pots: A/B; log/linear...

[andrei]

hello,

i've noticed that there are different types of pots with the same resistance.
first thing i noticed: it says on them 500kA/500kB.

now, snooping around on thomann (.de), i noticed in the descriptions, some say log (logarythmical?) and some linear.
the only thing i could connnnect these with are the functions with the same names.

is that so?
does it reffer to the way the pot is increasing/decreasing the resistence when turned?

i have 500kB pots with my duncans and they suck, cose when the volume is up, turning them doesn seam to have any effect at all and when turning it down, the sound cuts off abruptly.

also: how come emgs only need 25k pots (kindo a big difference from 25 to 500)?

i'll be verry grateful if someone with the technical knowledge could bring some light into this issue.

thanks!

[marauder]

Volume pots should have logarithmic (A, audio taper) pots, when these are at 5 then the volume is half what it is a 10. This doesnt give exactly half the output voltage at 5, but it works the same as the way your ears work in response to the change in volume.

Linear pots (B) pots, these give half the voltage output at 5 compared to 10, usually used for tone pots. When used as a volume pot, 5 will not be 50% volume, this is the problem you have with your volume control.

EMGs need 25k pots due to the output levels being totally different to passive pickups.

[metalhobo]

I think the EMG pots actually control preamp gain, not output volume.

[andrei]

thnks for the replys!

i've searched on wikipedia and here's what it says:



Linear potentiometers

A linear pot has a resistive element of constant cross-section, resulting in a device where the resistance between the contact (wiper) and one end terminal is proportional to the distance between them. Linear describes the electrical characteristic of the device, not the geometry of the resistive element.

Logarithmic potentiometers

A log pot has a resistive element that either 'tapers' in from one end to the other, or is made from a material whose resistivity varies from one end to the other. This results in a device where output voltage is a logarithmic (or inverse logarithmic depending on type) function of the mechanical angle of the pot.
Most (cheaper) "log" pots are actually not logarithmic, but use two regions of different, but constant, resistivity to approximate a logarithmic law. A log pot can also be simulated with a linear pot and an external resistor. True log pots are significantly more expensive.


now it's pretty clear.

[TheDevilHimself]

For guitar controls I highly recommend you use logarithmic ones. The linear ones are really quite useless IMO when used on a tone or volume.

[andrei]

i've wired my guitars with 2 independent volume controls.
pickup -> pot -> switch -> jack

on clean, everything works just fine, but
when distortion is on, if one pickup is selectet and it's pot is turned down, but the other is up, i still get a weak signal when hitting the strings.

can this phenomenon be avoided (when still using 2 independent vol knobs)?
is it cose of the switch (3way)?

on emgs this works better then on duncans.



another one:
i've seen in some whire diagrams the signal in the vol pot goes "in" through the middle and "out" on the right and in others its the other way around, "in" through the right and "out" through the middle.

whats the deal/difference?

[andrei]

is there a way to get 2 independent output signals out of the switch (in1 <=> out1, in2 <=> out2), so i may try wireing:
pickup -> switch -> pot -> out

i've tried separating the two "outs" of the 3way switch, that are usualy soldered together, hoping i'd achieve this, but it didn't work.

[Khali]

when distortion is on, if one pickup is selectet and it's pot is turned down, but the other is up, i still get a weak signal when hitting the strings.
Then your volumes are infact, not independent. From the input. ;)

can this phenomenon be avoided (when still using 2 independent vol knobs)?
Yes.
is it cose of the switch (3way)?
No.

You should provide a pic of what your wiring looks like, it'd be easier that way. But, since I have nothing, I'm gonna assume, you have it setup where the volumes are independent, they're just not masters. Basicly, if you turn the bridge pup to zero, then the neck to 10, and leave it in the middle position, because I'm assuming your trying to do this for the killswitch effect, in the middle position, you can still hear the neck pickup, and get a bit of tinniness from the bridge. If this is the case, you did it wrong.

The pickup leads should be going to the middle pole on the pot. Then you connect a wire from the left pole (assume your looking at the poles face up, and shaft for the pot face down) should then go from the pot to the switch position. Then the input connects to the middle of the switch. This means, the only connection for the input should be to the switch, and grounded at the switch. Then send a ground wire to the switch ground.

If you did it right, both volumes will act as a master. So if one volume is at 0, when you switch to the middle position, you won't hear anything, it'll be cut. But having the neck at 0 and bridge at 10 we'll say. You just hit a note, tap the switch down repeatedly, and it'll keep turning on and off. But as soon as you turn up the vol on the neck to 1, both pickups will be active in the middle position.

Basicly, your isolating the volumes from each other, and the input from anything else but the switch. The volumes should have 2 stops, potentiometer, then switch. Input just goes to switch. If the pups go to both pots, it won't work. If the input goes to either pot, it won't work.

[andrei]

thanks Khali,

this is what i want to do:

i want the volume knobs to act independent (not each as master) so each pickup may be set at it's own volume.
on the mid position, both pups are selected (the sound is mixed according to the percentage of each vol pot)
and that's also how it works but when (for instance) the bridge pickup is selected and its volume is turned down there should be no sound.
it's ok with cleans, but when distortion is on, i believe the higher input induces the signal from the other pickup to the out of the switch.

this is how it's done:

pup -> mid of the pot (in) -> right of the pot (out) (i'm looking at the bottom of the pot, the case, whrere the grounding is soldered) -> switch (in) (left) -> switch (out) (two thingies in the middel soldered together) -> jack

the other pup is wired the same way, it just goes in on the other side of the switch.

remember: it works fine on clean.

thanks!

[Khali]

remember: it works fine on clean.
Actually, your mislead! It isn't working fine on clean! There's just nothing to drive the other pickup, so you don't hear it as much, or if at all, but it is there. If you are using the neck with the bridge pup muted on clean right now for example, once you fix it and get it working right, you'll notice a noticeable decline in brightness/tinniness, with the bridge pup muted. That's because the bridge pup is still slightly on a tiny bit, and adding that bite. You would notice it a more with higher output pickups. But the point is, once its right, it'll be the full natural tone of the neck pup, without the tinniness its recieving from the bridge pickup.

Also, make sure both pots are linear, just so that's covered. The biggest difference with a linear pot, that might matter to you, is the fact, a linear will volume swell. A logarithmic will not. By that I mean, if your playing a song with a fade in or something, linear will fade, logarithmic will be like "BAM". You can still swell, it just takes practice, you have to turn the knob much slower. Very much slower.

Now, I assumed you were using EMGs, Iunno why. :lol But I guess that might not be the case. I'm not really sure. Anyway, here is traditional Gibson style wiring:
http://img.photobucket.com/albums/v731/Khali/emgvolvol.jpg
From what it sounds like, this is exactly what you have. I think for this wiring, it requires a capacitor to resist the other pot, but doing this, you lose natural tone.

Here is new Gibson style wiring, which I'm almost positive is what you want:
http://img.photobucket.com/albums/v731/Khali/emgvolvolmas.jpg

Now, if your using passive pickups, it works just the same, the differences are, erase the battery wire from the input jack, and draw on a wire grounding from the bridge to one of the pot casings, this is the only wire where it doesn't matter where it goes.

Keep in mind, with this setup, its direct vol, to pickup. Now what this means is, there's nothing even actually resisting the pickup anymore, its kind of like wiring on with no load at all. If you have a tone pot, you can have a slight loss in the natural tone of the pickup, so its not what it actually sounds like, but you get the advantage of a capacitor that'll modify the pickup's frequency. This way, it'll just be straight pickup, natural sound, all balls, nothing fighting it.

[andrei]

thanks a lot Khali,

i get what your saying.
you're right, now it's done like the first schematics shows.

please explain what the difference between wireing a vol pot like the first schematics shows (in: middle, out: right) and whireing it as showed in the second (in: right, out: middle) is.

and why is it essential to use linear pots? do they "cut" more?
(people where recomanding log in previous posts, cose they fade in smoother)

i was actualy hoping in the mid position to be able to mix the sounds of both pickups, but well...

hopefuly i'll get to mess with them tonight and see how it turnes out,
thank you!

[Khali]

and why is it essential to use linear pots? do they "cut" more?
(people where recomanding log in previous posts, cose they fade in smoother)
I meant logarithmic! :lol Sorry, I was tired when I wrote that and must've gotten the two confused. I've been living off sleeping pills at night, energy pills in the morning. Sorry for that confusion. :x

Logarithmic is the pot that swells, aslong as your pots say 500kA, your good to go. Don't use 500kB for vol/vol.

Really sorry about that!

please explain what the difference between wireing a vol pot like the first schematics shows (in: middle, out: right) and whireing it as showed in the second (in: right, out: middle) is.
The first one, makes it work so, when your in the mid position, if your turn up your bridge pickup vol, you will only hear the bridge pickup in the middle position. As you turn up the vol on the neck, it slowly mixes in the neck. Leaving the neck muted, flipping the switch to the neck repeatedly makes a kill switch effect, but its hard to do it that way.
The second one, makes it work so, when your in the middle position, if you max the bridge, and mute the neck, as soon as the neck vol pot hits zero, the middle position is muted entirely. It won't put out any volume, until the neck volume is set to 1-10. So, as soon as you turn it up, it'll bring the volume back. Basicly, both volumes have to be turned up for the middle position to work, if one is turned down, neither will work. This is prefered for the "killswitch" effect. This is how Tom Morello does it for example.
i was actualy hoping in the mid position to be able to mix the sounds of both pickups, but well...
You will be able to mix both volume outputs with either. Just like I said, the second one makes it so if either pot is muted in the mid position, it won't work. You can have both pots at 1, and it will work, both at 5, both at 10, and it will work. But if one of them is muted, both will be muted.

Its kind of hard to explain, I'm trying to eliminate confusion. Basicly, it only mutes the pickups in the middle position, so say you max the bridge volume and mute the neck. The neck pos will make no noise, the middle position will make no noise, the bridge position will be full volume. If the neck is full, bridge at zero, neck pos will be full vol, mid, nothing, bridge nothing. If both vols are turned up, regardless from 1 to 10. So we'll say...Neck 5, bridge full. Neck pos is neck at 5, mid pos is neck at 5, bridge at 10, bridge pos is bridge at 10.

So you have volume in the mid position as long as both pots are turned up to some extent, even the slightest, but if one is muted, your middle pos will be muted. This does not mute the bridge pos or neck pos if the opposite pot is muted though.

[andrei]

thanks a lot to all you guys for putting your knowledge into this and special thanks to Khali for his patience with this slow bastard.

i've wired the guitar as Khali described in the last post and it works just as it should.

allthou it's not exactly as i wanted it:
this way, in the mid position, each vol pot is master, so you get the same volume from both pickups, which is: min(vol1, vol2).
so, at all times the mix will be 50/50.

what i ment by "independent" is: in the mid position, each pup with it's own percentage, so if vol1=50% and vol2=100%, you get the output: 25% from pup1 and 75% from pup2.
this is how i had it set before, but this way i couldn't mute the damn thing by turning only one volume down, even if just it's pickup was selected.

so this is the compronise solution...

i've fryed some shitty new pots in the process and have ended up intsalling some the old ones from the EX (A/B)...
i'll order some decent (guitar) ones online since i can't find any suitable (big pots, long shaft) at the local electrics store.

thanks again and hopefuly this long thread will help someone thinking about rewireing his guitar, so he only has to do it once.