Lets talk about master volume controls.

[Sensei Tim]

two topics:

1.  which is the better approach:  a master volume as a voltage divider (board output to vol 3, vol 2 to output, vol 1 to ground) or as a variable resistor in the output path?

2. if using a voltage divider, does using a 100k pot just give you finer control over the volume because you go from (approximately) 100k:1R to 1R:100k vs using a 5k pot where you go fro 5k:1R to 1R:5k over the sweep?

[rocket88]

I think it depends on the circuit, but more commonly as you know, you'll see it as a voltage divider at the end of the circuit. As far as value, the way I understand it, which may be wrong, is that the higher the resistance of the pot and/or resistor to ground the more output you'll get. You have to keep in mind that you're also creating a high pass filter, since you have the output cap right before it. So changing either the pot value or the resistor to ground will change the tone of the effect.

[Frank_NH]

If you use the standard voltage divider at your output, you have to consider the impedance of the circuit upstream AND the device the pedal is connected to.  For the latter, if your pedal is connected to an amp, amps usually have about 1M ohm input impedance;  thus you can use 100K pot or lower (which is in parallel with the amp's input impedance) and not alter the output signal.  The upstream circuit feeding the output pot is in series with the pot and thus forms a voltage divider even when the pot is all the way up.  If the output circuit is low impedance (e.g. from an op amp stage), say 1K - 10K ohms, then again a 100K output pot will not appreciably lose any of the signal, but a 10K pot might.  It seems that 100K is about the right value for being higher than the circuit impedance but lower than the amp's impedance.  If you protect the circuit with input and output buffers a la Boss and Ibanez, then there is more flexibility in choosing the volume control value.

[induction]

Frank_NH wrote

If the output circuit is low impedance (e.g. from an op amp stage), say 1K - 10K ohms, then again a 100K output pot will not appreciably lose any of the signal, but a 10K pot might.  It seems that 100K is about the right value for being higher than the circuit impedance but lower than the amp's impedance.  If you protect the circuit with input and output buffers a la Boss and Ibanez, then there is more flexibility in choosing the volume control value.

I agree with most of what you said, but I want to point out that the output impedance of a real-world op-amp is in the tens of ohms. In other words, it's usually indistinguishable from a buffer. A 10k volume pot after an op-amp stage is very common.

A decent rule of thumb for master volume pot values is: 100k for low-impedance outputs, 10k for high-impedance outputs. Of course, unlike 'stupid' and 'clever' the line between low and high impedance outputs is fairly broad, so it pays to avoid too much reliance on rules of thumb. As with almost everything else in this hobby, the answer to almost every question is 'breadboard'.

Sensei Tim wrote

1.  which is the better approach:  a master volume as a voltage divider (board output to vol 3, vol 2 to output, vol 1 to ground) or as a variable resistor in the output path?

To the best of my recollection, I have never seen the latter in a pedal. Do you have any examples of this? The quantity affected by adjusting such a volume pot would be the output impedance, not the volume. The behavior of such a volume pot would be totally dependent on the input impedance of whatever comes next in the chain, which is generally bad practice. If you feed that pedal into a SHO, for example, adjusting that pot would likely have no noticeable effect on the volume at all.

2. if using a voltage divider, does using a 100k pot just give you finer control over the volume because you go from (approximately) 100k:1R to 1R:100k vs using a 5k pot where you go fro 5k:1R to 1R:5k over the sweep?

The output volume is governed by the voltage divider equation. The relevant quantity is the ratio between the bottom resistor (output to ground) and the sum of the top and bottom resistors, not the values of the resistors independently.

While the surrounding impedances can have an effect on the volume loss caused by a specific volume pot value, the pot value does not affect the amount of fine control or the taper. Referring to the voltage divider equation, in the ideal case Vout = Vin * 1/1 = Vin at maximum setting, regardless of the value of the pot. Likewise, at minimum setting, Vout = Vin * 0/1 = 0.

In other words, both pots go from full on at one end to full off at the other end. For convenience, let's assume both pots are linear. Mathematically there is no way to give finer-grained control when dumping a fixed quantity of something at a fixed rate between two endpoints that are a fixed distance apart. If that's what you want, you'd need to switch to a multi-turn pot, not a different value.

You can adjust the 'control-grain' by adjusting the taper, but that gives you finer control in part of the pot at the expense of coarser control in the rest of the pot, and again, the value of the pot itself doesn't play a part, assuming the surrounding impedances are reasonable.

Finally, note that while larger master volume pots do preserve volume in circuits with high output impedance, they do so at the cost of increasing the output impedance even more at settings less than max. Thus, while it's tempting to just slap a 1M master volume on every circuit, lower pot values do offer some advantage. Balancing the volume loss of a low value pot against the increased output impedance of a higher value pot is what gives rise to the rule of thumb given above.

[reddesert]

A problem with large-resistance volume pots is that they can yield a high output impedance and thus be loaded by a following stage with low-ish input impedance, as posters above have said. Here's an example:

Suppose I have an OCD-ish pedal with a 500K volume pot wired as the usual voltage divider, and I am getting my desired volume with it halfway up, so 250K below the wiper (pin 2 = output) and 250K above. Now suppose the next pedal in the chain has 100K input impedance, which is lower than desirable but not impossible.  I kick that pedal on, and now my voltage divider has 250 K above the wiper; and below the wiper is (250 K parallel with 100K) resistance to ground = 71K.  So now my voltage divider is 71K in series with 250K, and I'm getting 71K / 321K = 0.22 of the max volume, instead of 0.5.

See what happened? I kicked another pedal on and the volume dropped by half. I could turn up the volume on pedal 2, but what happens when I want to use pedal 2 without pedal 1, and so on?

[Frank_NH]

reddesert wrote

A problem with large-resistance volume pots is that they can yield a high output impedance and thus be loaded by a following stage with low-ish input impedance, as posters above have said. Here's an example:

Suppose I have an OCD-ish pedal with a 500K volume pot wired as the usual voltage divider, and I am getting my desired volume with it halfway up, so 250K below the wiper (pin 2 = output) and 250K above. Now suppose the next pedal in the chain has 100K input impedance, which is lower than desirable but not impossible.  I kick that pedal on, and now my voltage divider has 250 K above the wiper; and below the wiper is (250 K parallel with 100K) resistance to ground = 71K.  So now my voltage divider is 71K in series with 250K, and I'm getting 71K / 321K = 0.22 of the max volume, instead of 0.5.

See what happened? I kicked another pedal on and the volume dropped by half. I could turn up the volume on pedal 2, but what happens when I want to use pedal 2 without pedal 1, and so on?

That's one reason to build in a buffer at the input and output (like Boss and Ibanez).  It makes sense for a pedal manufacturer to do things that way, since player will purchase a variety of effects and then mix them in various ways.

As for the output impedance, it's not a simple matter to calculate a value (something like LTspice is helpful), but I agree that the output of an op amp or JFET buffer stage will permit a low output pot value like 10K.

 

[Sensei Tim]

induction wrote

To the best of my recollection, I have never seen the latter in a pedal. Do you have any examples of this? The quantity affected by adjusting such a volume pot would be the output impedance, not the volume. The behavior of such a volume pot would be totally dependent on the input impedance of whatever comes next in the chain, which is generally bad practice. If you feed that pedal into a SHO, for example, adjusting that pot would likely have no noticeable effect on the volume at all.

I'm rehousing a Danelectro Rocky Road leslie speaker simulator right now. It has no output volume control and has a pretty ridiculous amount of boost on tap.  I tried putting in a voltage divider to tame the output and it killed all of the high end in both the signal and bypass (buffered bypass, i guess?  not true bypass).  I then tried putting a variable resistor in line with the output wire and it worked almost perfectly.

The output volume is governed by the voltage divider equation. The relevant quantity is the ratio between the bottom resistor (output to ground) and the sum of the top and bottom resistors, not the values of the resistors independently.

that's what i said... using a 100k pot will give you ratios from 100,000/1 on one side to 1/100,000 on the other


thanks for the info. very enlightening!

[induction]

Sensei Tim wrote

I'm rehousing a Danelectro Rocky Road leslie speaker simulator right now.
<snip>

Interesting. If I get a chance, I'll look up the schematic and see what I can figure out.

that's what i said... using a 100k pot will give you ratios from 100,000/1 on one side to 1/100,000 on the other

Not quite. Look at the voltage divider equation again. The ratios are 100k/100k on one side and 0/100k on the other. These reduce to Vout = Vin and Vout = 0 no matter what the pot value is. The only difference between 100k and other pot values is the relative magnitude of the surrounding imperdances, which affect the min and max volumes only, not the sweep.