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  • #16
    Originally posted by Helmholtz View Post

    Well, you surely know that shorting kills voltage but may cause high currents. 1)
    So signal voltage between PI outputs will be zero2). But you may still have3) considerable common mode voltages wrt ground as Chuck H demonstrated in the other thread.
    Equal common mode grid signals cause plate dissipation but no output4).
    1) How high are the currents, if it is the bias voltage?
    2) If the signal voltage between PI outputs will be zero, how much is the signal voltage at g1 power tubes?
    3) In technique there is no "but you may still have", just have or not to have.
    4) I don't understand how there is plate dissipation when no output.

    Maybe I'll learn something I didn't know. I am especially interested:
    If the signal at point A as at the picture and level is U = 1V, what will graphic form of the signal at points ? and how much is his the level.
    It's All Over Now

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    • #17
      Originally posted by Helmholtz View Post
      A bright cap especially makes sense when a vol pot feeds a high gain triode stage, as there will be considerable Miller capacitance, typically >100pF.
      About Miller capacitance
      Miller capacitance is a story from the HF area (> 30MHz).
      Leo Fender, Jim Marshall and others probably haven't even heard of Miller capacitance, and they've made tons and tons of amps.
      It's All Over Now

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      • #18
        Originally posted by vintagekiki View Post

        1) How high are the currents, if it is the bias voltage?
        2) If the signal voltage between PI outputs will be zero, how much is the signal voltage at g1 power tubes?
        3) In technique there is no "but you may still have", just have or not to have.
        4) I don't understand how there is plate dissipation when no output.

        Maybe I'll learn something I didn't know. I am especially interested:
        If the signal at point A as at the picture and level is U = 1V, what will graphic form of the signal at points ? and how much is his the level.
        1) Measure yourself. I would estimate up to 4mA peak depending on circuit values.
        2) Study this thread, especially the simulations by Chuck H: https://music-electronics-forum.com/...olume-question
        3) Refrain from indoctrinating phrases if you want answers from me! I meant what I said.
        4) Common mode grid signals produce in-phase plate currents. In-phase plate currents don't produce OT induction, only see low DCR and don't produce OT output.
        Last edited by Helmholtz; 12-13-2020, 11:17 PM.
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        • #19
          Originally posted by pdf64 View Post
          Did you see the thread where Jon Snell raised the issue of a common mode positive bias caused by the type 3 / crossline master vol? https://music-electronics-forum.com/...olume-question
          I saw and read.
          #15
          I only know the theory as I have not recorded any results.
          Jon Snell fenced himself off correct.

          #13
          Consider drive of 10volts on each anode of the phase inverter, the drive is still there in this circuit wherever the control is set.
          By reducing the MV pot (1M), the drive on g1 output tubes still exist only smaller.
          When the MV pot (1M) at minimum, the g1 output tubes are short-circuited, and since anodes PI are also short-circuited, there is no drive at the g1 output tubes, only idle (bias) current will flow through the OT.

          Just to notice
          The whole story about CLMV comes down to interpreting various simulations.
          It is not difficult. It only takes 2 wires and a quality 1M pot to move from theory to practice.

          Simulation is just a simulation, personal practical experience is something completely different, which is worth discussing.


          It's All Over Now

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          • #20
            Originally posted by vintagekiki View Post
            About Miller capacitance
            Miller capacitance is a story from the HF area (> 30MHz).
            Leo Fender, Jim Marshall and others probably haven't even heard of Miller capacitance, and they've made tons and tons of amps.
            Another fallacy!

            From the Miller theorem it follows that any capacitance between plate and grid appears multiplied by the gain of the stage between grid and ground.
            The internal capacitance of a 12AX7 is 1.7pF and the wiring typically adds another 0.5pF.
            So the input capacitance of the stage is around 2.2pF x 60 = 132pF for a typical gain of 60.

            I actually measured values between 130pF and 150pF.

            If this gain stage is driven from a source impedance of e.g. 300k (1M vol. pot at some middle setting), the cutoff frequency will be 4kHz ( not MHz ! ) or even lower.

            I am quite sure that Leo Fender was aware of this effect as he used bright caps of suitable values to compensate the Miller loss.

            https://en.wikipedia.org/wiki/Miller_effect
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            • #21
              Originally posted by vintagekiki View Post

              By reducing the MV pot (1M), the drive on g1 output tubes still exist only smaller.
              When the MV pot (1M) at minimum, the g1 output tubes are short-circuited, and since anodes PI are also short-circuited, there is no drive at the g1 output tubes, only idle (bias) current will flow through the OT.
              Even with zero signal BETWEEN the grids you can have a common signal on both grids wrt ground.
              Just apply the same signal to both grids and measure what happens.


              Just to notice
              The whole story about CLMV comes down to interpreting various simulations.
              It is not difficult. It only takes 2 wires and a quality 1M pot to move from theory to practice.

              Simulation is just a simulation, personal practical experience is something completely different, which is worth discussing.
              If you don't trust the simulations I recommend to get out your 2-channel scope and measure yourself.
              That should be most instructive and practical enough .
              Last edited by Helmholtz; 12-14-2020, 03:38 PM.
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              • #22
                Originally posted by Helmholtz View Post
                ... ...
                I'm not really sure about the above math, only thing I don't understand is why Fender would make an guitar amps with the cutoff frequency < 4 kHz.

                (2nF capacitor between g1 and cathode OT)
                http://www.thetubestore.com/lib/thetubestore/schematics/Fender/Fender-Bandmaster-AD1269-Amp-Schematic.pdf
                http://www.thetubestore.com/lib/thetubestore/schematics/Fender/Fender-Bandmaster-Reverb-AA270-Schematic.pdf
                http://www.thetubestore.com/lib/thetubestore/schematics/Fender/Fender-30-Schematic.pdf
                http://www.thetubestore.com/lib/thetubestore/schematics/Fender/Fender-Dual-Showman-Reverb-AA769-Schematic.pdf

                (500pF capacitor between anode and cathode V1a / V1b / V2b / V4b)
                http://www.thetubestore.com/lib/thetubestore/schematics/Fender/Fender-140-Schematic.pdf
                http://www.thetubestore.com/lib/thetubestore/schematics/Fender/Fender-Twin-Reverb-II-Schematic.pdf

                I know that the road to hell has paved with good intentions, and so I refrain from comment about Miller and Fender, so that individuals are not found.
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                It is not difficult. It only takes 2 wires and capacitor between 130pF and 150pF. Connect a capacitor between the g1 (or anode) and cathode of the preamp tube and measure the cutoff frequency, and listen to guitar sound with and without 150pF.
                It's All Over Now

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                • #23
                  Originally posted by Helmholtz View Post
                  ... ...
                  I'm free to notice that you didn't understand
                  https://music-electronics-forum.com/forum/music-electronics/920070-master-volume-and-treble-loss?p=920246#post920246
                  CLMV Question
                  If the signal at point A as at the picture and level is U = 1V, what will be the signal at points ? (graphic form and how much is his the level).

                  About instructive and practical
                  Measured and practically checked everything I write.
                  I do not base my comments on retelling someone else's experience.
                  In order not to "heat hot water" (copy/ paste), I always list http where the original article can be downloaded.
                  It's All Over Now

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                  • #24
                    Originally posted by vintagekiki View Post
                    only thing I don't understand is why Fender would make an guitar amps with the cutoff frequency < 4 kHz.
                    That's why he used the bright cap at the vol. for compensation. But the frequency compensation is only perfect at mid setting of the pot.

                    2nF capacitor between g1 and cathode
                    In this example the source resistance is only around 10K (because of the low plate impedance of a 12AT7), so the cutoff frequency is around 8kHz. But as the LP filter is within the NFB loop, it's effect will be partly equalized.

                    Connect a capacitor between the g1 (or anode) and cathode of the preamp tube and measure the cutoff frequency, and listen to guitar sound with and without 150pF.
                    As said above I could verify the Miller theory by measuring input capacitances between 130pF and 150pF with my LCR meter.
                    LP cutoff frequency varies with source resistance. You should be able to calculate it yourself for a given circuit.

                    Effective source resistance between plate and cathode is typically only around 40k, so the cutoff frequency will be much higher than in my example with 300k.

                    And yes, I know how a 4kHz LP filter changes the sound.
                    Last edited by Helmholtz; 12-14-2020, 08:05 PM.
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                    • #25
                      Originally posted by vintagekiki View Post
                      I'm free to notice that you didn't understand
                      https://music-electronics-forum.com/forum/music-electronics/920070-master-volume-and-treble-loss?p=920246#post920246
                      CLMV Question
                      If the signal at point A as at the picture and level is U = 1V, what will be the signal at points ? (graphic form and how much is his the level).
                      I very well understood your question and was hoping that my above reply would answer it.

                      With a perfectly balanced PI the speaker output will be zero. But the signal wrt ground at the grids will depend on the common mode component of the PI output.

                      If you need real data use your scope and measure. Don't expect me to do it for you.

                      I don't like this CLMV so I don't use it.
                      Last edited by Helmholtz; 12-15-2020, 12:31 PM.
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                      • #26
                        Dear colleague
                        Obviously you either don’t understand, or you don’t want to understand about I’m writing. I don’t know, maybe the cause is in Mercury retrograde, or a difference in interlingual syntax.
                        To avoid unnecessary misunderstanding in communication as happened in post #19, I will refrain from comment until some better times.
                        It's All Over Now

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                        • #27
                          Originally posted by vintagekiki View Post
                          CLMV Question
                          If signal at point A as at the picture, what will be the signal at points ?
                          I have changed the picture and the question a bit. If there is signal at point 'X', what will be at the output?

                          Click image for larger version

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                          I think we can all agree the output will be zero, yet the power tubes are doing work.
                          So the question becomes "can there be signal appearing at point X" ?
                          The PI is not 100% perfect. How can we say it is not possible for some common mode signal to get through to X ?

                          Originally posted by vintagekiki View Post
                          I know that the road to hell has paved with good intentions,
                          I think the road to hell may be paved with google translator.

                          Attached Files
                          "Everything is better with a tube. I have a customer with an all-tube pacemaker. His heartbeat is steady, reassuring and dependable, not like a modern heartbeat. And if it goes wrong he can fix it himself. You can't do that with SMD." - Mick Bailey

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                          • #28
                            Originally posted by g1 View Post

                            So the question becomes "can there be signal appearing at point X" ?
                            The 2 grid signals with 2k between the grids in Chuck's simulation here:
                            https://music-electronics-forum.com/...679#post806679
                            are almost identical.
                            So that would essentially correspond to the common mode grid signal at point X with the CLMV adjusted to 2k.
                            As the CLMV can only reduce the differential voltage, I expect the common mode signal with the CLMV at zero to be about the same.

                            I have no reason to doubt the sims. Also I understand that nickb could confirm Chuck's results.

                            Last edited by Helmholtz; 12-18-2020, 02:51 PM.
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                            • #29
                              Originally posted by g1 View Post
                              I think we can all agree the output will be zero1), yet the power tubes are doing work.
                              So the question becomes "can there be signal appearing at point X"2) ?
                              That's right colleague. Output from PI at point ? (X) will be zero1) and graphic form at point ? (X) will be horizontal line, because equal half-periods are canceled.

                              When the output from PI is zero1), there is no drive at the g1 output tubes, through power tubes only idle (bias) current will flow.

                              2) I don't understand how you found the negative half-cycle at point X, when PI gives a symmetrical output voltage ???
                              It's All Over Now

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                              • #30
                                Originally posted by vintagekiki View Post
                                I don't understand how you found the negative half-cycle at point X, when PI gives a symmetrical output voltage ???
                                Because PI is not perfect.
                                Does not need to be complete half cycle, that was just an example. Can you guarantee no signal can possibly get to point 'X' ?

                                "Everything is better with a tube. I have a customer with an all-tube pacemaker. His heartbeat is steady, reassuring and dependable, not like a modern heartbeat. And if it goes wrong he can fix it himself. You can't do that with SMD." - Mick Bailey

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