0
\$\begingroup\$

The context here is simple volume control of a DAC generated audio signal with a potentiometer before an audio amplifier stage in a single supply, battery powered system.

Is there any advantage to attenuating only the AC component of a mid-supply biased audio signal, before buffering and feeding it into an AC coupled audio amplifier:

approach 1

As opposed to attenuating both AC and DC components, before buffering and feeding it into the AC coupled audio amplifier:

approach 2

Assuming the buffering op-amp is rail to rail, it seems that the two approaches should be theoretically equivalent.

The former approach seems like it might be more linear, avoiding the rails of the buffering op amp. The latter approach requires less parts, but there could be clipping near the rails due to non-idealities (albeit small) like input offset voltage in the buffering op-amp. Is one approach preferred over the other in practice?

Improve this question
\$\endgroup\$
  • \$\begingroup\$ C16 will block all DC. Not much use in blocking the bias with C12 and adding it back with U9B. \$\endgroup\$ – Aaron May 5 '20 at 23:59
  • \$\begingroup\$ Use the 1st schematic but replace U9B with a piece of wire. Then U9A can be a single opamp. \$\endgroup\$ – Audioguru May 6 '20 at 0:24
  • 1
    \$\begingroup\$ @Audioguru, I should add that I don't actually have a low impedance mid-supply reference available. The 1.65V labeled net is just a placeholder for that reference from somewhere. In my current schematic, I'm actually connecting pin 5 (BYPASS) of the audio amplifier, which is its internal mid-supply reference, to that buffer. I'm not sure I can safely load it by connecting it directly. \$\endgroup\$ – mountaingoat May 6 '20 at 1:07
0
\$\begingroup\$

Connect the bypass pin of the amplifier IC to the low end of the volume control with U9B removed. The 1uF bypass filter capacitor is a low impedance to audio frequencies and the bias current of the MCP6L01 or MCP6L02 opamp is extremely low.

Improve this answer
\$\endgroup\$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.