Dual rail to single rail supply

[rockaffe]

Hi everyone, I'm starting to make a pedal with the rockman x100 circuit and the scheme involves the use of a double-bar power supply (+6v/-6v).
For this reason I purchased two buck converters and therefore have the right power supply available.
However, if I wanted to modify the diagram to adapt it to a single power supply (+9v), would it be sufficient to connect pin 4 to ground and the voltage divider power supply to the input pin or are there other factors to take into consideration?
And for completeness, how to calculate the resistance that connects from the voltage divider to pin 3?
Thanks to anyone willing to waste some time with what is probably a newbie question

[amptramp]

If you want to go with a single supply, you have to bias the op amps into the active region where there is enough headroom for the signal to get through without clipping.  At the input, you need a capacitor in series with the signal input to isolate the 4.5 volt supply from the input.

You have a choice of adding an input resistor to the 9 volt supply that is the same 1 megohm as the existing resistor to ground or you can create a noiseless 4.5 volt source from a couple of 10 K resistors in series with one end grounded and the other resistor going to 9 VDC.  The junction of these resistors is at 4.5 VDC and you add an electrolytic capacitor of 22 µF to 100 µF from this midpoint to ground.  That gives you some isolation from any noise on the 9VDC line.

Anything in the signal that was referenced to ground through a resistive path now has to be referenced to the 4.5 volt supply.  If the connection to ground was through a capacitor, it can still go to ground.  You don't show the rest of the circuit but there is a second stage with a rectifier and JFET controlling the gain of the second stage.  This is quite a lot of current possibly flowing into the 4.5 volt divider, so it may have to be buffered using an op amp connected for unity gain but with a low output impedance to avoid having signals feed back through the 4.5 volt supply.  The alternative is to use separate voltage dividers for the 4.5 volts in different places in the circuit.

There is a third stage with a lowpass filter and a pair of antiparallel LED's to ground from the outputs.  This is a case where the LED's can be put in the feedback loop or use another voltage divider to generate this midpoint voltage.

[rockaffe]

If you want to go with a single supply, you have to bias the op amps into the active region where there is enough headroom for the signal to get through without clipping.  At the input, you need a capacitor in series with the signal input to isolate the 4.5 volt supply from the input.

You have a choice of adding an input resistor to the 9 volt supply that is the same 1 megohm as the existing resistor to ground or you can create a noiseless 4.5 volt source from a couple of 10 K resistors in series with one end grounded and the other resistor going to 9 VDC.  The junction of these resistors is at 4.5 VDC and you add an electrolytic capacitor of 22 µF to 100 µF from this midpoint to ground.  That gives you some isolation from any noise on the 9VDC line.

Anything in the signal that was referenced to ground through a resistive path now has to be referenced to the 4.5 volt supply.  If the connection to ground was through a capacitor, it can still go to ground.  You don't show the rest of the circuit but there is a second stage with a rectifier and JFET controlling the gain of the second stage.  This is quite a lot of current possibly flowing into the 4.5 volt divider, so it may have to be buffered using an op amp connected for unity gain but with a low output impedance to avoid having signals feed back through the 4.5 volt supply.  The alternative is to use separate voltage dividers for the 4.5 volts in different places in the circuit.

There is a third stage with a lowpass filter and a pair of antiparallel LED's to ground from the outputs.  This is a case where the LED's can be put in the feedback loop or use another voltage divider to generate this midpoint voltage.

That's a lot of work (and things to know) for a newbie like me, maybe it's better keep learning and then try. For the moment i'll go with the dual rail. Thank you very much for your time, it's very appreciated.

[antonis]

Just post the complete schematic and we'll try to help you..
(but always at your own risk..)

[Rob Strand]

The distortion/edge switch looks wrong to me.  The edge position doesn't make sense.
On the real unit, the edge position on the original pedal does actually do what is shown.

In isolation it doesn't make sense.  In the pedal it makes sense because they are using a particular type of slider switch which just happens to switch the 3k3 resistor across a dead network.

[antonis]

BTW, in the original circuit IC1A non-inverting input is "floating"..
(they're pretty sure for input signal absolute DC absence..)

[rockaffe]

Just post the complete schematic and we'll try to help you..
(but always at your own risk..)

Thanks Antonis!
Here is the full schematic:

[antonis]

Here you go..



P.S.1
Vcc/2 should be decoupled via a big value capacitor..!!
(to form an AC ground point)

P.S.2
There should be placed capacitors on some op-amp outputs but we hope for no significant DC offsets..

[rockaffe]

Thank you!
So, this is the first part of the schematic with the power section, is a 10uf caps ok?
And last question (I hope so): there's no need to connect the other ic directly to VA as for the first one?

[antonis]

So, this is the first part of the schematic with the power section, is a 10uf caps ok?

It's fine..

BUT that R30(1M) series resistor (part of R30/C21 supply LPF) value should be lowered by 10000 times..!!
Each amp draws about 2.3 mA typical (4.5mA max) supply current with no load (more when loaded) so you can calculate 6 X 0.0023 x 1000000 voltage drop across R30..
(you've probably misunderstood what amptramp told you above concering 1M resistor to +9V.. - Ron meant to retain 1M resistor to GND and connect another one 1M resistor from +9V to pin 3, effectively halving +9V at that junction point..)

there's no need to connect the other ic directly to VA as for the first one?

I'm not sure I get you..
IC1_B non-inverting input (pin 3) is connected to VA via R4 & R5..
(I reserve for JFET connection to VA for reasons explained by amptramp above.. - A seperate VA configuration should be advised there..)

In general, all amps non-inverting inputs should be connected to Vcc/2 (biased) either "directly" (when there is no signal going there) or "indirectly" (via a high value resistor to Vcc/2 or from previous stage output)..

P.S.
R1 serves as C19 anti-pop resistor and its lower end should be connected to GND (instead of VA)..

[rockaffe]

So, this is the first part of the schematic with the power section, is a 10uf caps ok?

It's fine..

BUT that R30(1M) series resistor (part of R30/C21 supply LPF) value should be lowered by 10000 times..!!
Each amp needs draws about 2.3 mA typical (4.5mA max) supply current with no load (more when loaded) so you can calculate 6 X 0.0023 x 1000000 voltage drop across R30..
(you've probably misunderstood what amptramp told you above concering 1M resistor to +9V.. - Ron meant to retain 1M resistor to GND and connect another one 1M resistor from +9V to pin 3, effectively halving +9V at that junction point..)

there's no need to connect the other ic directly to VA as for the first one?

I'm not sure I get you..
IC1_B non-inverting input (pin 3) is connected to VA via R4 & R5..
(I reserve for JFET connection to VA for reasons explained by amptramp above.. - A seperate VA configuration should be advised there..)

In general, all amps non-inverting inputs should be connected to Vcc/2 (biased) either "directly" (when there is no signal going there) or "indirectly" (via a high value resistor to Vcc/2 or from previous stage output)..

P.S.
R1 serves as C19 anti-pop resistor and its lower end should be connected to GND (instead of VA)..

Thanks again Antonis, as soon as I get some components I miss I'll breadboard the whole thing and see what happens.