Buffered Volume Pedal Schematic for Comment

[chumbox]

Hey All

Have put together a schematic to build a buffered volume pedal that has two outs (one out and one tuner).  It's my first attempt at splitting a signal and was hoping someone could cast an eye over it and make comment if necessary.  It's basically one signal in, splits off into two separate but identical buffers to avoid the usual signal loss.  Nothing miraculous just took a stab at how to draw it up.  Right at the end of the circuit I popped in a switch to 'potentially' change the pot taper, others may have comment on this as well.

Look forward to hearing what you think or if I've missed something.

Edit: It just occurred to me the tuner probably doesn't require the 100k and 560r resistors at the end?

[amptramp]

Delete R6, make the pot 100K and R8 changes to 20K and you have a pretty good design.

[chumbox]

Delete R6, make the pot 100K and R8 changes to 20K and you have a pretty good design.

Thanks amptramp.  Great suggestions and really pleased you like the design. Alright. Off to build.

[antonis]

If it isn't battery powered you may lower R1 & R2 5 to 10 times..

[merlinb]

Don't delete R6 -it is a good idea to have a small series resistor at the output of an opamp line driver.
You can delete R5 if you want, since the pot does the same job. Reduce the pot to 10k so your output impedance is as low as possible (and reduce R8 to 2.2k), and increase C2 to compensate (e.g. 10uF).
Add a small series resistor (e.g. 100R) where 9V power enters, to give you some filtering and emergency current limiting.

[chumbox]

Thanks so much everyone so far! Learning heaps along the way.

[T. B.]

I think it is a very good design.  However, I've never been a fan of the diode in parallel with the supply approach to handling a reversed battery.  The reason is a large current will flow through the diode and may burn it out.  I prefer to have a diode in series with the supply.  Its true you lose a bit of voltage but nothing will break if the battery is reversed.

Just my $0.02

[bool]

You don't need 47uF at C3. 4u7 or 10uF will be adequate; even 1uF would be "acceptable" ... for a battery-powered device.

I would place a protective resistor (1 - 10k) in series with C1.

As TB said, protective diode in series (1N5817 etc.) is a much better option.