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I'm new to analog electronics, so bear with me. Consider my guitar amp design:

enter image description here

What we have is a 12AX7 feeding into a cathode follower, going into a James tonestack, through another 12AX7 gain phase, then through another cathode follower.

So I'm kinda doing something cool here... I'm splitting the plate load resistors and adding the feedback in between the split resistors. According to Designing Tube Preamps for Guitar, this makes the net A/C across R5 and R13 near 0, making the gain of the stage very very high.

So problem #1: I'm unsure how to actually calculate the gain of those stages, and what my output Vac will be. My limited understanding calculations say 1Vrms coming in will be about 20Vac going out for the first stage... I think.

Problem #2: I want to be able to switch the amplifier from Carlos Santana mode to Hank Marvin mode. So if there is excess gain leftover from after the tone stack, I want to shunt it to clean up the amp. Normally a voltage divider pot does this between stages... but I'm trying to figure out how do that in this case. My thought is to put the voltage divider after C2, but I'm not sure how that'll affect the tonality of the stack.

Finally, bonus round... I want to use a dual gang pot between wherever I put the gain control and have RV3 be the inverse of the gain control. Any thoughts on that? So as I turn my gain control up, RV3 shunts more to ground, reducing the output signal. I couldn't find and designs using this type of idea, but I'm wondering if it's been done before.

Many thanks!

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The first tube is an anode gain stage bootstrapped with the signal taken on the cathode of the next stage. This is a positive feedback that tries to keep the current flowing trough R5 constant. This turns R5 to a virtually huge impedance load as seen by U1A. This is good because the ECC83 has a high internal impedance (rounded to 60k) so it needs a high load not to loose gain.

Since the current flowing through the tube becomes nearly constant, it compensates the fact R3 is not decoupled which results in a dramatic raise of the stage amplification. Using this bootstrap method you can bet it will give a gain between 80 and 100.

The direct coupling between the anode gain stage and the cathode follower is also a good thing in a guitar amp since, depending on the bias of the cathode follower, its grid may steal current from its neighbour causing nice H2 distortion.

Maybe a solution for being able to switch between high gain and low gain, would be a switch in the bootstrap loop of the second pair of ECC83. The filters behind the first couple of triodes will decide the output level of the amplifier.

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  • \$\begingroup\$ Ideally I'd like to be able to blend it gradually, not just an on/off. Hmm. \$\endgroup\$ – Jonathan S. Fisher Jun 22 '17 at 18:17
  • \$\begingroup\$ What do you think about a pot voltage divider right after c2? Would that change the source impedance of the cathode follower significantly? The cathode follower is extremely low impedance by itself, I'm curious if it can both drive the tone stack AND a volume control \$\endgroup\$ – Jonathan S. Fisher Jun 22 '17 at 18:21
  • \$\begingroup\$ Or what about a voltage divider after the tone stack, on the grid of V1b? I imagine that would greatly alter the tone stack properties :/ \$\endgroup\$ – Jonathan S. Fisher Jun 22 '17 at 18:23
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    \$\begingroup\$ The cathode follower have an output impedance of approx. 1/Gm hence it should be under 1kohm or so. If you use a reasonably high impedance pot just after C2, it should not change neither the CF load nor the tone stack behind. \$\endgroup\$ – greg Jun 23 '17 at 7:42
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This is a good design using a bootstrap unity gain positive feedback on U2A to the midpoint of the 47k pair making the 1st stage gain much greater. SO I might expect 60 dB gain with the Baxendall Filters adding some load to the high outpuit impedance of this method losing a few dB.

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  • \$\begingroup\$ You don't get much of a gain increase with triodes when you bootstrap the anode resistor because the triodes output resistance is in parallel. Using the quoted 60k output resistance with 100k resistor is 37.5k. Assuming perfect bootstrapping the output resistance is 60k - only a ratio of 1.6. It works much better with pentodes with their megohm output resistance. \$\endgroup\$ – Kevin White Jul 6 '17 at 0:02
  • \$\begingroup\$ Interesting but the low Zout of cathode follower on the 2nd stage bootstrap is AC feedback to high side of R5 and gate following the low side of R5 reduces the AC voltage drop near zero while still conducting DC current thus the AC impedance ratio of anode/cathode is raised significantly , the quoted transconductance in the specsheet S = 1,6 mA/V (milli-siemens) \$\endgroup\$ – Sunnyskyguy EE75 Jul 6 '17 at 1:32
  • \$\begingroup\$ Yes - but only to ~60k with bootstrapping vs 37.5k without. The resultant gain has only changed from 60 to 99. The triode's inherent output resistance is not affected. With a pentode it could change the gain an order of magnitude. \$\endgroup\$ – Kevin White Jul 6 '17 at 1:50
  • \$\begingroup\$ ah ok so only 40dB Gain per pair. I recall this bootstrap design was originally called a Williamson Circuit for pentodes and offered high gain and with good tubes not requiring negative feedback for low distortion of 0.1% \$\endgroup\$ – Sunnyskyguy EE75 Jul 6 '17 at 2:10
  • \$\begingroup\$ This circuit is similar except has a differential output with a gain of 260 angelfire.com/electronic/funwithtubes/… \$\endgroup\$ – Sunnyskyguy EE75 Jul 6 '17 at 5:17
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Don't forget the plate characteristics will be in parallel with the bootstrapped lumped load.

Thus your gain will be { rout / (1/gm + R3) }.

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  • \$\begingroup\$ its the gmR4 added to R3 that makes the big boost \$\endgroup\$ – Sunnyskyguy EE75 Jun 22 '17 at 4:29
  • \$\begingroup\$ gm2R4 that is////.... \$\endgroup\$ – Sunnyskyguy EE75 Jun 22 '17 at 5:22

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