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ECC83 - Plate Load Resistors, compression, Grid current limiting

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  • ECC83 - Plate Load Resistors, compression, Grid current limiting

    Hi Guys, been a while sine I last posted here but I am going through some ideas and this is probably the most knowledgeable group I know of, so..

    I am in the process of modifying the Load line plotter spread sheet that Merlin created, just want to get a bit more information off the sheet that can easily be included. Specifically I plan to add an input sign wave and map the output sine wave with the aim of viewing clipping and harmonic content. I know I can do this on LTspice already but this is for fun)

    Anyway one of the things discussed in Merlins book is how the plate resistor value affects the compression of the stage, higher plate loads have less compression as can be seen by the grid curves etc, but in practice (I think) I am finding higher plate loads add more compression not less, the grid curves don't show that but after a bit of thought and looking at the curves I remembered that with current JJ ECC83s valves grid current can start to flow around 100v and below which when using a higher plate load is a region (shown in blue) that the load lines cross into.

    The image below shows a MESA MKIIc stage (270k||3.3k), both DC and AC load lines, right now I have the final stage in an overdrive section set up like this and it is very compressed and small sounding, going to play with it next week to see if I can reduce the compression.

    My thought is that I want to avoid the load lines crossing grid curves under 100v to reduce grid current compression effect, but not so much that the signal is even more compressed by going too far the other way.

    Anybody thought about this before?

    Click image for larger version

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  • #2
    Your chart shows plate/anode curves not grid curves. The Vgk is the parameter.

    Grid current depends on your grid drive voltage. It will not start before the grid signal amplitude exceeds + 2.5V in your example.

    The negative sound effects of grid current distortion can be reduced by a large value grid stopper resistor.
    - Own Opinions Only -

    Comment


    • #3
      Originally posted by Helmholtz View Post
      Your chart shows plate/anode curves not grid curves. The Vgk is the parameter.

      Grid current depends on your grid drive voltage. It will not start before the grid signal amplitude exceeds + 2.5V in your example.

      The negative sound effects of grid current distortion can be reduced by a large value grid stopper resistor.

      Thanks, yes you are right of course, they are plate curves, that is what happens when you post too late at night.

      How did you determine the +2.5v off that graph?

      Reading the valvewizard book again, Merlin says that for ECC83 grid current may begins to flow at Vkg -0.9v, which equates to a plate voltage of Vp=130v on the load line. So I am really trying to understand if low plate voltages exacerbates the effect as it has been seen when you lower the supply voltage too much with variable voltage controls.

      Presumably I should be able to see the compression on a scope, and the effect of adding a grid stopper.

      Comment


      • #4
        When comparing the compression you need to compare at the same AC output voltage. If you do that you will see that the bigger resistor gives less compression.

        As Helm' says, no grid current will flow until the grid gets a few volts positive. Also, in practice due to the typical 35K ohm drive impedance, hardly any grid current will flow and so you will tend to get a rather abrupt cutoff rather than compression. To make matters even more complicated, the input is often coupled through a capacitor. When grid current flow the capacitor charges and the operating point shifts to more negative gird voltages. Thus you can see that the cutoff effect will change with input time constant.

        100K vs 270K:

        Click image for larger version

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        Time Constant effect. Top pane is grid voltage & current:

        Click image for larger version

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        Experience is something you get, just after you really needed it.

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        • #5
          Originally posted by nickb View Post
          When comparing the compression you need to compare at the same AC output voltage. If you do that you will see that the bigger resistor gives less compression.

          As Helm' says, no grid current will flow until the grid gets a few volts positive. Also, in practice due to the typical 35K ohm drive impedance, hardly any grid current will flow and so you will tend to get a rather abrupt cutoff rather than compression. To make matters even more complicated, the input is often coupled through a capacitor. When grid current flow the capacitor charges and the operating point shifts to more negative gird voltages. Thus you can see that the cutoff effect will change with input time constant.
          Thanks Nick, your top line is exactly what I have expected, but is not what I am hearing, in fact one customer asked me if I had a compressor built into the amp as it had "immense sustain". Maybe there is something else going on that I need to better understand, there is more compression than I would normally like in the amp.

          Comment


          • #6
            I should have just read the chapter more carefully

            "Additionally, the amount of grid current increases as anode voltage falls since the anode has less ability to draw electrons away from the grid at
            low voltages" http://www.valvewizard.co.uk/Common_Gain_Stage.pdf

            So larger plate resistors cause the anode voltage to fall lower (-45v with 270k Plate Load) and therefor grid current increases compared to 100k Plate load which falls to 110v

            Just need to prove it

            Comment


            • #7
              Originally posted by guitarmike2107 View Post
              Thanks Nick, your top line is exactly what I have expected, but is not what I am hearing, in fact one customer asked me if I had a compressor built into the amp as it had "immense sustain". Maybe there is something else going on that I need to better understand, there is more compression than I would normally like in the amp.
              You are (another) victim of the sloppy use of the term compression.

              What is going on in the common plate stage stage is really asymmetric distortion. True compression is a really different thing. It is a reduction in the amplitude of a signal as a function of it's original magnitude. So, for example, a compressor can be set up to give you nearly constant output amplitude for a very wide range of input magnitude, thus giving sustain. This a linear process and by that I mean the signal is not distorted.

              But you can get sustain as a consequence of clipping. You boost the signal and then clip. The output amplitude will still remain the same for a wide range of input but in this case the signal is distorted.

              So, what you're hearing is little to do with the asymmetric effect of the common plate stage but more to do with how it's over driven.

              What Merlin says about grid current it true but it's also a requirement for the grid to be positive with respect to the cathode before grid current can flow.
              Experience is something you get, just after you really needed it.

              Comment


              • #8
                Originally posted by nickb View Post
                What Merlin says about grid current it true but it's also a requirement for the grid to be positive with respect to the cathode before grid current can flow.
                Then how does grid leak bias work? You have grid current flowing and that current pulls the grid negative relative to the grounded cathode. Grid leak bias is just weird. You can change the grid resistor by double or half and the plate Voltage hardly moves, it self regulates. I tried to run a 12AX7 cascode with 100V plate Voltage. Couldn't make it work, too much grid current.

                The other fallacy I see in this thread is the idea that a high value plate resistor means low plate Voltage. Depends on how you bias the tube. Sure if you just shove in a 220K or 470K plate resistor and don't touch the 1.5K cathode resistor, you will have low plate Voltage. Re-bias with something like a 4.7K cathode resistor and see what you get.
                WARNING! Musical Instrument amplifiers contain lethal voltages and can retain them even when unplugged. Refer service to qualified personnel.
                REMEMBER: Everybody knows that smokin' ain't allowed in school !

                Comment


                • #9
                  Originally posted by loudthud View Post
                  Then how does grid leak bias work? You have grid current flowing and that current pulls the grid negative relative to the grounded cathode. Grid leak bias is just weird. You can change the grid resistor by double or half and the plate Voltage hardly moves, it self regulates. I tried to run a 12AX7 cascode with 100V plate Voltage. Couldn't make it work, too much grid current.

                  The other fallacy I see in this thread is the idea that a high value plate resistor means low plate Voltage. Depends on how you bias the tube. Sure if you just shove in a 220K or 470K plate resistor and don't touch the 1.5K cathode resistor, you will have low plate Voltage. Re-bias with something like a 4.7K cathode resistor and see what you get.
                  yip, from Merlins book

                  "Grid-current limiting does not happen instantly. Forward grid current actually begins to flow even before the grid reaches the cathode voltage, around Vgk = –1V, though
                  this varies considerably between valve types. The data sheet may even quote this as ‘Vgk (max)’ , indicating the point at which grid current will begin to exceed 0.3μA, and
                  for the ECC83 this is –0.9V."

                  It is not a fallacy that higher plate load means lower plate voltage, I am not referring to the quiescent state, but the dynamic state and I am working on the basis that the input signal is large enough to get full swing, just need to look at the load line, larger resistors have a larger voltage swing around the bias point, for a 270k load the voltage swings from 50v to 350v, for a 100k load the swing 110v to 350v (using the DC load line)

                  Comment


                  • #10
                    Originally posted by loudthud View Post
                    Then how does grid leak bias work? You have grid current flowing and that current pulls the grid negative relative to the grounded cathode. Grid leak bias is just weird. You can change the grid resistor by double or half and the plate Voltage hardly moves, it self regulates. I tried to run a 12AX7 cascode with 100V plate Voltage. Couldn't make it work, too much grid current.
                    The problem with this is that it's a question of magnitude. Negative grid current is in the low uA range. The grid current does change when sign when Vgk is negative but the current is tiny until you start to get to Vgk>>0 at which point you are dealing with orders of magnitude greater currents. That is why the effect is usually ignored. I'm pretty sure is not relevant to the OP's observations.

                    I'd be scoping the amp to see where the audible effect is coming from.
                    Last edited by nickb; 09-16-2018, 10:51 AM.
                    Experience is something you get, just after you really needed it.

                    Comment


                    • #11
                      In general, if you apply a large burst of sine wave signal to the grid of a 12AX7 stage through a coupling capacitor:
                      1. Grid current flows to change the charge on the capacitor. This can be large if the ac drive impedance is small.
                      2. Grid current can continue to flow through the dc impedance from the grid to ground.

                      The dc impedance can be high or low depending on the circuit, setting of a volume control, if any, at the grid, etc.

                      In general you have a transient effect (charging the capacitor) and a steady effect effect (involving the dc resistance from grid to ground). Both can effect the sound. This is a simplification of the actual effects, of course. You cannot determine very much from the tube curves alone, but it is necessary to do a nonlinear analysis of the whole circuit. Or just play with it until you get what you want.

                      Comment


                      • #12
                        "....current will begin to exceed 0.3μA, and
                        for the ECC83 this is –0.9V."
                        At Vgk = -0.9V the input impedance is still in the 100k range. By Vgk = 0V the average measured value has dropped to 5k to 10k (there is huge variation between individual ECC83/12AX7 tubes covering a spread of 1 to 10, as measured by Zollner).

                        IMO, compression is not only typical for overdrive circuits. In fact it is desirable. Instantaneous compression means distortion. And clipping means extreme compression. If you want less compression, go for less distortion and reduce the drive signal.
                        - Own Opinions Only -

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                        • #13
                          Well tonight I tried a 120k plate load on the last stage and it sounded much less compressed, better picking dynamics for sure. This is the last stage feeding the tone stack.
                          Tomorrow I want to try changing the plate load on the previous stage, currently 100k, so will put it up to 270k and re-bias to see the effect.

                          Ultimately I will put a test rig together to see if I can understand this better by measurement and will share my results if I gain some understanding that I can evidence

                          @Nickb, regardless of semantics, by compression I am referring to the sound you get when you drive a compressor into an overdrive pedal/amp, great fluid lead sound but you lose picking dynamics, I want to turn off the compressor

                          Comment


                          • #14
                            Originally posted by Helmholtz View Post
                            At Vgk = -0.9V the input impedance is still in the 100k range. By Vgk = 0V the average measured value has dropped to 5k to 10k (there is huge variation between individual ECC83/12AX7 tubes covering a spread of 1 to 10, as measured by Zollner).

                            IMO, compression is not only typical for overdrive circuits. In fact it is desirable. Instantaneous compression means distortion. And clipping means extreme compression. If you want less compression, go for less distortion and reduce the drive signal.
                            thanks, will look up Zollner

                            You can have distortion without lots of compression and visa versa, it is about finding the balance that make a sound good for rhythm and lead, that is all I am trying to do with this.

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                            • #15
                              Originally posted by guitarmike2107 View Post
                              Ultimately I will put a test rig together to see if I can understand this better by measurement and will share my results if I gain some understanding that I can evidence
                              Excellent

                              I think it would help enormously if you were to post a complete schematic and approx settings of controls.
                              Experience is something you get, just after you really needed it.

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