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Thread: Antilog/ reverse pots

  1. #1
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    Antilog/ reverse pots

    Cant find a 500k revrse log pot for a MXR microamp clone, so Im using a std 500k with obvious 'last 10th' of dial where the action is.. annoying.. so where do I find one (Im in uk)? or is there a way another pot can be made into one? cheers Capt

  2. #2
    Noodle of Reality Steve Conner's Avatar
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    I've never had much luck finding reverse log pots either. The obvious trick is to use a standard log pot but swap the two outer lugs over. You now have a reverse log taper, but you have to turn the knob anticlockwise to increase

    Alternately, you can use a linear pot, which is a better match to a reverse log characteristic than a log pot.

    Finally if you're really keen, you can get a 500k log dual gang pot in the mini Alpha style and dismantle it. The rear section is a reverse log, and you can swap it with the front one to get a 500k dual gang reverse log, or install it in the body of a single gang Alpha pot. I could never be bothered doing this though.
    "Enzo, I see that you replied parasitic oscillations. Is that a hypothesis? Or is that your amazing metal band I should check out?"

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    Read "The Secret Life of Pots" at GEOFEX.

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    Quote Originally Posted by R.G. View Post
    Read "The Secret Life of Pots" at GEOFEX.
    Thanks for both replies- elusive little b'stards to source.. I dont see any prob with having the knob 'wrong way round' so I'll try just swapping lugs 3 and 1. Useful link too- I told my wife I was reading the 'secret life of pots' and she just let out a big sigh! I think I have become a proper nerd.

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    There is NO need, I repeat NO NEED, to use a reverse-log pot on the micro-amp and Distortion+. Either will work just fine with a widde range of other values/tapers. Where taper/value starts to matter is if there is some specific range where your desired gain setting becomes hard to nail.

    Do the math for yourself and isolate where your preferred gain settings are. For the Micro-amp, the gain is given by the ratio of the feedback resistor (56k), plus the pot resistance and 2k7 fixed resistor, to the sum of the pot plus 2k7. So, if the pot is set to zero resistance, then the gain is [56k + 0k +2k7] / [0k + 2k7] = 21.7. If the pot is set to maximum resistance (500k), the gain is then [56k+500k+2k7]/[500k+2k7] = 1.11.

    The reason why the anti-log pot is needed is to get you through the first couple of hundred K of pot resistance in a hurry. For instance, you will note that if the pot is set to half-resistance, the gain is 56k+250k+2k7/250k+2k7 = 1.22, meaning that half the resistance of the pot actually changes the gain factor by a little over 1%. Drop the pot resistance to 100k and the gain factor is still only 1.55. At 50k pot resistance, the gain is 2.06. So, with 90% of the pot resistance used up, we have still only budged from a gain of 1.11 to a gain of 2.06.

    My advice? Unless you plan on using gain factors less than x2, use a linear OR log 50k pot and be done with it. Alternatively, along with the 50k pot, change the 2k7 resistor to 3k3, change the 56k feedback resistor to 68k, replace the 10k output terminating resistor with a 10 log volume pot, and you will be able to nail ANY preset level from less than unity up to x26, with no sacrifice in tonal quality or bandwidth. Plus, with the added versatility of the output level pot, you'll have far more precision than the original permitted. Keep in mind that 70's pedals tended to opt for fewer knobs/switches, so the requirements of the pots they did use were sometimes outlandishly specific because of that.

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    Great help as always there Mark, Im going to spend time on this reply tnite with a lin 50k and do some maths etc. Muchos gratias, Capt

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    Custom tapers for pots

    FWIW you can custom-make an antilog taper on your average linear or log taper pot, by wiring in a fixed resistor (of whatever nominal value you have worked out by running the numbers) between the pot input and the wiper. (just as you can custom-make a log taper from a linear pot, or extend a log taper pot, by wiring a resistor in from the wiper to the pot ground).

    Attachment courtesy of Steve A.
    Attached Files Attached Files

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    Pickup Maker David Schwab's Avatar
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    I dislike log pots. I use linear for just about everything.
    It would be possible to describe everything scientifically, but it would make no sense; it would be without meaning, as if you described a Beethoven symphony as a variation of wave pressure. ó Albert Einstein

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    Smallbear have them.

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    Will this work for balance control pots of an audio preamplifier. I have a McIntosh MX110 tube preamp with a failed balance control. The OEM control consists of 2 ganged 68K pots. I think one is log and 1 is anti log but not sure. I can only find linear pots to replace. Can I use the 20% of the total pot resistor value from the center tap out to give me what I need.

    Thanks

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    Senior Member Enzo's Avatar
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    If you are not sure, take the control out and check with an ohm meter to find out.

    However, do you operate the amp with the balance control not centered very often? In other words, does it matter much what the taper is if it balances in the middle?
    Education is what you're left with after you have forgotten what you have learned.

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    Noodle of Reality Steve Conner's Avatar
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    Have you tried contacting McIntosh? They seem like the kind of company that might send you a replacement.

    McIntosh | McIntosh Owners

    If not, I guess a dual-gang linear pot would do. The original control would probably have been log and antilog.
    "Enzo, I see that you replied parasitic oscillations. Is that a hypothesis? Or is that your amazing metal band I should check out?"

  13. #13
    Senior Member Enzo's Avatar
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    Mac will sell parts to individuals, I just had to do that a couple months ago.
    Education is what you're left with after you have forgotten what you have learned.

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    Quote Originally Posted by kaplang View Post
    Will this work for balance control pots of an audio preamplifier. I have a McIntosh MX110 tube preamp with a failed balance control. The OEM control consists of 2 ganged 68K pots. I think one is log and 1 is anti log but not sure. I can only find linear pots to replace. Can I use the 20% of the total pot resistor value from the center tap out to give me what I need.

    Thanks

    Balance pots are typically ganged D and W tapers a variant of the A & C Log and Anti-Log tapers. Pretty specific for the job. MacIntosh can supply these. They aren't cheap and QC is often not to the old standard, but they got 'em and they work.

  15. #15
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    The AMZ trick "works", sort of, *only* on pots connected as voltage dividers (1 pin hot, other Gnd and wiper out) by loading down one of the voltage divider halves.
    What he does not mention, is that your pot becomes a much lower resistance one, when on "10"
    Example: in a classical Fender tone stack followed by a volume pot, the tone stack sees that typical 1M *always*, no matter what the volume setting.
    If you "simulate" it with a 1M linear pot plus a 220K resistor from wiper to ground, everything is fine from 0 to 5, but on 10 your pot becomes a 1M//220K resistor, around 160K, *much* lower than intended and throwing out of whack the equalization.
    And the curve is not that smooth either, if used on a real world amplifier.
    If the pot is used as a variable resistor , not as voltage divider (as in MicroAmp, Dist+ and countless others) the "mod" suggested does not work at all; when on 5 the pot will still have 500K, not the 100K expected on a true antilog.
    By the same token, in a Tube Screamer it will have 250K, not the 50K expected from a true log, *or* it will only get to a maximum value given by the padding resistor in parallel with the 500K track.
    Where the designer specifies a Log or AntiLog pot, he has his reasons to do so, please respect him.

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