In the circuit below, where would I place an audio taper potentiometer to modulate the output volume level? Which of the resistors should get replaced with the pot and why?

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If understand correctly, I should place the pot between the input decoupling capacitor and the buffer amp is this correct? Additionally, this circuit was designed for +-12V, if I am only capable of providing half of the +-12V, around +-6.2V, what effects will I observe, will the amp still function, just at a lower output?

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    \$\begingroup\$ I would replace R1 with a 500k audio taper pot. Input signal goes to the top end of the pot, the wiper connects to C1, the bottom end of the pot goes to ground. \$\endgroup\$ – Dwayne Reid Apr 21 '15 at 4:01
  1. Since this circuit requires both positive and negative power, you can place a pot either before of after capacitor. If you had a virtual ground (for example made using two resistors), you would want to place the pot before the capacitor.

  2. Reduced power will reduce output, but halving the supply voltage will decrease the output power by more than 50%. Both devices are may operate from +5/-5V power supplies; however Figure 6 in TPA6120 datasheet shows that distortion-free power is 0.5W for +-15V, and 0.05W for +-5V (into 32ohm speaker).

  • \$\begingroup\$ Reduced voltage will lower the clipping point. It won't change the gain of the circuit. \$\endgroup\$ – user207421 May 23 '15 at 9:01
  • \$\begingroup\$ Yes, that is why the comment speaks about 'distortion-free power'. In many audio amps, the maximum gain does not matter -- you do not run with 100% volume, instead you increase the volume until you hear distortion due to clipping. This is what Figure 6 describes, and this is what will be the important limitation with +/-5V supplies. \$\endgroup\$ – theamk May 27 '15 at 19:19

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