Searched and read but still need a clear explanation on volume pot value choice?

[chumbox]

Hi All

Assuming the gain and output of a circuit remains constant what difference should I be finding placing eg an 10K, 100K and 500K at the end of a circuit?  And why? Let's ignore taper and assume they are all audio, I understand that bit.

Why this has become so elusive to my brain is beyond me 

Thanks in advance

[mistahead]

Have a glance here - might help, I'm skimming it:

http://www.beavisaudio.com/techpages/Pots/
"This causes more of the signal to be dumped to ground. This dumping essentially sends the signal into oblivion, thereby lowering the overall output level.

"So if you think about it, you are never really turning the volume up! The volume or level in the circuit is always running at full tilt. What you are doing the above "volume" control is actually attenuating (making smaller) the full volume that was there to begin with."

With a bouncy AC (audio) signal you have what would happen if you decreased the width of pipe in the above? I imagine it would start chopping into the signal, not sure what would be the outcome of going too large however...

[chumbox]

Thanks mistahead.

So in essence a 500K pot should have a slightly higher output volume than a 100k when fully open?  This is where I get confused.  Fully open I assumed all pots would be the same and then the speed at which it dumps to ground (as you turn it counter-clockwise) would change with pot value, in which case I struggle to see why people use so many different options.

[mistahead]

Ummm maybe, sorry brain odd right now... the summary you have about the speed/responsiveness is right, but you can also hit the tone by going the right/wrong way... I think its too low a width (1K type thing) where 500K and 100K are both over the min required width so dont make much difference at all.

Most of these volume controls are Voltage Divider configs, so its more about ensuring there is enough bandwidth through the pot to not eat into the signal (10K or 100K) from there its just proportions - one third on the left, two thirds on the right, etc.

I've only every seen 500K pots (or 1M) when they're wired as true variable resistors, so not the voltage divider config, and I only RECALL seeing the 10K voltage dividers on the low-fi fuzz-az type stuff.

Happy to be corrected here - could be spinning misinformation, been dragging through a cold-mush-head-thing and VBA script, so yeah no brain means something something....

[chumbox]

Makes sense. 

In reference to the values used in actual pedals, two examples that started half my confusion are why a Lovepedal Woodrow runs a 500K and a Timmy runs a 10K on the output.  Granted I think the Timmy is linear but the outcome shouldn't change at full tilt.

I'm gonna go back and read 'Secret Life of Pots' a few more times and see where I end up.

Thanks again mistahead.

[mistahead]

Good question, I'd have to go learn those two up real quick which I'm not going to today... It'd be interesting to swap them both in a 100K and listen.

Nothing against the Lovepedal gear - but I've always had a "feeling" that they believed they were more nuanced than they really were, maybe its big sweepin' knobs that I'm missing.

[ashcat_lt]

I really don't like that thing from the Beavis site.  There's really nothing being "dumped to ground", and in this config, the source doesn't really get any closer to ground no matter where the wiper sits.

The assertion that it's a voltage divider is the hint, but the resistances in the pot are not the only ones to consider when trying to figure what gets divided and by how much. 

There is always a source impedance which is almost always frequency dependent - the output impedance of whatever the active device was, any series resistor after it, the coupling cap...  This is always added to the "top" of the pot, even when it's at 10 and the pot itself contributes 0, for frequencies for which this value is very large compared to the "bottom" of the pot, there will be noticeable attenuation. 

There is also always a load, which is almost always also frequency dependent.  This is sits parallel to the "bottom" of the pot, and can do nothing but reduce that part of the pot's value.  For frequencies where this impedance is very small compared to whatever is on the "top" of the pot (including all that other crap we said we have to add in there), there will be noticeable attenuation.

So, if the pot's too small you will lose bass, or use much larger coupling caps, and you'll lose volume.  If it's way small, you'll also start to demand too much current from the active device and either stuff will distort or stuff will get hot.

If the pot's too big everything's generally cool until you turn it down from 10, at which point you end up with very large values at the "top" of the pot, which works with cable capacitance to divide down treble frequencies much more quickly than the broadband signal.

This is kind of why there's no definitive answer coming up in your searches.  It always depends on what's on either side!  A lot of times we have to compromise depending on our goals for the performance of the control.

[earthtonesaudio]

As ashcat said, the potentiometer does not exist in a vacuum.  There is a source impedance and a load impedance.

The source impedance (R1) forms a voltage divider with the total resistance of the pot (R3).  So if your source impedance is high and the pot value is low you will get a large voltage division and therefore a lower total volume output.  Note that this is with no load (infinite R2).

The second part to consider is the load impedance (R2).  At maximum volume R2 is in parallel with the pot value and together these form a voltage divider with R1.  Again, if R2 is much smaller than R1 you'll get a big attenuation.

Finally consider the ratio of R3 to R2.  If it is large then turning R3 down a small amount will result in a large attenuation. 


This ends up being a trade-off.  An idealized voltage divider will have R1 = 0 and R2 = infinite.  Then the value of R3 is completely irrelevant.
In the real world you might strive for a set of ratios like R2 >= 10*R3 and R3 >= 10*R1.

[R.G.]

Yes - what ETA said.

It is a classical mistake to fail to consider that every circuit has something driving it, and something it drives. The impedances and signal levels of what is driving the circuit (signal through R1 in ETA's example)  and what it drives (R3) always matter to how well the circuit performs.

There is another set of things to consider. There are always circuit components that do not appear in the schematic. Mother Nature puts them there. There are capacitances from every conductor to ever other conductor literally in the universe. Every wire has inductance and resistance. The whole idea of schematics is to idealize and ignore these parasitic capacitances, inductances, and resistances. This is justified in many cases because they are quite small, and may safely be ignored at low currents, low voltages, and low frequencies, like most pedals work with.

Where you get into trouble ignoring them is when you start using big resistances - at or over 1M - and high gains, like the distortion pedals we use. The big resistances, and the special case of the large inductance of a guitar pickup, mean that a few pF of capacitance *can* cause unwanted coupling or tone changes. And high gains can cause issues with those same small capacitances.  So the bigger the value of a pot, up in the 1M range, the more you have to worry about the wiring to and from it changing the perceived tone of the audio through it.

[gjcamann]

So RG is saying a smaller pot value (within reason) is better than a larger one. (Good to know - thanks)
For the benefit of noobs...
Your pot will appear as a resistor to ground with current flowing through it (the value of the resistor for this consideration is the full pot value).  So you will need to make sure that whatever tranny or op-amp is driving the pot, can comfortably source the current through the volume pot (and any other tone circuitry paralleling the volume pot).

The tranny in the woodrow is probably comfortable with the 500k load - or Mr LP is lazy and used it because 500k is a pretty typical value.
But the timmy has an op-amp that can source alot of current, so a smaller pot could be used.

Stand by for someone to correct me. 

[R.G.]

In most cases, a pot value around 10K is good for generic audio work - which is why a lot of non-guitar gear tends to use similar values.

Lower value pots are not as susceptible to noise pickup and crosstalk, and thermal noise effects are smaller, if the signal level is such that it makes a difference. When you get below 2K or so, IC opamps tend to have difficulties driving the load. A 500 ohm pot is very hard for a generic opamp to drive. Between 10K and 100K, you're fine. You see a lot of those. More than 100k, you need to ask yourself whether you have a reason for going that high.

1M was the de facto standard in tube circuits, but that was usually driving a tube grid with an even higher input impedance. Even then, parasitic capacitance crept in. I read one audio magazine article from the 50s on choosing volume control values and it spent several pages on the idea that simply flipping in a 1M was not necessarily the best thing to do.

The loading on the source from the pot and its wiper load varies with rotation, as does the source impedance to the pot load. When the wiper is full up, the source sees the entire pot resistance in parallel with the wiper load. When the pot is full down, the source sees only the pot load.  In between, the source sees some portion of the pot resistance in series with the rest of the pot resistance paralleled by the load.

The wiper load sees the pot as a source resistance in series with the fractional-voltage of the setting. If you drive a 10K pot with a 1V signal, and set the pot to half its electrical rotation, 5K above and 5K below the wiper, then any load on the wiper thinks this looks just like a 0.5V signal in series with the upper and lower pot fractions in parallel, or about 2.5K. In fact, the mid point electrically is the highest source impedance for any pot, and this is 1/4 of the pot's end-to-end resistance.

If the load on the wiper of the pot is greater than 2.5 time the pot end-to-end value, it will not be very affected by the resistance of the pot and the voltage divider approximation will hold true. If the source impedance driving the whole pot is less than about 1/10 of the pot value, the pot will not significantly load it down.

Often the exact value of a pot **when used as a voltage divider or volume control** is not very critical, and the pot value can be somewhat flexible or used for other things, like damping a transformer output, treble rolloff, etc. When used as a two-terminal variable resistor, of course the exact value of the pot may make a difference.

It's worth thinking about using fixed resistors to limit the pot's range a lot of times. A resistor at the top of a pot limits how far the output can be turned up electrically when it's at full mechanical motion. A resistor at the bottom limits its range to turn things down.

The wiper of a pot is the weak point usually. It's a fragile sliding contact, and may have several hundred ohms of resistance just in the contact itself in some pots. A resistor in series with the wiper adds to the source impedance seen by any wiper load. Sometimes this can keep you from burning a pot out.

[chumbox]

Now that I think I have got my head around this one last question.

If I have A100K pot on the end of a boost circuit and I am getting unity gain close to 3 o'clock (3/4 clockwise turn) is there a way using a bridging resistor to move this closer to 12 o'clock or should I consider maybe a B100K based on the secret life of pots article?

Thanks everyone this has been a great explanation.  Really enjoyed the read and makes a whole lot more sense now. 

[ashcat_lt]

I find myself wondering why you would want a "boost" pedal to run at less than unity???  Isn't that kind of the opposite of what the name impliles?

[chumbox]

I understand what you mean ashcat_it but I don't want it to run at less than unity, I just want unity to be at 12 o'clock on the dial.  I understand this may/may not be possible because the max volume of the pedal is the what it is based on the circuit. 

I guess a better way to put it.  How does a company ensure unity gain is at 12 o'clock generally when they release a pedal?  Is it purely based on the circuit or can bridging resistors adjust where this is on the dial?

[ashcat_lt]

ummm...  install the knob in a different position or physically rotate the pot body so that the marker points to 12 at unity?

Okay, but really I hear you saying that you want it at unity when it's at the half-way point in its rotation.  Changing to an audio taper pot or installing an appropriate tapering resistor will likely get you closer if not exactly there.

But it still leaves the question, what do you get when you turn it down below 12 oclock?  Less than unity.  So why?

[earthtonesaudio]

If the volume pot is allowed to interact with the load that follows it, that load then helps determine at what "o'clock" position unity gain is.  So unity might be 3:00 for one amp but 12:00 for a different amp.  The way to make it consistent and not care what you plug in to is to put a buffer between the volume pot and the output.  See for example the stock Tube Screamer schematic.

And as mentioned the taper of the pot will also make a difference.

[chumbox]

Thanks all.  Appreciate the advice.

and ashcat_It now I fully understand what you are saying and I guess less than unity is a little absurd on this type of pedal.  It's probably an aesthetic and mental thing. 

[PRR]

> 1M was the de facto standard in tube circuits

For *AM band* work.

You find lower values in utility PA, and in Hi-Fi you rarely find much over 250K as classic volume pot.

______________________________________

> there a way using a bridging resistor to move this closer to 12 o'clock or should I consider maybe a B100K

Not a good way.

Increase the gain of the booster, you will "have" to turn-down more for the same overall gain. (But it may distort before the pot, not-good if you want a clean boost.)

A resistor wiper to ground will suck-off more signal at part-way than at full-up. But to make a real difference you need heavy loading which will change the full-up condition.

An Audio-taper has 5:1 loss at half-way, so 2:1 loss around the 3:00 setting. A Linear taper has 2:1 loss at mid-way. Sounds like you already found your path.

[chumbox]

Hey PRR

Thanks for the ratios 5:1 and 2:1.  That really is super useful info.