Looper to remove looper

[cheeb]

I know bypass boxes have been beaten to death, but I can't find anything in the search that answers my question, or I'm just not recognizing my findings as my answer.

I have a Digitech Jamman loop recorder. It's a magical piece of machinery, but it sucks tone. I need all my pedals to run into it so I can use them in the loops I record, but when I'm not using it, I need a box just to remove it from the chain but keep the chain up to the Jamman going into the amp at all times.

Is this just a regular loop? I've never used one so i really don't understand them so much.

Any ideas? Thanks!

[Paul Perry (Frostwave)]

That's an interesting problem, if you are intending to 1. run stuff into it & record a loop and 2. then play back the loop AND 3. play the guitar at the same time as the loop plays.

I guess in an ideal case you wnat the looper to be completely bypassed when not recording, but the output from it to go to a small mixer, so that the guitar (and whatever fx are following it) can be added in.

It's certainly doable, but a standard bypass box isn't going to be right.

[Tuemmueh]

I had the same problem with my Digitech DigiDelay, which I use for short atmosphere-loops. I solved the problem by using an A/B-Box and a simple Mixer.
With the A/B-Box you select between Guitar -> FX Mixer -> Amp and Guitar -> FX -> LOOPER -> Mixer -> Amp. Using the Mixer you can control the level of both paths etc.

If you want to bypass the "A/B -> Looper -> Mixer"-Thing, you may use an extra TB-Box. I don't think it's necessary if you have a good mixer. I'm using a Nobels MIX-41C Mixer but there should be some more small mixers out there.

[cheeb]

I was hoping it was doable with a box that in normal mode passes the signal from the FX to one output jack which would be connected to the Jamman, and then when switched passes the signal to another output which would be an ABY or a mixer, I suppose. Kinda like this:

[Tuemmueh]

Allright, why not? Just Buffer the input & output and it should work

[cheeb]

So basically a buffered A/B and then a Y-splitter is my answer?

[Tuemmueh]

yes.

input -> buffer -> switch (a: send -> Jamman -> return) / (b: bypass) -> buffer with mixing stage -> output

maybe you could place the first buffer behind the switch.

[cheeb]

Maybe this is a stupid question (read: probably) but why the need for a buffer?