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How does it all work?

Discussion in 'Pickups & Electronics [BG]' started by mjolnir, Feb 27, 2008.


  1. mjolnir

    mjolnir Thor's Hammer 2.1.3beta

    Jun 15, 2006
    Houston, TX
    I've been on a HUGE modding streak lately, yanked the electronics on every bass I have and mixed, matched, bought new stuff, and generally made frankensteins out of every bit of electronics I can get my hands on, and I've (so far...) been very successful in getting the sounds I want with my random tinkerings. The thing is... I don't really know how it all works. I know what goes where, and I know it's a good idea to solder this wire to this spot and not that one, but I have no idea why...

    I got curious and opened up a volume pot just now, that's what brought all this on.

    So. Why is it that when you turn a volume pot, the volume increases or decreases? How does that little capacitor on the tone pot color the sound going through it? Just in general... Why do the electronics in a bass do what they do?
     
  2. Rickett Customs

    Rickett Customs

    Jul 30, 2007
    Southern Maryland
    Luthier: Rickett Customs...........www.rickettcustomguitars.com
  3. the volume knob perhaps is a potentiometer, it just decreases the amount of current flowing thourgh the pickup hence the volume when you turn it down
     
  4. mjolnir

    mjolnir Thor's Hammer 2.1.3beta

    Jun 15, 2006
    Houston, TX
    Absolutely fascinating...

    I feel really nerdy saying that, but after reading through "The Secret Life of Pots" I'm so hungry for more...

    Thanks for the responses, guys!
     
  5. Things to check out:

    1) Ohms law (this will basically answer your volume pot question) - http://en.wikipedia.org/wiki/Ohm's_law

    2) Basic RC filter analysis (this will answer your tone pot questions) - http://en.wikipedia.org/wiki/RC_circuit
     
  6. the volume pot is in fact a variable resistor in series with the pickup. In the completly off position all the signal is shorted to ground, if you turn it up a little part of the signal is ''lost'' in the pot since the voltage is split between the pot and the pickups. As you turn the volume up, the resistance gets lower and lower till it is completly bypassed in the fully open position then, all the voltage comming from the pickup gets to the jack.

    As for the tone pot, if you put a capacitor in series with a resistor, you have a low pass filter. what it does is it lets all the lower frequency that passes through it and send all the high frequency down to a certain point to ground, this point is decided by the value of the resistor. So bassicly, when the tone is all the way off, it cuts the maximum high frequencys, as you move that point up, more and more high frequencys are allowed through up to the full on point where almost all highs are there, I said almost, because there is always a small number lost, the only way to counter-act this is bypassing the tone knob.

    I hope this helps I tried to be the least technical I could.
     
  7. mjolnir

    mjolnir Thor's Hammer 2.1.3beta

    Jun 15, 2006
    Houston, TX
    ...Actually I don't mind you getting technical at all. I'm an IT technician that is forced to do electronics work at my job all the time, so I'm not entirely clueless. Hit me with yer best shot, guys, I'll let you know when you go completely over my head. :D
     
  8. The volume pot is a typical V=IR Ohms law problem.

    You have a variable resistor with a tap -


    Vin+
    |
    \
    / Ra
    \
    |
    |----- Vout+
    |
    \
    / Rb
    \
    |
    |
    |
    GND


    Where Ra/Rb are the volume pot, and Vout+ is connected to the pot tap.

    Ra + Rb = pot resistance. As Ra rises Rb falls and vice versa.

    Your pickup generates some voltage across the equivalent resistance, and from that you can determine what voltage is dropped across each resistance.

    Lets assume a 250k ohm voltage pot, and a voltage around .7 volts (a typical output voltage from a passive pickup).

    We know Vin+=IR. We know Vin+=.7V. We know R=Ra+Rb=250k ohm.

    I=.7V/250k ohm = 2.8 x10^-6 amps.

    Vout+ = I*Rb = (2.8x10^-6)*Rb

    When the volume is at "zero":

    Rb= 0 ohms.
    Ra = (250k-0) = 250k ohms.
    Vout+ = (2.8x10^-6)*0 = 0V


    When the volume is full on:

    Ra=0 ohms.
    Rb = (250k-0) = 250k ohms.
    Vout+ = (2.8x10^-6)*250k = .7V


    When the volume control is at midpoint:

    Ra=125k ohms
    Rb = (250k - 125k) = 125k ohms
    Vout+ = (2.8x10^-6)*125k = .35V
     
  9. mjolnir

    mjolnir Thor's Hammer 2.1.3beta

    Jun 15, 2006
    Houston, TX
    Wow, very thorough, aborgman. I think I get the gist of it now. I had to read over it a couple times, but I got it. :D
     

  10. Tone pot:

    General passive tone circuit:

    Vin+
    |
    |
    \
    / Rpickup
    \
    |
    |
    \
    / Rtonea
    \
    |
    |---- Vout+
    |
    \
    / Rtoneb
    \

    |
    = C
    |
    |
    GND

    Vin+ is the input voltage to the tone circuit, and Vout+ is the connection to the next circuit. This circuit is a basic first order RC lowpass filter.

    This filter functions for the following reason - the capacitor appears as an open circuit to DC and "low" frequency signals, thus not passing DC/"low" frequency to ground. The capacitor acts as a short circuit at "high frequency", thus being the equivalent of connecting Vout+ to ground at high frequencies.

    The frequency at which a filter "turns over" is related to the time constant of the RC network.
     
  11. mjolnir

    mjolnir Thor's Hammer 2.1.3beta

    Jun 15, 2006
    Houston, TX
    Here's a question:

    What makes a 250k pot 250k? How does one go about making a 500k pot or a 50k pot and so on and so forth?
     

  12. Depends on the type of pot - first it helps to understand what makes a 50k or 500k resistor etc. without it being a potentiometer.

    Resistance is proportional to length and cross sectional area.

    Resistance = (resistivity * length)/cross sectional area

    Resistivity is specific to the material ( Copper: 1.68x10^-8 ohm m, Carbon: 60 x10^-5 ohm m, etc.)

    Think of a wire "resistor".

    If you make your "resistor" longer without changing any other parameters - your resistance will go up proportionally.

    If you increase the cross sectional area (go from 20 gauge to 12 gauge wire) - your resistance will go down.

    If you change materials you can make the resistance go either up or down - higher resistivity will give you higher resistance.

    Tailor the options available to generate the resistor you want.
    Want a 50k ohm silicon resistor that is .1 meter long?

    50,000 ohm = ((60 ohm m)*(.1m))/cross sectional area

    It'll need to have a cross sectional area of .12mm, or say a "wire" with a diameter of about .4mm

    There is a basic resistor.

    Now imagine you have a third terminal, that can slide along the length of the "wire" - there you have a pot.
     
  13. This thread is a complete +1. I'm bringing this thing back because I have a question along these lines. So for tone control, what I've come to believe from reading online recources is that the capacitor rating will not affect the tone of your bass if the tone knob is in the full open position, but affects the tone as you turn the knob counter clockwise to raise the resistance. Is this true?

    Or, given that having a tone knob does eliminate some high frequencies as opposed to not having a tone pot, could one capacitor eliminate more or less high frequencies than another if used in the full open position? For example, comparing a .1 MFD to a .047 MFD capitor. Would these roughly eliminate about the same amount of high frequencies in the full open position? I do realize the tone will be different as the resistance is raised on these two set ups. Thanks.
     
  14. RyreInc

    RyreInc

    May 11, 2006
    Kalamazoo, MI
    Great advice/knowledge here; I just wanted to add that Kirchoff's Voltage and Current Laws (KVL & KCL) are the two most basic laws of electric circuits and are fundamental to the understanding of those circuits.

    http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

    I like to compare electricity to water in a pipe or stream. Voltage is like a waterfall--the higher the voltage, the higher the waterfall, and the more potential energy. Current is like the water itself, the more water the more current. Power is the product of the two--if you have a lot of voltage but little current you won't have much power, the same if you have a lot of current but little voltage. Like in a hydroelectric dam, the power its capable of producing is a product of the water height and amount of water flow. Electricity is also like water in the sense that when you turn the tap on, the water is there instantly--it doesn't have to travel all the way from the source.
     
  15. mjolnir

    mjolnir Thor's Hammer 2.1.3beta

    Jun 15, 2006
    Houston, TX
    Yes, to a point. Only a no-load pot will be truly unaffected when full open, and the rest are technically affected a tiny bit. And though it seems to work backwards, you can actually reverse it if you want it the other way around. It all just depends on how you wire it.

    The difference between .1MFD and .047MFD when the tone pot is full open is negligible, or at least it is to my ears.
     
  16. Bassamatic

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

    WOW - What a great thread. It is so refreshing to see so many smart people out there.

    Like RyreInc, I also use water as a simile to try to explain electricity to non-techies.

    GREAT JOB!
     
  17. Cool. Thanks dude.
     

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