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[jetbase]

Using a console, what's the best way to connect guitar fx pedals to use as outboard fx in a mix, achieving the best s/n ratio & tone? I would think possibly you could simply connect an aux send directly to the pedal & keep the level low, fine tuning the level with the pedal's input level (if it has one). Or maybe, like reamping a guitar, use a passive DI in reverse (or a reamp box - but I don't have one). For returns I guess just DI into mic pre into fx returns or into mic input on a channel. Is there a better or simpler way?

Thanks,
Glenn

[MagnetoSound]

That is exactly the way I would do it, although I wouldn't bother with the reverse DI in front of the pedal.

[0dbfs]

I've used both methods and even if I "felt better" withe the re-amp/reverse-DI method, it's still an fx pedal and acomplished the desired result (it fucked up the $ignal).

Best,
j

[jetbase]

Thanks guys!

[jonoaustin]

these things are fantastic:  http://littlelabs.com/pcp.html

Actually, all his stuff is fantastic.  You can occasionally find PCP's used...

[jetbase]

Thanks Jono, it likes like a pretty handy little unit. Nice to know what's out there.

[Ned Ward]

Radial's Pro RMP is $99, licensed from John Cuniberti, and is great for exactly this. I use it running an out from my 003 into this into a pedalboard when I want to reamp, but you could easily use this and a Radial JDI at the other end. Given that you're starting with a line-level signal, it's really what FX units you have - good ones will do well with a range of signals; cheaper ones will yield "interesting" results.

[drknob]

Lots of good advice here.
Remember - with guitars, it's not that the level is low. The problem is the source impedance is high, usually in the 100k+ area. When you plug your guitar into a device, the source impedance of the guitar and the input impedance of the device you've plugged into create a voltage divider. This makes it impossible to drive anything but a high impedance device without changing the sound and level of the guitar itself. Think of a direct box as an impedance converter. This is why a passive direct box, in order to lower source impedance must also lower the signal level. An active DI does not have this problem. Ohm's Law strikes again!
Nerd alert.