Is this a feasible distortion with blend?

[nchauhan890]

It’s definitely possible to do what you're trying to do in a 1590A with only two pots but the reasons for the limitations on this build seem kind of arbitrary. Especially with bass there is an advantage to having gain and volume controls as max gain isn't necessarily what will sound best.

On the clean side you would just make that buffer into an inverting amplifier stage (to flip the phase to match the fuzz side) and then bring the two signals together with some mixing resistors into the output buffer.

I found a link to this on Elliot sound products:


https://sound-au.com/articles/audio-mixing.htm

This looks like it would work well and I can definitely incorporate this. Would you agree that having a separate volume control per section would work better than a blend/panner? It also reverts the phase back to what it was at the start of the circuit too, which is nice.

My other bit of confusion is why this inverting amp uses ground as the non-inverting input, but the one from AMZ effects uses the reference voltage of 4.5V on this pin 

[bluebunny]

My other bit of confusion is why this inverting amp uses ground as the non-inverting input, but the one from AMZ effects uses the reference voltage of 4.5V on this pin 

It's intended that this circuit uses a bipolar power supply, so there's no need for a "virtual" ground like we have in 9V pedals - we have "real" ground!

Edit: I should say that my use of the term "virtual ground" might be confusing.  The "mix point" in the schematic is a virtual ground.  (The non-inverting + input of the opamp is ground, and the opamp will do everything it can to make the two inputs the same.  Therefore the inverting - input is also ground: virtual ground.)  We more often call our virtual ground "Vb", or something similar.

[nchauhan890]

It's intended that this circuit uses a bipolar power supply, so there's no need for a "virtual" ground like we have in 9V pedals - we have "real" ground!

Thanks for the brief yet clear explanation - I did some further research into inverting amps last night and I'm starting to get it now.

I've revised this schematic a little bit - it now uses an inverting stage to match the distortion circuit and another inverting stage to perform active mixing. I've read that the high resistor values to produce gain could add some noise, so does anyone think these values are appropriate? I'm not sure how much gain I should add to the clean side's inverting stage so it matches that of the transistor/distortion section.

I also wasn't sure about using the 1M resistors in the summing recovery stage since the input impedance isn't so much of an issue there... I'm now thinking I remove R10 and make R11 about 4.7k to 20k.

[marcelomd]

Looks great!

I'd remove R10 and use a capacitor between the volume pots and summing resistors. R11 can be something like 4k7, if you want unity gain.

For the inverting buffer, maybe 4M7 will be an issue. Maybe not. Suggestion: You can make it a unity gain (1M/1M) and attenuate a little the wet signal (adjust R12 and/or R13). Then recover the volume in the mixing stage (increase R11).

[nchauhan890]

I'd remove R10 and use a capacitor between the volume pots and summing resistors. R11 can be something like 4k7, if you want unity gain.

For the inverting buffer, maybe 4M7 will be an issue. Maybe not. Suggestion: You can make it a unity gain (1M/1M) and attenuate a little the wet signal (adjust R12 and/or R13). Then recover the volume in the mixing stage (increase R11).

Cheers! It's thanks to people like you and everyone else who's helped me put this together that beginners like me continue to love making effects. I'm excited to see how it turns out.

[antonis]

Well done..!!

(just some minor notes..)

There is no nedd for R7 to be of such a large value..
It ideally shoud be of value equal to Q1 equivalent input impedance for equal signal attenuation on both input stages (and equal tone sucking..)

If you wish absolutely equal mixing for equal mixing pots setting (e.g. for both R5 & R6 set at 50%, wet/dry also 50-50..) you'll have to "equalize" Q1 & U1A output impedances..

C7 value is 100 times C5 one, hence dry signal stage HPF cut-off frequency is 100 times lower than upper stage one..

[nchauhan890]

There is no nedd for R7 to be of such a large value..
It ideally shoud be of value equal to Q1 equivalent input impedance for equal signal attenuation on both input stages (and equal tone sucking..)

If you wish absolutely equal mixing for equal mixing pots setting (e.g. for both R5 & R6 set at 50%, wet/dry also 50-50..) you'll have to "equalize" Q1 & U1A output impedances..

C7 value is 100 times C5 one, hence dry signal stage HPF cut-off frequency is 100 times lower than upper stage one..

With C7 and C5, I don't mind the HPF cut-off frequency because I would like more lows on the clean side anyway. (The cut-off on the other side is still really low so it shouldn't make much difference) With the in/out impedances, how can I measure calculate them? All I know at the moment is that R7 gives the U1A section 1M impedance input, but I am fairly clueless as far as Q1 goes.

[antonis]

Don't mind a lot about imput impedances..

The problem rises from output impedances mismatch, especially when R5 & R6 are set at their upper setting..
Signals "fight" at mixing point having to "face" equal value resistors R12/R13 but coming from different voltage sources output impedances..
(For U1A it's about a hundred of ohms, give or take, but for Q1 stage is (R3+R9) // [(R2/(1+Gain)]

[nchauhan890]

The problem rises from output impedances mismatch, especially when R5 & R6 are set at their upper setting..
Signals "fight" at mixing point having to "face" equal value resistors R12/R13 but coming from different voltage sources output impedances..
(For U1A it's about a hundred of ohms, give or take, but for Q1 stage is (R3+R9) // [(R2/(1+Gain)]

Im considering putting trimpots of about 20k for the mixing resistors R12/13. What exactly is the 'gain' - Do you mean something like HFE?

[antonis]

I refered on Q1 stage Gain..
(it's late here to type a complex formula of open loop gain divided with feedback factor times open loop gain plus unity, e.t.c.)

Feedback resistor (4M7) "effective" value is divided with Q1 stage open loop Gain plus unity either "seen" from Input or from Output..

P.S.
Never mind.. Just take into account that signal looking back from mixing point doesn't see equal impedances..