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    I can usually sort this sort of stuff out, and have read up on a coupe different Tube Screamer analysis articles. But I really want to build a a pair of Tube Screaming Ultra into one enclosure with only three physical potentiometer housings – using three dual-gang concentric pots. The problem is that I cannot locate an Alpha 9mm sized dual-gang concentric in a B25K value. My only easily available options are B10K or B50K. So, which of these two values could best be used in place of the B25K Tone pot (P2) in the Tube Screaming Ultra? And I’m reasonably sure that a value change in C5 and\or C6 would  be needed to compensate the resistive value change within that low-pass filter configuration. I’m taking a guesstimate of 47nF in place of the 220nF value(s) for C5 & C6 if using a B50K pot. (And yes, I’m aware the original TS circuit uses a 20K pot with a “‘G’ or ‘W’ taper”.)

    I also know I could use the B50K dual-gang concentric pot and hang a 50K resistor across lugs #1 & #3 of each range to achieve the original 25K pot value, but doing so would drastically change the taper of the Tone pot. To some, a taper change in the Tone pot would be no big deal, but I’d like to preserve the linear sweep if possible. (This seems the most practical approach because the taper change, (using the added parallel resistance), can be somewhat controlled and more closely approximate the “G taper” were I to use a 25K resistor across both pot lugs 1 & 2 and 2 & 3. That configuration would would achieve a total of a 25K pot, but the taper would approximate more of a “w” shape taper vaguely mimicking the traditional “‘G’ or ‘W’ taper”. Strapping a 50K resistor in parallel across pot lugs 1 & 3 of a linear taper pot typically changes the pot’s taper to more closely approximate a “C taper” (reverse log\audio).

    I just thought I’d ask the group before experimenting with the “socket & see” approach. Thanks for reading.


    The “B” pots are a linear taper.  Adding a resistor across lugs 1 and 3 is adding a linear resistance to a linear taper.  Therefore, the resulting taper is unchanged.  Only the resistance value changes to @25K.

    If you have a log pot; either “A” taper, “C” taper or some other, adding a linear resistance in the form of a resistor will definitely change the taper by reducing the slope of the log curve.

    I’m not sure about the result of adding 2 25K resistors as mentioned.  I would like to try that and plot a resistance curve as the wiper is rotated through the 300 degree (or so) rotation.


    Another question would be, “Can you hear the difference?” or would you just notice that the pot knob was in a different position for a perceived tone or level?  Interesting questions.


    Wilkie – Thanks! “The “B” pots are a linear taper.  Adding a resistor across lugs 1 and 3 is adding a linear resistance to a linear taper.  Therefore, the resulting taper is unchanged.  Only the resistance value changes to @25K.” When I tried this with a linear pot, my measurements seemed to indicate a different taper than linear. Perhaps my plotting was off.

    I haven’t gotten to the point of firing it up much less populating the PCBs. I’ still in the planning stages and before ordering the parts, (I don’t stock concentric dual-gang pots), I wanted to answer the questions floating around in my head.

    And yes, the resultant difference is audible between “B”, “A” and “C” tapers as tested in other tone circuits. In my experience, “A” tapered pots will reverse the curve when a resistor is laid across lugs #1 & #3. Again, just going thru discussions in my head before committing to a specific build direction. And as always, will socket and see before boxing is completed.


    I’ll look around to see if there is more info on this subject.  An interesting issue for sure.


    Well, my intuitive logic that says if it looks like a duck and quacks like a duck…it must be a duck!  WRONG!

    My 1963 Electronics 101 must have lost this amidst the Kirchhoff Law discussions.

    Here is a link to some equations that will assist you.  You are correct in using 2 resistors instead of one.  Now, I have some new info to use in my quest for more exotic pedal circuits.

    I’ll wipe the mud from my face as I thank you for broaching this subject.




    I took a B50K pot (measured 50K5 Ω) and a C50K pot (measured 47K8) and measured them at the following positions:

    Position     “B” Value     “C” Value

    • Full Left         0 Ω              0 Ω
    •  8:00            3K1 Ω          0K2 Ω
    •  9:00            8K2 Ω         11K1 Ω
    • 10:00         14K8 Ω         31K3 Ω
    • 11:00         20K8 Ω         39K6 Ω
    • 12:00          25K Ω          41K1 Ω
    •  1:00          30K9 Ω         43K4 Ω
    •  2:00          36K8 Ω         44K8 Ω
    •  3:00          43K8 Ω         46K6 Ω
    •  4:00          48K7 Ω         47K2 Ω
    • Full Rt        50K5 Ω         47K8 Ω

    Then I tied 51K (2 x 25K5) Ω across their lugs #1 & #3 respectively

    The “B” then measured @ 25K2 Ω and the “C” @ 24K8 Ω across lugs #1 & #3  . . . .

    Position     “B” Value     “C” Value

    • Full Left         0 Ω               0 Ω
    •  8:00            3K3 Ω          0K6 Ω
    •  9:00            9K2 Ω         11K7 Ω
    • 10:00         15K1 Ω         20K1 Ω
    • 11:00         17K2 Ω         23K8 Ω
    • 12:00         19K4 Ω         24K1 Ω
    •  1:00          21K9 Ω         24K4 Ω
    •  2:00          23K5 Ω         24K8 Ω
    •  3:00          24K7 Ω         24K6 Ω
    •  4:00          25K4 Ω         24K2 Ω
    • Full Rt        25K2 Ω         24K8 Ω

    Sorry for sloppy list format, I couldn’t find my graph paper to do a proper graph. Still, it tends to support the concept that adding a resistor across lugs #1 & #3 of a pot does change its taper.

    And after doing all of this, it occurs to me that the “sweep” behavior (taper) of a Tone pot is not as critical as Volume, Gain or Drive.

    So, I shall go with the B50K dual-gang concentric pot and strap a 50K resistor across lugs #1 & #3 for the Tone pot. I’ll just live with the tiny bit of weirdness toward the extreme full-right position.

    Thanks for the feedback Wilkie. Your initial counter-point prompted me to do the manual measurements and re-acquaint myself with the “mathematic visual” aspect of pot tapes. Good link too, by the way. Thanks for that. I gotta new bookmark!


    But to really get funky with the original question, (and here’s where “socket-and-see” will really expose what may or may not transpire in discussion), how would it affect the circuit were I to use a B50K pot in place of the B25K pot for P2 (the Tone pot) in the Tube Screaming Ultra?????


    From your results above, I guess you would not have much tonal change from 12 o’clock to full CW.  Interesting.


    Right. I’ve yet to apply two 25K Ω resistors, using them one each across lugs #1 & #2 and #2 & #3 respectively. I’ve not plotted that variant yet. If the dual resistor method doesn’t deliver a somewhat more linear response, perhaps EMH’s “Tapered Pot” calculator can help compensate for the constrained linearity. More reading and measuring to do.


    Please share your results.  Curious minds await.


    Had few minutes this morning to tied those 25K5 Ω resistors to lugs #1 & #2 AND #2 & #3 respectively and got the following plot . . . . .

    Position       “B” Value     

    • Full Left         0 Ω
    •  8:00            2K9 Ω
    •  9:00            7K1 Ω
    • 10:00           9K1 Ω
    • 11:00         10K1 Ω
    • 12:00         12K8 Ω
    •  1:00          14K3 Ω
    •  2:00          15K1 Ω
    •  3:00          16K2 Ω
    •  4:00          16K8 Ω
    • Full Rt        16K8 Ω

    I find it odd that the max is only 16K8 Ω when, according to Ohm’s Law, should be right about 25K Ω. My best guess is that Kirchhoff’s Law has something to say about it, but I only ever dove into those rules in the 80’s and I’ve slept since then. At least this result more closely approximates a linear response, with a of pinching at far-right extreme.

    Next I shall play with some different values across lugs #1&#2 AND  #2&#3 to see how EMH’s “Tapered Pot” calculator may help compensate to bring the resultant total closer to a total of 25K Ω with closer proximity to a linear 0 Ω to 25K Ω.


    That appears to act like a “C” taper with more changes taking place in the first half of rotation and more linear in the last half.



    Agreed. It seems the 2 equal value resistor method renders the pot to a reverse-analog taper.

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