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Better Low Z driver stage design based on our Bootstrapped Gain Stage thread

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  • Better Low Z driver stage design based on our Bootstrapped Gain Stage thread

    I made a couple of changes to the driver stage in my amp based on some of the findings we discussed the "Bootstrapped Gain Stage Theory" thread posted here a while back:
    https://music-electronics-forum.com/...n-stage-theory

    Here is how the stage was previously designed incorporating cathode biased followers. The goal was to use the drivers as a transparent buffer for a DC coupled output stage.
    Click image for larger version

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    However, one of the more revelatory things to me was the effect bootstrapping has not only on the grid impedance, but also the cathode impedance as well. Therefore, the output impedance of a cathode biased cathode follower (as was the case in my earlier design) would be dramatically increased by the bootstrapping effect of this configuration. Interestingly, I've never seen this mentioned in any of the literature I've read on cathode follower operation or design before. This seems like an important thing to mention, particularly because cathode followers are often used as low impedance buffers.
    Did we ever come up with a formula to determine the output impedance of a cathode biased follower? I couldn't find it in the other thread.

    In any case, here is the modified circuit incorporating a fixed bias adjustment to the driver stage.

    Click image for larger version

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    I think this improves it's function as a transparent buffer in a couple of ways. The current biasing more changes the input impedance, dropping it to around 213k (set by the 422k in parallel with 433k). While this might seem to be a drawback from the previous 5M input impedance, 213k more closely approximates the usual 220k (or so) loading at the output of the LTP in a normal AC coupled stage. This also drops the output impedance around 322Ω (≈1/gm) making it a much better buffer.
    If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

  • #2
    I understand that you want to avoid blocking distortion by DC coupling. But why do you need a very low driver output impedance? Do you really want to "over-saturate" the ouput tubes by lots of grid current in class AB2 operation? Remember that pentodes like the EL84 have lots of gain and easily saturate in class AB1 without any grid current.
    - Own Opinions Only -

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    • #3
      Click image for larger version

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      As shown, the output impedance is about 850 ohms, ignoring the effect of source impedance. If you take the output from the cathode, the impedance goes down to 370 ohms as you no longer have the 500 ohm resistor in series with the output. The source impedance does have an effect so for Z src = 50K the cathode output impedance goes from 370 to 400 ohms. All this seems insignificant compared to the grid stopper of likely 1.5K or so.

      In addition to what was said above is that a lower drive impedance will improve the HF response. That might not be desirable in a guitar amp.
      Last edited by nickb; 10-02-2020, 07:54 AM.
      Experience is something you get, just after you really needed it.

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      • #4
        The source impedance does have an effect so for Z src = 50K the cathode output impedance goes from 370 to 400 ohms.
        Do I interprete you correctly that the bootstrapping causes only a very small increase in output impedance probably because the source impedance is relatively small compared to the bootstrapped resistor?
        I seem to remember from earlier discussion that high source impedance increases the cathode output impedance with bootstrapping because it increases the positive feedback signal at the grid.
        Last edited by Helmholtz; 10-01-2020, 11:48 PM.
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        • #5
          nickb that attached pic in post #3 is 'invalid file specified'. You will need to re-size it or zip it.
          "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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          • #6
            Originally posted by g1 View Post
            nickb that attached pic in post #3 is 'invalid file specified'. You will need to re-size it or zip it.
            Well that was weird. It would show for me sometimes. A bugette.
            Experience is something you get, just after you really needed it.

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            • #7
              Originally posted by Helmholtz View Post

              Do I interprete you correctly that the bootstrapping causes only a very small increase in output impedance probably because the source impedance is relatively small compared to the bootstrapped resistor?
              I seem to remember from earlier discussion that high source impedance increases the cathode output impedance with bootstrapping because it increases the positive feedback signal at the grid.
              Change in feedback is fairly small for Zsrc= 50k because of the 470K grid leak resistor. For infinity (quite useless) it's something like 3k for this cct with a 12AU7.

              Another thought. The bias voltage is very uncertain as you rely on the tube to set it. It would be more predictable if you were to feed a reference to the grid. That would lower the input impedance and couple power supply noise. This first is not an issue an the second just requires attention to filtering.
              Experience is something you get, just after you really needed it.

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              • #8
                Some equations to answer the OP.

                Gain = Vcathode / Vgrid = Ra.(R1+R2).Gm/(Ra + R1 + R2 + Ra.(R1 + R2).Gm) ; Ra = plate resistance, Gm = transconductance;

                Output Z = 1/(1/(R1+r2) + Gm + 1/Ra) for zero source impedance

                Output Z = 1/(1/(R1+r2) + Gm.k + 1/Ra) for non zero Zsrc where k = R3/(R3+Zrc). Assumes R1<<R2 and R2<<R3


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                I don't doubt HH will check my math soon enough
                Last edited by nickb; 10-02-2020, 10:02 PM.
                Experience is something you get, just after you really needed it.

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                • #9
                  Originally posted by nickb View Post
                  Some equations to answer the OP.

                  Gain = Vgrid / Vcathode = Ra.(R1+R2).Gm/(Ra + R1 + R2 + Ra.(R1 + R2).Gm) ; Ra = plate resistance, Gm = transconductance;

                  Output Z = 1/(1/(R1+r2) + Gm + 1/Ra) for zero source impedance

                  Output Z = 1/(1/(R1+r2) + Gm + 1/Ra) for non zero Zsrc where k = R3/(R3+Zrc). Assumes R1<<R2 and R2<<R3

                  I don't doubt HH will check my math soon enough
                  Right you are: I am confused.

                  Gain should be Vcathode/Vgrid. And of course Ra is zero for the cathode follower, so 1/Ra means trouble (singularity).
                  The second Output Z formula is identical to the one above for zero source impedance and doesn't contain the parameter k.
                  Last edited by Helmholtz; 10-02-2020, 09:43 PM.
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                  • #10
                    Originally posted by Helmholtz View Post

                    Right you are: I am confused.

                    Gain should be Vcathode/Vgrid. And of course Ra is zero for the cathode follower, so 1/Ra means trouble (singularity).
                    The second Output Z formula is identical to the one above for zero source impedance and doesn't contain the parameter k.
                    I knew you'd check!

                    Vcathode /Vgrid is just a typo, fixed (edited).

                    No, Ra is not zero, it's the internal anode (plate )resistance of the tube, about 5.7k for a 12AU7 as shown.

                    Yes, the second formua is almost the same, it should be. I did it like that to illustrate the effect of the feedback factor, k.
                    Last edited by nickb; 10-03-2020, 09:52 AM.
                    Experience is something you get, just after you really needed it.

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                    • #11
                      As an experienced confuser, I have to add...

                      You can use power MOSFETs instead of cathode followers and get an even larger decrease in output impedance. There's an illustration of this in the MOSFET Follies on geofex.com.
                      Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!

                      Oh, wait! That sounds familiar, somehow.

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                      • #12
                        Originally posted by nickb View Post

                        No, Ra is not zero, it's the internal anode (plate )resistance of the tube, about 5.7k for a 12AU7 as shown.
                        Sorry, I thought you meant the external plate resistor. I'm used to see ra for internal and Ra for external plate resistance.

                        Yes, the second formula is almost the same, it should be. I did it like that to illustrate the effect of the feedback factor, k
                        But the 2 formulae are exactly the same. No k contained in either. But the Zsrc should have some small effect, right ?

                        - Own Opinions Only -

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                        • #13
                          Originally posted by Helmholtz View Post

                          Sorry, I thought you meant the external plate resistor. I'm used to see ra for internal and Ra for external plate resistance.



                          But the 2 formulae are exactly the same. No k contained in either. But the Zsrc should have some small effect, right ?
                          Ah, another typo.I left out the "k". Fixed. Thx for checking
                          Experience is something you get, just after you really needed it.

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                          • #14
                            FWIW, source or output impedance can be determined from the signal reduction caused by adding a load resistor.
                            Or more easily by using an LCR meter that provides AC resistance measurement (also works for input resistance). To block DC wire a cap in series with the meter.
                            Last edited by Helmholtz; 10-03-2020, 12:58 AM.
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                            • #15
                              I'd love to have a decent LCR meter or rather impedance analyzer. The cheap ones have about 1% accuracy which means the real part is wildly ( i.e several hundred percent) off when the phase is close to +/-90 degrees.
                              Experience is something you get, just after you really needed it.

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