Choosing a resistance for volume (input side) potentiometer

I have an amp that delivers 30W to each of the two (2) channels. Each channel contains a crossover composed of a 6 ohm and a 4 ohm speaker in parallel with each other.

That's all irrelevant other than you have a stereo power-amplifer. The potentiometer will be going to the input. The output power doesn't matter.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Your setup.

To add a simple volume knob between the source (a phone or iPod) and the input jack of the amp, what is the logical process for determining a resistance value, and type (Lin/Log)?

  • The phone and iPod are designed to drive earbuds in the range 8 to 32 Ω. Using the 10:1 rule any pot over 80 Ω would not load the output significantly.
  • The amplifier input is typically >= 100 kΩ (but check the specifications). To prevent it loading the potentiometer too much we'll drive it with a source impedance of <= 10 kΩ.
  • Due to the logarithmic response of the human ear to sound pressure levels a logarithmic potentiometer is required to get a perceived even increase in volume with rotation.

A 1 kΩ to 10 kΩ pot should be fine.

If you use too high a value for the potentiometer the amplifier input impedance will load it excessively and distort the volume curve. See Electrically loaded resistive Potentiometer! for more on the subject.


You need to describe the characteristics at the iPod output (possibly 16 to 64 Ohms) and the input of the amp (perhaps 600 to 10000 Ohms) to be able to match the pot at that location.

I would look at a 1kOhm pot for testing and do further measurements and research if planning for production.

Logarithmic is popular for human characteristic volume management.