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Is there a typical impedance needed to properly load passive pickups?

Discussion in 'Pickups & Electronics [BG]' started by Flux Jetson, Aug 9, 2012.


  1. Is there some typical amount of input impedance that your average passive pickups work best with?

    This is part of setting up my http://www.talkbass.com/forum/f36/experimental-fully-modular-bass-rig-902681/ ... I'll be adding a Squier Vintage Modified Jazz bass to the works in a day or so (as soon as I generate some cash-hish to finish paying off the lay away on it). I've been using an active Cort A4 up to this point.

    Using the passive Jazz made me think about input impedance levels and if I need to add a DI or something (a mic pre?) to place the proper load on those passive electronics that Jazz has.

    I'll be modifying the wiring, so that the pickups are given their own outputs straight from the volume controls (no tone control). So each pickup will have it's own channel in my modular preamp. Should I concern myself with using some sortof pickup loading circuit, like a DI on each pickup? Keeping in mind the way I'm setting this up.

    This experimental rig is part of a research and development effort. But it's also the way I want to use my bass. Yea, I know I'm a total square wheel.

    Thanks.
     
  2. After some sniffin' around my guess is that it's roughly 1meg/ohm?
     
  3. 1M amp input impedance, parallel to anything from one 500k volume, to three 250k volumes.

    Z[SUB]Total[/SUB]= 1/([1/Z[SUB]1[/SUB]]+[1/Z[SUB]2[/SUB]]+...[1/Z[SUB]n[/SUB]])

    So, with one 500k volume, 333.33k. Three 250k volumes, 76.92k.
     
  4. What will the passive Jazz pickups need (as far as the volume controls' values). Keeping in mind each pickup will go to it's own volume control, then straight out it's own output jack to separate signal processing? In other words:

    Neck pickup volume pot value = ??
    Input impedance for that output = ??

    Bridge pickup volume pot value = ??
    Input impedance for that output = ??

    Thanks. I'm not as smart as you are so that algorithm is above me.

    The pickups will be mixed/panned/blnded outboard using outboard gear for that purpose (in case that makes any difference).
     
  5. bongomania

    bongomania Gold Supporting Member Commercial User

    Oct 17, 2005
    PDX, OR
    owner, OVNIFX and OVNILabs
    Yes, that's the standard all-purpose good answer for passive basses. Many other numbers can be chosen, either for personal tone tastes or based on practical limitations; but the simple answer that never fails is 1 Meg.
     
  6. But that does not account for pots that decrease the input impedance seen by the pickups. As mentioned above, throwing some pots in will often drop the impedance by an order of magnitude.
     
  7. The math is very simple, if you just plug in numbers for the "Z"s and use a calculator. Or you can simplify to (Z[SUB]1[/SUB]*Z[SUB]2[/SUB])/(Z[SUB]1[/SUB]+Z[SUB]2[/SUB]) for two impedances at a time.

    Remember that resistive loading is a personal preference. Some prefer the darker tone of low value pots, others like very little loading.
     
  8. bongomania

    bongomania Gold Supporting Member Commercial User

    Oct 17, 2005
    PDX, OR
    owner, OVNIFX and OVNILabs
    If I have understood the OP correctly, he is giving each pickup its own separate output jack, and its own volume pot. The pots are not in circuit with each other. So yes, 1 meg certainly does account for a single volume pot.

    If I misunderstood and there will be multiple vol pots in circuit with each other, then maybe the math is required.
     
  9. Nope, you got it right. :) Your understanding of my intentions is spot on.
     
  10. So he's trying to provide an input impedance for each signal path that would be typical of a Jazz bass, correct? A 1M Ohms input impedance is typical for amps and effects and such, but what the pickup coils see is actually much lower, because volume and blend pots decrease that impedance. One 250k or 500k volume pot through a 1M input impedance should be fine, however, there would be no need for extra resistors in the circuit.
     
  11. No .. not a typical Jazz bass. There's no blend pot. Only one volume pot for the neck pickup, then out a jack. And one other volume pot for the bridge pickup, then out it's own jack. Two pickups, two discrete volume pots, two discrete output jacks. Both connected to separate inputs on two separate preamps.

    So "no" not anything like a typical Jazz bass. Just in case any of that makes a difference.
     
  12. The only difference between that and a typical Jazz bass, when pickups are soloed, is that the resistance of the volume pots' load doubles, since you have only one pot. (Assuming your traditional Jazz bass didn't have mixed pot values.) With two 500k volumes, that's a 250k load on the pickups, so you can use a 250k pot. Or, with two 250k pots, that's a 125k load. You can drop a 220k resistor parallel to a 250k volume, if you want that.
     
  13. Thanks. :) Anotated in my notebook (yes, I actually keep a notebook).
     
  14. khutch

    khutch Praise Harp

    Aug 20, 2011
    suburban Chicago
    No. If you are experimenting you should load them to suit your tastes. Most amplifiers you would use will have a 1 MegOhm input resistance so the total load on your pickups can never be any higher than that but how closely you approach that number is up to you.

    Ken
     
  15. Understood, but I needed a starting point. I also needed to know - for my notes - what nominal was.
     
  16. bongomania

    bongomania Gold Supporting Member Commercial User

    Oct 17, 2005
    PDX, OR
    owner, OVNIFX and OVNILabs
    No, more like a P bass with no tone pot. :)
     
  17. Badda-friggin-BING!
     
  18. Then don't worry about it. Typical 1M input impedance and your preference of a 250k or 500k volume pot. No further thought and planning needed.
     
  19. Thanks.

    Like I said, just needed a starting point.
     

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