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6SL7 Phase Inverter Loading Problem

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  • #16
    Originally posted by GregS View Post
    Max clean signal at amp output = 84V p-p
    @ 8 ohms isn't that 880 watts!?! I think I must be confused about the formula.
    "Take two placebos, works twice as well." Enzo

    "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

    "If you're not interested in opinions and the experience of others, why even start a thread?
    You can't just expect consent." Helmholtz

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    • #17
      Yes, I did a double take, but -

      V peak = V p-p/2
      Vrms = Vpeak x 0.7071

      84V p-p = 29.7V rms, which provides 110 watts to an 8 ohm load

      Pete.
      My band:- http://www.youtube.com/user/RedwingBand

      Comment


      • #18
        Gingertube: thanks for your informative post. Good information there. In my case I'm limited in what solutions I can use. An extra tube or one that draws more heater current isn't possible. There is no physical room and the heater winding is already 22-25% over it's rating with the current tubes. The MOSFET solution sounds like it involves significant modifications to the power supply and a number of extra parts which I don't think can realistically be done at this point.

        What is the bias voltage splitter resistors?
        What is the DC voltage on the 6SL7 between the plate and the cathode?
        I may be able to describe this in words as it is not very complicated. Each tube has it's own bias pot for a total of four. Each plate of the PI connects to a junction of 2 coupling caps. The other side of each coupling cap connects to a 120K gridleak which connects to the bias pot. There is a grid stopper of 1.5K at each tube. So there is a minimum of 120K impedance at each tube grid as that is the only path to the bias circuit. The grids of each pair of tubes are isolated from each other by the coupling caps. In this case the PI plate sees the grid leaks in parallel for 60K as I understand

        If only one bias pot were to be used per side then there would be one coupling cap connected to the PI plate. On the other side of the coupling cap would be the connection to the 120 grid leak and the grids of both tubes. In this case the PI plate sees at least 120K impedance. However, both tubes grids are sharing the 120K gridleak. Would it not be equivalent (as far as draining the grids of positive charge) to use gridleaks of twice that in the 4 bias pot case since each grid has its own separate grid leak?

        In my test #4 the volate between plate and cathode is 298V - 88.6V = 209V at no signal.

        Originally Posted by GregS
        Max clean signal at amp output = 84V p-p
        @ 8 ohms isn't that 880 watts!?! I think I must be confused about the formula.
        The way I calculated the formula is to take the peak voltage of 42V and divide that by 1.414 to get the peak rms voltage. The resulting value is then squared and then divided by 8. I'm assuming that's the correct way to do it but could be wrong.

        Looking at the plate curves for a 6SL7 it seems to me likely that my signal swing is being limited on the HT side by the bias voltage sitting at -2V. My bias resistor value is kind of high at 750R. It seems to me, in theory anyway, that I could get significantly more signal swing by reducing the bias resistor to centre bias. I'll try this and see what happens.

        My approach to solving this problem is to first see how much clean signal I can squeeze out of the PI and then make up the rest by upping the grid leaks accordingly. If that ruins my power tubes then so be it. I may have to resort to EL34's to handle the increased gridleaks. The amp should be compatible with them.

        Greg

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        • #19
          Update:

          Reducing the bias of the phase inverter actually reduced the headroom. I also tried increasing the HT to 459V but got about the same amount of headroom. No more signal swing to squeeze out of this 6SL7.

          Next I tried changing the 120K gridleaks for 220K and used the phase inverter setup with an HT of 436V. I got a signal swing of 116V p-p at the output of the phase inverter and a 96 - 100V p-p signal at the output of the amp across the 8 Ohm load. I could see the start of power amp clipping on the scope this time. Anyway, with that measurement I'm getting anywhere from 145 - 156W rms output into 8 Ohms which is what it should be.

          I know the 220K gridleaks seem high for 6550/KT88 but each tube has it's own gridleak without any sharing. To me this seems about the equivalent of two tubes sharing a single 100K gridleak in terms of draining positive charge from the grid but I'm not sure of this so correct me if I'm wrong.

          Also, I suspect that how critical the gridleak value is depends somewhat on the application. Perhaps if maximum power is being demanded of the tubes where they're setup with a very high plate voltage and a significantly lower screen voltage from a separate supply, the 50K maximum gridleak value is more important.

          The power amp I'm using is actually based on the Traynor Custom Special - the main differences being the 6550/KT88 output tubes instead of EL34's and the 6SL7 phase inverter instead of the 12AX7. I have approximately 580V on the plate and 570 on the screens (well above the 6550 max rating). The plate and screen supplies are separated by a choke. I have 1K screen resistors for each tube.

          Anyway, I'm going to go with the 220K gridleaks to solve this problem as it is by far the least painful. If I have any power tube problems I'll report back on that.

          Thanks to all who participated in this thread. Much appreciated.

          Greg

          Comment


          • #20
            Good work.
            I would'nt be too concerned about the 220K grid leaks, not ideal BUT - Max Rg1 values for output tubes is the single most abused thing in guitar amp design. Fender, Marshall etc. have been happily abusing these specifications for years.

            Of-course HiFI guys have been known to abuse it too. One of the local HIFi nutters brings me his French Jadis JA80 (80W Class A with 4 x KT88) monoblocks to fix after a "blow up". This happens every couple of years. It uses a single 510K Rg1 for the pair of KT88 on each side. I have to specially select KT88s for low grid current.

            For your info, I found that the Gold Lion re-issues are much better in this respect (low grid current) than JJs but then they are a lot more expensive. I haven't tried KT88 from any other manufacturer.

            Cheers,
            Ian

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            • #21
              Thanks, I got one set of Electro Harmonix 6550's which look like KT88's and a set of Sovtek KT88's for this amp. I've had good experience with the EH tubes and they're very reasonably priced also.

              Greg

              Comment


              • #22
                Greg,
                For future reference - have a squiz at the Mosfet Follies clarfication thread above.
                Cheers,
                Ian

                Comment


                • #23
                  I think you have a PI biasing problem here.
                  Don't know why you chose those values.
                  HT (at PI node) = 343V
                  Plates = 234V, 235V

                  Max clean signal at PI output = 74V p-p
                  Max clean signal at amp output = 70V p-p
                  You also said you have +67V at the cathodes.
                  Also that the
                  KT88 bias voltage of -62V
                  Let´s see where that leads us to.
                  For an initial analysis let´s imagine perfect PI tubes, which can swing end to end and power tubes which have infinite input impedance.
                  Also remember that power tube grids idle at bias voltage and must be driven "up" to pass max. current; so the positive swing is the important one.
                  The negative one is not that important, since power tubes are being driven more into cutoff anyway.
                  And what do we see?
                  (Perfect) positive swing: 343-234V=109V peak.
                  (Perfect) negative swing= 234-67V=167V peak.
                  In the real world, you lose, say, around 40 to 60V each way, plus loaded voltage swing is much lower than unloaded one.
                  So you see that a PI stage fed over 300V rails which "should" have no problem putting 62V peak out (into the next grid) in fact is struggling to do so.
                  You have somewhat solved it by going the "brute force" method of rising +B to the PI.
                  Not bad, but I would change its biasing too so its plates rest at a lower idle voltage, so they have more headroom to go "up".
                  In a nutshell: besides the "current" problem, which is real, you also have a "voltage" problem, in part caused by the PI bias point you chose.
                  In a nutshell 2: *if* the Traynor "original" worked well (I trust them, good no nonsense folks) , why don't you just build it in full, as is?
                  I do not doubt you can use an octal instead of a 12AX7, for example, but the development effort may not compensate building "just one" (or a couple).
                  Good luck.

                  PS: why use 6SL7's anyway?
                  And EL34 biased at, say. -38V are much easier to drive than any other power tube biased at -62V, that's for sure.
                  Everything adds up
                  Juan Manuel Fahey

                  Comment


                  • #24
                    I think you have a PI biasing problem here.
                    Don't know why you chose those values.

                    HT (at PI node) = 343V
                    Plates = 234V, 235V
                    Max clean signal at PI output = 74V p-p
                    Max clean signal at amp output = 70V p-p
                    JM, Thanks for your reply.
                    What's in the quote above was the first of a series of tests I did with increasing the HT to the PI by reducing the value of the dropping resistor.

                    4. Changed PS dropping resistor to 22K
                    HT (at PI node) = 436V
                    Plates = 297V, 298V
                    Max clean signal at PI output = 96V p-p
                    Max clean signal at amp output = 84V p-p
                    Cathodes = 88.6V
                    Above tail resistor = 86.6V
                    Bias = 2V
                    With this one I was able to get a 48V peak signal at the output of the PI - still not enough to fully drive the power amp. Going higher than this with the HT seemed to not make much difference. I thought I could get more signal swing by biasing more to the centre (-2V seemed to be off centre on the cold side). However, I got less doing this.

                    Using the 22K dropping resistor and increasing the power tube grid leaks from 120K to 220K got me to 58V peak at the output of the PI. With this I could reach the headroom of the power amp as it runs out about the same time as the PI - a little sooner actually. The goal here was simply to get the max clean output out of the amp which was achieved. I'm not interested in overdriving the power amp as this is a bass amp and I want it clean.

                    I realize the 220K gridleaks are not ideal for 6550/kt88 tubes. However, I'm not going to assume this is a problem until it actually is. I'd rather not make drastic changes if not necessary. If it proves to be a problem I'll look to other solutions then. For now it's working well.

                    Greg

                    Comment


                    • #25
                      Fine.
                      I would decrease slightly (say, 20 to 30% lower for starters) the PI cathode resistor, so they pass a little more current and more voltage drops across plate resistors.
                      This will increase the possible positive swing, which is what´s needed .
                      Juan Manuel Fahey

                      Comment


                      • #26
                        Hi,

                        I recently ran into the same problem with my 6SL7 PI. I have this box of NOS russian tubes I got from Ebay for pennies and I finally decided to make use of them so it's actually the russian analog to 6SL7 - 6H9C running two 6П6С (6V6). It looks like the PI is starting to clip at around 16V (measured with DMM). I have no real scope only PC-software via sound card but what I can see is the amp starts clipping at same time with the PI. According to datasheets the amp shoudd produce 14 Watts with 300V anode voltage but in reality it's only around 10W. I'm running the 6П6С (6V6) at around -22V and 26mA at idle.
                        The PI has the standard Marshall values typical for 12AX7, 50k FB resistor from the 8 Ohm tap, the usual Presence control.
                        Anode voltage at the PI node is 300V (82k and 100k plate resistors). Unfortunately I have no option of increasing the voltage in this amp. The plate to cathode voltage is 150V with the plates at around 190V. As I'm not very good in theory so I would like to ask how and which resistors values to change in order to optimize the PI performance as it is if possible at all. I tried to tweak the 470 Ohm and 10 k resistors without any success. Increased the 220k grid leak resistors to 470k, no change.

                        Comment


                        • #27
                          I tried 12AT7 and 12AX7 in place of 6SL7 but it's still clipping at the same level, so I assume this is how much the power amp can take.

                          Comment


                          • #28
                            It's still helpful to know exactly where the clipping is. If you don't have a scope you can put a 1M pot across the power tube grids as a temporary master volume. If turning down the pot allows for more swing at the PI prior to clipping then it's the power tubes. If not you can remove the 1M pot and use it to attenuate the signal at the input to the PI. If you still have clipping at same volume pot setting then it's the preamp.
                            "Take two placebos, works twice as well." Enzo

                            "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

                            "If you're not interested in opinions and the experience of others, why even start a thread?
                            You can't just expect consent." Helmholtz

                            Comment


                            • #29
                              I replaced the 220k grid leak resistors to 470k without any effect. With all the three tubes I tested the clipping occurs at ~2V (PI input).
                              I get ~16V at the PI caps/power tubes' grids.

                              Comment


                              • #30
                                That's pretty low. What's your Vp on the PI?

                                And can you post a schematic?

                                AND... A pet peev of mine here is when the issue is unclear, someone asks for a schematic and the OP replies with a very long description of the circuit. It takes a long time to digest and something relevant is always left out. If you can't post a schem, please at least keep the description short and expect to answer follow up questions.
                                "Take two placebos, works twice as well." Enzo

                                "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

                                "If you're not interested in opinions and the experience of others, why even start a thread?
                                You can't just expect consent." Helmholtz

                                Comment

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