Buffer/splitter trouble - for parallel fx use

[Coriolis]

Long time since I posted here, but here goes:

I'm trying to build a 3-way splitter to use with a parallel fx system. IE split the signal to 3 different fx chains (or 2 chains + 1 tuner, etc) and then mix these chains back together.
Sorta like the ROG splitter-blend, but with opamps - no fets.

So I figured a unity-gain buffer (voltage follower) should do the trick. In and old Craig Anderton book I have, he does a simple voltage follower, and then just sticks 3 output jacks after it.
My reasoning tells me I would have to buffer each output, like in the AMZ splitter here: http://www.muzique.com/lab/splitter.htm
...but what do I know.
So I tried Craigs idea.
Well...it loads my signal down (sounds duller anyway). Some hum too, but that's probably from the lack of enclosure.
  So - my first question: Is Craig wrong on this one? Do I need to build it more like the AMZ splitter? Or do I have the values wrong?

Another thing is the way I generate the Vref for the opamp. I don't know much about this subject, so just stole a solution from the Boss OS-2 schematic.
After reading up a bit, it seems there are several ways to skin this one, so...
  Next question: Do these values look right? Because when messing with them, I get results varying from no sound  - to almost no sound when strumming lightly - and a sudden, loud, distorted  roar when whacking the strings (this happened with a divider made up of 100k resistors with a jumper going to the non-inv input on the opamp).

Here is a pic of the buffer and the psu:


Any takers?   

C

[Mark Hammer]

Check page 15 here: http://www.jaycar.com.au/images_uploaded/TL071074.PDF

You'll probably want to tack an electrolytic cap on the output of each send.  I think I stuck a 4.7uf and a 10k log pot on each output of the one I made.  Note that the circuit shown provides a gain of 11, given the 1M feedback resistor and 100k ground leg resistor.  You may want to  drop that gain a bit.

The only thing I don't like about the circuit is that, for some reason or other, mine seems to take a little while (seconds, actually) to turn on when power is applied.

[grapefruit]

Coriolis,

For your circuit to work you need a resistor between the two VCC/2 points. For guitar about 1M or 2.2M is suitable. As it is the non inverting input has an ac path to ground via the 100uF capacitor.

You can do it with one op amp but it would be a good idea to have a series resistor of 1k on each output. One of the more suitable op amps for this is the LM833 that can drive 600 ohm and capacitive loads, however since the circuits you are driving will generally have a high input impedance and be close together you could use something like a TL072.

Regards,
Stewart.

[Coriolis]

Thanks both for your suggestions.

Actually, in the original Anderton circuit, there is a 470k resistor from the non-inv input to ground - to "provide a high input impedance" when using the circuit with an input cap.
I did not put this resistor in, but maybe I should. OTOH, isn't the point of using a jfet-input opamp like the TL074 (which I'm using) that it has very high input impedance? Do I need this resistor unless I'm using the 833 - or 5532 or other?

Thanks for the pointer on the Vref resistor - which I left out. What confuses me, is that some designs don't seem to use them - like the GGG mini mixer:
http://www.generalguitargadgets.com/diagrams/mixer_sc.gif

But let's say I put in a 1M Vref resistor. What values do I then put in as the divider? I've read R.G's paper on this subject, as well as a few threads. I understand that the divider resistors are chosen depending on the current draw of the opamp. So I guess the Vref resistor is used to reduce that current draw. But if 1M is a "good" value here, what do I put in the divider? Reading the 074 datasheet, I look under "input bias current" and see something like 65-200 pA. That current is then reduced even further towards nothing by that 1M resistor, right? So it hardly matters what the divider resistors are, or? In some circuits, I see 10k's in the divider, and 10k as Vref resistor. So what gives?

Confusing.

Anyway, I want this buffer to be fairly versatile (drive long cables with minimal loss, etc), but I'd still prefer low parts count, as I'll be adding a mixer, and possibly phase-reversal circuits.

Thanks for the help sofar.

C

Edit: I added the 470k resistor to ground, and the 1M as vref, and now it sounds as it should. Still puzzled by the whole vref-issue though....

[grapefruit]

Edit: I added the 470k resistor to ground, and the 1M as vref, and now it sounds as it should. Still puzzled by the whole vref-issue though....

Is the 470k resistor to ground on the op amp side of the input cap? If so, it shouldn't be. The 1M resistor going to Vref is the only one you want there. That will set the input impedance to 1M and the voltage at the non inverting input to Vref. If you put a resistor to ground it will reduce Vref at the non inverting input. Maybe the Anderton design was using a split supply.

10k is still fine for the Vref voltage divider resistors.

You could put another 1M resistor on the input side of the input cap to pull it to ground to avoid popping, but it sounds like you'll have another circuit in front of it anyway....

Regards,
Stewart.

[Coriolis]

Is the 470k resistor to ground on the op amp side of the input cap?

Yes!

Maybe the Anderton design was using a split supply.

Yes, it was. Now that you mention it, I can see how that 470k could mess with things in a single supply situation.
Sounds fine though. Maybe I should measure Vref before and after pulling it out, to see for myself what it does.

I have another question actually - regarding the mixer circuit I linked to previously:
Say I want to simplify it even further, and cut off the second inverting stage, power-indicator led, etc.
Would I just put in a 2.2uf cap and a 1k resistor in series with the output of the opamp? And if I wanted a master volume (connected as a voltage divider to ground after the output) would I keep the 1k resistor in there?

C