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volume pot subbing question

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  • volume pot subbing question

    I have a Hartke HA3500 head in with a broken volume pot VR1. It is a long shaft 50K linear pot with the wiper tied to one leg. Full Compass has this pot in a 100K linear, can I get away with connecting a 100K resistor across the legs to make it act like a 50K pot? I realize the taper won't be the same, but it shouldn't matter to the customer, he bought it this way.
    It's weird, because it WAS working fine.....

  • #2
    In a pinch, paralleling a resistor across the control will 'work'.

    You are correct that the taper will be off.

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    • #3
      No, it should be exactly the same. You have a linear pot with the wiper tied to the end, 100k across it now it's a 50k linear.

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      • #4
        Oops.
        I was thinking audio taper.

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        • #5
          Originally posted by mozz View Post
          No, it should be exactly the same. You have a linear pot with the wiper tied to the end, 100k across it now it's a 50k linear.
          Not quite. A linear 50k pot with the wiper tied to the end gives 25k in middle position. A linear 100k pot in middle position paralled with a 100k resistor means 50k parallel to 100k resulting in 33.3 k.
          Last edited by Helmholtz; 08-15-2019, 07:27 PM.
          - Own Opinions Only -

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          • #6
            Is the pot either VR1 or VR2 in the schematic below? (Both in the feedback loop of an opamp.)

            http://schems.com/bmampscom/hartke/h...lifier_sch.pdf

            If so, then you can probably get away with replacing the 50Kpot with 100K if you also double the value of either R121 or R122. (10K will work fine; you won't find 9.98K. )

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            • #7
              Close enough. How else would you get a close result without overly complicating the repair?

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              • #8
                Close enough.
                That may well be.
                My comment was meant to support the statement that taper will change and no longer be linear with a parallel resistor. In fact resulting resistance will change faster in the lower resistance range.
                Last edited by Helmholtz; 08-15-2019, 05:50 PM.
                - Own Opinions Only -

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                • #9
                  "Is the pot either VR1 or VR2 in the schematic below? (Both in the feedback loop of an opamp.)"

                  Yes, it is VR1.

                  "Not quite. A linear 50k pot with the wiper tied to the end gives 25k in middle position. A linear 100k pot in middle position paralled with a 100k resistor means 50k parallel to 100k resulting in 33.3 Ohm."

                  Did you mean to say 33.3K Ohm?
                  It's weird, because it WAS working fine.....

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                  • #10
                    Did you mean to say 33.3K Ohm?
                    Yes, thanks.
                    - Own Opinions Only -

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                    • #11
                      Originally posted by Randall View Post
                      "Is the pot either VR1 or VR2 in the schematic below? (Both in the feedback loop of an opamp.)"

                      Yes, it is VR1.
                      So, if you replace the pot with one with twice the resistance, then the gain will double at all settings of the pot. Doubling R122 (4.99K -> 10K) will restore normal operation.

                      C115 (0.68uF) puts the low frequency cutoff at 46Hz in the stock circuit. Doubling R122 will cut that frequency in half (23Hz). If you really wanted to do it up, you would replace C115 with half the capacitance (0.33uF) but I doubt it will matter much in the end. There is some low frequency attenuation up into the frequencies of interest with the factory circuit, especially with bass guitar. This will change if you don't also replace C115.
                      Last edited by Tony Bones; 08-15-2019, 07:46 PM.

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