Which is wrong? If either?

[samhay]

To get unity gain, you just have to short the op-amp output to the negative input - i.e. across the feedback loop - so you can do this with an SPST.

[slashandburn]

Ah! So with the switch open it's a unity gain buffer and the pot does nothing, and with it closed the distortion circuit comes into play? 

I love over complicating things, me.

[R.G.]

Ah cool I think I follow roughly what that was about!   So for basically, you can put the feedback network to either ground or Vref (or even the positive rail!) but from a novice point of view it's probably simplest to put it to ground?

Careful with terms. I think you mean the correct thing, but it's easy to get parts misnamed. "Feedback network" can sometimes be used for just the parts between the opamp output pin and the (-) input pin. Sometimes and some people mean it to also include the parts from the (-) input to either a signal or to an AC ground point. It's best in terms of not being confused to separate them carefully. In this circuit, the 3.3K and 100nF in series serve to set the gain. The other junk between the output pin and (-) pin is what I'd call the feedback network.

With that as background, the opamp itself, internally, always considers the voltage of the (+) pin to be the real, no-fooling reference. It is the *difference* between the (+) and (-) pins that make the output move around. That leads immediately to the questions of whether the Vref, or ground at the Vref network, or the ground of the signal voltage is the best reference, and you're asking one or two variations of that question.

The REAL answer is that you want to connect the 3.3K/100nF gain setting network to the quietest place in terms of other circuit noise. In opamp circuits with +/- voltages this the middle "ground" of the power supply - and that is the "Vref" that the opamp uses. It's expensive to make a +/- power supply, so effects circuits use resistors or an active circuit to split the single power supply and make a fake "AC ground" in the middle of the available single voltage.  However, a generated "AC ground" can never be quiet as quiet as the real ground, which in these circuits is the negative power supply terminal (and let's not get off into positive ground circuits until the OP gets the concepts here). So as a practical matter, it is quieter to connect the 3.3K/100nF gain setting parts to the minus rail, as that acts as ground for the other parts of this circuit and for the signals in and out. Note that the 100nF does double duty in keeping the DC voltage difference from mattering.  So - in circuits like this, the negative voltage DC ground is a bit quieter.

It also can introduce a few other issues which are not pertinent to this issue.

So yes, from a novice point of view, take it to the DC ground for circuits like this. It works when connected to Vref, too, but there are other minor issues.

Another quick question on the same circuit then, could I expect any issues putting a dpdt "bypass" switch on that feedback network? To bypass the whole gain network and turn it into a simple buffer? Or would it be better to use a dual opamp and use the other half to provide the switchable buffer?   I hope I'm explaining that thought okay.  Basically a switch to toggle from buffer to TS style distortion?

The simple thing to do is to use an SPST (or one subsection of any more complex switch) to "short out" the feedback network from output pin to (-) pin. This forces the opamp to have a gain of unity from the (+) pin, and the 3.3K/100nF gain setting network now just looks like a load to the output.

[ashcat_lt]

Ah! So with the switch open it's a unity gain buffer and the pot does nothing, and with it closed the distortion circuit comes into play?  

I love over complicating things, me.

Other way around.  With the switch closed it acts like straight wire from output to negative input.  With the switch open, it has to go through the gain pot et al.

Course, you need at least two poles if you want an LED.

[slashandburn]

Brilliant. Thanks everyone.  Very well explanied (i think, maybe!) R.G.

Just to triple check though I'm not still misunderstanding the switch issue, basically if using the circuit linked in the original post, I'd add the switch straight across the output and the negative pin?   With the switch open the signal the circuit works as normal and when closed I essentially bypass the pot and diodes etc by creating this direct route between the negative pin and the output? 

[ashcat_lt]

Just to triple check though I'm not still misunderstanding the switch issue, basically if using the circuit linked in the original post, I'd add the switch straight across the output and the negative pin?   With the switch open the signal the circuit works as normal and when closed I essentially bypass the pot and diodes etc by creating this direct route between the negative pin and the output? 

Yep!