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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 for it.

The best I could find was this link below.

http://techchannel.radioshack.com/difference-audio-linear-potentiometers-2409.html

There they said this:

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 does the log changes in the pot resistence relates to sound waves and how the human ear works?
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maybe helpful en.wikipedia.org/wiki/Psychoacoustics –  kenny Feb 27 at 17:43
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This question appears to be off-topic because it is not related to electronic design, it's rather about sound waves and how the human ear perceives them. –  alexan_e Feb 27 at 18:02
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@alexan_e Although the question does require some physiology knowledge (which is off-topic), it ultimately asks, "Why use logarithmic instead of linear taper?" That is an electronics design question, it's just informed by biology. Electronics design is often about how to engineer something that is usable by a human, and for better or worse, that requires input from other fields of study. –  JYelton Feb 27 at 18:18
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@alexan_e I'm not saying we should accept questions that ask exclusively about physiology, but that questions that ask "How do I accomplish X in electronics design?" where X requires external fields of study, should be on-topic. The bold questions are indeed off-topic, and the OP should ideally ask them on a suitable site. However, you can't be a good electronics engineer if you work in a black box. Some insight on the bold questions and their answers are intrinsically part of component selection and the performance of the resulting device. –  JYelton Feb 27 at 18:36
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This is a relevant question and should not be closed –  Andy aka Feb 27 at 19:29

6 Answers 6

up vote 15 down vote accepted

Consider this: -

enter image description here

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: -

enter image description here

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.

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Additionally, professional linear faders use an "audio taper" that is neither log nor lin, to give you more control close to "nominal zero" where fine tuning is needed. –  Jon Watte Feb 28 at 17:17
    
@Jon: Could that be because the described relationship of perception, paraphrased as "tenfold increase in intensity is a doubling of perceived loudness" is actually a power curve, not exponential-logarithmic? Which is to say, $$loudness \propto intensity^{0.3}$$ not $$loudness \propto \log(intensity)$$ –  Ben Voigt Feb 28 at 23:40
    
@BenVoigt, if this is all that your fuss is about, we could have cleared this up well before now. See the 2nd addendum to my answer. –  Alfred Centauri Mar 1 at 5:04
    
No, it's because mixing engineers like getting the level roughly right early in the chain, and then fine tune the most prominent stems with better resolution. It's entirely an ergonomics thing AFAIK! –  Jon Watte Mar 2 at 3:01

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...


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.


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.

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A side question: Is "mmw" preferable over μw for microwatts? I've not seen this convention before. –  JYelton Feb 27 at 21:08
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@JYelton, "mmw" was preferable in this case because I was too lazy to type the extra characters. Remember the old style \$\mu \mu F = pF\$? –  Alfred Centauri Feb 27 at 21:21
    
What you have described is a power relationship, not a logarithmic one. i.e. loudness = intensity<sup>0.3</sup> It is a straight line on a log-log plot, whereas a log relationship forms a straight line on a log-linear plot. –  Ben Voigt Feb 28 at 19:03
    
@BenVoigt, in my answer, I make two points: (1) the relationship between perceived loudness and sound intensity is not linear (10x intensity is perceived as 2x loud) and (2) thus, a linear volume control would be quite useless while a logarithmic volume makes sense. I don't get the connection between your points and my answer. –  Alfred Centauri Feb 28 at 19:17
    
No, the logarithmic control does not cancel (or "account for") a non-linearity of the type you describe. Are you really claiming that people want a logarithmic range of loudness, and the logarithmic potentiometer provides this? Maybe, since your final sentence describes an exponential progression. 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. –  Ben Voigt Feb 28 at 19:24

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):

enter image description 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).

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Perceptions of loudness are not quite logarithmic, especially in a noisy environment. A 3dB change in the volume of a signal which is barely audible over ambient noise can be huge. Further, a 3dB change in a signal level which is loud enough to be distorted somewhat can have a huge effect on the level of distortion. On the premise that most people don't care about fine adjustments between "nothing" and "clearly audible", it makes sense to condense that range. On the premise that people often want things as loud as possible without too much distortion, it makes sense to expand that range. –  supercat Feb 28 at 0:28

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.

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This would be easier to read if it used quote blocks instead of code blocks. –  TRiG Feb 28 at 11:23
    
@TRiG: fixed. Thanks for the suggestion. –  Bjorn Roche Feb 28 at 15:17
    
Assuming the other answers are correct about a decade of intensity corresponding to an octave of loudness, then "When we move the pot twice as high, we want to perceive twice as much volume, and log pots give us that" is wrong. –  Ben Voigt Feb 28 at 19:26
    
@BenVoigt: oops, fixed. Thanks. –  Bjorn Roche Mar 1 at 0:29

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.

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If human hearing were linear, the superposition principle would apply, and every sound would be heard the same way, regardless of background noise.

You can easily hear a pin drop in a silent room. You can't hear a pin drop next to a running jet engine. So as the noise level rises, you need more and more volume to make a noticable difference.

So a log pot effectively turns up the volume of the "pin drop" as the ambient level of the sound increases.

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This is not right - "log pot effectively turns up the volume of the "pin drop" as the ambient level of the sound increases" - how can a pot do that? –  Andy aka Feb 27 at 19:06
    
I was speaking metaphorically, of course the pot doesn't turn up the sound of the pin, it makes the delta in sound power higher per degree of pot movement. –  John D Feb 27 at 19:50
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None of this handwaving explains why the required pot is logarithmic. –  EJP Feb 27 at 22:35
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Superposition does hold: sound is additive in a linear fashion. The phenomenon you are referring to is called "masking", and has nothing to do with our perception of sound intensity, which is called "Loudness". –  Bjorn Roche Feb 27 at 23:06
    
@Bjorn, yes, of course superposition holds for actual sound, I was referring to the transducer/processing that is the ear and brain. If it were a completely linear system then I still believe that the a pin drop would be just as audible with a jet engine running as it was in a quiet room. Of course we're not linear and perception plays a big role, which is kind of my point.... –  John D Feb 28 at 1:50

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