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Electric Bass Tone Filter/Volume circuit

  1. Sep 17, 2010 #1
    Hello all,

    I'm starting an undergraduate research project at my University regarding guitar pickups and their geometry. To begin, I'm first trying to understand the pickups and the circuits they are connected to, pre-amplifier.

    Please understand I have very limited knowledge in circuits and almost everything I've done has been basic DC circuits involving resistors and capacitors. I'm doing the project to expand my knowledge in the subject independently (and because I'm building an electric guitar from scratch, so it ties in nicely).

    The circuit I'm looking at now is off of a simple Fender style Bass. I found a schematic on the internet.

    [PLAIN]http://www.axiomatic-music.co.uk/acatalog/PBassSchematic_compressed.jpg [Broken]

    Can someone help explain to me how the tone control works when it is in parrallel with the volume knob. It seems like some frequencies would escape volume change.

    While typing this, I thought of an error I made when using the function generator, so that might be the cause of my confusion. Regardless, I wanted to post the question to get some input from more experienced people. Sorry if I didn't word it clearly, some things got deleted when I realized the mistake I had made.

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 17, 2010 #2


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    That is a simple treble bleed circuit. The lower the resistance offered by tone pot, the more the high frequencies will get shunted to ground.

    Look at the input lugs of the two pots. They are tied together. If the tone pot is set to its highest resistance setting, most of the input signal feeding the two pots will exit the "hot" lug of the volume pot and make it to the jack. If, instead, you set the tone pot to a lower resistance level, more and more of the high frequencies will be dumped to ground (and the effected frequencies can be tuned by your selection of the capacitor value). Even simple passive circuits like this are interactive, though it's not always intuitive.
  4. Sep 17, 2010 #3
    Right on. That helps a lot. Could you explain the idea of "dumped to ground" a bit more? Right now it's my understanding that the circuit isn't grounded to the bridge so much as the bridge is grounded through the circuit and therefore, through the amplifier (to help keep away noise).

    I guess I'm just asking how grounding works. I'll be reading on it after posting this, but you did an excellent job explaining in your first answer, so I'm hoping for another great reply.

  5. Sep 17, 2010 #4


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    The output signal from the guitar to the preamp is the difference between the hot and the ground legs of that circuit. If you you bleed some of that input signal off to ground (via the tone circuit), then the total signal out of the guitar is diminished. It's not that the signal is non-existent - it is diminished to the extent that the difference between the "hot" and the "ground" in the guitar circuit are diminished. I hope that helps.
  6. Sep 17, 2010 #5
    Yes, it does indeed. Now, the way I understand it, the filter (at a certain setting) makes certain high frequencies consider it to be a low resistance path and those same high frequencies consider the volume knob to be a high resistance path. Therefore, despite the voltages the same at the leads to the V and T pot, the current flowing through them will vary greatly, and the T (tone) pot will have almost all of the certain high frequencies passing through it. Since it leads to the ground terminal and not the "hot" (I believe that's the right word) terminal, they won't be nearly as prominent in the output.

    I believe this is pretty much just what you said in the first post, but the second post helped further my understanding. I don't *completely* get it, as in, I'm not sure what really happens to the high frequencies bypassed to the ground, but I imagine some of it needs to come with experience.

    If anything I said trying to summarize what is going on in the circuit was wrong, please correct me. Thanks again!
  7. Sep 18, 2010 #6


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    I'd say you're pretty close. As an exercise, draw the same circuit without the tone pot. You can see that the "can" of the volume pot, the ground side of the pickup lead, the bridge and the ring of the jack are all at one potential. The magnitude of the signal coming from the output (center lug) of the volume pot is dependent on the setting of the volume pot. There is a wiper inside the pot that is in contact with a resistive strip, so that pot acts like a variable resister, depending on where the wiper touches the strip. At the highest resistance, very little signal gets to the center lug (output) of the pot.

    Now look at the original drawing that you posted here. The capacitor could have been attached to any of the grounds in the schematic, and it would still provide a path to bleed treble frequencies to ground. For purposes of keeping the circuit simple and rugged, the capacitor is soldered to the "can" of the volume pot. If there was a simple capacitor in the circuit and NO tone pot, you would only have one (rather dark) tone available. In this case, though, you have another variable "resistor" - the tone pot - that regulates how much of the treble signal that you will bleed to ground.

    Some simple guitar wiring years ago consisted of a pickup selector switch and a volume pot. The tone control was a simple 3-position toggle switch that retained all the treble in the center position and allowed you to choose between two capacitors in the other two positions. Gretsch guitar-owners used to call the tone selector the "mud switch". Gretsch guitars were very popular with country-western players who tended to be pretty fond of the "twang" and didn't use the tone switch much.
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