Quick impedance question

Discussion in 'Pickups & Electronics [BG]' started by eukatheude, Jan 31, 2013.


  1. eukatheude

    eukatheude

    Joined:
    Apr 2, 2012
    Location:
    Italy, Brescia
    So, i was wondering. Since having 500K pots instead of 250K on, for instance, a J, can give you a brighter sound, wouldn't it be the same if i put a 750K resistor between the output wire and the output jack? Thanks
  2. Crater

    Crater

    Joined:
    Oct 12, 2011
    Location:
    Dallas, TX
    No, it wouldn't be the same. It would reduce your output (loudness) a lot, and would reduce some treble too.

    You could replace the standard 250K or 500K with a 1 megohm (1,000 K ohms) pot, or use a Fender "no-load" pot that removes the resistance to ground entirely when the the control's on 10.
  3. line6man

    line6man

    Joined:
    Jun 20, 2008
    Location:
    Close to Los Angeles, CA
    I'm not sure where you got 750k from, but for the average pickup impedance, that's enough to kill most of the output.

    What you would do to go from 500k to 250k is place a 500k resistance parallel to the signal. Realistically, 470k or 510k resistors will be easiest to find.
  4. eukatheude

    eukatheude

    Joined:
    Apr 2, 2012
    Location:
    Italy, Brescia
    (500*3) - (250*3)

    So if you were looking at a J wiring, where would that be? (don't waste time drawing, it's just a curiosity)
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  6. line6man

    line6man

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    Jun 20, 2008
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    Close to Los Angeles, CA
    Where did the number 3 come from?

    When dealing with resistances and impedances, just remember the series and parallel formulas.
    Series: R[SUB]Total[/SUB]=R[SUB]1[/SUB]+R[SUB]2[/SUB]+...R[SUB]n[/SUB]
    Parallel: R[SUB]Total[/SUB]=1/([1/R[SUB]1[/SUB]]+[1/R[SUB]2[/SUB]]+...[1/R[SUB]n[/SUB]])

    The resistor can just run parallel to the jack.
  7. eukatheude

    eukatheude

    Joined:
    Apr 2, 2012
    Location:
    Italy, Brescia
    3 pots.

    How would i wire it parallel to the jack?

    Not sure about parallel, if i had say a 100k and a 200k resistors, total resistance would be 1/[(1/100)+(1/200)] = 1/(0.01+0.005) = 1/0.015 = 66.666?
  8. line6man

    line6man

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    Jun 20, 2008
    Location:
    Close to Los Angeles, CA
    The tone pot in a VVT setup presents a resistance that varies with frequency. At high frequencies, it approaches the value you have dialed in. At low frequencies, it approaches infinity.

    One side of resistor goes to the tip, and the other goes to the sleeve.

    100k parallel to 200k is 66.667k. Be mindful of the "k" suffix, which means "*1000."
  9. SGD Lutherie

    SGD Lutherie Supporting Member

    Joined:
    Aug 21, 2008
    Location:
    Bloomfield, NJ
    Disclosures:
    Owner, SGD Music Products
    The series resistance would increase the impedance and dull the tone and reduce the output. The control pots are parallel resistance.

    You can use a larger value pot, like a 1M if you want a brighter tone, or go active, which accomplishes the same thing since it isolates the pickup (buffers) from the cable and controls.

    As much as some people say that active sounds artificial, it's the true sound of the pickup with minimal loading.
  10. eukatheude

    eukatheude

    Joined:
    Apr 2, 2012
    Location:
    Italy, Brescia
    Thank you. I'll try putting a spare resistor and see what it does to my tone.
  11. wcriley

    wcriley

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    Apr 5, 2010
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    Western PA
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    Uncompensated endorsing user: fEARful
    Repeating for emphasis.

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