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Fixed resistor to replace volume control?

Discussion in 'Pickups & Electronics [BG]' started by fourstringbliss, Feb 19, 2014.


  1. fourstringbliss

    fourstringbliss Supporting Member

    Joined:
    Oct 5, 2003
    Location:
    Puyallup, WA
    I'm going without a volume control on my bass, but don't know about wiring my pups straight to the jack once past the blend control. Can I wire in a 500k resistor between the blend output and the wire to the jack? Would that simulate a 500k pot wide open?
     
  2. ngh

    ngh

    Joined:
    Feb 6, 2013
    Location:
    brooklyn, ny
    no that would simulate a 500k pot rolled off all the way. a 500K pot wide open has virtually no resistance.
    in short your over thinking it. it might be a good idea to have a kill switch on there though.
     
  3. fourstringbliss

    fourstringbliss Supporting Member

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    Oct 5, 2003
    Location:
    Puyallup, WA
    So, a 500k pot wide open sounds pretty much like the pickup wired straight to the jack?
     
  4. ngh

    ngh

    Joined:
    Feb 6, 2013
    Location:
    brooklyn, ny
    pretty much, the vary fact you are running through a component means you are going to get some resistance but you would be hard pressed to notice the difference in any practical application. also if you are running through a blend pot you are already running it through two volume pots (one for each pickup). so your pickups are actually not running straight to the jack.
     
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  6. aphexafx

    aphexafx A mind is a terrible thing. Supporting Member

    Joined:
    Dec 10, 2013
    Location:
    Denver, Colorado
    No.

    To mimic a 500K pot wide open you would add a 500K resistor between the output hot and ground. On your bass with a blend, the blend output is the output hot. That goes straight to the output jack tip and the 500K resistor would go between the output jack tip and the output jack sleeve (ground).
     
  7. fourstringbliss

    fourstringbliss Supporting Member

    Joined:
    Oct 5, 2003
    Location:
    Puyallup, WA
    I'm wiring up a Bartolini NTMB and am going without a volume pot because I never use it. I'm working out the active/passive switch and have the active side of it thought out, but wanted the passive side of it to sound like it would through the blend and a 500k volume wide open. You're saying it will sound that way if it goes through the blend and then to the output (routed through switch terminals), right?
     
  8. Crater

    Crater

    Joined:
    Oct 12, 2011
    Location:
    Dallas, TX
    That is completely wrong. A 500K pot wide open is


    ...500K.

    When it's rolled all the way down the resistance is 0 ohms.
     
  9. Bassamatic

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

    Joined:
    Dec 7, 2007
    Location:
    Studio City, SoCal, USA
    As mentioned above, you would put the 500K resistor from hot to ground to simulate the pot being full ON. However, since the 500K pots are chosen to have minimal effect on the sound, there seems to be little reason to do it.
     
  10. Phalex

    Phalex Semper Gumby Supporting Member

    Joined:
    Oct 3, 2006
    Location:
    G.R. MI
    I thought I wanted a toggle switch on my bass for "on" and "off", but as I was considering this easy mod, I noticed that I seem to use the volume knob a fair amount. Not so much for setting the volume of my bass, but for swells and such.
     
  11. garfoid

    garfoid

    Joined:
    Apr 10, 2010
    If you don't want a volume pot, just wire directly to the jack.Yet, don't forget that the Blend pot is putting a load on the signal, so, why not leaving the volume pot there anyway!?
     
  12. garfoid

    garfoid

    Joined:
    Apr 10, 2010
    I believe that if you do it that way, you'll kill the signal.When you use a pot, when(on) the resistance is "zero", when you turne it (off) the pot is at max resistance shunting the signal to ground.Anyway, I might not be right, But I'm sure that if you just send the blend straight to the jack tehre you have your sound open wide.The 500k Why put a resistor if you already have your sound open?!
     
  13. line6man

    line6man

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    Jun 20, 2008
    Location:
    Close to Los Angeles, CA
    When a volume pot is turned all the way up, there is a resistance equal to it's rated value, plus or minus the tolerance, parallel to the signal path, and zero Ohms in series with the signal path.

    To simulate a volume pot on full, you would place a resistor parallel to the signal path.
     
  14. fourstringbliss

    fourstringbliss Supporting Member

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    Oct 5, 2003
    Location:
    Puyallup, WA
    So, if I put a 500k resistor between the bottom right and middle right switch lug in your a/p switch would that work?
     
  15. bassbenj

    bassbenj

    Joined:
    Aug 11, 2009
    Depends on how it's wired. Sometimes the pot shorts the pickups and others it acts like a voltage divider.

    However, point is that pickups (or blend control) wired directly to the jack is the same as volume up max. Yes, in reality the volume control has a tiny series resistance and 500k in parallel with the output. But that really makes no difference because the impedance of the pickups is only a couple KOhm. Thus for all practical purposes a direct connection is identical to volume up max. (difference between a few thousand ohms and a few thousand ohms in parallel with 500 thousand ohms)

    And remember the input resistance of the amp is in parallel too which can be from maybe 50k to 1 meg depending on the amp. Which also won't matter.
     
  16. Chainsaw Willie

    Chainsaw Willie

    Joined:
    Nov 6, 2011
    Location:
    Redmond, Washington
    Volume pots are wired as a voltage divider. There is no "series resistance". All resistances are in parallel to the input or the ouput.

    The pickup always sees 500 ohms. it is wired to the outer lugs.

    The input jack is what sees zero ohms, 500k ohms or something in between depending on knob position. The input jack is wired: tip to the center lug of the pot (the variable resistance), the outer ring is wired to the same grounded outer lug of the pot that the grounded side of the pickup is.

    So, the pickup is always loaded by 500K ohms acoss it, no matter what the volume setting is.

    If you constantly want to run full volume:
    1. Wire grounded side of pickup to ground and one side of 500K resistor and to the outer ring of the output jack.

    2. Wire hot side of pickup to the other side of the 500K resistor and the tip of the output jack.

    The word "pickup" above can be replaced by "output of the blend pot".


    Edit for example of simplified wiring:
    http://cdo.seymourduncan.com/blog/wp-content/uploads/101_Diag_4.png
    http://fenderguru.com/wp-content/uploads/2011/06/PM-guitar-simple.png
     
  17. line6man

    line6man

    Joined:
    Jun 20, 2008
    Location:
    Close to Los Angeles, CA
    This is untrue. Look at the physical manner in which a potentiometer does what it does. As you turn down from full volume, the wiper terminal moves away from one side, causing the resistance between the two terminals to increase. At the same time, the wiper moves closer to the other terminal, and the resistance between those two terminals decreases. This means that there is a resistance between the input and the output, and a resistance between either the input or output (depending on wiring), and ground.

    The pickup does not see a constant input impedance.
    Let us assume that the first terminal of the pot is wired to the pickup, the second terminal of the pot is wired to the output of the bass, and the third terminal of the pot is grounded. Parallel to the pickup, there will be the resistance of the rating of the pot, plus or minus its tolerance. However, there will also be the input impedance of whatever the bass is plugged into, in series with the resistance between terminals one and two. The sum of the input impedance from what the bass is plugged into, and the resistance from terminals one to two appears parallel to the pickup. Since the input impedance of whatever the bass is plugged into is not infinite, the pickups will always see a resistive load that is less than the rating of the pot, plus or minus its tolerance. This is because the total resistance when two or more resistances are placed parallel to each other is less than the lowest of the resistances. Whatever the total resistance comes out to be, it is also not constant, because the resistance between terminals one and two of the volume pot is variable.
     
  18. Chainsaw Willie

    Chainsaw Willie

    Joined:
    Nov 6, 2011
    Location:
    Redmond, Washington
    Indeed,

    but we talk too much!

    The OP just needs to put 500K in parallel with the output jack, or just wire direct to the ouput jack (with no resistor) and let the amp imput impedance be the load.
     
  19. RobbieK

    RobbieK

    Joined:
    Jun 14, 2003
    Umm, I think the OP only wants the resistor there in passive mode, and I think was also asking how to wire this with their active/passive switch. But to clarify, wiring this across the jack means it will be there in both passive and active modes, but because it's after the preamp, it will have no effect in active mode due to the nice low impedence of the preamp, so there's no real reason to wire it to the switch. In fact if it's a typical preamp made with modern opamps, you could have a 10K or even 5K resistor across there before you notice much in active mode!

    As others have said, frankly it will have only a very small effect in passive mode, but if it gets you to your tone and/or you learn a little bit about guitar wiring, then go for it.

    Oh, and if you go to a hobby electronics shop, your resistor will be probably have to be a 470K. 500K is not a standard size.
     
  20. line6man

    line6man

    Joined:
    Jun 20, 2008
    Location:
    Close to Los Angeles, CA
    +1 to all of this.
     
  21. Munjibunga

    Munjibunga Total Hyper-Elite Member Gold Supporting Member

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    Location:
    Groom Lake, NV
    Disclosures:
    Independent Contractor to Bass San Diego
    Note to OP: Good luck deciding on whom you're going to rely.
     

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