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animalfarm

3 way switch in middle, turn one volume down, BOTH pup outputs go down. Wanna know why?

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This question has been asked before, I'm doing this as an information

reference, and as a way of showing a solution called "Independent

Volume Control". Thanks to teegar for idea!

 

Any thoughts are welcome as well as clarification on any thoughts I may

have incorrectly addressed.

 

Most Folks wire up their pups using the standard Semour Duncan Les Paul

wiring diagram (I did), which will cause both pup volumes to drop if either

volume knob to be turned down when 3 way is in middle position.

(Note Volume Knob/pot wiring).

 

4000_04.gif

 

2h_2v_2t_3w.jpg

 

Why does this happen?

The FULL electrical output of each pup is wired to the input of each volume pot,

then off to switch via the wiper arm after signal goes thru the resistive strip in the pot.

The way this is wired makes it a PARALLEL resistance electrical circuit. Huh?

 

In a parallel circuit, both "legs" mathematically combine to produce an output

LESS than the value of the lowest leg's resistance.

 

If I have 2 500K ohm resistors in parallel, the output of the circuit will be 250K ohms.

The lower the resistance, the more signal sent to ground, meaning less signal being

sent out to amp. Turning down either volume pot decreases the resistance in one "leg",

lowering the total resistance of the circuit.

 

In the S.D. diagram, if you drop the volume of either pot with switch in MIDDLE position,

the overall resistance ouput value of circuit DROPS, sending more of BOTH pot signals to

ground, causing volume of BOTH pots to DROP.

-------------------------------------------------------------------------------------------------

 

INDEPENDENT VOLUME CONTROL (3 way still in middle position)

 

4000_04b.gif

smwiring.jpg

 

4000_03.gif

 

The way it's wired into the WIPER allows you to vary the OUTPUT OF THE PICKUP at

the point it's wired into the circuit!!!!! The INDIVIDUAL VOLUME CONTROL comes from

the fact that you're controlling how much of the pickup signal is making it into the circuit

using the resistive strip inside the pot.

The less resistance (lower volume) more pup signal goes to ground, the higher resistance

(higher volume), less signal to ground, more to amp.

 

By controlling each individual pup's input INTO the circuit/shorting to ground before making it

out to the amp, you're seperately controlling each pup's VOLUME!!!!!!!!!

 

The individual pickup is shorted to ground instead, rather than the total output of the circuit

with both pups if it would be if wired like the Semour Duncan diagram.

 

There's nothing wrong with the SD wiring, I use it, I occasionally use the middle position

of my 3 way switch, but generally just for getting a different TONE rather than volume blending.

 

StewMac link/Different explanation:

http://www.stewmac.com/freeinfo/Electronics/Misc/i-4000/i-4000_4.html

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I just assumed it was just someones law "Nope' date=' can't do That..":-$ [/quote']

 

[biggrin] [biggrin] [biggrin] You mean "OHM's Law"!!!!

I tried to stay away from an explanation that would require

having to drag out TEXTBOOKS just to understand....

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Once again, thanks animalfarm, this was really helpful. I hope to have the wiring completed in a couple of days, life keeps interrupting and refusing a couple of hours of free time, but I'll let you all know how the independent wiring job went.

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I always thought the best, if oversimplified, way of explaining it is this:

 

----------

Volume pots are potential dividers. They basically take the middle lug and chose how to divide that signal between the other 2.

 

e.g.

At one end, middle is 100% connected to A and 0% connected to B

At other end, middle is 0% connected to A and 100% connected to B

At middle (on linear pots), middle is 50% connected to A and 50% connected to B.

 

----------

 

Independent volume is choosing how much of the pickup (middle lug) is connected to the hot line on the jack (lug A) and how much to the ground of the jack (lug :-. This could mean the pickup is essentially connected directly to the hot at one extreme, or that the pickup is connected directly to the ground at the other.

 

If you turn the independent one to 0, that pickup is shorted to earth and doesn't go to the hot line of the jack. The part of the circuit that is shorted to earth, and therefore bypassed, is the pickup; so the rest of the circuit is not affected.

When the pickup is bypassed, only that pickup is muted.

 

----------

 

Traditional volume is choosing how much of the hot line on the jack (middle lug) is connected to the pickup (lug A) and how much to the ground of the jack (lug [thumbup]. This could mean the hot is essentially connected directly to the pickup at one extreme, or that the hot is connected directly to the ground at the other.

 

If you turn the traditional one to 0, the hot line of the jack is shorted to earth and doesn't go to the pickup. There is a short circuit between the hot of the jack and earth of the jack.

 

You can think of this as meaning the whole guitar circuit is bypassed, or that the signal is grounded. Either way, what an amplifier input needs is a potential difference (voltage difference) between the 2 wires in the guitar cable; and if they are connected to each other, there isn't going to be a difference.

 

When the jack is bypassed, the whole guitar circuit is muted.

 

If both volume pots have the ability to short circuit the jack, both have the ability to mute the whole guitar, no matter what the other one is doing. This is the case in the traditional volume; whereas in the independent volume, each pot only has the ability to bypass its respective pickup.

 

Hope people can understand my logic. I USED TO be good at expressing it.

 

----------

 

P.S. (this is a bit off topic, but still vaguely related, read it if you're especially interested)

In a traditional volume, as both pots are acting on the jack rather than individual pickups, they follow much more complicated mathematics as they are in parallel [ R = (R1 × R2)/(R1 + R2) ] And bearing in mind that there are essentially 4 different resistances changing with each other, 2 in each potentiometer, this gets confusing, but the basic practical concept is this:

 

One pot affects the other as if it were changing the taper of the pot.

When one pot is near 0 (but not on 0 because that would mute them all), the other pot is made to be more exponential, with more of the effect of the pot up between 8-10 and less of it near the bottom.

When one pot is near 10, the other pot is made to be more logarithmic, with more of the effect of the pot down between 0-3 and less of it near the top.

When one pot is at 50% resistance (this won't be at 5 on an audio taper pot), the other pot is acting at its natural taper, unaffected by the other pot.

 

If you have traditional volume controls, try turning them both around and having a little experiment in the middle position which puts them in parallel. It's often very hard to predict how the volume will change and you find yourself just finding the volume by ear rather than turning one knob and expecting it to change a certain amount.

This is also the reason why some people will say a linear taper pot sounds more natural and a log taper has all its range in a small part of the turn and some people say the exact opposite. They most likely have the OTHER knob in a different position.

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Thanks, KX - was wondering if anyone else was going

to "weigh in" in this topic.

:-[biggrin][thumbup]

 

Next - Ohm's Law of Capacitance (Xc)

Ohms' Law for an capacitor: Vc = IcXc

 

Never mind......[thumbup] [thumbup] [thumbup]

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Thanks' date=' KX - was wondering if anyone else was going

to "weigh in" in this topic.

:-k[biggrin'] [biggrin]

 

Next - Ohm's Law of Capacitance (Xc)

Ohms' Law for an capacitor: Vc = IcXc

 

Never mind......[blink] [blink] [blink]

For a resistor, V=IR. if you ignore the subscript "c" which is just showing it's for a capacitor, V=IX. X is the symbol for a capacitor's reactance, a measure of its internal resistance, so that's just the same as V=IR

 

V=IR is flawed though, it doesn't take temperature into account. I once got a mock GCSE question that said

 

"V=IR; but why is V not equal to I x R"

 

My answer was something like this:

"[blink] Err, cos Mr Ohm was an idiot who lived a long time ago"

to which I got a "see me" when I got the paper back. The actual answer was because it didn't account for temperature... a really mean question for a GCSE.

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