Definitive value for volume pedal pot?

[SISKO]

Hi, im making a volume pedal out of an expression pedal ive been given. Its the classic pinion-rack system like in a crybaby. It is known that this kind of system dont fully rotate the potentiometer, so i either you end up with not fully closed or not fully open, but if the pot is high enough in value, the diference is not noticeable (assuming a next stage of a high impedance, and the previous of a low impedance.)

I wanted to know how high is high. In other words, what pot value do i need?. Ill be using the pedal between two buffered pedals, so the scenario is good, but ill be happy if i could use it directly between my guitar and amp too.

[Mark Hammer]

If you know that it will always receive its signal from a nicely-buffered low-impedance source, then 50k should be just fine.  If there is every chance that it could be seeing the direct feed from the guitar, then you will want at least 500k to make sure you aren't loading things down too much.

As a historical aside, most console organs that come with a volume foot-pedal use 50k.  When the Roland/Boss CE-1 was released, it had a 50k attenuator pot on the input.  Why?  Because it was more or less assumed that the user would be feeding a keybpard into the unit, instead of lugging around a Leslie speaker.

The 50k input attenuator loaded down the guitar signal like crazy, and many complain about its tone-sucking.  The replacement of that with something MUCH better suited to the higher impedance of guitars is one of the things that people liked about the Retro-Sonic  CE-1 clone.

[PRR]

> It is known that this kind of system dont fully rotate the potentiometer, so i either you end up with not fully closed or not fully open, but if the pot is high enough in value, the diference is not noticeable

I do not understand this.

[SISKO]

> It is known that this kind of system dont fully rotate the potentiometer, so i either you end up with not fully closed or not fully open, but if the pot is high enough in value, the diference is not noticeable

I do not understand this.

The pot dont rotate its full 270º.

I am aware of the pedagogical risk of explaning one fact with many examples, but ill take my chances:
If the pot dont fully rotates its 270º you can set it up in a way that when the pedal fully backward, the pot is fully shorted to ground,letting no signal pass thru, but on the other (heel?) position, the pot is not fully open, ther is some resistances between the in and the out, plus a resistance to ground makin it a voltage divider.
The other case is the oposite, having no resistance between in and out on the (heell?) position, but when it is backwards, the signal is not fully shorted to ground, letting some audio pass thru.

I know a schematic would be useful, but its just a voltage divider.

[armdnrdy]

You got that now Paul?

[induction]

> It is known that this kind of system dont fully rotate the potentiometer, so i either you end up with not fully closed or not fully open, but if the pot is high enough in value, the diference is not noticeable

I do not understand this.

The pot dont rotate its full 270º.

I am aware of the pedagogical risk of explaning one fact with many examples, but ill take my chances:
If the pot dont fully rotates its 270º you can set it up in a way that when the pedal fully backward, the pot is fully shorted to ground,letting no signal pass thru, but on the other (heel?) position, the pot is not fully open, ther is some resistances between the in and the out, plus a resistance to ground makin it a voltage divider.
The other case is the oposite, having no resistance between in and out on the (heell?) position, but when it is backwards, the signal is not fully shorted to ground, letting some audio pass thru.

I know a schematic would be useful, but its just a voltage divider.

How does any of that change with a high value pot?

[GibsonGM]

It kind of sounds like he's concerned about being "stuck" with his settings rolling from ex: 45K on one end to 120K on the other one, as you would when you 'tune a wha'?

Except - for a volume pedal, you'd by definition be trying to operate from right about ZERO and roll in resistance to lower your signal (unless you often play with axe at 1/2 volume??)...so really you only need to look at what Mark wrote.  If you will be able to rotate it back and get 100's of K of resistance between amp & guitar, you are ok.   If not, you should use an input buffer.     That is what I would do so as to make it most versatile, useful for many guitars and all that...so it plays well with others.

[induction]

It kind of sounds like he's concerned about being "stuck" with his settings rolling from ex: 45K on one end to 120K on the other one, as you would when you 'tune a wha'?

Except - for a volume pedal, you'd by definition be trying to operate from right about ZERO and roll in resistance to lower your signal (unless you often play with axe at 1/2 volume??)...so really you only need to look at what Mark wrote.  If you will be able to rotate it back and get 100's of K of resistance between amp & guitar, you are ok.   If not, you should use an input buffer.     That is what I would do so as to make it most versatile, useful for many guitars and all that...so it plays well with others.

That makes sense for an expression pedal wired as a variable resistor, but SISKO soecified a volume pedal wired as a voltage divider.

Mark was talking about putting lots of resistance between signal and ground (not between signal and amp), to avoid loading. But SISKO isn't asking about loading, he's talking about not getting the full volume range from a voltage divider because of limited range of motion. He's asking what value of pot to use to avoid this problem.

And, of course, the value of the pot is not relevant to this question. There is no pot value that will give you the full range of volume without a full range of motion. That's not how voltage dividers work.

I suspect it would be better to lose volume in favor of being able to completely silence the signal. You could wire a slight boost in front if the pot to make up for the lost volume, or you could just live with it.

Maybe someone else has a better solution? How do most volume pedals handle this?

[slacker]

How do most volume pedals handle this?

Some use custom pots so the full range matches the rotation available in the pedal.

You could wire the pot as a variable resistor and make it the bottom of a voltage divider with fixed resistor for the top. Mount it so the pot goes to zero at the heel down end of the travel then sweep it to toe down measure the resistance and size the resistor so the voltage drop is acceptably low. Might need some tweaking to get the sweep to feel right and probably wouldn't work too well with straight guitar as it would load the output at low settings.

[SISKO]

You got that now Paul?

I know sometimes i can be rude, even rude-er when trying to explain myself in english (my native language is spanish).
While i was writing the explanantion, i always had in mind that Pual surely knew what i was trying to say (thus the "it just a voltage divider", i know Paul knows a voltage divider very well), and the problem was me not explaning myself correctly in the first instance.
Paul i respect you, dont get me wrong!!

How does any of that change with a high value pot?

I started writing some equations, but a picture is worth a thousand words


Where the 10k resistor is the guitar output impedance.
With a buffer in front of the volume pedal, i think there will be no noticeable effect with a 500k pot or a 50k pot.

My intention is to make it passive and with the possibility of using it between the guitar and amp. Of course, im not aiming for a prefect volume pedal, but to get the best result of it.

[ashcat_lt]

...That's not how voltage dividers work...

...Because a voltage divider is ratiometric.  The actual value of the resistances involved doesn't actually matter, only the ratios...

Except that we're never dealing with ideal sources and loads.  Given pot value and available rotation angle, a higher source Z and/or lower load Z will get closer to silence at the low end, but further from full on at the top.  Lower source and/or higher load will have the opposite effect.  It's still about ratios, though, and in this case it's the relationship of the outside impedances to the pot itself.  Given both source and load as constant, a smaller pot will get closer to 0, while a bigger pot will get closer to 10.

Your idea to set the pot to hit 0 and then add makeup gain is probably the best (easiest) way to go.  It compromises S/N a bit, but shouldn't be that bad.  Here's another "but", though:  These things are designed not to hit the ends for a pretty good reason.  That rack and pinion offers a whole lot of torque, and will allow you to push that pot beyond its physical limits pretty easily.  Then it's borked.

I think the very best is to use a larger than necessary pot and modify the internals so that the limited rotation actually works.  I tried to do it with conductive paint on either end of the track, and if I'd had a steadier hand it might have worked, but I don't think that's the best way to go.  You could use that "rivet tap" trick, or maybe if you had a really really fine conductive foil...

[GibsonGM]

Yes, I 'got' the idea, Induction...as noted, with the limitations of the rack & pinion, risk of destroying the pot if you hit its rotation limits...far and away, the easiest method to achieve the desired results is (what I was getting at) to use Mark's comments re. a buffered input as a starting point.   By recognizing the possible need to boost after, you're also seeing this but from the other end.   Losing signal and then re-boosting, vs. using buffers to not lose signal...

I know we sometimes 'really want to' make something passive, but this problem has been solved for us by the 'good' volume pedals that use buffering and, if necessary, gain compensation.    Just seems that to reproduce exactly what we do on the guitar using a pot, and trying to now make it happen via a (rotation limited) expression pedal involves a lot more work than what has already been designed for us.  But that's just my 2 cents, maybe there is more to the story than I am seeing.  *shrug*

It's a good question as it is a bit outside the usual scope of volume pots and impedance/loading issues... 

[induction]

as noted, with the limitations of the rack & pinion, risk of destroying the pot if you hit its rotation limits...far and away, the easiest method to achieve the desired results is (what I was getting at) to use Mark's comments re. a buffered input as a starting point.   By recognizing the possible need to boost after, you're also seeing this but from the other end.   Losing signal and then re-boosting, vs. using buffers to not lose signal...

I was actually recommending putting the gain before the volume pot. A non-inverting op-amp stage with a little gain will solve both the loading problem and the incomplete rotation problem. With just a buffer, you'll still lose a little signal because of the rotation problem. Pre-gain should also improve the S/N slightly (over post-gain) because the noise added by the gain stage will be reduced along with the signal from the incomplete rotation. Though if the gain was small, the noise wouldn't be too bad anyway. You could even make the gain variable with an internal trimpot or side-mounted pot to allow a real boost option in the toe-down position.

[GibsonGM]

Nice solution     Things like this can be aggravating, yes, LOL.    Sometimes experimentation is the simplest way, IMO. 

[PRR]

Sorry, my hasty question was un-clear.

> don't fully rotate the potentiometer

This part I understand. Your ankle does not rotate 270 degrees. Some ankle-pivot pedals gear-up the rotation, but going from 20 deg(?) of the ankle to the 270 deg of the standard rotary potentiometer is heavy gearing, cost, wear, and alignment trouble.

> if the pot is high enough in value, the difference is not noticeable

This is the part which I do not understand.

In *voltage divider* connection, it seems clear to me that if you do not go one end or the other, you do not go to 100% or to zero %. A "pot ...high enough in value" does not change the end conditions, only the loading for mid-rotation. Since in pedal work the loading is often "unknown", this is unpredictable, so any solution is very specific to that situation.

In shunt connection (shorting-out the guitar) we have to go over 100K to give small-loss for the treble peak, yet under 5K to get large loss for the bass/body of the sound. If the source is not a guitar but another box, there's no set of values which will work for "all" the boxes we find on stage (from 47 Ohm opamp outputs to 470K out-volume pots).

In series connection you are entirely at the mercy of whatever the next stage is. If it is one of the old-old Ampegs with 3.6Meg inputs, even a 1Meg pot won't cause much drop; OTOH if it is the ~~20K input of some old fuzzes (which loaded-down raw guitar treble to reduce ice-pick sound) then much lower series resistance is needed for a reasonable ankle/loss relation.

> pinion-rack system

It is "possible" for such a mechanism to give "full rotation". Lift the front of your modern sedan, push the front wheels left and right, the steering wheel will turn about 720 degrees. You "just" need a rack much longer than your pinion. Of course there is more room side-side under a car than up-down under a pedal. And more force in a steering system than you may care to put into a pedal while plucking a fancy musical passage.

A rotary pot does not "have" to be 270 degree rotation. 90deg and 360deg pots exist for very special purposes (and prices). In 270deg models, if they print the track 1/3rd silver, 1/3rd carbon, 1/3rd silver, then all the action happens in 90 degrees. It is not expensive. Problem is that today's pot-makers only want to deal in ship-load quantities of the same part, and the global demand for center-action pots is only a crate or two. Still, I know some custom-taper pots are on the market.

You "can" open-up a standard pot of ~~3X desired value (you want 250K-500K effective, so start with a 1Meg), get some rear window defroster repair paint, and glop-up the ends of the carbon track. Would be best to try it mechanically and mark the actual ends of rotation in the pedal before you glop it up. There's also PCB-repair paint. If I get a dry day I can go up on my tractor's roof and see if standard high-Aluminum roof paint conducts well.

You "can" do fancy things with opamp, addition and subtraction, so a given pot variation produces a large gain variation. But any active solution adds power complication.

[garcho]

Why make a passive volume pedal?

[ashcat_lt]

Why make a passive volume pedal?

In my case it was because I don't have V pots on my guitar.  In fact, my pickups are completely unloaded.  The passive pedal makes it sound a little more "normal".  Also, I only have two hands, and they're usually busy doing other things.

[garcho]

The passive pedal makes it sound a little more "normal".

why wire the guitar that way? or maybe i should ask, why play a guitar wired that way? just curious, not criticizing!

[SISKO]

This is the part which I do not understand.

See the picture above. The higher the load impedance, the greater the differencie is between a high resistance and a low resistance pot. Of course, this is true if the volume pedal is driven by a high impedance source (directly from the guitar)

You "can" open-up a standard pot of ~~3X desired value (you want 250K-500K effective, so start with a 1Meg), get some rear window defroster repair paint, and glop-up the ends of the carbon track. Would be best to try it mechanically and mark the actual ends of rotation in the pedal before you glop it up. There's also PCB-repair paint. If I get a dry day I can go up on my tractor's roof and see if standard high-Aluminum roof paint conducts well.

This is a good option that some people suggest. Ill keep it in mind.

Why make a passive volume pedal?

In my case it was for two reasons: 1) noise: I want the volume pedal to replace my volume pot (i suck at change my volume knob while playing), so the pedal would be first or second in the chain and always on, so any noise it may have, will be amplified by the next distortion,fuzz and OD stages.
2) Fuzz faces dont like low 'z' sources, and active *üsually* means low 'z'. Not always, but very often.

BUT

Given the situation and your opinions, i think that the best way it is to have a preboost stage (carefully desingned so as to keep noise at minimum) and a post volume pot as suggested by induction.
The FuzzFace problem can be fixed by placing a 10k resistor in series with the output of the volume pedal in order to increase the output impedance if needed (i think it may work).

[anotherjim]

Just suppose...
You fit a pot of higher value than the normal value vol control for your pickups. You wire it in the pedal so the wiper faces the pickups. You set it on the gear so the wiper position is grounding the guitar out at whatever end of the pedal tilt you decide for zero volume. You connect the non-ground end of the pot to the output jack. Also connect the wiper to output, or, if you like, via a cap - give some treble bleed.
The value of the pot depends on how much rotation of it your pedal provides. So if 50% rotation then a 1Meg pedal pot stands in for a  500k guitar pot.