Converting a V-, 0, V+ schematic to 0, V/2, V circuit help

[deadlyshart]

Hi guys, today I thought I'd just put together a really simple fuzz pedal from Craig Anderton's Electronic Projects for Musicians. I had all the parts on hand and wanted to try some simple fuzz, so I thought I'd go for it. However, he always draws his circuit diagrams assuming you're using two 9V batteries in series and center tapping them so you have an easy -9V, 0V, +9V.

Of course, most people making pedals these days use just a 9V supply and do a voltage divider, and provide the rail pins of the op amp with 0, +9V and give +4.5V to any pin that would usually require ground, which is what I want to do.

So I'm trying to convert the schematic from the book into the 0V, 4.5V, 9V format. A couple of the replacements are obvious to me, but there are two I'm unsure about, so I wanted to check here.

Here is the schematic:



(The red numbers are just the correct pins for the 4558 chip I'm using; ignore them.)

The blue are the connections that I have to change, plus the ones I'm not sure about.  For example, I"m pretty sure I have to change pin 3 (using the 4558 notation in red) to +4.5V instead of GND. I know the shield of the input and output jacks connect to the chassis, so they're necessarily GND. However, the two I'm unsure about are the point labeled "G" and the first pin of the R5 pot. My natural inclination is to change them to 4.5V (because that's the "new 0V" in my circuit), but I'm unsure...

Here's my current logic for each:

for the G one, in the original circuit it's between the rails voltages, so in my circuit, that would mean it's 4.5V (in between 0 and 9V).

For the pin 1 of R5, I think that one might actually still be chassis, because it's after the output cap C6, so it's "outside" of the +4.5V biased section of the circuit (the wave is with respect to 0V there, right?). Therefore, I'd guess that that one should actually be 0V still in my new circuit (connected to the chassis).

Is my reasoning right here? Thanks for any advice.

[M.A.P]

Hi deadlyshart,

if you connect 9V to the "+"pin in front of D1 and 0V to the "-"pin after D2 then G should be half the voltage. In this case 4,5V.
I'm not 100% sure about what to put on R5, but I would give it a try with 0V.

Greetings from Germany
Marcus

[antonis]

Keep GNDs for IN, OUT & R5, disconnect GND from G and wire G to pin3..

Your IC will be feeded with 8.4V and 0.6V respectively so you MAYBE have to delete D1 & D2 for a little more headroom..

[ElectricDruid]

+1 agree with Antonis.

The two on the RHS have to be connected to the same potential (gnd) because there's no DC-blocking cap between the pot and the output. Having one at 4.5V and the other at ground would cause all sorts of mayhem.

Similarly, the input on a pedal is usually ground-referenced, so the input needs to be to ground.

That only leaves pin G, which needs to be mid-supply (4.5V nominal).

The only other thing I'd say is that you could get rid of D2. D1 and D2 are protection diodes for the batteries, but you'll only have one battery, so you only need one diode. Two is perfectly safe, but also perfectly unnecessary, and the circuit will be happier for longer without the extra diode drop stealing supply voltage.

HTH,
Tom

[balkanizeyou]

If I were to redraw this schematic to single supply, I'd probably do it like this:

doing usual offboard connections (jack sleeves to ground etc).
As a side note, I once made Anderton's comparator fuzz (very similar to this one, but without the input gain stage) and, to my surprise, it sounded absolutely awful with a dedicated comparator chip (lm311). It was alright with usual TL072 though.

[Cozybuilder]

Isn't the signal inverted at the output?

[deadlyshart]

Keep GNDs for IN, OUT & R5, disconnect GND from G and wire G to pin3..

G shouldn't actually be at 4.5V but at about 3.3V, because of 2 diodes forward voltage drop on each side.

Your IC will be feeded with 8.4V and 0.6V respectively so you MAYBE have to delete D3 & D4 for a more equal swinging..

Hi, thanks for the advice! I think I actually need D3 and D4 though, because they're kind of what gives rise to the distortion. I think G will still be at 4.5V though, right? If it goes: (9V, 10k res, diode, G, diode, 10k res, 0V), then G is still exactly in the middle of 0 and 9V and should be 4.5V just due to symmetry.

[deadlyshart]

+1 agree with Antonis.

The two on the RHS have to be connected to the same potential (gnd) because there's no DC-blocking cap between the pot and the output. Having one at 4.5V and the other at ground would cause all sorts of mayhem.

Similarly, the input on a pedal is usually ground-referenced, so the input needs to be to ground.

That only leaves pin G, which needs to be mid-supply (4.5V nominal).

The only other thing I'd say is that you could get rid of D2. D1 and D2 are protection diodes for the batteries, but you'll only have one battery, so you only need one diode. Two is perfectly safe, but also perfectly unnecessary, and the circuit will be happier for longer without the extra diode drop stealing supply voltage.

HTH,
Tom

Hi, perfect, this is exactly what I ended up doing (hoping I wouldn't wake up, read something else here, and have to go back and change it!).

I actually took out both of the protection diodes (D1 and D2). He's crazy about them in this book, but they just add components...

[deadlyshart]

If I were to redraw this schematic to single supply, I'd probably do it like this:

doing usual offboard connections (jack sleeves to ground etc).
As a side note, I once made Anderton's comparator fuzz (very similar to this one, but without the input gain stage) and, to my surprise, it sounded absolutely awful with a dedicated comparator chip (lm311). It was alright with usual TL072 though.

Thank you for the schematic! That's really useful.

Is that circuit in his book, or just from elsewhere? Was it using the 4558, and if so, do you think it was due to that? Do you think I should use a TL072 instead? thanks.

[balkanizeyou]

you can read about it here: http://www.diystompboxes.com/smfforum/index.php?topic=67377.0

As for the op-amp - sorry, I just use tl072 symbol for all my Eagle schematics where the circuit calls for a dual op-amp. But you may try all dual op-amps you have at hand for this effect - the sound should be very dependent on the choice of op-amp, because there's no negative feedback in the output stage.

[antonis]

Keep GNDs for IN, OUT & R5, disconnect GND from G and wire G to pin3..

G shouldn't actually be at 4.5V but at about 3.3V, because of 2 diodes forward voltage drop on each side.

Your IC will be feeded with 8.4V and 0.6V respectively so you MAYBE have to delete D3 & D4 for a more equal swinging..

Hi, thanks for the advice! I think I actually need D3 and D4 though, because they're kind of what gives rise to the distortion. I think G will still be at 4.5V though, right? If it goes: (9V, 10k res, diode, G, diode, 10k res, 0V), then G is still exactly in the middle of 0 and 9V and should be 4.5V just due to symmetry.

You're absolutely right...!!!

It's remarkable the way (my)mind works without enough coffee..