Impedance buffer with single-supplied opamp, how does that work?

[WickedBlade]

Hello,

this is my first post so I'm hoping I'm not trespassing some rules here.

I'm a noob with a very ambitious project that includes an impedance buffer. I've decided to use Jack Orman's superbuffer design as detailed in this page:
http://www.muzique.com/lab/superbuff.htm
Only I've decided to go with only one opamp, which shouldn't be a problem as far as functionnality is concerned (tone is another thing, right?).
I'm going to use an OPA2134 (I'll actually make 2 buffers, one for each opamp in the OPA2134). Power supply will be 5V.

Anyway, I was on the verge of etching the PCB I've made when suddenly something occured to me, so here is my question.

In this design as in many other buffer designs, the opamp is, I believe, single-supplied (V- pin is supplied with ground). This doesn't show in the page but it's the case with the other buffers on his site, and on many other designs somewhat similar (i.e. using an opamp as a unity gain device). Please correct me if I'm wrong on this, as the whole point turns around that. I'm somewhat comforted in this belief by the fact that the schematics use a reference point, which is generally built for that purpose.

So I was thinking, allright, it looks like my signal (from the guitar) will also be connected to ground, and this is the same ground everywhere (that's the point of labelling it ground). What I don't understand is how that can work when my oscillating guitar signal is negative? I've read what I could about "reference point" and buffers and still don't quite understand.
In particular I've read this page that is very informative:
http://www.geocities.com/thetonegod/opamp/opamp.html
And also I tried the search function on the forum and perused the results with varying degrees of satisfaction...

So OK, the point of the reference voltage is to make the "V+/ground" pair behave as "V2+/V2-" as far as the opamp is concerned, where V2 is the half of V+. I just don't understand how.

I'll be using Jack's schematics for this discussion, so here goes. If I got it right, C1 is there for decoupling, so if my input signal is negative, it still is after C1. Which means that the opamp + input also sees a negative value, that is outside of its power supply range, so how can that work? I don't see how the Vref point can be helping there...

Can someone shed some light?

Thanks a lot!

[Andi]

The junction of R6, R8 and R9 is at half the supply voltage. R6 therefore pulls the signal level up so that it's wiggling around half the supply voltage, and hence in the middle of the op-amp's range.

R5 and C1 make sure that this offset (or bias) voltage doesn't affect whatever is connected before the buffer, and R7 and C3 do the same at the output.

[WickedBlade]

Thanks for your reply.

So if I understand it right, I was actually reading the function of C1 backwards? It's not here to filter out DC from the signal (which would be strange, as far as I know the guitar signal is centered aroung the ground) but to filter out DC coming from the Vref point.
Is that it?

Am I correct then in assuming that C2/C3 are there to filter out the DC so that the output signal goes back to being centered around ground?

[brett]

Hi
capacitors block/filter DC, irrespective of the direction and purpose.  This is important when "stepping up" the background DC voltage.  A guitar signal is usually modulated around 0 V DC in the guitar lead, but needs to be modulated around a higher voltage (a few volts) at the input to a single-supply op-amp.  Similarly, a step-down is required from op-amp output to stompbox output.  So there's an input cap and an output cap.  (Likewise in almost every circuit - inputs to FETs and valves being the most common exceptions). 

The capacitor at the Vref point stabilises the Vref voltage.  To some extent it also conducts some of the stray signal and unavoidable noise to ground.  Because there is only a little noise or signal at this node anyway, it is an unecessary cap in a buffer.  However, for a high gain op-amp it is very desirable to include this cap.  That's my crude understanding.
cheers

[Andi]

So if I understand it right, I was actually reading the function of C1 backwards? It's not here to filter out DC from the signal (which would be strange, as far as I know the guitar signal is centered aroung the ground) but to filter out DC coming from the Vref point.
Is that it?

It's both - it floats the input so that the buffer isn't affected by a DC bias from a previous stage, and it also prevents the internal DC bias getting out and causing pops when switching.

[aziltz]

lots of good answers here.

the "dc blocking" cap as its called is used to separate the opamp from the circuit before.   it allows the op-amp to have a fixed Reference or Bias, regardless of what kind of DC signal is coming at it before.  This is probably most useful in circuits that have multiple op-amp stages and different Vref's.  It's not always needed, but its good practice.

With many designs for op-amps, discretes, tubes, you have to set the right DC conditions to get the desired AC function from the device.  With op-amps, that usually means centering the Reference between the rails.  Sometimes you want it skewed in one direction or the other.  With transistors, it means maintaining the .6 drop (or whatever it is) from base to emitter, etc...

[WickedBlade]

Thanks a lot for your answers guys, I feel I have a much clearer grasp of what is happening there.

It turns out that it's my understanding of the workings of a capacitor that was lacking