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This circuit should work as part of DC servo circuit in audio amplifier with single power supply, maintaining power amplifier stage working point at Vcc/2 (output is AC coupled).

Intuitive approach is to get generic non inverting integrator and replace all ground connections with stable low impedance reference voltage source of Vcc/2.

enter image description here

But the network at op amp non-inverting input is just an RC circuit, so it can sit on any low impedance source, regardless of its level. So I returned it back to the ground. This gave me a freedom to replace voltage reference with simple divider. Am I right?

enter image description here

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  • \$\begingroup\$ It isn't a "generic non inverting integrator" and it may cause your amplifier to go unstable. \$\endgroup\$
    – Andy aka
    Commented Mar 24, 2017 at 17:10
  • \$\begingroup\$ There's no negative DC feedback shown; hopefully, there's some OTHER feedback? \$\endgroup\$
    – Whit3rd
    Commented Mar 24, 2017 at 20:09
  • \$\begingroup\$ This is a feedback network itself :) \$\endgroup\$
    – e_asphyx
    Commented Mar 25, 2017 at 18:09

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I think what you are asking is if the two circuits in your diagram are equivalent.

If so, the answer is yes. The Thevanin equivalent of the bottom network attached to the non-inverting input of the amplifier is exactly the top network. So they are identical in operation, at least ideally.

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  • \$\begingroup\$ Not exactly. These two circuits are obviously equivalent. I'm asking if my decision to put RC network to ground is right. I provided the second circuit just as an illustration how ground referenced RC network can help me to simplify this circuit (I don't need low impedance Vcc/2 source anymore). \$\endgroup\$
    – e_asphyx
    Commented Mar 24, 2017 at 17:13
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    \$\begingroup\$ Sure, you can put the RC circuit on the non-inverting terminal to ground. You essentially have a differential integrator at that point (Vout is the difference between VCC/2 and Vin integrated), so the input has to be referenced to VCC/2 as well or you will integrate the difference. Common mode rejection won't be great over frequency as it depends on matching of the capacitors. \$\endgroup\$
    – John D
    Commented Mar 24, 2017 at 18:19
  • \$\begingroup\$ Expanding a little on my comment above, the nominal initial integrator output will be referenced to ground, not VCC/2, but even if that's not desired if it's part of some overall feedback loop the integrator will eventually get to the level that it needs to. Be sure to check that the common mode input range and output range vs. supply is OK. \$\endgroup\$
    – John D
    Commented Mar 24, 2017 at 20:32

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