Can imbalance in the rail values of a dual-rail op amp circuit cause problems?

[west portion]

For clarity, say I've ended up with +9V on the positive and -8.5V on the negative. Can this contribute to poor circuit behaviour? I think I understand that it could cause asymmetrical clipping, but what about something like switch pop? I've tried many different anti-pop techniques and it has gotten much better, but I'm now wondering if this could potentially be the last mild cause.

Little bit of back story.
I'm using the 7660S to generate negative voltage for my op amp distortion circuit. Whether on the breadboard or soldered to a prototype board I can never seem to end up with almost equal opposite value rails. It shouldn't be a problem with a few mV, but half a volt, which it frequently is (or more), seems to be a bit much. I've followed the datasheets recommendations by trying a 100nF capacitor in various configurations. Is this just due to poor current capabilities of the chip? Has anyone had better results with any others? ex. MAX1044 or LT1054?

Thanks so much for any input, I @#$%ing love this community!

[ElectricDruid]

Yeah, not your fault. While those chips are pretty amazing at what they do (Turn +9V into -9V! Woah! Is it just me that thinks that's like magic?!) there are inevitably some losses in the process, and that tends to get worse as the current drawn goes up. So, yep, the negative rail tends to be a smaller voltage than the positive rail.

As to whether that's a problem? I'd say not. Pretty much *all* of the problems people have around here with these chips come from them creating noise on the supply, not the fact that the voltages end up slightly asymmetrical.

[Rob Strand]

For clarity, say I've ended up with +9V on the positive and -8.5V on the negative. Can this contribute to poor circuit behaviour? I

Usually not.

FYI, some opamps don't clip symmetrically in the first place.

It's also normal to expect some drop on the output of a converter.   The +9V is fairly low impedance so loading it down doesn't cause much drop.   The charge-pumps have a much higher output impedance so when you place a load on the output it is normal to see a drop.   Most charge-pump datasheets give some sort of output voltage vs output current curve.    They also quote "on resistances"  for the switches.  Lower on resistances will produce a lower output impedance.  The output impedance is always higher than the on-resistances by a significant factor.

As for the drop.  You could have an issue with you circuit pulling too much current.  It might have problem.    If your filter capacitors on charge-pump or on the input DC rail are too small you can see higher than normal voltage drops under load.   Charge pumps that use diodes instead of the chip along will have more voltage drops, use Schottky diodes to reduce the drop.
(You an only know the expected drop for a charge-pump from the manufacturers data and the current drawn by the circuit.)

[PRR]

If your circuit is unhappy about 0.5V, it is too fussy.

[amptramp]

If you are running ±9 volt rails, your bias point is going to be zero volts with a resistor directly to the ground and any anti-pop resistor is also going to go to ground.  Since there is no voltage across the input coupling capacitor, you are unlikely to get a switch pop, in theory.  Similarly, the output may have a nominal zero volt resting state but if you are using an electrolytic coupling capacitor for the output, it would have to be a non-polar variety unless the output is biased in one direction only.  In a high-gain circuit, the op amp output may deviate from zero at rest as it will multiply its input offset voltage by the (gain +1).  None of this has anything to do with the voltage supply.  Your input and output circuit voltages determine the amount of switch pop, not the supply voltages to the devices.