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  • Bootstrapped Gain Stage Theory

    I have a couple of questions about Bootstrapped stages. First, is there a "universal boot" equation? Where input impedance can be calculated whether the stage uses a FET, BJT, Tube, or op amp?
    Also, using the example of a simple triode gain stage; I know that bootstrapping can increase the input impedance of the grid(inverting input) substantially . Would the same be true of the cathode(non-inverting input), as in a grounded grid stage?
    That's all.
    If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

  • #2
    Too sleepy but basically:
    1) you are mixing very different gain stages, differentb inherent/"built in" impedances.
    2) we are not talking the stage *on/internal* input impedance by itself 8which as said above varies a lot) but the input impedance degradation caused by the usually inevitable input resistor.
    That is what can be improved by bootstrapping.

    Now back into a nice cozy warm bed, here itīs 1AM of a cold, polar wind swept Winter night.
    Juan Manuel Fahey

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    • #3
      Originally posted by J M Fahey View Post
      Too sleepy but basically:
      1) you are mixing very different gain stages, differentb inherent/"built in" impedances.
      2) we are not talking the stage *on/internal* input impedance by itself 8which as said above varies a lot) but the input impedance degradation caused by the usually inevitable input resistor.
      That is what can be improved by bootstrapping.

      Now back into a nice cozy warm bed, here itīs 1AM of a cold, polar wind swept Winter night.
      Oh yeah, I sometimes forget that the seasons are opposite us down where the water circles the drain the wrong way
      If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

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      • #4
        Bootstrapping to increase the input impedance of a circuit works by reducing the signal current through the (external) input resistor. This is accomplished by connecting the far end of the resistor to a circuit point where the signal is in phase with the input and amplitude is just a little lower (instead of connecting to signal ground), while still providing the necessary DC path. The now much smaller signal voltage across the resistor means lower input current. The resistor appears increased by the ratio of full input voltage/actual voltage across the resistor.
        To take it to the extreme, if signal voltages at both ends of the resistor were the same, there would be no signal current through the resistor - thus no input loading (like infinite resistance).

        As JMF indicated this method only works for the external resistor but not for tube internal impedance. A basic grounded grid circuit has a low cathode input impedance of roughly 1/gm (typically some hundred Ohms) and bootstrapping the external input resistor seems useless.
        Last edited by Helmholtz; 07-15-2019, 04:59 PM.
        - Own Opinions Only -

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        • #5
          Yes.
          Bootstrapping was necessary (so popular) in simple bipolar transistor gain stages.
          Base impedance could be made *relatively* high value, think 470k or so, but biasing resistors were typically 1/10th of that (say 47k to 22k) so they keep bias stable relatively independent from transistor variations.

          Now if you can bootstrap that low resistance offender and make it "look" , as helmholtz says, 10X or 20X larger, then you can still sell your cake and eat it.

          Not really neded in, say, Fet or Fet input Op Amp since bias currents are minuscule, think picoamperes, so internal impedance is already very high, and reference resistor can be made 1M or larger.

          I have used 10M reference resistors innPiezo preamps with no ill effects, so there bootstrapping isnīt needed.
          Juan Manuel Fahey

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          • #6
            Originally posted by Helmholtz View Post
            Bootstrapping to increase the input impedance of a circuit works by reducing the signal current through the (external) input resistor. This is accomplished by connecting the far end of the resistor to a circuit point where the signal is in phase with the input and amplitude is just a little lower (instead of connecting to signal ground), while still providing the necessary DC path. The now much smaller signal voltage across the resistor means lower input current. The resistor appears increased by the ratio of full input voltage/actual voltage across the resistor.
            To take it to the extreme, if signal voltages at both ends of the resistor were the same, there would be no signal current through the resistor - thus no input loading (like infinite resistance).
            Right, I understand the principle when applied to a traditional grounded cathode/emitter stage. Would the cathode impedance remain the same in this configuration?:
            Click image for larger version

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            This is similar to what you might see in a long tail pair. In this case the voltage at the top of the long tail resistor (where the grid and cathode are "grounded") follows the voltage at the cathode. Although, I can see how you would still need to contend with the internal cathode impedance to drive the current in the tube? So I guess that's where I'm unsure.
            If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

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            • #7
              Would the cathode impedance remain the same in this configuration?:
              Click image for larger version. Name: example.jpg Views: 4 Size: 69.3 KB ID: 54333
              I would think so, as the internal cathode impedance acts between cathode and signal ground, not between cathode and grid (grid current must be zero).

              Also your circuit will have very low gain, as the grid - instead being (AC) grounded - is connected to a voltage divider which reduces the actual input voltage for the tube (= difference between cathode and grid voltage) by a factor > 40.

              This is similar to what you might see in a long tail pair.
              In a LTP the non-inverting triode is driven from its cathode in grounded grid mode. But its grid is essentially grounded by the low resistance feedback circuit (or directly grounded via a capacitor in non-NFB amps). The high value grid reference resistor connecting to the top of the tail resistor won't change much, as the signal from the cathode circuit gets divided by a huge factor before it reaches the grid.
              Last edited by Helmholtz; 07-16-2019, 03:08 PM.
              - Own Opinions Only -

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              • #8
                One not-so-common use of bootstrapping is to raise the gain of a pentode stage. Most find that an input pentode stage has too much gain already, but I wanted to see if I could get a minimalist 3 valve guitar amp going with cheap throw away valves.

                The key to the 3 valve guitar amp I had in mind was to try out the Jeffery pentode/triode input gain and PI circuit from WW Aug 1947, which effectively uses bootstrapping of the input pentode stage anode load to substantially increase the loading impedance on the pentode. It works quite nicely to achieve enough gain for a guitar input (circa 20mVrms input) as well as drive a substantial input voltage swing (circa 100Vpp) for a PP output stage with just one valve, and can directly couple the pentode to the following cathodyne PI stage for simplicity.

                Jeffery WW Aug 1947
                https://www.americanradiohistory.com...ld-1947-08.pdf

                Also a section in RDH4 on page 523, and in Briggs & Garner, Amplifiers, 1952, p.102.

                Operating prototype circuit - but still going through guitar testing:
                https://www.dalmura.com.au/static/6C...0Schematic.pdf

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                • #9
                  How effective is the 500k tone pot across the 43R cathode resistor?
                  I wouldn't think it would have any effect until it hit the CW end stop.

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                  • #10
                    Yeh a bit off topic, but I had a few spare dual gang pots so was wondering how I could use a spare gang to 'enhance' the tone function. My evil thought was to use the CW end of travel to act as effectively a switched Presence effect to increase treble gain by suppressing feedback.

                    I also tried an 'enhancement' of the volume pot, whereby lowering the PPIMV would also progressively suppress global feedback. That is a bit quirky, as max volume has ended up not at 100% CW. Anyway, I'll see what my son thinks of it when he has had some play time.

                    I was also using the amp to demonstrate how switchmode supplies could be used, and just got a 12V 8A plugpack in as a comparison to the 12V 5A plupack I have had - it has had to supply about 5A at max overdrive but hasn't hiccuped so far, so not too bad a performance.

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                    • #11
                      Originally posted by SoulFetish View Post
                      Right, I understand the principle when applied to a traditional grounded cathode/emitter stage. Would the cathode impedance remain the same in this configuration?:
                      [ATTACH=CONFIG]54333[/ATTACH]

                      This is similar to what you might see in a long tail pair. In this case the voltage at the top of the long tail resistor (where the grid and cathode are "grounded") follows the voltage at the cathode. Although, I can see how you would still need to contend with the internal cathode impedance to drive the current in the tube? So I guess that's where I'm unsure.
                      I gave this uncommon circuit some more thought and now think that connecting the grid to the top of the tail resistor as shown actually increases cathode input impedance by what I would call feedback. But as cathode impedance acts between cathode and ground, the upper limit for the input impedance is the tail resistance which shunts the cathode impedance.
                      Needs to be verified by measuring.
                      - Own Opinions Only -

                      Comment


                      • #12
                        Originally posted by trobbins View Post
                        Yeh a bit off topic, but I had a few spare dual gang pots so was wondering how I could use a spare gang to 'enhance' the tone function. My evil thought was to use the CW end of travel to act as effectively a switched Presence effect to increase treble gain by suppressing feedback.

                        I also tried an 'enhancement' of the volume pot, whereby lowering the PPIMV would also progressively suppress global feedback. That is a bit quirky, as max volume has ended up not at 100% CW. Anyway, I'll see what my son thinks of it when he has had some play time.

                        I was also using the amp to demonstrate how switchmode supplies could be used, and just got a 12V 8A plugpack in as a comparison to the 12V 5A plupack I have had - it has had to supply about 5A at max overdrive but hasn't hiccuped so far, so not too bad a performance.
                        How did the non-steady state operation of class AB factor into your choice of boost converter?
                        If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

                        Comment


                        • #13
                          That particular unregulated dc/dc step-up module has only minor sag with B+ load variation. But its B+ output also varies directly with input DCV, so any minor sag on the 12V supply effectively passes through to the B+ level. So the outcome is still a small amount of sag on B+ (and bias) when cranked - which imho is a better outcome for a guitar amp than a regulated output with no sag.

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                          • #14
                            Originally posted by Helmholtz View Post
                            I gave this uncommon circuit some more thought and now think that connecting the grid to the top of the tail resistor as shown actually increases cathode input impedance by what I would call feedback. But as cathode impedance acts between cathode and ground, the upper limit for the input impedance is the tail resistance which shunts the cathode impedance.
                            Needs to be verified by measuring.
                            Interesting. This circuit was only an experimental exercise for the most part. I'm trying to better understand how bootstrapping works, and see how the concept can be expanded. I'm not sure the circuit is really all that useful the way it's drawn, so I wasn't giving much consideration to gain.
                            But the question had it's genesis in the mixing circuit I decided to use in my submini reverb project. I needed a way to mix the signals in phase, without the need for much gain. So, I decided to use a non-inverting cascode (which I first learned from a Mike Sulzer circuit), combined with a traditional technique of mixing the dry signal into the cathode.

                            in the following two variations, it's obvious that the grid of the first triode is bootstrapped (ie cathode follower), but how would we analyze the cathode input impedance of this example? (again, I'm not trying to increase the voltage gain really. I would need to pad the output down anyway)
                            Click image for larger version

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                            If you're correct about the "feedback effect" by tying the grid to the top of the tail resistor, maybe it might be useful in this circuit. If I tie the grid of the second triode to the top of the tail, could we not take advantage of the presumed effects of feedback? If it does increase the cathode input impedance, would it also not decrease the output impedance at the plate?
                            Click image for larger version

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ID:	854323
                            If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

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                            • #15
                              Just thinking aloud here;...about the impedance at the plate, per my last post.
                              In the second schematic, the second triode is a kind of grounded grid stage. But the grid voltage follows the cathode voltage because it is “grounded” to the top of the tail resistor - (grid degenerative feedback?) Almost like the effect of the cathode degenerate feedback you see from an unbypassed cathode resistor. So I am wondering if it has the same affect and would increase the output impedance.
                              If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

                              Comment

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