tone knob became volume knob?

[frobro808]

long time no see GNUTZ!!!
so last rehearsal i've discovered that when turning down my tone knob the volume drops to an inaudible decible. when i run my dirties and turn down the tone knob, it acts like a treble bleed and makes it clean. i have a super-strat style jackson with one MV and one MT both have p/p switches that work fine. i got my schematics from yall and everything works fine. i used my old tone cap for this project. did something come loose or should i switch out the old cap? thanks in advance!

-frobro808

[sumgai]

bro,

Take a resistance reading across that tone cap. It does sound like it might be bad (shorted), or else there could be a short around the cap somewheres, such as a bent over pot terminal, or a drop of solder where there shouldn't be any..... That's the first thing I'd check closely.

HTH



sumgai

[Deleted]

Most of cheap pots suck. I had to stay up till 01:00 last night to finally discover i had a defect pot. So your VOM meter is your best friend in this one.

If you see the tone pots have only two pins connected, while the volume pot all three.
Now with a VOM you should measure the following :

a) Tone pots : Place/grip the pins of your VOM to the two connected terminals of the tone pot.

a1) With the pot dial turned to 10, you should get the highest resistance figure of this pot (250Kohm or 500 Kohm). With the dial turned to 0, you should get 0 Ohms.

a2) Now take the VOM pin off the outer pot terminal and connect it to the unconnected (free) pot terminal. You should get the reverse results : Dialed to 10 it should read 0 Ohm, and dialed to 0 it should read 250 (or 500) Kohm.

Thats the way to test the two tone pots.

b) volume pot : Here you must do two kind of measurements
b1) Disconnect all pins from the vol pot (or just the pin coming from the 5-way switch)
Do the tests a1) and a2) your pot should exhibit a similar behavior
b2) Connect correctly all pins , that you might have disconnected in the previous step. Then plug the cable into your jack and dial the vol pot to 10. Switch to the bridge pup (or any pup that you already know its resistance). Placing the pins of your VOM to the two conducts of the other side of your guitar cable it should read smth slightly less than the Ohm value of your pup. Now gradually turn the vol dial from 10 to 0. It should start increasing ...20 kohm->30Kohm->60Kohm till a value around 70-100 Kohm where it will start decreasing again. With the vol at 0, it should show 0.

[Deleted]

To the best of my understanding the vol pot is a variable resistance in parallel with the guitars circuit (or just the resistance of one pickup if only one pickup is selected from the switch). If we assume max 250Kohm pot and a 8Kohm pup, then the final resistance is

1/R = (1/Rpot + 1/Rpup) ==> R = (Rpup*Rpot)/(Rpup+Rpot) ==> R = (8Kohm*Rpot)/(8Kohm+Rpot), where Rpot varies from 250Kohm (10 dial position, open, where most signal reaches the amp) to 0Kohm (0 dial position where the hot actually is grounded, and no signal reaches the amp).

However i have, one comment , if the pup resistance is too high (like the dimarzio hs-3, or similarly overwound pups) then the pot figure must be increased. (i think)

Also, i have a big question : Making a graph for the function above shows that this function
for Rpot=0 is 0 and for Rpot=250 is 7.75 Kohm, also this function is a monotone function, so it can only increase.

Here is the graph :



So the question is, how is the reading getting higher and higher from 7.7 to 50, to 70 and then finally getting to 0.
How does it get above its limit? (which is 7.7)

[newey]

pyrros-

Your formula is correct to find the combined resistance of 2 components in parallel- but that number doesn't tell us much.

A guitar pickup has a certain resistance, but it also generates a certain current having a certain voltage. That current carries the signal to the amp.

A potentiometer works by shunting a varying amount of that signal to ground. As the knob is turned down from 10, resistance to ground decreases as resistance between the input and the output increases- which sends more signal to ground.

When at maximum, the pot provides only very little resistance between in and out, while the resistance to ground is at maximum. More of the current (the signal) then flows to the output.

So, your formula (and graph) aren't representing the situation.

Plotting the action of a pot can be done- and has been done around here somewhere. If it's a linear taper pot, the graph is more or less a straight diagonal line; with a log taper pot it's a parabola.

[Deleted]

Hmm,
my question is the following : If we are having 2 resistors in parallel (note that when the pup is not sensing anything it is a mere resistor), then the equivalent resistor of those two (the result) is *ALWAYS* less than the minimum of them.

How on earth do we have readings on the pot greater than 50, 60, 70 Kohm, before it starts dropping to zero as we close the volume knob all the way to 0?

This defies the laws of physics!! so somewhere i am wrong, but i want the exact reason, in terms of maths!!

[Deleted]

maybe i got the logic wrong, i gotta read more Trying strict high school physics and maths dont seem to help much. Maybe i got the concept wrong.

[Deleted]

Hmm maybe its a series-parallel combination...

[Deleted]

I got it!!!
for pot = 250 kohm and pup 8 kohm the formula is:

R = (Rpot*258 - Rpot^2)/258!! and the graph is here:



So this explains it all!

[sumgai]

I don't know whether to laugh or cry.

Rather than scribing a 5,000 word essay, and coincidentally re-inventing the wheel, let's just all of us troop on over to another website, and have a look at an essay on....

The Secret Life Of Pots

The pertinent material, relevant to the current discussion, is almost all the way down to the bottom of that page.

HTH


But pyrros, keep it up. You are pitching in and helping, and you're very close to a +1. I predict that it won't be much longer now. ;D





sumgai

[Deleted]

I don't know if you followed your own link, but what it explains is about building your own audio (Logarithmic) pot. Not explain the resistance figure as sensed by the VOM pins on the terminals of the guitar's output jack, as the pot dial is turned from 0 to 10.
I don't know if i just re-invented any wheel (apparently i have, all students across the globe everyday re-invent some kind of wheel) , but this specific "wheel" you presented above (although a useful wheel) seems to be a different kind of wheel.

[ashcatlt]

I could have sworn we had just talked about this.

When the volume pot is at 10, all of its resistance is parallel to the pickup. As it's turned down, some of its resistance is in series with the pickup and the rest is parallel to that series resistance. The total resistance will increase until the pickup plus the series portion equals the parallel portion and then it goes back down.

It's a little early for me to work out the formula for this.
Edit - nevermind, you got it right with that last graph, assuming that RPot is the resistance between lugs 1 and 2 of the volume pot.

[sumgai]

Feb 11, 2011 8:34:03 GMT -5 @pyrros said:

I don't know if you followed your own link, but what it explains is about building your own audio (Logarithmic) pot. Not explain the resistance figure as sensed by the VOM pins on the terminals of the guitar's output jack, as the pot dial is turned from 0 to 10.

Hmmmm, it might be that I'm serruptitiously channeling ChrisK here..... Suppose I try it one more time:

Brain Scanning Through A Nostril, Strat Edition


Similarities are.

Hidden similarities aren't.

[Deleted]

Feb 11, 2011 11:39:59 GMT -5 ashcatlt said:

I could have sworn we had just talked about this.

When the volume pot is at 10, all of its resistance is parallel to the pickup. As it's turned down, some of its resistance is in series with the pickup and the rest is parallel to that series resistance. The total resistance will increase until the pickup plus the series portion equals the parallel portion and then it goes back down.

It's a little early for me to work out the formula for this.
Edit - nevermind, you got it right with that last graph, assuming that RPot is the resistance between lugs 1 and 2 of the volume pot.

Thnx! It seems we are absolutely correct!

The formula was constructed as follows:
Let us denote with Rpotmax the resistance spec (max resistance)
of the pot, and Rpup the resistance of the pickup.
Lets us denote with Rpot the variable resistance of the pot between the middle and ground terminals. For any Rpot, the resistance between the middle and the other terminal (between the middle and hot signal of the pup) is Rpotmax-Rpot.

For any given Rpot the resistance R between the middle and the ground, (which is what the VOM is sensing) is a parallel resistance of the following two resistances, which denote the two paths from the middle terminal of the pot to the ground:

1) Rpot
2) Rpotmax-Rpot+Rpup

So, applying the parallel resistance formula, we have:

1/R = 1/Rpot + 1/(Rpup+(Rpotmax-Rpot)) =
1/Rpot + 1/(Rpup+Rpotmax-Rpot)

Multiplying both side of the equation with Rpot*(Rpup+Rpotmax-Rpot) he have :

Rpot*(Rpup+Rpotmax-Rpot)/R = Rpot*(Rpup+Rpotmax-Rpot)/Rpot + Rpot*(Rpup+Rpotmax-Rpot)/(Rpup+Rpotmax-Rpot) =
(Rpup+Rpotmax-Rpot) + Rot = Rpup+Rpotmax
==>
Rpot*(Rpup+Rpotmax-Rpot)/R = Rpup+Rpotmax ==>
R = Rpot*(Rpup+Rpotmax-Rpot)/(Rpup+Rptomax) =
((Rpup+Rpotmax) * Rpot - Rpot^2) / (Rpup+Rpotmax)

[ashcatlt]

The Strat version of the Brain Scan Through a Nostril doesn't involve turning the volume pot. The LP version does, though, and is has more to do with this discussion.

Extra Credit - Where'd ChrisK get the multiple of 4?

[Deleted]

^^^
Nice article, this one is worth reading.

[ashcatlt]

Course, I got frustrated trying to fiddle with unreliable pots and come up with an Rmax with which I could be confident, so I figured out a way to do it without all that messing around here

[sumgai]

Now we're starting to cook with gas! ;D