matt youwin
pumping the mids
Yes, unless there is a pad on the focusrite pres, you'll need to buy a suitable DI. You will probably get better results out of a good DI anyway.
Hmzzz i just wanted to use my Focusrite Pro 26 as a clean signal for the reamping. Now i got instrument level inputs i can use. However the instrument level inputs totally clip the hell out me with EMG's. Even if i bring the gain knob from the focusrite to like 0, like 7o'clock. Now if i put my other guitar in, there's no problem using the instrument level input. I even have to bring up the gain a bit. So does this mean i just can't use the instrument level inputs of my focusrite? Do i have to buy a DI anyway.
Did found this info on the instrument level inputs of the focusrite, but to be honest. I lost you guys on all the calculations it's probably easy as hell but still XD
Instrument Inputs
Frequency Response: 20Hz - 20kHz +/- 0.1dB
THD+N: 0.004% (measured with 0dBu input and 20Hz/22kHz bandpass filter)
Noise: -87dBu (20Hz/22kHz bandpass filter)
The formula:
voltage% = di box input impedance/(source impedance+di box input impedance)*100
So lets see what % of your signal will you get if you connect your passive humbucker (37403ohm) directly to:
Bellari RP220 with 600ohm input impedance:
600/(37403+600)*100
=1,578822725%
One and a half mothersucking % !!!
J48: (220000ohm)
220000/(37403+220000)*100
=85,46908933%
OK but not great
Type 48: (10000000ohm)
10000000/(37403+10000000)*100
=99,62736377%
No comment ! Numbers don't lie .
Demeter VTDB-2B: (27000000ohm)
27000000/(37403+27000000)*100
=99,86166201%
The thing with DIing for reamping is not that you get the absolute maximum from your pickups. Guitar tone is generally as much about what you lose as what you keep.
The guy talks in percentages but fails to factor in for example the resistence of the pots in the guitar circuit. These are there as much for the effect they have on the sound as anything else. If more signal was "better" you could just get rid of them completely.
Take a humbucker equipped Gibson-style guitar: generally they have 1 x 500 k volume and 1 x 500 k tone wired in parallel giving a total resistence of 250 k Ohm. Change over to 1 meg pots or just wire the pickups directly to the output and see do you like the difference. You will certainly be getting more of the signal from you pickups but does this sound better? For high-gain stuff, maybe yes, for cleaner stuff probably no.
Get a DI like the Radial JDV which has variable input impedance from 3.9 M down to 200 k and experiment. Install push-pull pots to your guitar circuit to allow you to take the volume and tone pots completely out of the circuit.
Fully open at 3.9 M Ohms and using the formula you get
3900000/(37403+3900000)*100 = 99.05% or so.
This would be rated as good, and I'm sure in the case of an active bass or an acoustic guitar pickup it probably is, but I can tell you that for reamping this sounds like complete ass in my experience. You are trying to get your guitar to behave like it does when plugged into an amp, you are not trying to get the maximum signal from it. You are trying to get it sounding musical.
Just applying the formula to the same pickup into your average amp it would be
1000000/[37403+250000(volume and tone pots)+1000000(amp input)*100 = 77.68%
And in my experience I have found that even a bit less than 1 MOhm for the amp input works better in a lot of cases (which when worked out according to the formula would score even less)
And of course these percentages are failing to take into the account pickup inductance and cable capacitance and how they interact with one another and the resistence to create the sound