miles from here in Wensleydale, and was never allowed to leave. There was never an Iver, AFAIK. http://www.ivesonaudio.pwp.blueyonder.co.uk/carrot.GIF So this circuit also confirms that if CCS and equal RL are used, and gM, Ra and u are both quite differnt, you get equual VO? Hope so. It is also the circuit you were just describing, so you can compare. Don't know where you got the 0.35 from but it's about right. Incidentally, a signal to the resistor side's output results in a gain of exactly 0.5 at the other output. The value of the resistor doesn't matter for balance until it is so small that the valve turns off. Then the circuit works via the interelectrode capacitance to give small, in-phase and unequal outputs. The electrical life of a carrot is quite interesting. Resistance seems to fall as it ages at first, as does its apparent capacitance. Then the resistance rises again but the capacitance keeps falling. Throughout this, the LTP remains balanced. Anyway, more interesting questions are about how the output impedances behave, and how it reacts to unequal loads and power supply hum. cheers, Ian |
#61
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On Sat, 26 Mar 2005 20:13:22 GMT, "Ian Iveson"
wrote: http://www.ivesonaudio.pwp.blueyonder.co.uk/carrot.GIF Ian, I hadn't seen this before now. You're right of course and I was wrong. Thanks for the correction, and your patience. Chris Hornbeck |
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