Powering op-amp directly with splitted/buffered 9V supply - will it work?

[marg]

I've been having some struggles with getting schematic that has dual supply powered op-amps to work with single supply because at some point it requires negative voltage (if I'm understanding it correctly). It is not typical situation where you can bias opamp to Vcc/2 and decouple it with caps on input and output.

So, I've read a bit on the subject and came up with an idea of "simulating" dual supply with buffered resistor voltage splitter.

This is my idea:


The problem is that I've only seen it being done when you need to bias an op-amp input, but not to power the whole op amp. I've simulated it and it seems ok, but I wonder, for example, will there be problems in connecting it to another circuits (pedals) whose signals are referenced to "real" ground (the negative pole of battery)? Or are there some other drawbacks?

I understand that certainly a problem will be op-amp's maximum output current, but I don't need a lot of current from that supply for schematic that I'm working on (cca. 20 mA) so I can assume that won't be a problem here.

Every response is aprecciated

[PRR]

> It is not typical situation where you can bias opamp to Vcc/2 and decouple it with caps on input and output.

Why not? Are you sure?

The plan you post has two different points marked "ground". Such systems usually do not not play well with other gear.

Maybe post your original problem, not a possible solution? Guys here have smoked some really strange stuff. (Equipment, that is.) Maybe someone knows a better path to try?

[bool]

Just google up "virtual ground", "ground channel", etc. (in context of headphone amplifiers)

Lots of ready made examples, usable in all sorts of other applications.

Those headamp guys have really outdone themselves.

[Ice-9]

The circuit you have drawn out is simply a buffered voltage divider and not a split rail psu. Although it is marked wrongly with the gnd and -Vcc symbols it is only dividing the 9v supply into 9v / 4.5v and 0v

Using your markings for the output gives
Vcc =9v
gnd =4.5v   (Vref)
-Vcc = 0v

But be aware as mentioned by PRR you have both the gnd and -Vcc marked as grounds.

As suggested post a schematic and someone will most likely be able to point you in the correct direction.