I salvaged a motorized pot from a stereo receiver which I want to use between a CD player and a power amplifier (both devices sit on the same shelf at my workplace). Currently, the line out jacks feed a volume pot through 30+ feet of (cheap) cable, and then another 30+ foot run back to the amp. Nobody at work thinks there is anything wrong with the sound (except for the ground hum), but I was nonetheless able to convince my boss to let me "have at" the sound system.

I like the motorized pot idea because the audio cables will be very short, and noisy environments shouldn't be a problem when driving the motor remotely over long cables (plus multiple control points are possible).

My question concerns the pot's value which is 50k, versus what the vast majority of the audio circuits I've found use, which is 10k. In How do I determine if I can safely substitute potentiometers?, general guidelines are given and I seem to meet them all. All except for: try to stay within a factor of 2 (or 1/2) from the ideal. But this doesn't satisfy; I'm after a more definite answer for my specific application. Then, in What potentiometer should I choose?, I was dismayed to read, "As a rule of thumb, for audio, pots generally range from a resistance of about 10K to around 1 megohm." (I thought we were talking 10k to 50k?)

Since I've found 10k to be the "consensus" value, and 50k to be the "exception" value, it seems reasonable (to me) to make this post. So I ask, "What's the deal with the 10k and the 50k audio pots?" I guess my question comes down to one of degree: how much difference will this make, and in what circumstances? Sound quality is my top priority. If there is a trade-off, I don't want to give up more than 2% to 5%.



simulate this circuit – Schematic created using CircuitLab

Figure 1. A potentiometer between a pre-amp and a power amplifier.

  1. A lower pot resistance will load the output of the previous stage. When DC blocking capacitors are used the combination of the capacitor and pot (C1 - R1) forms a high-pass filter. In most cases the higher the pot resistance the better as the cutoff frequency will fall. (Cut off frequency = \$ \frac {1}{2 \pi \cdot R C} \$.
  2. On the input to the next stage a lower source resistance is better as it will better drive the next high-pass filter formed by the input DC blocking capacitor and input impedance. Note also that R2 in parallel with R1 loads the audio taper somewhat so as R1 increases I would expect the taper to be somewhat skewed. This would only be significant if trying to make a linear fade-in or fade-out while doing DJ at your workplace.

In the end the designer makes a trade-off and 10k is a reasonable value.

I would be surprised if you can hear any difference other than, perhaps, a slight bass roll-off.

| improve this answer | |
  • \$\begingroup\$ So if my power amp has a low input impedance, using 50k R1 might load down the source output too much? What are the symptoms of an overloaded output stage? Is that like an unbalanced -10dB output trying to drive 100 yards of cable? Or does that degradation mainly come from excessive capacitance? Maybe I should hone my question: {The manufacturer's specs for the power amp are: "Input Impedance (nominal): 20 kilohms balanced, 10 kilohms unbalanced." In this case, would a 50k pot be risky (and make the 10k the best choice)? Or would either 10k or 50k not make much of a difference?} \$\endgroup\$ – user160409 Nov 3 '17 at 8:18
  • \$\begingroup\$ Have a look at en.wikipedia.org/wiki/Potentiometer#Theory_of_operation and then create a spreadsheet to calculate what the output voltage will be at 10%, 20%, etc., position of the volume control for a 10k and 50k pot feeding a 10k load. \$\endgroup\$ – Transistor Nov 3 '17 at 19:32
  • \$\begingroup\$ OK, I made tables for 10K and 50K into a 10K load using Python so I could play with # of steps, load etc. I repeated this many times to get it comfortable. For Vs, I always used 1Volt. I was quite surprised you said plot voltage because any value pot could form a divider to bring the 1V down to, say, 0.5V which then drives the input. The 50K would need less rotation to reach 0.5V than the 10K, but as long as the input felt 0.5V, why would rotation matter? I had already done this exercise, but looked at the wiper resistance to gnd, since I thought that was how loading occurred. \$\endgroup\$ – user160409 Nov 5 '17 at 2:47
  • \$\begingroup\$ Loading generally refers to the effect the next stage input has on the previous stage's output. The pot loads the pre-amplifier so increasing it's value would reduce the load and, generally, be a good thing. The amplifier input loads the pot so, as you should see in your spreadsheet, it still works but distorts the normal curve of the pot (so rotation only matters as covered in my point 2). \$\endgroup\$ – Transistor Nov 5 '17 at 8:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy