Big thought experiment

[zionstrat]

Guys you've been kind enough to walk me through some electronics that I couldn't get my head around in the past, but I've reached a point where I really need to understand more.

My problem is I think of electronics like plumbing and I think that is what is messing up my thinking.

The high end question is why doesn't a tone control only affect half of the signal?

So here's my thought experiment... A single pickup connected directly to a jack. The signal forms a circuit and we'll hear sound through the amplifier.

Now let's split the output to a second jack B. Just to be clear, all we did was splice the input and output wires into two wires so we now have 50% signal at jack A and B.

So far I think everything works the way I'm expecting it to.

But now let's put a volume control in the circuit for jack A. If we turn it down, does only jack A output go down or does Jack B output go down as well?

Either way, please explain why?

Assume now that we put a tone control on the Jack b side of the circuit. Does it only affect the jack be output or does it affect the A output as well?

Same why as before.

As you've probably already guessed, My thought experiment is clearly separating tone and volume which which both go to ground. So they are parallel, but that is also true of the normal single jack version that we're used to.

So it seems in my mind that the tone control should only be controlling half of the signal tone.

I'm pretty sure my fallacy has to do with the outputs described above.. I think you guys are going to tell me that when Jack a volume goes down it also turns down Jack b volume.. same for tone and that would explain why a regular circuit isn't half and half.

But the why completely confuses me... If this was plumbing, the two outputs could be affected on their own.

I hope this makes sense and also hoping I might learn something from it :-)

[beaubrummels]

But now let's put a volume control in the circuit for jack A. If we turn it down, does only jack A output go down or does Jack B output go down as well?

Either way, please explain why?

Assume now that we put a tone control on the Jack b side of the circuit. Does it only affect the jack be output or does it affect the A output as well?

Same why as before.

If the controls are each after the split, I think they should work independently.

[zionstrat]

Thanks Beau... As previously mentioned, that seems logical to me.

But here's where it falls apart in my mind... If we go to standard wiring, a signal goes to the volume with a parallel wire going to tone. Doesn't that represent a split essentially the same as I have described with two outputs?

Which brings me back to why doesn't the tone control only affect half of the signal.

I think you may be pointing me in the correct direction by saying "after the split."

But the split is just creating two parallel paths and why is that different than the two parallel paths with a regular volume and tone control?

Again I apologize for not understanding what may be obvious.

[ImmortalSix]

Well in a regular guitar, the whole signal gets back together at a single input jack.

[Chistopher]

I think your question is the type that's easier solved by just sitting down and thinking about it until something clicks rather than trying to have someone explain it to you in words.

Consider your plumbing idea. if you have a pump that puts water out into two pipes they will each have half the water pressure. If you close the valve on one pipe so no water reaches the "amplifier" the other pipe will have 100% pressure. If you put a leak in the pipe so no water reaches the amplifier the other pipe will still be at 50% pressure. This isn't completely accurate as far as circuits are concerned, but the water metaphor breaks down after a certain level of detail.

Another interesting thing to get you thinking is that even though the bridge of a Strat doesn't have a tone control, the middle tone control still effects it in position 2.

[freefrog]

I think you guys are going to tell me that when Jack a volume goes down it also turns down Jack b volume.. same for tone and that would explain why a regular circuit isn't half and half.

Yes.

But in fact, when a tone pot is paired to a volume control of the same value, this tone control effectively causes half of the resistive load taming the resonant peak(s) . If you short its tone cap, this tone control becomes just another volume pot.

Tone capacitors limit this phenomenon by behaving like frequency dependent resistors: in parallel with pickup(s), they keep bass frequencies intact but don't "resist" to high frequencies going to ground. The larger the value of a tone capacitor, the lower is the corner frequency - but only at the end of the curve: from 10/10 to approximatively 3.5/10, a typical tone control is mostly resistive, squashing/flattening progressively the resonance of your pickup(s).

More info here: https://guitarnuts2.proboards.com/th...one-capacitors

And I'll humbly recommend you the "Guitarfreak" Excel sheet generously shared on the same website... It's a fantastic tool IMHO/IME - and normally safe to download...

https://guitarnuts2.proboards.com/thread/3627/guitarfreak-guitar-frequency-response-calculator

HTH. :-)

[beaubrummels]

Yes.

Hmm. If both volumes join to the same jack, for sure yes. If the volumes are after a split in the wire and going to separate jacks, is that still true?

[freefrog]

Hmm. If both volumes join to the same jack, for sure yes. If the volumes are after a split in the wire and going to separate jacks, is that still true?

Well, the volume of jack A wouldn't work as a standard volume control on jack B but it would still affect the response of the pickup "heard" through jack B (by changing the frequency and height of its resonant peak).

Tone pot on jack B would affect both outputs anyway. That's why my admittedly simplifying answer was focused on tone controls. :-)


Side note - putting a diode in series with the split wire going to a tone pot should immunize the volume pot of jack A against tone control on jack B. But it would also make this tone pot unuseable on the jack B side, with a weak output level and a strange scooped response...

[GuitarDoc]

You can conveniently think of electricity as water, but remember it has to go somewhere to have an effect. Just adding another jack doesn't do ANYTHING if nothing is plugged into it. Just like adding an additional drain pipe doesn't do anything if that pipe is capped off.

If you split the output wire to two jacks and put a vol pot to jack A (after the split), it will only affect jack A since the wire to jack B is still continuous to the source (the pickup). Similarly, if you put a tone pot with cap to jack B (again, after the split) it will not effect jack A. The electrons will not go through the tone pot to jack B then come back to the split and down the other wire to jack A and its vol pot. Electrons don't go back and forth through the wires, just like water doesn't go back and forth through pipes (thank godness, or we'd have toilet water coming through our sink faucets).

Electrons will always follow the shortest, most direct, and easiest route from a source (pickup) to the end point (jack). This is related to conservation of energy.

Another consideration equating electrons to water: If you have a large utility sink with two drain holes cut into the bottom (instead of just one) and the sink is full of water, if you open drain A you'll get a certain flow of water through that drain. Now if you also open drain B, it won't change the flow you've been getting through drain A. You won't get 1/2 of the original water flow through drain A and 1/2 through drain B. In fact, if you open both drains, now you'll get twice the amount of drain water flowing out of the sink. And if you have a valve on drain B that you turn 1/2 closed, it will also not affect the amount of water through drain A even though it cuts the flow through drain B in half.

[freefrog]

Let's share below a 5spice sim of what happens precisely (simplifications aside) if the signal of a passive pickup goes in parallel to a jack output plug A with volume control and to another jack B output plug with tone control.
The accuracy of this sim should be easy to check with real components if needed.

[zionstrat]

Yes.

But in fact, when a tone pot is paired to a volume control of the same value, this tone control effectively causes half of the resistive load taming the resonant peak(s) . If you short its tone cap, this tone control becomes just another volume pot.

Tone capacitors limit this phenomenon by behaving like frequency dependent resistors: in parallel with pickup(s), they keep bass frequencies intact but don't "resist" to high frequencies going to ground. The larger the value of a tone capacitor, the lower is the corner frequency - but only at the end of the curve: from 10/10 to approximatively 3.5/10, a typical tone control is mostly resistive, squashing/flattening progressively the resonance of your pickup(s).

More info here: https://guitarnuts2.proboards.com/th...one-capacitors

And I'll humbly recommend you the "Guitarfreak" Excel sheet generously shared on the same website... It's a fantastic tool IMHO/IME - and normally safe to download...

https://guitarnuts2.proboards.com/thread/3627/guitarfreak-guitar-frequency-response-calculator

HTH. :-)

Thanks as always my friend!

[jeremy]

good stuff! always good to have that type of empirical knowledge