Keeley parallel mixer as serial to parallel amp loop converter

[neurino]

I'm building the Keeley parallel mixer and I'd like to use it to convert my amp effects loop from serial to parallel (side chain).

I breadboarded it and, as far as I understand, all I have to do is connect Send 1 to Return 1 like this:


What then if I want to keep also the second send / return for later use?

Do I only need to add another 1uF cap + 100K res like C7 - R8 and C8 - R9?

Thanks for your support.

[Ben N]

Correct. Or you could use a switching jack as your Return 1--that way, with nothing plugged in it is a jumper, as you have it now, but you retain the option to put something in that loop.

[neurino]

Thanks.

The switching jack is a good idea too but I always want dry signal along with effects so I'll add another send/return

[neurino]

I resume this for another question: I used a MAX1044 to get +9V and -9V a la geofex.

Now what about if I want to build another one just using half the voltage like this:



Is that too low or will it work?

Would be better using splitter blend for phase inversion?

[PRR]

To run on one 9V supply do this:

[neurino]

Well, what more can I ask?

Thanks a lot Paul, I'll report the build

[Ben N]

If your second unit is also going in an fx loop, you might want the extra headroom that the +-9v power supply gives you, depending on how hot your amps fx send is.

[neurino]

Ben you're right, this new one is supposed to work with bass and some effects before going into the amp to blend some dry signal.

I reserve to later power it with 18V, with or without charge pump, if necessary.

[neurino]

I breadboarded the circuit and it works  , haven't had chances to see if it suffers of lack of those 9V.

Anyway I noticed R6, the 100k send level pot, which I bought as log, acts only in the very last part of the action, between 95% and 100%.

Maybe a linear pot would have been better but I don't want to order a new one and wait more time again.

There's some little turnaround to linearize a pot? Of course if this could fix the problem, I read time ago about the contrary i.e. simulating logarithmic with linear and a resistor.

Anyway what I'd like to achieve is a better send level excursion.

Any advice?

[neurino]

Another question (the one above ended up being unfounded, the log pot is fine):

this is the adapted schematic of what Paul suggested above (also in case some one needs it):



what if I want to place a blend control where where dry and wet join?



Are these reasonable values for R8, R9 and the blend pot R12?

[PRR]

Try it.

Or look at it.

In center, each side has gain of 220K/51K or about 4.

At either extreme, the "less" side has gain of 220K/101K or about 2, the "more" side has gain of apparently 220K/1K or 200.

Yes, you do get "more". But you get it with a HUGE boost of the "more" side, way out of line with the gain in the "equal blend" setting.

It is actually not a trivial problem, even after 80 years (talking movies needed to smoothly fade from one projector to the other every 20-min reel; cross-fade was a key tool of DJs and disco).

The most direct path is to convert U3 to a non-inverting (high impedance input) stage, and use your pot network.

BTW: now you may omit the 1K resistors. They simply limit the depth of the "less" input to 1%. It is often useful to get to 0%, and it saves two parts, a whole dime.

[neurino]

The most direct path is to convert U3 to a non-inverting (high impedance input) stage, and use your pot network.

BTW: now you may omit the 1K resistors. They simply limit the depth of the "less" input to 1%. It is often useful to get to 0%, and it saves two parts, a whole dime.

So it should look more like this? (adapted from the moosapotamus paralooper)

Again I guess I can omit 1k resistors