Guitar pedal - biasing - why?

[ribolovec2]

Hello fellows,

I am designing my own distortion pedal (based around the Tube Screamer). But I have this question about biasing: In the original schematic of the TS (as well as any other guitar pedal whose schematic I found), the op-amp feedback network, the input transistor's base and the output pot are biased to 1/2 U.. which seems reasonable.. but the input and output signal grounds, as well as any other ground in the schematic (filters, RC networks, etc.) are at 0V... Why is that? Shouldn't all grounds be connected to a common point with a potential of U /2.. That's what I've seen on countless battery powered headphones amps and other devices.... Please explain why is that.

Regards

[Minion]

The grounds don"t change , 0v is still ground and +9v is still +9v , but the input signal has to be biased to 1/2 supply so it has a voltage referance so the signal can swing positive and negitive in relation to it .....

Don"t think of the 1/2 supply as a ground , think of it as a voltage referance ....

Most all Single supply opamp circuits use a Voltage referance ....

I hope I explained that right .....

[Nigel Goodwin]

It makes no difference whatsoever, the negative supply, the positive supply, and the split middle supply are all at the exact same AC potential (and are exactly thre same point as far as AC is concerned).

[gain-wire]

Originally Posted by Minion

The grounds don"t change , 0v is still ground and +9v is still +9v , but the input signal has to be biased to 1/2 supply so it has a voltage referance so the signal can swing positive and negitive in relation to it .....

Don"t think of the 1/2 supply as a ground , think of it as a voltage referance ....

Most all Single supply opamp circuits use a Voltage referance ....

I hope I explained that right .....

I think you did. But it's important to see that it is in fact a Unipolar supply: we aren't trying to go for a bipolar supply where the middle of the divider is ground, it's the other way around, we build a unipolar circuit and therefore must bias the opamps (and transistors) to a more positive potential.

Now, I hope I haven't confused you. Funny because I was understanding the same as you at first.

[wakibaki]

Originally Posted by ribolovec2

the input transistor's base and the output pot are biased to 1/2 U..

OK, for a start the output pot is NOT biased to 1/2 U. It's decoupled by a cap, the DC voltage is at ground.

The input transistor's base is biased to 4.5V so that there's some standing current in the transistor otherwise at best there'd only be conduction in the transistor when the input signal exceeded ~0.6V

Originally Posted by ribolovec2

but the input and output signal grounds, as well as any other ground in the schematic (filters, RC networks, etc.) are at 0V... Why is that? Shouldn't all grounds be connected to a common point with a potential of U /2..

No, the grounds should all be at 0V, if they were at 4.5V they'd all be at the same potential as the signal path, which would defeat the point of biasing IT to 4.5V.

Originally Posted by ribolovec2

That's what I've seen on countless battery powered headphones amps and other devices....

No, you haven't.

w

[sreten]

Hi,

Ground is ground for signals with split supply.

With a single supply all signals are referenced to 1/2 the supply.
Ground is still ground and capacitor coupling removes this offset.

You have not seen "countless examples" as you describe, however
the 1/2 supply point is effectively the same as the split ground point.

For the split case coupling capacitor capacitors are optional.

rgds, sreten.

In your diagram only two of the ground points need to be ground,
the other three could be connected to the 1/2 supply point, but
its better to have a a strong DC bias across electrolytics.