What is behavior of this virtual ground circuit when sourcing/sinking current?

Question

^{simulate this circuit – Schematic created using CircuitLab}

This is the virtual ground circuit from a battery-powered headphone amplifier(direct link to schematic PDF). What is the behavior of the pseudo-ground as the opamp attempts to sink/source current to/from it?

A suggestion for improving performance of the amplifier is to switch from this passive circuit to an active one (using opamps, rail splitter ICs, etc.) to provide a "lower-impedance" virtual ground. Before going down that road I'm curious what my starting point so I can know what sort of improvements I might see.

EDIT TO ADD: Another way of asking the question is: does this circuit have any benefits over grounding the load to V- (or V+) and decoupling it from the opamp with a 470uF capacitor?

Answer 1

It's amazing what passes for circuit design in some of the DIY forums.

Your link goes to this schematic:

The center point created is limited by the charge on the 220 uF capacitors, so particularly at high amplitude/low frequencies (as the amp draws longer current and perhaps asymmetrical pulses) the center point will wander.

This is just the same as if the amplifier output had been capacitor coupled which negates what was probably thought of an advantage (DC coupled output).

To fix up this amplifier, I'd simply use a 2 pole supply switch and ground the center point of the batteries ...Voila ....dual supplies and proper DC coupling.