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#1
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Cathode Follower
I have a Super It Phono Preamp that has plenty of gain (70db). This is more
than enough to drive my UL P/P amp. My question is how do I add a Cathode Follower to the IT? Are there any good circuits out there that I can copy? Thanks for your opinions in advance. Cordially, west |
#2
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west wrote:
I have a Super It Phono Preamp that has plenty of gain (70db). This is more than enough to drive my UL P/P amp. My question is how do I add a Cathode Follower to the IT? Are there any good circuits out there that I can copy? Do you really need it? If the preamp drives the amp with room to spare, then there's no real need. A cathode follower has slightly less than unity voltage gain, but its output impedance is lower than most other tube circuit outputs. If you have a lack of bass, a larger coupling cap would help that. We will need more discussion. |
#3
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Robert Casey wrote: west wrote: I have a Super It Phono Preamp that has plenty of gain (70db). This is more than enough to drive my UL P/P amp. My question is how do I add a Cathode Follower to the IT? Are there any good circuits out there that I can copy? Do you really need it? If the preamp drives the amp with room to spare, then there's no real need. A cathode follower has slightly less than unity voltage gain, but its output impedance is lower than most other tube circuit outputs. If you have a lack of bass, a larger coupling cap would help that. We will need more discussion. The only reason to add a CF would be to reduce the Ro of the preamp. We are not told about the existing circuit, yet asked for an opinion. I'd be a good idea if he had quite long cables between preamp and power amp. Patrick Turner. |
#4
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A common rule of thumb says that the input impedance of a stage
should be no less than 10 times the output impedance of the preceding stage. Hence if your valve power amplifier has the typical 100k or more input impedance, then it is OK to drive it with 10k. A cathode follower would not be necessary. If, OTOH, you were driving an SS power amp with 10k input, you would need 1k out of the preamp, for which a CF or transformer would be necessary. Have you thought of using a transformer, BTW? A natural way to go because it trades gain for impedance. Why do long cables make a difference, as Patrick says? I guess you should consider a long cable as a stage in itself. As an extreme example, if the cable itself has an input impedance of 10k, then the preamp output needs to be 1k or less, no matter what the input of the power amplifier is. cheers, Ian "west" wrote in message . .. I have a Super It Phono Preamp that has plenty of gain (70db). This is more than enough to drive my UL P/P amp. My question is how do I add a Cathode Follower to the IT? Are there any good circuits out there that I can copy? Thanks for your opinions in advance. Cordially, west |
#5
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west wrote:
I have a Super It Phono Preamp that has plenty of gain (70db). This is more than enough to drive my UL P/P amp. My question is how do I add a Cathode Follower to the IT? Are there any good circuits out there that I can copy? Thanks for your opinions in advance. Cordially, west In order to drive a long cable properly you will need a White Cathode Follower. That will drive the cable in both +ve & -ve directions. An ordinary CF is good only for +ve going changes unless its cathode resistor is chosen quite low resistance. That results in very low gain Cheers, John Stewart |
#6
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IAN IVESON wrote:
A common rule of thumb says that the input impedance of a stage should be no less than 10 times the output impedance of the preceding stage. Hence if your valve power amplifier has the typical 100k or more input impedance, then it is OK to drive it with 10k. A cathode follower would not be necessary. If, OTOH, you were driving an SS power amp with 10k input, you would need 1k out of the preamp, for which a CF or transformer would be necessary. Have you thought of using a transformer, BTW? A natural way to go because it trades gain for impedance. Why do long cables make a difference, as Patrick says? I guess you should consider a long cable as a stage in itself. As an extreme example, if the cable itself has an input impedance of 10k, then the preamp output needs to be 1k or less, no matter what the input of the power amplifier is. cheers, Ian "west" wrote in message . .. I have a Super It Phono Preamp that has plenty of gain (70db). This is more than enough to drive my UL P/P amp. My question is how do I add a Cathode Follower to the IT? Are there any good circuits out there that I can copy? Thanks for your opinions in advance. Cordially, west Good to see you back Ian. I thought you had stepped off the planet. Cheers, John Stewart |
#7
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nice to read ya
-- .................................................. ........................ Choky Prodanovic Aleksandar YU "don't use force, "don't use force, use a larger hammer" use a larger tube - Choky and IST" - ZM .................................................. ........................... "IAN IVESON" wrote in message . .. A common rule of thumb says that the input impedance of a stage |
#8
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IAN IVESON wrote: A common rule of thumb says that the input impedance of a stage should be no less than 10 times the output impedance of the preceding stage. Hence if your valve power amplifier has the typical 100k or more input impedance, then it is OK to drive it with 10k. A cathode follower would not be necessary. If, OTOH, you were driving an SS power amp with 10k input, you would need 1k out of the preamp, for which a CF or transformer would be necessary. Have you thought of using a transformer, BTW? A natural way to go because it trades gain for impedance. Why do long cables make a difference, as Patrick says? I guess you should consider a long cable as a stage in itself. As an extreme example, if the cable itself has an input impedance of 10k, then the preamp output needs to be 1k or less, no matter what the input of the power amplifier is. And the top of the morning to you Ian, you ain't been around for sooo long.... A friend has a pair of Leak 20s for a biamped system. His gutted and rebuilt-to-different-design Quad22 pre amp sits beside the TT and Denon cart, and there is about 6 metres of cable to the power amps, running under the floor, and up where the power amps sit. If the capacitance of the cable is 100pF per metre, he'd have 600 pF of shunt C just from the cable, and if the Ro of the preamp was 10k, it causes a pole at 26 kHz. but other losses in the amp would probably result in a final pole at 20 kHz. If the signal from the pre was from the anode circuit of a 6SN7 triode and the cathode R was unbypassed in the interest of sonic purity, and dislike for electro bypass caps, (even the good Oscon types,) and if the tube was a 6SN7, then the Ro would be maybe 25kohms, and the response would be far more dissappointing. But with a gain triode with Ro = 10k, it might not seem too bad, but I have always found that if the Ro was 10 times lower, the signal seemed to have more crispness and speed. Ah, but then we enter tha land of a cathode follower to get the Ro to less than 1k, and I already hear howls of dismay. Well, it depends how you do the follower. I myself find the use of a simple CCS load on the CF allows for a healthy current flow, but removes most of the R load that would other wise be there, and thus the only load the CF sees is the cap coupled R at the power amp. CF produce very low thd, about 1/15 of that produced by the same 6SN7 creating the same voltage in an anode circuit. With a CCS tail, there is an extra 10 dB thd reduction since the load the tube sees is so high, where the thd is so low for a triode. At a volt, 0.01% is possible with a CF..... The other benefit of the follower is that it acts as a buffer between tbe gain control pot which needs to be about 100k value to be driven by a preceding gain triode. The gain pot has maximum Ro of 25k, so you don't want much C at its output, and the CF has very low C at its input. Allen Wright has a couple of circuits for "super cathode followers", which use 3 triodes. See http://www.vacuumstate.com Whether they sound any better is a moot point. The use of a transformer is a nice idea, but it needs to be a real good one to reap the benefits. Say the Ro from 1/2 6SN7 was 10k, and you had a 4:1 step down transformer. This gives a 16:1 impedance transformation, so 10k is measured as 625 ohms at the secondary, then output Ro is as low as the CF, which relies on the NFB to transform the plate resistance of the triode. But then you have to produce 4 times the output voltage from the gain triode connected to the transformer, but this isn't difficult, ( just expensive.) The transformer adds distortion to the output signal which can be greater than the tube distortion, and mostly 3H. Having said that, transformer coupling can sound very nice. Patrick Turner. |
#9
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John Stewart wrote: west wrote: I have a Super It Phono Preamp that has plenty of gain (70db). This is more than enough to drive my UL P/P amp. My question is how do I add a Cathode Follower to the IT? Are there any good circuits out there that I can copy? Thanks for your opinions in advance. Cordially, west In order to drive a long cable properly you will need a White Cathode Follower. That will drive the cable in both +ve & -ve directions. An ordinary CF is good only for +ve going changes unless its cathode resistor is chosen quite low resistance. That results in very low gain Cheers, John Stewart An ordinary CF will display premature cut off when the cap couple load is a lowish value compared to the DC supply R. A CF also tends to suffer from slew distortion with a capacitance load because the discharge of the capacitance is due to the current flow via the DC load R, and the time constant for this is longer than the charge time constant when the tube is turned on. BUT, its not a big deal at preamp voltages and the slew distortion is only a bother when C is large, or at high voltage swings at HF. Usually a 6SN7 CF has a banwidth of maybe a Mhz at a volt, and with 600 pF, the slew distortion is negligible, and cut off won't occur until maybe 15 vrms of output..... I have tried two 1/2 triodes as used in the White, and its very nice, but the same two triodes paralleled will do almost the same job because there is more current flow, and more current to discharge the C... The White can be RF un-stable if you are not careful.... Patrick Turner. |
#10
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"John Stewart" wrote
Good to see you back Ian. I thought you had stepped off the planet. Thanks, John. Yes, I just disappeared. Phone cut off, no money, squabble, etc. Now I have a broadband cable connection, and a new ISP. Dunno why I'm in capital letters though. I'll change that when I find out how. Just now I feel disoriented coz of a change in OS to 'doze XP. cheers, Ian |
#11
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Patrick Turner wrote:
IAN IVESON wrote: A common rule of thumb says that the input impedance of a stage should be no less than 10 times the output impedance of the preceding stage. Hence if your valve power amplifier has the typical 100k or more input impedance, then it is OK to drive it with 10k. A cathode follower would not be necessary. If, OTOH, you were driving an SS power amp with 10k input, you would need 1k out of the preamp, for which a CF or transformer would be necessary. Have you thought of using a transformer, BTW? A natural way to go because it trades gain for impedance. Why do long cables make a difference, as Patrick says? I guess you should consider a long cable as a stage in itself. As an extreme example, if the cable itself has an input impedance of 10k, then the preamp output needs to be 1k or less, no matter what the input of the power amplifier is. And the top of the morning to you Ian, you ain't been around for sooo long.... A friend has a pair of Leak 20s for a biamped system. His gutted and rebuilt-to-different-design Quad22 pre amp sits beside the TT and Denon cart, and there is about 6 metres of cable to the power amps, running under the floor, and up where the power amps sit. If the capacitance of the cable is 100pF per metre, he'd have 600 pF of shunt C just from the cable, and if the Ro of the preamp was 10k, it causes a pole at 26 kHz. but other losses in the amp would probably result in a final pole at 20 kHz. If the signal from the pre was from the anode circuit of a 6SN7 triode and the cathode R was unbypassed in the interest of sonic purity, and dislike for electro bypass caps, (even the good Oscon types,) and if the tube was a 6SN7, then the Ro would be maybe 25kohms, and the response would be far more dissappointing. Assuming that the preamp's output is just a plate resistor and coupling cap, and that he has more then enough gain, just make the plate resistor less resistance. I also assume that any RIAA curve shaping circuits are not included in the plate circuit, but done elsewhere upstream. That should make the Ro lower, equal to the tube's plate resistance in parallel with the new value of the plate resistor. |
#12
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"Anonymous" wrote
The amp has to drive their capacitance. Right. I was trying to keep to the theme of impedance matching, partly because I wonder where the 10:1 rule comes from, and how far it stretches. For example, if the capacitance of the cable is such that, combined with its inductance and resistance, it appears as a 10k load at, say 20kHz even when open-ended, then driving with 1k or less would be necessary even if the power amp had 100k input impedance. I wonder therefore if the 10:1 rule is universally applicable, as long as you use the *minimum* input impedance, and the *maximum* output impedance of the stages, across the bandwidth and dynamic range of interest. That is, the ratio should be at least 10:1 at all combinations of amplitude and frequency. That should automatically keep you well clear of poles, zeros, and as John points out (I think), slew-rate limiting. cheers, Ian |
#13
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Robert Casey wrote: Patrick Turner wrote: IAN IVESON wrote: A common rule of thumb says that the input impedance of a stage should be no less than 10 times the output impedance of the preceding stage. Hence if your valve power amplifier has the typical 100k or more input impedance, then it is OK to drive it with 10k. A cathode follower would not be necessary. If, OTOH, you were driving an SS power amp with 10k input, you would need 1k out of the preamp, for which a CF or transformer would be necessary. Have you thought of using a transformer, BTW? A natural way to go because it trades gain for impedance. Why do long cables make a difference, as Patrick says? I guess you should consider a long cable as a stage in itself. As an extreme example, if the cable itself has an input impedance of 10k, then the preamp output needs to be 1k or less, no matter what the input of the power amplifier is. And the top of the morning to you Ian, you ain't been around for sooo long.... A friend has a pair of Leak 20s for a biamped system. His gutted and rebuilt-to-different-design Quad22 pre amp sits beside the TT and Denon cart, and there is about 6 metres of cable to the power amps, running under the floor, and up where the power amps sit. If the capacitance of the cable is 100pF per metre, he'd have 600 pF of shunt C just from the cable, and if the Ro of the preamp was 10k, it causes a pole at 26 kHz. but other losses in the amp would probably result in a final pole at 20 kHz. If the signal from the pre was from the anode circuit of a 6SN7 triode and the cathode R was unbypassed in the interest of sonic purity, and dislike for electro bypass caps, (even the good Oscon types,) and if the tube was a 6SN7, then the Ro would be maybe 25kohms, and the response would be far more dissappointing. Assuming that the preamp's output is just a plate resistor and coupling cap, and that he has more then enough gain, just make the plate resistor less resistance. But the plate resistor needs to be several times Ra, to get low thd. I also assume that any RIAA curve shaping circuits are not included in the plate circuit, but done elsewhere upstream. That should make the Ro lower, equal to the tube's plate resistance in parallel with the new value of the plate resistor. We dunno what the original poster has in his amp. Patrick Turner. |
#14
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IAN IVESON wrote: "Anonymous" wrote The amp has to drive their capacitance. Right. I was trying to keep to the theme of impedance matching, partly because I wonder where the 10:1 rule comes from, and how far it stretches. For example, if the capacitance of the cable is such that, combined with its inductance and resistance, it appears as a 10k load at, say 20kHz even when open-ended, then driving with 1k or less would be necessary even if the power amp had 100k input impedance. The capacitance whould have to be 800 pF before the cable presented 10kohms of Z at 20 kHz. The cable inductance will be tiny. I wonder therefore if the 10:1 rule is universally applicable, as long as you use the *minimum* input impedance, and the *maximum* output impedance of the stages, across the bandwidth and dynamic range of interest. That is, the ratio should be at least 10:1 at all combinations of amplitude and frequency. That should automatically keep you well clear of poles, zeros, and as John points out (I think), slew-rate limiting. Slew rate limiting also depends on signal voltage. Patrick Turner. cheers, Ian |
#15
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"Patrick Turner" wrote
The capacitance whould have to be 800 pF before the cable presented 10kohms of Z at 20 kHz. I was trying not to get hung up on actual values. Also, if you wish to establish a mathematical relationship, extreme values are usually a better test. Looking at your own example in an earlier post to this thread, I note you cite a cable of 600pF, which is not far off from my example. You also demonstrate that, combined with a 10k output impedance, the resulting pole will be unacceptably low. Now, how far *should* the this pole be from the usual 20kHz limit? Since the power amp will add its own poles after, and the preamp already has added its own before, presumably you would wish the cable connection to extend to 80kHz or further? If you work it out, you will see that the 10:1 rule fits well with your own example. That is, the output impedance of the preamp would need to be about 1 tenth of the impedance presented by your 600pF cable at 20kHz. The cable inductance will be tiny. Compared to even a smallish planet, so is the capacitance. Slew rate limiting also depends on signal voltage. I included dynamic range. Does that not cover signal voltage? If not, then I meant the full range of signal voltage that the amp is expected to handle. What does dynamic range mean, then? A cathode follower has an output impedance of roughly 1/gm, which may be fine at the operating point. But as the signal goes low and Vgk approaches zero, gm quickly falls, and so Ro rapidly increases. Unacceptable distortion may result if the 10:1 ratio is not maintained under these conditions. The effect that folk call "slew rate limiting" occurs when a capacitance in parallel with the load is unable to discharge quickly enough because of the high Ro at the time the discharging needs to happen. The effect is exacerbated if the CF is within a closed loop circuit. But, if the 10:1 ratio is applied using the worst-case Ro at max negative signal voltage, then that problem will not occur. Cathode followers are by no means the only offenders. If you were to get into simulating, you would find how easy it is to check Ro for all sorts of common circuits and find they don't stand up to scrutiny. Measuring the frequency response of Ro at full amplitude is very revealing. Mu-followers can be terrible if you are not careful. I'm not happy with the term "slew rate limiting", BTW, and wish I hadn't mentioned it. I see amps with quoted slew rates of so many volts per microsecond. What does that mean? No mention of amplitude, so are they just talking about upper frequency limit? Or do they mean at full power? Perhaps, since the effect depends not just on slew rate, but also slew magnitude, it should be called "slew slew rate limiting". cheers, Ian in message ... IAN IVESON wrote: "Anonymous" wrote The amp has to drive their capacitance. Right. I was trying to keep to the theme of impedance matching, partly because I wonder where the 10:1 rule comes from, and how far it stretches. For example, if the capacitance of the cable is such that, combined with its inductance and resistance, it appears as a 10k load at, say 20kHz even when open-ended, then driving with 1k or less would be necessary even if the power amp had 100k input impedance. I wonder therefore if the 10:1 rule is universally applicable, as long as you use the *minimum* input impedance, and the *maximum* output impedance of the stages, across the bandwidth and dynamic range of interest. That is, the ratio should be at least 10:1 at all combinations of amplitude and frequency. That should automatically keep you well clear of poles, zeros, and as John points out (I think), slew-rate limiting. Patrick Turner. cheers, Ian |
#16
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IAN IVESON wrote: "Patrick Turner" wrote The capacitance whould have to be 800 pF before the cable presented 10kohms of Z at 20 kHz. I was trying not to get hung up on actual values. Also, if you wish to establish a mathematical relationship, extreme values are usually a better test. Looking at your own example in an earlier post to this thread, I note you cite a cable of 600pF, which is not far off from my example. You also demonstrate that, combined with a 10k output impedance, the resulting pole will be unacceptably low. Now, how far *should* the this pole be from the usual 20kHz limit? Since the power amp will add its own poles after, and the preamp already has added its own before, presumably you would wish the cable connection to extend to 80kHz or further? Well if you use a cathode follower, and one set up to be able to charge ans discharge capacitance without slewing, or tending to act as a detector at HF, then the Ro can easily be less than 1/10 of the 10k from the anode of a 6SN7. Thus the pole caused by 600 pF moves out to over 200 kHz. If you work it out, you will see that the 10:1 rule fits well with your own example. That is, the output impedance of the preamp would need to be about 1 tenth of the impedance presented by your 600pF cable at 20kHz. Wiser minds than I suggest wider bandwidth than just 20 Hz to 20 kHz will give the better music. a CF has Ro = 1 / Gm. So if you use two paralleled halves of a 12AT7, you get Ro = about 125 ohms. The use of something even more gutsy allows the transfer of video signals down quite long cables. But then you move into the area of signal transmission down transmission lines... The cable inductance will be tiny. Compared to even a smallish planet, so is the capacitance. Slew rate limiting also depends on signal voltage. I included dynamic range. Does that not cover signal voltage? If not, then I meant the full range of signal voltage that the amp is expected to handle. What does dynamic range mean, then? I agree. One needs about 0.5v/us per volt of output to get wide bw. If you want 250 kHz bw at 300 watts into 8 ohms, you need about 300v/us But if you want only 10v of signal, then a lot lower slew rate is needed. A cathode follower has an output impedance of roughly 1/gm, which may be fine at the operating point. But as the signal goes low and Vgk approaches zero, gm quickly falls, and so Ro rapidly increases. But its all subject to the series voltage NFB. The Ro does not change greatly during any part of the voltage cycle providing there is enough current... A CF will rapidly cut off if the output voltage is large enough because the DC load R and the cap coupled R load of the following amp form a divider. The lower the cap coupled load, the earlier the voltage cut off threshold is reached. Its easily seen when you have a 1/2 6SN7 set up as CF with 47k DC loadR, and the cap coupled load is only say 10k. If the idle current in the CF is only 2 mA, then the maximum negative going voltage will be 20 peak volts. Just because the CF has low Ro, it doesn't mean you can use loads about equal to the Ro. So CF should only ever be coupled to the same sort of load as a gain tube, say 3 x Ra at least. Unacceptable distortion may result if the 10:1 ratio is not maintained under these conditions. The effect that folk call "slew rate limiting" occurs when a capacitance in parallel with the load is unable to discharge quickly enough because of the high Ro at the time the discharging needs to happen. The effect is exacerbated if the CF is within a closed loop circuit. That's the CF acting like a detector. There are CF detectors used at 455 kHz in radios, typically using 12AU7. But, if the 10:1 ratio is applied using the worst-case Ro at max negative signal voltage, then that problem will not occur. Cathode followers are by no means the only offenders. If you were to get into simulating, you would find how easy it is to check Ro for all sorts of common circuits and find they don't stand up to scrutiny. Measuring the frequency response of Ro at full amplitude is very revealing. Mu-followers can be terrible if you are not careful. And any tube stage. Tubes can turn on with more current than they can supply when they turn off. I'm not happy with the term "slew rate limiting", BTW, and wish I hadn't mentioned it. I see amps with quoted slew rates of so many volts per microsecond. What does that mean? No mention of amplitude, so are they just talking about upper frequency limit? Or do they mean at full power? Perhaps, since the effect depends not just on slew rate, but also slew magnitude, it should be called "slew slew rate limiting". Most amps quote the volts per u-sec at full power. As the power increases, so does the voltage required, and to get a constant bandwidth, the V/us must also increase, so twice the voltage needs twice the slew rate, which is 4 times the power. 10V/us might be OK for a Williamson at 16 watts, but not much chop if its a 300 watt amp. If you sketch a 10 vrms wave form at 250 kHz, and then draw a straight line parallel to the wave line where it crosses 0V, you can see what rise time is required to describe the HF sine wave without distortion occuring. Once than straight line becomes less steep, the wave form starts to become triangular. The fundemental F is attenuated, and higher harmonics are generated. This may not matter much if F is way above the audio band, but some power amps are into triangular waves at say 16 khz at the full rated power level, and it often is associated with gross over loading of the input stages by the HF signals. And in a real world where amplifiers have to cope with a mighty *BASH* of cymbals, along with all the other frequencies, the cymbals come out sounding unlike cymbals. Som to my mind, the better amps mange clean sine waves without any stage overloading right up to 65 khz into the rated resistor load. Using capacitor loads tests the amp sorely, and as the Z of the C falls at 6 dB/octave, the amplifier current needed to maintain the voltage into the declining ZC rises to impossible levels. People place LR zobel networks on an amp's output to prevent low value R loads, or C loads ever forcing an amp to try to make absurd currents and voltage at F above 10 kHz. Nearly all SS amps have say 10 uH plus 6 ohms in parallel, placed as a network between the emitter outputs and load terminals. I serviced an ARC SP8 preamp today. It was able to pump out 56 vrms from its line stage amp, unloaded. It has two cascaded gain tubes, the second one direct coupled to a CF, and its DC load R taken to the Rk of V1, and there is considerable global NFB around the 3 stages. So instead of the usual 300 ohms Ro of a 1/2 6DJ8, the Ro is more like 30 ohms or less. Despite all this NFB, the cut off distortion would still occur way below 56vrms if the load coupled were really low, like a pair of heaphones. But maybe at 0.1 vrms into those 32 ohm phones, the sound would be OK. 0.14 peak volts needs only +/-4.3 mA current change, and the CF would provide that, and the thd wouldn't be excessive, since so much NFB is applied. A CF using a power tube like an EL84 with 40 mA will give better performance into low value RLs and C loads. Some say that's the best sounding output from a preamp. There is a point where increasing current ability is useless. No need to feel so insecure about the interface betwen preamp and power amp that one acts obsessively; we have all seen examples where someone has used a 300B as the preamp output tube. I don't see the point, and I sure don't hear the point. Even a 1/2 12AX7 CF with 0.7 mA of idle current will have an extraordinary bandwidth at a volt into 100kohms and 68 pF in a metre of decent shielded interconnect cable. Last but not least, we should also consider the gain amp reduced to unity gain by a shunt FB R network, and such a set up is an "anode follower", and will have similar attributes of low td, low Ro, and widened bw as the CF. CF don't invert the signal phase. Anode folls do, and this upsets some audio enthusiasts. Patrick Turner. cheers, Ian in message ... IAN IVESON wrote: "Anonymous" wrote The amp has to drive their capacitance. Right. I was trying to keep to the theme of impedance matching, partly because I wonder where the 10:1 rule comes from, and how far it stretches. For example, if the capacitance of the cable is such that, combined with its inductance and resistance, it appears as a 10k load at, say 20kHz even when open-ended, then driving with 1k or less would be necessary even if the power amp had 100k input impedance. I wonder therefore if the 10:1 rule is universally applicable, as long as you use the *minimum* input impedance, and the *maximum* output impedance of the stages, across the bandwidth and dynamic range of interest. That is, the ratio should be at least 10:1 at all combinations of amplitude and frequency. That should automatically keep you well clear of poles, zeros, and as John points out (I think), slew-rate limiting. Patrick Turner. cheers, Ian |
#17
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On Wed, 10 Nov 2004 14:14:40 GMT, "IAN IVESON"
wrote: I'm not happy with the term "slew rate limiting", BTW, and wish I hadn't mentioned it. I see amps with quoted slew rates of so many volts per microsecond. What does that mean? No mention of amplitude, so are they just talking about upper frequency limit? Or do they mean at full power? Perhaps, since the effect depends not just on slew rate, but also slew magnitude, it should be called "slew slew rate limiting". Slew rate limiting is only of relevance at full power. There is no such thing as 'slew magnitude', the slew rate simply determines the frequency at which a full-power sinewave will begin to triangulate, it has no other relevance. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
#18
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Please feel comfortable in opening my Super IT Phono Preamp Gif file. It's
clean. Hope this help. As you can see the output is a hybrid, making it difficult for me to articulate its configuration. But I was told that a Cathode Follower increase the highs & is much more open by taking the load of the RIAA network out of the gain stage. Makes sense? Please give me your opinions. Thank you in advance. Cordially, west "west" wrote in message ... Please feel comfortable in opening my Super IT Phono Preamp Gif file. It's clean. Hope this help. As you can see the output is a hybrid, making it difficult for me to articulate its configuration. Cordially, west "west" wrote in message . .. I have a Super It Phono Preamp that has plenty of gain (70db). This is more than enough to drive my UL P/P amp. My question is how do I add a Cathode Follower to the IT? Are there any good circuits out there that I can copy? Thanks for your opinions in advance. Cordially, west |
#19
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Assuming that the preamp's output is just a plate resistor and coupling cap, and that he has more then enough gain, just make the plate resistor less resistance. But the plate resistor needs to be several times Ra, to get low thd. Is that because I would reduce the internal feedback one sees with triodes? Because of the differing voltage of the plate over the course of an audio waveform? What if I take the output signal off a tap on the plate resistor? That is, a voltage divider between the plate and B+. That would reduce the output impedance (and gain) without much changing what the plate sees (the destination load will have some effect, but one gets that if the plate drives thru a cap that load anyway). |
#20
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"Stewart Pinkerton" wrote
Slew rate limiting is only of relevance at full power. There is no such thing as 'slew magnitude', the slew rate simply determines the frequency at which a full-power sinewave will begin to triangulate, it has no other relevance. Hi Stewart! Thanks. Full power defines the magnitude of the slew perfectly. As long as "full power" is also adequately defined. But they never say "at full power" on the box, hence the usual quotation is meaningless on its own. OK, *we* know what they mean but, like the equally meaningless "rms power", it must confuse heaps of regular folk. Also, I take issue with your "only relevant at". If an amp has a slew rate limit at full power, then its slew rate will only be slightly less limited at slightly less than full power. Just because it is only quoted at full power, doesn't mean it is only relevant at that point. But perhaps you meant that the value given only applies at full power? Fair enough, I'm being picky. Anyway, this highlights the reason I would prefer not to use the term. If it is only quoted at one point, then it only applies to an infinitesimally small proportion of an amplifier's output. It is only a guide if you know something of the architecture of the amp, so you can make an educated guess at the profile of the limit, surely? Audio art is music engineering. cheers, Ian ps I feel slightly bemused to find you here Stewart, but I can't remember why. I'm glad. Hooray. Really, I remember that you know some things. I still entertain an imagined vision of your porch and courtyard, with a little path leading down a gentle slope to a well. I must know you have a well...I can't see how I would have made that bit up. Whoo...now I see fairies. I know what's happened...I saw a TV programme, or film perhaps, in which a pair of girls saw fairies and took a photo of them with their fathers' camera...or was that a little brook, rather than a well? Somewhere in my mind you became a tragic Edwardian gentleman. |
#21
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An ordinary CF will display premature cut off when the cap couple load
is a lowish value compared to the DC supply R. A CF also tends to suffer from slew distortion with a capacitance load because the discharge of the capacitance is due to the current flow via the DC load R, and the time constant for this is longer than the charge time constant when the tube is turned on. BUT, its not a big deal at preamp voltages and the slew distortion is only a bother when C is large, or at high voltage swings at HF. Usually a 6SN7 CF has a banwidth of maybe a Mhz at a volt, and with 600 pF, the slew distortion is negligible, and cut off won't occur until maybe 15 vrms of output..... I have tried two 1/2 triodes as used in the White, and its very nice, but the same two triodes paralleled will do almost the same job because there is more current flow, and more current to discharge the C... The White can be RF un-stable if you are not careful.... Patrick Turner. The White CF like any other audio circuit may suffer from parasitic oscillations if one is not careful with lead dress. A large current loop coupled with some stray C & mutual coupling & an RF oscillator is a sure thing. A very effective insurance against such results can be easily applied thru the use of 1/2 watt resistors of one to ten K connected as close as convenient to the control grid leads of every tube. When I worked with fast rise time pulse generators I found I could make 50 ohm attenuators out to about 2 GHZ using 1/2 watt R's. Two watt R's were not much good above one GHZ because of skin effect. In an audio circuit these R's in circuit work to reduce the Q of any potential oscillatory circuit, since Q = ( 2*PI*f * L ) / R. Increasing the RF resistance reduces the Q. Notice I say RF resistance, which as frequency rises is not the same as the DC resistance. The lower triode of the White CF acts as an active current source for the upper triode, just as the upper triode in a Mu Follower acts as an active current source for the lower triode. The White CF needs only a single polarity supply. Other resistor only solutions often need a 2nd negative supply. Go to ABSE to see a stable, wide band White CF, complete with its RF swamping resistors. The input signal is 10 volts peak to peak at 50 KHz, driving into a 10 nF load. The screen shots are offset above & below zero signal so that you can see the output at the top while the drive signal to the lower triode is the bottom trace. You can see the effect of the cap load as the upper triode switches the lower triode to more or less conducting in response to the current required. That way, charge & discharge of the transmission line are both active. Cheers, John Stewart |
#22
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Robert Casey wrote: Assuming that the preamp's output is just a plate resistor and coupling cap, and that he has more then enough gain, just make the plate resistor less resistance. But the plate resistor needs to be several times Ra, to get low thd. Is that because I would reduce the internal feedback one sees with triodes? Because of the differing voltage of the plate over the course of an audio waveform? As the load on a triode is reduced, the distortionm is increased. Its best seen when one draws a load line across set of plate resistance data curves. The more horizontal the load line is, the less thd, ( high value RL ), and the when the LL is close to vertical, the thd is worst, along with low gain. There are 3 parameters in any tube which affect its linearity, U, Gm, and Ra. U is the most constant, being dependant mainly on the electrode structure dimensions, which are fixed, unless designed on purpose to allow variable U like a remote cut off pentode where the U reduces with bias current. Gm and Ra vary considerably. Gain for all tubes = ( U x RL ) / ( Ra + RL ) U = Gm x Ra, so in the gain equation there are two variables, and gain varies at different Ea, and Ia conditions. A CF has all its output fed back in series with the input, and if a load of 1k is used instead of say 50k for a 6SN7 CF, then the " open loop " tube gain will be ( 20 x 1k ) / ( 1k + 8k ) = 2.2, and so for 2 volts output, we need 0.9v between G1 and k and we need 3.1v input, so the overall gain is 2 / 3.2 = 0.625, and the gain reduction is 2.2 / 0.625 = 3.38 times, so the distortion is only reduced this much with the CF, and because of the 1 k load, the open loop thd might be 5 times higher than with a 50k load, so the CF with 1 k does not achieve lower thd than what would be achievable with a plate loaded tube with 50k load. As internal gain reduces in triodes, there is less internal electrostatic feedback. What if I take the output signal off a tap on the plate resistor? That is, a voltage divider between the plate and B+. That would reduce the output impedance (and gain) without much changing what the plate sees (the destination load will have some effect, but one gets that if the plate drives thru a cap that load anyway). Yes, you could do that indeed. But is it worth the loss of gain? Is it as good as a CF? Consider the 1/2 6SN7 with a bypassed Rk, RL = 50k, and 4 mA of anode current. Ra will be about 10k, and the Ro from the anode will be 10k with the RL in parallel, so we get Ro = 8.33 k. If we tapped up the RL towards the B+ we could get a point where Ro was also = 8.33k, and at about 10k down from the B+, because below the tap you have 40k + Ra, and above the tap you have 10k. The gain would reduce to 1/5 of what it is from the anode, so about 16.6 / 5 = 3.3. If we went further up RL to say 1k down from the B+, the Ro = 1k // 49k = about 980 ohms, gain is reduced to 0.33 times, so the performance cannot be as good as the CF, where Ro = 1 / Gm = 1 / 0.002 = 500 ohms, and gain is just below unity, and the signal is low thd since lots of NFB applies. A CF can be used purely as a class A current source with an attrocious load match, ie, a 1 k load on the 6SN7, and as explained, it doesn't matter whether its CF or plate loaded, the tube with 4 mA of standing current can only make a max of 2.82vrms before its cuts off or runs into grid current, or both. the thd will be high even at this low signal. But if there is a gain tube ahead og the CF or plate loaded tube, say a 12AX7, and there is loop FB of series or shunt, and gain is reduced to unity even for the 1 k load, then thd is reduced by at least the gain of the 12AX7, or perhaps 75 times, so if it was 5% at 2.8 vrms with no FB it will be about 0.07% with the AX7 used ahead of the SN7. Its not what I would do, but it can be done. But if the load on the follower was 50k, the thd at 2.8v would be maybe 0.01% without the 12AX7 in the FB loop, but with the 12AX7 in the loop, the thd = 0.01 / 75 = 0.000133% in theory, but in practice the noise would prevent its accurate measurement. The ARC line stage in the SP8 uses such techniques to be able to claim the thd of their amplifiers is rather low indeed. They use 3 tubes, where I would use one, or maybe two, at most. Whether that translates into the best sound is a moot point. Patrick Turner. |
#23
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"Patrick Turner" wrote in message ... ssssssssssssssssssssssssssssssssssssnip But if the load on the follower was 50k, the thd at 2.8v would be maybe 0.01% without the 12AX7 in the FB loop, but with the 12AX7 in the loop, the thd = 0.01 / 75 = 0.000133% in theory, but in practice the noise would prevent its accurate measurement. The ARC line stage in the SP8 uses such techniques to be able to claim the thd of their amplifiers is rather low indeed. They use 3 tubes, where I would use one, or maybe two, at most. Whether that translates into the best sound is a moot point. Patrick Turner. amongst other reasons,that 's why SP8 sound like expensive crap. -- .................................................. ........................ Choky Prodanovic Aleksandar YU "don't use force, "don't use force, use a larger hammer" use a larger tube - Choky and IST" - ZM .................................................. ........................... |
#24
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Choky wrote: "Patrick Turner" wrote in message ... ssssssssssssssssssssssssssssssssssssnip But if the load on the follower was 50k, the thd at 2.8v would be maybe 0.01% without the 12AX7 in the FB loop, but with the 12AX7 in the loop, the thd = 0.01 / 75 = 0.000133% in theory, but in practice the noise would prevent its accurate measurement. The ARC line stage in the SP8 uses such techniques to be able to claim the thd of their amplifiers is rather low indeed. They use 3 tubes, where I would use one, or maybe two, at most. Whether that translates into the best sound is a moot point. Patrick Turner. amongst other reasons,that 's why SP8 sound like expensive crap. I only fix the bloomin things. I was going to start a service where I could sent the amp to the God of Triodes Himself, but He dictated terms which were too difficult to accept; I had to be a resident of Heaven first, and who knows, I may not even make it there. Patrick Turner. .................................................. ....................... Choky Prodanovic Aleksandar YU "don't use force, "don't use force, use a larger hammer" use a larger tube - Choky and IST" - ZM .................................................. .......................... |
#25
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What if I take the output signal off a tap on the plate resistor? That is, a voltage divider between the plate and B+. That would reduce the output impedance (and gain) without much changing what the plate sees (the destination load will have some effect, but one gets that if the plate drives thru a cap that load anyway). Yes, you could do that indeed. But is it worth the loss of gain? Is it as good as a CF? Consider the 1/2 6SN7 with a bypassed Rk, RL = 50k, and 4 mA of anode current. Ra will be about 10k, and the Ro from the anode will be 10k with the RL in parallel, so we get Ro = 8.33 k. If we tapped up the RL towards the B+ we could get a point where Ro was also = 8.33k, and at about 10k down from the B+, because below the tap you have 40k + Ra, and above the tap you have 10k. The gain would reduce to 1/5 of what it is from the anode, so about 16.6 / 5 = 3.3. If we went further up RL to say 1k down from the B+, the Ro = 1k // 49k = about 980 ohms, gain is reduced to 0.33 times, so the performance cannot be as good as the CF, where Ro = 1 / Gm = 1 / 0.002 = 500 ohms, and gain is just below unity, and the signal is low thd since lots of NFB applies. My motivation was that the tapped plate resistor would not require the installation of an extra tube in the equipment the CF would require. The tapped plate resistor would be cheap and easy to try. He may not need 500 ohms impedance. If he does, then he will need the CF. But if his cables are not long, he shouldn't need low impedance. |
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