Bipolar supply and op amp bias?

[Lizardmn]

Hey guys, it's my first post so firstly thanks for having me.
So I've got this basic little op amp boost circuit. Due to some stuff going on in the rest of the pedal I'm wanting to use a bipolar power supply +/- 14V instead of the usual 9V, 4.5Vb, 0V deal. My question is regarding Vb with a bipolar supply, I assume this should be 0V aka ground? If so, is there any redundancy with the two 2M resistors if they're both now connected to ground? Could I get rid of one and make the other a 1M? And finally would it matter which one?
Thanks

[strungout]

Yes, Vb is at 0V.

Both resistors are useful, but for a different purpose. And both go to ground.

The 2M at the input is a pull-down resistor, to keep it from going 'pop!' when you plug in your instrument and your amp is on.
The other is for biasing the non-inverting (+) input to 1/2 the supply (Vb = 0V).

[Lizardmn]

Great, thank you!

[GibsonGM]

To follow up on this - does the 2Meg on the Vb add noise?  You generally see them smaller in value except in 'starved' designs like Distortion Plus, I'm just curious.

[duck_arse]

and of course, welcome to the forum.

[antonis]

and of course, welcome to the forum.

This..!! 

And a comment about the utter disutility of IC1A..

@ Sir Mike: You're right about noise but Vbias resistor value is considered in relation with previous stage output impedance..
(in "noisless" applications it's bypassed with a some pF cap..)

[Lizardmn]

And a comment about the utter disutility of IC1A..

@ Sir Mike: You're right about noise but Vbias resistor value is considered in relation with previous stage output impedance..
(in "noisless" applications it's bypassed with a some pF cap..)

The circuit I started with originally was starting at IC1b onwards and had a really low input inpedance, so since I had a whole side of the op amp going spare I decided to add the klon's input buffer. I got the 2M values from the aion schematic, which (I hope) should set the input inpedance at a nice 1M.

It seemed like a not totally useless thing to do

[antonis]

It seemed like a not totally useless thing to do

IC1B  with 2M bias resistor has exactly the same input impedance with IC1A with 2M bias resistor..

What I'm saying is that if you insist to use the spare amp, better implement it as an output buffer 'cause as input buffer provides nothing..
(in case of single supply, you could use it as Vbias "stabilizer"..)

[Lizardmn]

IC1B  with 2M bias resistor has exactly the same input impedance with IC1A with 2M bias resistor..

Yeah, that makes sense.

How about if I threw in this switchable transistor booster in front of the op amp boost that I totally could have mentioned earlier? Going off of how a ds1 works my line of thinking was input buffer -> transistor boost -> op amp gain stage.
(Thank for all these answers btw, this is really helpful)

[strungout]

Oh, yes, welcome! I have no real manners...

[antonis]

How about if I threw in this switchable transistor booster in front of the op amp boost that I totally could have mentioned earlier?

Good thought - Bad implementation..
(I let you figure out the reason..)

hint1: IC1A output impedance..
hint2: Quiescent voltage of transistor booster and IC1B amp..


edit: There is a more simple way for selective boost/gain..

[Lizardmn]

Damn, my knowledge has run out fast.
Output inpedance of the ic is meant to be as low as possible, basically 0, I think? I don't fully understand inpedances yet so not sure what impact this has further in the circuit.
Quiescent voltage is when no signal is applied, so does that mean that I've not biased something correctly?
I'm gonna refer to this bit of the ds1 schematic from electrosmash I've been looking at. In my case since I'm using the bipolar supply, instead of 4.5V I'm to use ground as strungout said, so it's not jumping out at me where I've gone wrong.






How much more knowledge do I need?

[antonis]

Output inpedance of the ic is meant to be as low as possible, basically 0, I think? I don't fully understand inpedances yet so not sure what impact this has further in the circuit.

Q1 Collector is effectively AC grounded..
(no signal goes into IC1B non-inverting input..)

Quiescent voltage is when no signal is applied, so does that mean that I've not biased something correctly?

You need to bias BOTH Q1 & IC1B at the same quiescent voltage for equal fully symmetrical signal swing..
Q1 output signal swings from +9V to 0V where IC1B can swing from +9V to -9V (both swing margins are considered ideal..)
You see that Q1 output "restricts" IC1B output to +/- 4.5V headroom, hence +/-9V bipolar supply becomes useless..

I'm gonna refer to this bit of the ds1 schematic from electrosmash I've been looking at. In my case since I'm using the bipolar supply, instead of 4.5V I'm to use ground as strungout said, so it's not jumping out at me where I've gone wrong.

Quite right for IC1A/B but not for Q1 Booster..
(see above..)

For a complete Q1 & IC1A/B cooperation, Q1 should be fed with +/- 9V bipolar supply with its Collector biased around 0V..

[Lizardmn]

Ahhhhhh I think I get it. The transistor needs Vb, V+ and V- the same as the op amp. However on usual pedal schematics they don't bother to call it V- around the transistor because it's usually just ground. But here with a bipolar supply I need to use my acual -9V somewhere with the transistor as well. Correct?

You guys are so smart btw.

[antonis]

Ahhhhhh I think I get it. The transistor needs Vb, V+ and V- the same as the op amp. However on usual pedal schematics they don't bother to call it V- around the transistor because it's usually just ground. But here with a bipolar supply I need to use my acual -9V somewhere with the transistor as well. Correct?

"Somewhere" is where presently lies GND..

You guys are so smart btw.

We would be utterly burk if there were not some other smart guys before us..

[Lizardmn]

"Somewhere" is where presently lies GND..

[/quote]

So cryptic, I like it  Put's me on the spot tho.
It's NPN so the base terminal should be postive with respect to the emitter, so I'm gonna say (on the ds1 schematic) R9 on the emitter should go to -9V instead of ground?
I've got a 1 in 3 shot here so I feel good about this.

[antonis]

It's NPN so the base terminal should be postive with respect to the emitter, so I'm gonna say (on the ds1 schematic) R9 on the emitter should go to -9V instead of ground?

Quite right..

It can be modified by deleting C3, R5, R7 & C4, connecting R4 to -7.5V (instead of +4.5V) and make R22 1k..
(-7.5V can be obtained via a resistive voltage divider between -9V & GND)

[Lizardmn]

Q1 output signal swings from +9V to 0V where IC1B can swing from +9V to -9V (both swing margins are considered ideal..)
You see that Q1 output "restricts" IC1B output to +/- 4.5V headroom, hence +/-9V bipolar supply becomes useless..

Since I've been trying to fix up this bit of the CE1 preamp, I assume that this also applies to that original circuit?



If I've understood correctly, with the transistor boost not engaged in the circuit, the op amp swings between -13 and +13V for 26V of headroom. But when the transistor, which only swings between 0 and +13V, is in the circuit the op amp's clean headroom is restricted by half?

The op amp here is the ancient TA7136, so part of my problem is I'm not sure if it works the same as say a 4558, as there's very little information abut them and it has it's own thing going on with how it's powered.

[antonis]

If I've understood correctly, with the transistor boost not engaged in the circuit, the op amp swings between -13 and +13V for 26V of headroom. But when the transistor, which only swings between 0 and +13V, is in the circuit the op amp's clean headroom is restricted by half?

Quite right..

[Lizardmn]

How odd. It seems then that, when used with guitar instead of keyboards as intended for the CE1, potentially forcing the op amp into some clipping with this transistor boost is a key feature of this circuit. How about that