I kinda already asked the question in title, but I will reiterate —

In the case of audio, would it be accurate to think of a potentiometer with, for example, three lugs, as a fader between two inputs?

Please tell me why that's dumb or not dumb or something!


A pot can be used as a fader between two inputs, but not at constant power, so it's probably not a particularly useful way to think of it.

Here's the problem: suppose you have a 10k pot with an input attached to each end and the output coming out the wiper. When the wiper is moved to either end of its range, you get full volume out of one channel and a severely attenuated contribution from the other. When the pot is set to the middle of the range, both inputs are attenuated equally, but they're not at half-volume-- they're much quieter.

In the case of audio, I think it would be better to think of a pot as a single-channel fader that has an extra lug on the end.

(I don't actually know whether in real audio equipment the extra lug has some use that I haven't thought of. Anyone know?)

  • 2
    \$\begingroup\$ Just a bit to clarify pingswept. The pot is dividing the voltage. You need to look at power to get a better idea of how much the audio is cut. In addition, the dB scale is used to determine the perceived audio drop by a human. \$\endgroup\$ – Kellenjb Sep 14 '10 at 16:01
  • \$\begingroup\$ The relationship between voltage, power, and the loudness perceived by a human ear is nontrivial. I really like the chart on this page gcaudio.com/resources/howtos/voltageloudness.html because it highlights how 3dB = 2x power, 6dB = 2x voltage, and 10dB = 2x loudness \$\endgroup\$ – ajs410 Sep 14 '10 at 21:14
  • \$\begingroup\$ be careful of this chart. It is based on a linear load. This is not always true, and thus directly measuring voltage may not allow you to determine power. The dB scale is probably one of the more important things you should learn. i would suggest you spend some time getting a feel for it. \$\endgroup\$ – Kortuk Sep 15 '10 at 14:00
  • \$\begingroup\$ Also, please remember that dB scale is relative. Audio is often dBA(en.wikipedia.org/wiki/…). Power is often dBm(power relative to 1mW). \$\endgroup\$ – Kortuk Sep 15 '10 at 14:11
  • \$\begingroup\$ Another way of putting it is that a pot may be used as a fader between two inputs with near-zero impedance, driving an output with near-infinite impedance, but (1) each input will see the other connected by the pot's resistance; if the devices driving the inputs have any significant resistance, and if they also feed any other devices, this may produce unwanted coupling; (2) if the pot is driving a device whose input impedance is too far short of infinite, the pot itself will add a series resistance which varies with its position. \$\endgroup\$ – supercat Oct 24 '11 at 16:17

You could use it that way, as a blend or balance control, by connecting each leg to an input and the wiper to a high input impedance amp.

linear pot used as a balance control

More commonly, you have two 3-terminal pots, and each acts as a volume control (one leg to the input, the other to ground), and are then fed to a summing amplifier to mix them. That way you can control the level of each instead of just their relative difference, and you can use log-taper pots so that the change in loudness is proportional to the change in angle of the knob.

simple mixer circuit with two volume controls http://www.circuitdb.com/downloadimg.php?fileID=130

simple inverting op-amp mixer circuit with 3 volume controls
(source: aaroncake.net)

  • \$\begingroup\$ The two potentiometers could be physically attached to one another so that they both turn at the same time. \$\endgroup\$ – Brad Gilbert Sep 15 '10 at 13:58
  • \$\begingroup\$ Yes, that would make it a blend control instead of a mixer. \$\endgroup\$ – endolith Sep 15 '10 at 14:14
  • \$\begingroup\$ R4,5,6 are wired oddly \$\endgroup\$ – DarenW Oct 15 '10 at 17:41
  • \$\begingroup\$ They are wired so that the noise gain decreases with the signal gain. The signal gain as a function of pot position is the same, but the signal-to-noise is better. \$\endgroup\$ – endolith Oct 15 '10 at 17:55
  • \$\begingroup\$ The emitter-followers have an output impedance of 10k || (50k / beta). 10k is huge, so let's call it 50k/beta. If the transistors have a beta of 200, then the output impedance is 250 ohms. So maximum attenuation is about 250/50k, or -46dB. This is OK but not great performance. Use op-amp buffers instead of the emitter-followers if fidelity is important. \$\endgroup\$ – markrages Feb 27 '11 at 5:09

Maybe it will help some of you guys to see a simulation of this.


Full size can be found at http://www.flickr.com/photos/kellenjb/4991010642/lightbox/

The reason I wouldn't do it this way is because you might have sources with different resistance. Or be driving something with a much lower resistance. Or... well lots of things. It is just much cleaner to use an op-amp adder.

ADDITION: 1) If you make the 50K pot larger, lets say 500K, you get a much smaller signal out when you are at 50% on the pot and the same signal out as the 50K when you change it to the 0% and 100%.

2) If you keep the 50K pot but change R2 and R3 to be bigger, you aren't able to get purely just the signal from 1 source over the other and the effect of the percentage that the pot is at becomes less and less important.

3) If you keep the pot and the source resistances the same but change the output to be 8ohms (such as a speaker might be) you get an effect similar to #1.

  • \$\begingroup\$ I can understand this, but I could before, this needs a bit more detail of what you have done. I see two sources with 1ohm output resistance fed through a 25k resistance to the same line which has a load resistance of 101k. \$\endgroup\$ – Kortuk Sep 14 '10 at 19:41

It's better to think of a potentiometer as a volume control(*) for a single audio channel.

Your + and - audio input signal connect to the non-wiper pins of the potentiometer. The attenuated output + signal connects to the wiper pin, while the attenuated output - signal connects to the same pin you connected the - audio input signal to.

That application actually works, and is how you actually make volume controls. To think of a single, unaided pot as a fader really stretches the analogy into inaccuracy land.

(*) - For pots that have a audio/log taper, or a simulated log taper.


I know this question is old, but I just wanted to add that I've done exactly what the OP asked about as a volunteer sound guy with a full board. We had two click tracks (electronic metronomes, basically) that needed to be mixed in with the musicians' headphones at the same volume each. So I did this:


simulate this circuit – Schematic created using CircuitLab

I also tied all of the grounds together, not shown here.

I had the drummer start both tracks and adjust the pot until they were equal to each other in his headphones, then I mixed the combined signal into all the headphones as usual. Worked perfectly.


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