Why should I use a logarithmic pot for audio applications?

Question

Just got really curious about it reading this answer from Spehro Pefhany. There Spehro comments that one should use a logarithmic pot for audio applications. So I googled it.

The best article I could find was one titled "Difference between Audio and Linear Potentiometers"^[1] which now seems to have been removed from the original website.

There they said the following:

##Linear vs. Audio

Potentiometers, or "pots" to electronics enthusiasts, are differentiated by how quickly their resistance changes. In linear pots, the amount of resistance changes in a direct pattern. If you turn or slide it halfway, its resistance will be halfway between its minimum and maximum settings. That's ideal for controlling lights or a fan, but not for audio controls. Volume controls have to cater to the human ear, which isn't linear. Instead, logarithmic pots increase their resistance on a curve. At the halfway point volume will still be moderate, but it will increase sharply as you keep turning up the volume. This corresponds to how the human ear hears.

Well, I'm not satisfied.

• What does it mean that the human ear isn't linear?
• How do the log changes in the pot resistance relate to sound waves and how the human ear works?

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^{[1] Original (now broken) link was http://techchannel.radioshack.com/difference-audio-linear-potentiometers-2409.html.}

Answer 1

Consider this: -

Sound level is measured in dB and, a 10 dB increase/decrease in signal equates to a doubling/halving of loudness as perceived by the ear/brain.

Look at the picture above and ask yourself which is the better choice for smooth (coupled with extensive) volume controller. Below are the Fletcher Munson curves showing the full range of decibels that a human can comfortably hear. Note, that unless your stereo system is very powerful, a range of 100 dB is "about right" for volume control. The Fletcher Munson curves also relate loudness to the pitch of a sound. Note also that the curves are all normalized to 1kHz in 10 db steps: -

Approximately every 10% of travel of the wiper on the LOG potentiometer can reduce/increase the volume by 10 dB whereas a LIN pot will need to move all the way down to its middle position before it's reduced the volume by only 6 dB! When a linear pot is near the bottom end of its travel (sub 1% of movement left) it will be making massive jumps in dB attenuation for just a tiny movement hence it would become very difficult to set the volume accurately at a low level.

It's also worth pointing out that a LOG pot is only able to cope with so much dynamic range of adjustment before it does the same (below -100 dB) but, the point is, this will hardly be noticeable at the tiny, quiet end of its travel.

You might also note that the markings on a pot such as CW and CCW tell you which end of a pot is the ground end and the high-volume end. CW = clock wise and CCW is counter clock wise end points for the wiper.

Answer 2

What does it mean that the human ear isn't linear?

In this context, if the human ear were linear, a sound wave with twice the power of another would sound twice as loud.

However, the fact is that a sound wave must have 10 times the power of another to sound twice as loud.

How does the log changes in the pot resistence relates to sound waves and how the human ear works?

Assume the potentiometer (volume control) varies the signal power applied to the loudspeaker and assume the amplifier can produce a maximum of 100W.

Assume the pot is linear, the control is evenly marked from 1 to 100 and we start with the control set to 100 - there is 100W of power sent to the loudspeaker.

To halve the volume, we would reduce the output to 10W which would require turning the volume control 90% CCW to the "10" mark.

To halve the volume again, we would want just 1W which would require turning the volume control to the "1" mark.

To halve the volume again, we would want just 0.1W and... do you see the problem?

If however, the pot were logarithmic, the spacing on the knob between 0.1W and 1W, 1W and 10W, and 10W and 100W would all be the same. If there were ten marks, evenly spaced, we'd have something like:

0, 1mmw, 10mmw 100mmw, 1mW, 10mW, 100mW, 1W, 10W, 100W

So we go from no sound to barely audible, double that, double that, double that, double that, etc...

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This addendum is to address a question raised in the rather longish comment thread. According to @BenVoigt, the hypothetical attenuator proposed above does not adjust the sound level evenly.

@Alfred: I'll repeat my previous comment, since clearly you glossed over it: "your dial has "loudness 1, 2, 4, 8, 16, 32 ... 1024" as its equally-spaced ticks. One click at the bottom is a change of 1 loudness unit. One click at the top is a change of 512 loudness units." 1 and 512 are vastly different changes.

Since I haven't been able to convince Ben of his error nor has Ben been able to convince me of mine in the comment thread, I'd like to address this dispute in this addendum.

According to this source, the just noticeable difference in sound intensity is about 1dB:

about 1 decibel is the just noticeable difference (JND) in sound intensity for the normal human ear.

If the sound intensity changes by 1dB, we just notice the change in loudness.

Thus, it follows that if our hypothetical stepped attenuator adjusted the attenuation by 1dB increments, adjusting the control by 1 step would make the sound just noticeably louder or softer to the human ear.

In other words, this attenuator would smoothly adjust the loudness of the sound, in just noticeable increments, over the entire range.

So, rather than 10 evenly spaced steps as I gave above, imagine 100 evenly spaced steps on the control.

Each step changes the power by 1dB; turning the control CW 1 step increases the power by a factor of 1.2589...; turning the control CCW 1 step decreases the power by a factor of 0.79433...

For example, if the control were set to 1W output, turning the control 10 steps would increase the power by \$(1.2589...)^{10} = 10\$ to 10W. Tuning the control CW another 10 steps would increase the power by another factor of 10 to 100W.

But this differs from the previous attenuator only in resolution, i.e., we've only increased the number of (evenly spaced) marks in between the original marks.

Also, questioned in the thread is whether this is a logarithmic attenuator.

I explicitly said the relationship you describe is not linear, and not logarithmic, it is a power.

Recalling that the relationship \$y = \log(x)\$ implies \$x = 10^y\$, if a pot is logarithmic, there is necessarily a related power (or exponential) relationship implied.

That fact is, we can say that in the above attenuator, the number of steps required to change the power by some factor is proportional to the logarithm of that factor.

For example, to change the power by a factor of 5, e.g., to increase the power from 1W to 5W, requires turning the control

$$10 \log(5) \approx 7$$

7 steps.

So, the number of steps (or change in angle of a pot) is logarithmic in the power.

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2nd addendum to address further comments.

According to @BenVoigt, the answers given here are misleading or plain wrong:

But I get the general impression from reading any of these answers that the logarithmic resistance inverts the biological response, and then look closer at the math described and realize that isn't true.

I wish to demonstrate that a logarithmic pot is what is desired but not because it inverts the biological response (which I don't believe anyone has claimed nor is it what's desired as I shall show below.).

Starting with the well-known (and approximate) "rule of thumb" that 10 times the intensity is perceived as 2 times the loudness, let us write the following relationship between relative loudness \$l\$ and relative intensity \$k\$:

$$l = 2^{\log k}$$

Clearly, if the relative intensity \$k\$ is 10 then the relative loudness \$l\$ is 2 as desired.

For our 1dB stepped attenuator, the relative power is given by:

$$k = 10^{n/10}$$

Combining the previous two equations, we have that the relative loudness is

$$l = 2^{n/10}$$

Thus, for each step, the loudness increases by a factor of 1.0718... or decreases by a factor of 0.93303...

But this is what we want. We don't want the loudness to increase by a fixed amount each step, we want the relative loudness to increase by a fixed amount each step.

Thus the need for a logarithmic attenuator.

Answer 3

Andy has answered this, and he hinted at the end that A-taper (log) pots are not perfect. Here is a comparison between an ideal log response and what a real commercial log pot actually does (taken from here):

It's a two-segment piecewise linear approximation to the ideal log taper (dashed line). Crude, but it does the job well enough in many cases.

Note also the flat bits at the end of even the linear (B-taper) pot curve. That's when the wiper gets near the ends of travel in either direction.

Oftentimes these days, electronic volume control is implemented which has constant dB steps of attenuation or gain.

Here is an example datasheet for the PGA2320. It has gain adjustable from +31.5dB to −95.5dB in 0.5dB Steps. A step of 0.5dB is considered to be just perceptible. That's an 8-bit number to select the volume level (255 levels plus mute). If you were to try to simulate that with a linear multiplying DAC (MDAC), you'd need something like \$4\cdot10^6\$ steps to get 0.5dB resolution at the low end (about a 22-bit DAC).

Answer 4

Although this question has been adequately answered, I found some of the answers confusing, and this something of a specialty for me, so here's an attempt at a simpler answer:

What does it mean that the human ear isn't linear?

The human ear perceives the intensity differently from how the world actually is. In the world, sound has a property called "Volume," (or sound intensity) which we perceive as "Loudness." A doubling in volume does not produce a doubling in loudness, and this is what is referred to as "non-linear."

How does the log changes in the pot resistence relates to sound waves and how the human ear works?

The idea of using log-taper pots is that they more closely copy the human ear's perception of reality: when we move the pot by a fixed amount, we want to perceive the same amount of change, regardless of where the pot started. (incidentally, the human ear is not the only thing to perceive things this way: Most of human perception is ruled by the so-called Weber-Fechner Law, but hearing is particularly sensitive because the loudest sound we can comfortably listen to is about 1 million times louder than the quietest sound we can hear.)

This works well for gain controls (including gain controls as part of an EQ or other circuit), but not everything in audio should be log-taper: balance/pan controls for example.

Answer 5

Regarding the perceptual aspect of hearing: It is a fact that sounds seem louder in proportion to the log of the actual sound intensity, and not linearly directly proportional. This is a very common aspect of all animal and human perception of the environment. For example if you have two weights, one that weighs 1 ounce and another that weighs 2 ounces, you can use both hands and tell that the 2 ounce weight is heavier. However if you have a 1 pound weight and another which weighs 1 pound plus 1 ounce, you will be very hard pressed to discriminate the difference.

In general the neurological processes in perception are set up to discriminate ratios between stimuli intensity and not subtractive differences. This means that really you are sensitive to subtractive differences in the log of the stimuli intensity. This also includes vision, where the eye and brain normalize for the average background brightness and contrast. And when we perceive differences, these are differences in ratio relative to the normalized average. This involves a fundamentally log transfer characteristic of the sense organs plus temporal adaptation processes in human sense organs, and also involves relational renormalization and adaptation responses in the many layers of interconnected neurons that process information in the nervous system.

In vision, the eye has to be able to cope with light levels that go from 10^{-4} to 10^6 candelas per square meter from an environment with a starry night to one at noon on a sunny day. So given this 10 orders of magnitude scale, representing the visual signal at the retina using a linear system would be unreasonable. (That's like a camera requiring more than 32 bits of binary representation per pixel just for brightness without considering color.)

The field of psychophysics studies aspects related to the perception of stimuli relative to the actual measured stimuli. Two important concepts are the just noticeable difference (JND) curves, which describe how threshold intensity awareness for change relates to the background intensity, and the Weber-Fechner law which basically just states that most perceptual processes are sensitive to ratios between the intensity of stimuli.

One can see that living organisms have to have the ability to adapt to the average level of environmental stimuli - visual, auditory or other sensory inputs (e.g. in a loud environment not to be constantly triggered by minor changes) - but at the same time be aware of important significant changes that could be relevant to survival.

In addition, every sense organ and neural process has a limited dynamic range of representation, and also a background internal noise level (typical aspects of any communication channel). It makes sense that the brain attempts to re-normalize the sensory input signals in order to constantly optimize the internal representation's signal to noise ratio, so that the probability of detecting relevant changes is highest. It is similar to the problem of representing audio signals in only 8 bits - if you can accurately represent the quiet signals, then the loud ones will saturate the range. This is why A-law was invented.

Anyway this is the biological and perceptual rationale behind the fact that we judge sound intensity on a log scale.

Ref 1: Just noticeable difference concept.

Ref 2: Weber-Fechner law

Ref 3: A-law

Answer 6

Many others have explained why a lin pot is not much use, as it stands, as a volume control, and discussed the various pot laws available.

What hasn't been mentioned is the effect on reliability of the log law. Basically the pot is a carbon or conductive plastic track, and the whole thing is mechanical. Non-linear pots have a thinner track at one end, and so tend to deteriorate more over time.

There is a common "hack" used in pro-audio gear to get around this, and allow the use of a linear pot. A resistor from the wiper to ground of a lin pot "fakes" the log law well enough.

If you think about it - what people want with a volume control is that they get "loud" with it full up (or nearly), "medium" in the middle, and "quiet" at the bottom. Noone at all worries about whether each segment of 10dB has the same angular rotation.

In practice, if you have a 10k linear pot and put a resistor to ground on the wiper, you get a circuit like this:

^{simulate this circuit – Schematic created using CircuitLab}

Now Ra + Rb = 10k, and a spreadsheet is handy to see the law (rotation is 0 for anti-clockwise and 1 for full up - Rb is just 10 * rotation. I leave out the "k" as everything is just normalised here.)

From experience, it turns out that something around -15dB in the middle (it is not that precise) feels right - and saves you from waiting around for those special pots to arrive (also reduces the lines in your BOM), and gets you a more reliable product. (For that you want Rp =~ 1k3 with a 10k lin pot.)

Given that the accuracy of most "log" pots is terrible anyway, this is just fine. If you are making a stereo volume pot and care about imaging (you should) then this might also be a bit more accurate - or perhaps you are better with a switched attenuator.

Answer 7

Sounds is pressure. Like a balloon. You're blasting More Than a Feeling on about volume '1' on your radio, and you're 10 feet away, then you move to 20 feet away, you need to turn the dial up. The radio is the centre of the balloon, you want a 5 foot balloon to become a 10 foot balloon? The volume of air required doesn't just double right? It's way more. Actually, for a balloon its about 8 times. But our brains don't work like that. Changing your radio dial from 1 to 8, just coz you moved 10 feet would seem 'wrong'. So, use a log pot, then change it from 1 to approx 2, and you've got the sweet sounds of Boston ringing in your ears at just the 'right' volume.