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Question of pots and ohms

Discussion in 'Pickups & Electronics [BG]' started by Toastfuzz, Aug 22, 2012.


  1. Toastfuzz

    Toastfuzz

    Jul 20, 2007
    Pittsburgh, PA
    Uh oh, this may be getting into dangerous territory... this isn't a bass question, its for my guitar.

    Adding a Seymour Duncan SD-1 mini humbucker to my G&L ASAT (Tele clone) neck. The SD wants a 500k ohm load to not sound muddy, but the bridge needs 250k ohm to not be too bright. Guitar has 1 vol, 1 ton, 3 way switch.

    Do I need to consider the tone pot ohms in the following plan? One solution I found out on the web was putting in a 500k ohm volume pot for the humbucker, then running a 470k ohm resistor between the bridge pup hot and ground, to reduce the ohms for the bridge only to around 250k. But I wasn't sure if the tone pot would have an affect.

    Reading a bit more its sounding like if I had a 500k vol and 250k ton, the total load would be 375k, right in the middle. But could I then, instead, replace both vol and ton with 500k, and run that resistor on the bridge pup's hot line (470k to cut it in half) then I'd essentially have the bright 500k effect on the HB and darker 250k effect on the bridge single coil. Is that right?

    Hopefully picking up parts tonight, so any help today is much appreciated!
     
  2. A higher value tone pot functions identically to a lower value pot throughout part of the rotation, and will be brighter when set beyond the lower value's resistance. For example, aside from taper differences, a 500k tone pot will function identically to a 250k tone pot from 0 to 250k Ohms, then further remove the capacitor from the circuit from 250k to 500k Ohms.

    166.66k Ohms.

    Also, you're referring to "resistance," not simply "Ohms." The Ohm is a unit of measure, and can be used to describe resistance, reactance or impedance.
     
  3. Just to explain that number, when resistances are in parallel they add up like this:

    1/Rt = 1/R1 + 1/R2

    So

    1/Rt = 1/500k + 1/250k

    1/Rt = 3/500k

    Rt = 500k/3 = 166.66k
     
  4. Bassamatic

    Bassamatic keepin' the beat since the 60's Supporting Member

    Just FYI - The sum resistance of loads in parallel as always less than the lowest resistance.
     
  5. Unless there is a theoretical true zero Ohm load, in which case, an infinite number of resistances can be added without changing the total.

    Then again, things tend to burst into flames when you divide by zero. Wouldn't recommend it.:bag:
     

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