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Questions about Ampeg Reverberocket R-12-R reverb

Discussion in 'Shock Brother's DIY Amps' started by chas.wahl, Dec 4, 2020.

  1. Dec 4, 2020 #1
    chas.wahl

    chas.wahl Tele-Meister

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    I'm having trouble understanding how the Ampeg R-12-R preamp works. It's a push-pull 6V6-based design with a paraphase phase inverter and all-octal tubeset, incorporating on-board tremolo and reverb. I'm pretty sure I get how the tremolo works: a triple RC network between anode and grid of the 2nd triode V1b, and output fed into the circuit between the PI and grids of the power tubes. But the reverb mystifies me. I've traced the signal path from input on the attached image (whole schematic included as an attachment) from input jack through V1a, then through V2a and V3a to the reverb tank (so far in green). After the tank (highlighted yellow) the signal path (now pink) goes through V3b and the "dimension" pot, then back to V2b, and thence to the PI.

    upload_2020-12-4_20-8-2.png

    What I don't understand is that the reverb tank seems to sit squarely in the signal path, without any circuit "fork" where dry and wet signal intersect. If the reverb is turned off, then how does signal bypass it and make its way "dry" to the PI? Maybe there's just something about how a reverb circuit (or tank) works that I don't understand. I'm not even sure whether the reverb is "on" when the switch is open, or when it's closed.

    Any help will be greatly appreciated.
     

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  2. Dec 4, 2020 #2
    Lowerleftcoast

    Lowerleftcoast Friend of Leo's

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    Look at R15. Signal traveling through the cathode path.
     
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  3. Dec 4, 2020 #3
    chas.wahl

    chas.wahl Tele-Meister

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    Thanks, but . . . How does the signal end up on the cathode? And traveling by what path to the PI? Is the reverb on or off when the footswitch is closed? And when reverb is off, are any of the triodes taken out of the signal path?
     
  4. Dec 4, 2020 #4
    Lowerleftcoast

    Lowerleftcoast Friend of Leo's

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    There is always signal on the cathode when there is signal going through the tube.
    Draw a pink line on that R15 resistor path. The signal will travel through the tube to the anode and continue on the pink signal path to the PI
    The switch 2 footswitch when connected dumps the signal on the green line to ground, just ahead of the reverb tank.
    V3a and V3b still function but there is no signal for V3b to amplify.
    .
     
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  5. Dec 4, 2020 #5
    Lowerleftcoast

    Lowerleftcoast Friend of Leo's

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    A Phase Inverter works in a similar manner. The signal travels to the other side through the cathode. A Cathode Follower works in a similar manner. Have a look at a 5F6A schematic. It has a Cathode Follower before the tone stack. The signal flows from the Cathode to the tone stack.
     
  6. Dec 4, 2020 #6
    King Fan

    King Fan Poster Extraordinaire Ad Free Member

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    This is way above my paygrade, but I'm thinking @Lowerleftcoast is on it. Although it starts out talking about the trem, I wonder if there's some insight in this thread -- maybe from post #24 onward about cathode coupling...
     
  7. Dec 4, 2020 #7
    Lowerleftcoast

    Lowerleftcoast Friend of Leo's

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    I should mention the phase of the cathode signal is 180 degrees off. Confused yet?:confused:
     
  8. Dec 4, 2020 #8
    King Fan

    King Fan Poster Extraordinaire Ad Free Member

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    ^^^ No, that's the only part I understand. :) My 1.5% insight on this includes the 1% I know about cathode followers and the 0.5% I know about LTPs...
     
  9. Dec 4, 2020 #9
    chas.wahl

    chas.wahl Tele-Meister

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    Thanks for the explanation(s). Actually, I think I knew that the phase inverted, subliminally, from casual reading about cathode followers. Let me see if I've got this straight:

    When the footswitch is open, the signal path is as I've shown in the markup above, through the reverb tank. When it's closed, the signal developed through the anode of V2a and C17 is dumped to ground after V3a, so V3b has no signal on its grid, and no effect -- but the signal on the cathode through R15 goes to V2b's cathode, and (here's where my grasp may be shaky) through V2b to its anode, through C18 (which is the same value as C17 on V2a's anode), and straight on to the PI.

    Now I have to ask: is the signal re-inverted by the transmission through V2b, or is the non-reverb signal arriving at the PI out of phase with the signal that reaches the PI when reverb is active? Also, does the signal through R15 flow all the time to the PI, or only when the reverb is inactivated? That is, should my signal path markup for "reverb on" include signal from both anode and cathode of V2a, or only from anode? I'm suspecting that the 22k resistor in the cathode path creates greater impedance on that signal path than the cathode path which has no such resistor -- but maybe I'm still out of my depth!

    Well I guess I am, since I'm still mystified about how the signal can make its way through V2b (the only path to PI being through that triode's anode) when there's no signal being applied to V2b's grid!

    Also, I'm wondering if there's a significant difference in signal gain when reverb is active compared to the signal when reverb is inactive.
     
    Last edited: Dec 4, 2020
  10. Dec 5, 2020 #10
    mrriggs

    mrriggs Tele-Meister

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    V2 is a differential amplifier, just like a long-tail phase inverter. It looks a little different because the "tail" is split in two with R15 going between them. That resistor is local feedback which lowers the gain of the amp. You see this kind of setup in tube oscilloscopes, only the "R15" resistor is a pot that acts as the gain control.

    The pre-amp is fed into the diff-amp, the inverted output goes to the reverb driver, the non-inverted output goes to the phase inverter. With the reverb turned down, that is all that is happening. Turning up the reverb will feed the reverb signal into the second grid input of the diff-amp which will output to the phase inverter.

    Interestingly, though, it will also output to the reverb input. Perhaps this was done to increase dwell? As long as the signal is smaller on it's return, it'll eventually die out. You would have to be careful with the circuit design because if the signal is too great then you will end up with a positive feedback loop.

    You can totally ignore "phase" when dealing with reverb circuits. Once the signal goes through jiggly springs all phase is lost. It's just random, chaotic short-delays all mish-mashed together.
     
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  11. Dec 5, 2020 #11
    chas.wahl

    chas.wahl Tele-Meister

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    OK! I think I'm getting somewhere with this now, between @Lowerleftcoast's and @mrriggs's explanations. I've looked up @robrob's explanation of the voltage gain through a LTPI in his "How Amps Work" page, which helps to visualize this. Thanks to all of you! This amp has several circuit features that I haven't run into previously -- very interesting to me, at least.
     
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  12. Dec 5, 2020 #12
    theprofessor

    theprofessor Poster Extraordinaire

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    This is why I love TDPRI. I've learned a TON in the past several years, but I'm still in elementary school when it comes to this stuff. And yet the high school students and the college profs still let me sit in on their discussion, just to see what I may be able to pick up. @King Fan is several grades ahead of me.
     
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  13. Dec 5, 2020 #13
    King Fan

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    *Not!* You’ve aced the AP speaker and tube classes; I just sat in on one session of each by mistake. It’s only that I love the 5F6a and try to understand it, and I built a 5G9 with the LTP... @Lowerleftcoast and @chas.wahl and @mrriggs deserve all the credit for a smart question and smart answers.
     
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  14. Dec 5, 2020 #14
    Lowerleftcoast

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    You may have a grasp on this already, but just to be clear, the signal through V3a and V3b is only responsible for the *wet* reverb signal. Through the *Dimension Pot* the wet signal will be added to the dry signal in V2b.
    The *Dimension Pot* allows the user to vary the amount of wet signal that is available to V2b. I doubt the additional signal gain would be noticed any more than in other reverb circuits.

    Even if the wet signal were perceived to be equal to the the dry, adding the two together does not register as a significant difference to the human ear. Think of two singers vs one. Two are twice as loud as one.
     
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  15. Dec 5, 2020 #15
    chas.wahl

    chas.wahl Tele-Meister

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    Yes, I believe I understand that -- basically whatever I've shown in the markup from the anode of V2a through the reverb circuit and back to the grid of V2b is all "wet". And the "dry" is from V2a through cathode, and V2b from cathode to anode. Right? Or am I still on the wrong side of the river?

    Good to know, and somewhat reassuring.

    Yeah, if they're singing in unison maybe. But if they're singing different parts, probably not. A guitar doesn't sound so much louder to me when a chord is played, compared to single-note melody work. Thanks a lot for your help with this.
     
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  16. Dec 5, 2020 #16
    Lowerleftcoast

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    Right.
    The *twice as loud* concept was relayed to me years ago when learning about FOH PA gear. The example was two jet engines. Two are twice as loud as one, yet we do not perceive it as twice as loud. It is arguably about 3dB difference, which is just noticeable. This carries over to twice the power in an amp as well. Twice as loud but just noticeable. The wife tells you to turn down no matter which one you use.;)
     
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