B500K pot down to a B100k Pot

Discussion in 'The Projects Forum' started by dragonrose, Oct 14, 2009.

  1. dragonrose

    Thread Starter New Member

    Oct 14, 2009
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    Hi all, currently working on a guitar were space is an issue, so im having to swap the pots for concentric pots, the issue is the active EQ in this guitar has B100K pots, and i cant find anywhere selling concentric B100K pots.

    I do how ever have a 500k/500k concentric pot so what size resistor would i need to use to bring a 500k pot down to 100k or as close as possible?

    On the multi-meter readings from my concentric pot which is rated 500/500, are 462/482, in case thats important.

    Thanks, Dave

    EDIT: For future reference in case anybody else gets the same EQ circuit, the circuits in question are the MT2 and MT3 circuits from artec sounds, for which the official wiring diagram is this:
    [​IMG]

    As not everyone is comfortable making new holes in there guitars and not all guitars are suitable for it anyway (quilted top guitars etc) here is how to wire one by replacing the pots with concentric pots:

    [​IMG]

    Also of note is the official diagram is rubbish, and next to useless, these EQ boxes have to be put int he circuit before the volume control not after the volume, as most guitars are wired.
     
    Last edited: Oct 16, 2009
  2. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    Dave,

    You need to have a 125kΩ in parallel with the 500k pot. You can see an example in the attachment. Whenever you need to do something like this again, maybe this will help:

    First, according to the equation for adding resistors in parallel, \frac{1}{R1} + \frac{1}{R2} = Rout So substituting algebra:

    \frac{1}{500k} + \frac{1}{x} = \frac{1}{100k}

    Therefore, x or the resistor you need is 125kΩ.
     
  3. dragonrose

    Thread Starter New Member

    Oct 14, 2009
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    0
    Hi, thanks for the help :)
     
  4. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    dragonrose,

    No problem!

    Notice that the wiper is completely irrelevant to changing the resistance value, but it focuses on both sides of the resistor. Keep in mind that whenever you have a resistor and you want to increase the resistance, place it in series; but when you want to have a fraction of that value, place it in parallel. Do you understand the math correctly?
     
  5. dragonrose

    Thread Starter New Member

    Oct 14, 2009
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    Yeah the math is sinking in somewhat :) Thanks, how does the 'sweep' of the pot react after havign the resistor on it, does the sweep stay the same or does it become 'all or nothing' effect? With it been an EQ say the treble the sweep has to stay steady not all or nothing, so it maintains the usability it was meant for :)
     
  6. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    The 'sweep' or the 'wiper' moves over the 100k pot you now have. Since you added that 125kΩ parallel resistor, it becomes a 100k pot. Imagine the wiper moving left and right of the resistor, one direction will give you less resistance and the other direction will give you more resistance. So the wiper simply selects what resistance you want and moves over that resistor to get the correct selection. Note that the wiper is in the middle, that would mean we have half of 125k which is 62.5k. That also means we have half of the magnitude it's capable of. When you move the wiper to the maximum value (125k) it will be theoretically "nothing" as you put it. Although, it actually means there is no current going through that point. But, if you move the wiper so there is no resistance, it will be at the most magnitude or "all" as you put it.
     
  7. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    Might I add that when I said the resistance would be 62.5k if the wiper was in the middle, this would be true, except for one thing. The attachement just shows a regular schematic symbol for a pot so whenever you see the wiper in the middle, don't automatically assume it would be half of the total pots value. It's just a shematic symbol for a pot.
     
  8. dragonrose

    Thread Starter New Member

    Oct 14, 2009
    6
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    Got my parts order they should be with me tomorrow, then i can put all this into action, thanks for all the help!
     
  9. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    Great! If you have a multimeter, I would advise you to check the resistance just to make sure it's 100kΩ. So going back to the diagram in the attachement, measure the resistance from point A to point B. Don't pay any attention to the wiper for this test.
     
  10. dragonrose

    Thread Starter New Member

    Oct 14, 2009
    6
    0
    Yeah i have my meter here been doing some tests on some pots i bought out of curiosity, the variation between actual resistance in some cases is surprising, out of 4 pots i bought all of which were 500k 1 is only 418k whilst another is 534k. And these were CTS pots which are meant to be good quality.

    On a side note, ive added my schematic (still got to add the resistor in) to the 1st post just in case somebody else with the same or similar EQ circuit can get some use from it.
     
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