Can someone explain the purpose of this buffer stage? I understand how it works in the sense that I understand the cause of each reading on the oscilloscope but I don't understand why the buffer stage is designed this way. The circuit is a five band equalizer.

I thought the purpose of putting a buffer at the beginning of an audio circuit was generally to amplify the signal (I don't think my output is load sensitive) as explained here:

What is the reason for adding a buffer stage?

Here is the design of the buffer stage: enter image description here

The corresponding voltage readings are simulated here (red is audio signal, green is amp input, blue is output of the entire stage): enter image description here

To be clear, the audio input contributes a negligible amount of current so the 12V supply effectively goes through one giant voltage divider which reduces it to 2.4V which it adds as a DC offset to the audio signal which is needed because this is a single supply op amp.

But why would they design the buffer circuit so that the output is REDUCED?

The original design is taken from here where you can see the entire circuit:


  • 1
    \$\begingroup\$ Circuits on the web can have dozens of reasons for being designed this way or that that and, without some written spec about performance it’s virtually impossible, in some situations, to 2nd guess those initial preferences. the buffer is clearly there to be able to handle the low impedance from the 5 parallel filters so maybe that answers that part. As for the slight gain reduction, it might be arbitrary so leave a comment on that site and ask them. \$\endgroup\$
    – Andy aka
    Aug 25, 2019 at 11:24
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    \$\begingroup\$ It is a power (in this case current) amplifier. the voltage may be nearly the same but it can generate a lot more current into the next stage(s). This prevents the input signal from collapsing under the load. (Low input impedance of the next stage) \$\endgroup\$
    – Oldfart
    Aug 25, 2019 at 11:25
  • \$\begingroup\$ Oh wow I hadn't thought of the 5 filters as effectively being in parallel and thus low impedance. Durp, that makes sense. However, I'm sorry if this is an ignorant question, but how does this op amp act as a current amplifier/supplier? It's easy for me to see it as a voltage amplifier. \$\endgroup\$
    – Andrew
    Aug 25, 2019 at 11:47
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    \$\begingroup\$ Simplified: There is a feedback that detects the output voltage dropping and then corrects it by supplying more power. This works until the OPAMP output stage reaches its maximum power, then it starts collappsing too. \$\endgroup\$
    – Oldfart
    Aug 25, 2019 at 11:52
  • \$\begingroup\$ To check my understanding: So say, when the OPAMP input is decreasing (because let's say its sinusoidal and the input is on the falling part of the oscillation), then the output stage doesn't need to provide as much voltage and so current actually moves the opposite direction away from the feedback loop into the filter stage? And if so, wouldn't this provide sporadic power to the filters? Wouldn't that be an issue? Sorry for the many questions, thank you for your help \$\endgroup\$
    – Andrew
    Aug 25, 2019 at 11:58

1 Answer 1


It is a differential non inverting amplifier with a gain of ~1, (all 4 legs have the same resistance if you ignore R18)

R18 is acting as a 6V bias to keep the output and input between the 0 to 12V supply rails of the op amp, this mainly seems like they did not want to use symmetrical supplies for the op amp,

So the output voltage of the op amp is roughly signal in +6V

C3 is the input coupling capacitor, which removes any bias outside, and C4/R15 are to remove the DC offset this buffer has,

  • \$\begingroup\$ I know you answered this a while ago but could you tell me under what circumstances would an audio input signal have a bias as you mentioned when explaining C3? \$\endgroup\$
    – Andrew
    Sep 6, 2019 at 22:56
  • \$\begingroup\$ @andrew You rarely get to control what currents may flow through a signal ground, let say something isn't quite right, so some fixed current ends up flowing down the signal ground, this creates an offset voltage or bias to the signal, you only want to amplify the signal, not this offset, so you use an AC coupling cap like C3 to remove it. \$\endgroup\$
    – Reroute
    Sep 7, 2019 at 8:50
  • \$\begingroup\$ But why would a fixed current occur in the first place? If I understand you correctly, a fixed current would correspond to a fixed voltage which would essentially lost across the capacitor C3 (or another way of saying it, the constant current is stopped by the capacitor which only allows varying currents to pass) \$\endgroup\$
    – Andrew
    Sep 8, 2019 at 0:51

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